DK Jain M3

S h re e R am

F ra k a sh a n

NGINEERING MATHEMATICS-III !.

tf'ri; I 1 n . t h i h u s

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h'i'iin U il! I

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SHREE RAM PRAKASHAN

Published by

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Patankar Bazar. Gwalior (M.P.) Ph: 0751-2626678.4079P I

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k oteshw ar Road. Gwalior (M. P) Ph 0751-4060657.2491877 c-inai I:roh it2007_swami@rediffmail.com Website . www.shreeramprakashan.com

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I Al l r i g h t s r e s e r v e d . N o p a r i (it thi> b o o k m a \ b e r e p r o d u c e d o r t r a n s m i t t e d in a n \ f o r m or by j

m e a n s , e l e c t r o n i c o r m e c h a n i c a l , i n c l u d i n g p h o t o c o p y i n g , r e c o r d i n g , o r a n y i n f o r m a t i o n storage; r e t r i e v a l s y s t e m w i t h o u t p e r m i s s i o n , in w r i t i n g . f r o m t h e p u b l i s h e r

REFACE OF THE FlRST EDITION We are very happy to present to our readers, students, and teachers alike, this book on Engineering Mathematics-Ill. The present book is to meet the requirements of the students of B.E. Second Year, the need of which was being felt very anxiously. In the treatment of subject matter, we have tried to Maintain die same style as used in the other books of Engineering Mathematics (namely I, II and Itecrete Structure). f All the topics have been covered comprehensibly not only with clarity but in a lucid and easy way d be grasped. There are a good number of fully solved examples with exercises to be worked out at the Hid of each unit. B esid es, th ere are fu lly so lved q u estio n papers o f previo u s year RG PV exam ination a t th e en d o f th e b o o k. We are sure that this book will receive the same warm welcome by its users, as in the case of our previous volumes. We shall feel highly obliged for any suggestions and constructive criticism for the improvement of he book. The book is fully on accordance to the syllabi provided for the Bachelor of Engineerging Fourth semester students of Rajiv Gandhi Prodyogiki Vishwavidyalaya, Bhopal. Lastly, we thank our Publisher, M/s Shree Ram Prakashan, Gwalior for publishing this valuable wok. AUTHOR For suggestions, write to Dr. D .K . Ja in, B-3 Dwarika Puri, Behind Prem Nagar, Gwalior (M.P.) 474002 Phone:9826086788 Email:jain_dkj@yahoo.co.in ........................................ .......... — —

P reface

o f th e

............

11

...................

F o u r th E d it io n

We are pleased to bring out the fourth edition of this book. This new edition is throughly revised and updated in accordance with the demands of the modified syllabi and that of the technical students and the interdisciplinary applications that they need. The new revised edition includes modified units dealing with statistics adding up topics namely, correlation and regression, theoretical distribution and testing of hypotehsis, with particular references to mathematical statistics. Suggestion for the further improvement and upgrading of the book are heartily welcome and the editors and the authors will be thankful to the readers for responding with the same. AUTHOR Dr. D. K. Jain Gwalior January 2012

SYLLABUS U nit- •I

Functions of Complex variables: Analytic functions, Harmonic Conjugate, CauchyRiemann Equations, Line Integral, Cauchy’s Theorem, Cauchy’s Integral Fomiula, Singular Points, Poles & Residues, Residue Theorem, Application of Residues theorem for evaluation of real integrals. '

i Unit- -II

Errors & Approximations, Solution of Algebraic & Trancedental Equations (Regula Falsi, Newton-Raphson, Iterative, Secant Method), Solution of simultaneous linear equations by Gauss Elimination, Gauss Jordan, Crout’s methods, Jacobi’s and Gauss- t Siedel Iterative methods. i

'

Unit -I II

Difference Operators, Interpolation ( Newton Forward & Backward Formulae, Central j Interpolation Formulae, Lagrange’s and divided difference formulae ), Numerical Differentiation and Numerical Integration. 1 1

U nit- -TV

Solution of Ordinary Differential Equations(Taylor’s Series, Picard’s Method, Modified Euler’s Method, Runge-Kutta Method, Milne’s Predictor & Corrector method ), , Correlation -and Regression, Curve Fitting (Method of Least Square). 1

Unit- -V

Concept of Probability : Probability Mass function, Probability density function, f, Discrete Distribution: Binomial, Poisson’s, Continuous Distribution: Normal Distribution, j Exponential Distribution,Gamma Distribution ,Beta Distribution,Testing of Hypothesis | |:Students t-test, Fisher’s z-test, Chi-Square Method.

I V

C ontents U N IT -1 FUNCTIONS OF COMPLEX VARIABLES 1.1 *2 1.3 04 1.5 1.6 1.7 J.8 •+.9 1.10 -t.ll 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 -120 421 H-22 123 J24 125 >+26 127 128 129

Introduction Complex Variables Polar Forms ofACom plex Number Functions o f A Complex Variable Limit o f A Function Continuity o f /( z ) Differentiability o f f(z) Analytic Function C^auchy-Riemann Equations Cauchy-Riemann Equations In Polar Coordinates Harmonic Function Orthogonal System Methods for Constructing an Analytic Function Bilinear Transformation Conformal Transformation Some Standard Transformations Some Definitions Complete Integration Some Definitions Complex Line Integral Cauchy's Integral Theorem or Cauchy's Theorem Cauchy's Integral Formula Residues o f /( z ) at Pole M ethods o f Finding Out Residues o f /( z ) at A Pole Cauchy's Residue Theorem Application o f Residues To Evaluate Real Integrals Taylor's Series Laurent's Series Circle and Radius o f convergence Practice Questions

~

l 1 I 2 2 2 2 3 4 6 7 7 15 32 40 40 48 52 53 54 58 59

(A (A 65 79 99 99 105 106

U N IT -2 NUMERICAL ANALYSIS-I 2.1 22

23 2.4 2.5 2.6 2.7

Introduction Approximation And Errors Some Important Rules For Absolute and Relative Errors Algebraic and Transcendental Equations Bisection or Bolzano Method Iteration M ethod or Successive A pproxim ation M ethod Method o f False Position or Regula Falsi Method

115 115 118 119 119 123 128

2.8 2.9 2.10 2.11 2.12 . 2.13 2.14 2.15 2.16 2.17 2.18 2.19 220 221 222 223 224 225 226

Secant Method or Chord Method , Newton-raphson Method Geometrical Interpretation of Newton-Raphson Method Order of Convergence of an Iterative Method Convergence of Bisection Method Convergence of Secant Method Convergence of Regula Falsi Method Order of Convergence of Newton-Raphson Method Comparison Study of Iterative Method Graeffe’s Root Squaring Method Bairstow’s Method or Method for Complex Root Solution of Simultaneous Linear Algebraic Equations Gauss-elimination Method (Direct Method) Gauss-Jordan Elimination Method (Direct Method) Crout’s Triangularization Method or Choleskey’s Method or Lu Factorization Method (Direct Method) Iterative Method Gauss-Jacobi Method Gauss-Seide! Method Relaxation Method Practice Questions

134 137 138 150 150 “ 151 152 152 154 154 157 159 159 166 169 172 172 178 186 189

U N IT -3 NUMERICAL ANALYSIS-II

3,1 '■dl 33 3.4 3J J.6 3.7 y.% 39 •^.10 3.11 3.12 3.13 3.14 3.15 3.16 >~3fl7 3.18 3.19 320 321 322 \_^23

Calculus of Finite Differences Different Operators Relation Between The Operators Difference ofA Polynomial Factorial Polynomial Interpolation with Equal Intervals Gregory-Newton's Forward Interpolation Formula for Equal Intervals Gregory-Newton’s Backward Interpolation Formula for Equal Intervals Error Iri Newton's Interpolation Formulae r Gauss’s Central Difference Formula for Equal Intervals Gauss's Forward Interpolation Formula for Equal Intervals Gauss’s Backward Interpolation Formula Stirling’s Formula for Equal Interval Bessel’s Formula for Equal Interval Everett’s Formula for Equal Intervals Interpolation with Unequal Intervals Divided Differences Relation between Divided Differences and Forward Differences Newton’s Divided Difference Interpolation Formula for Unequal Intervals Lagrange’s Interpolation Formula for Unequal Intervals Inverse Interpolation Lagrange’s Inverse Interpolation Formula Numerical Differentiation *

195 195 198 200 208 217 217 218 219 233 234 235 236 238 241 242 242 243 243 244 254 254 257

H 32

Derivatives ofy =f(x) Based on Newton’s Forward Interpolation Formula Derivatives ofy = /(x) Based on Newton’s Backward Interpolation Formula Derivatives ofy =f(x) Based on Stirlings or Central Difference Formula Numerical Integration Newton-Cote’s Quadrature Formula Trapezoidal Rule Simpson’s One-third Rule (or 1/3) Rule Simpson’s Three-Eighths (3/8) Rule Weddle’s Rule Practice Questions

257 258 “ 258 269 269 270 270 271 272 290

UNIT- 4 NUMERICAL ANALYSIS-III, CORRELATION AND REGRESSION

^

4# ill .12 S3

Numerical Solution of Ordinary Differential Equations Picard’s Method Taylor Series Method Euler’s Method Improved Euler’s Method Modified Euler’s Method Runge-Kutta Method Milne’s Predictor-corrector Formulae Working Rule for Milne’s Predictor Corrector Method Concept of Correlation Coefficient of Correlation Types of Correlation Properties of Coefficient of Correlation Covariance Methods to Study Correlation Scatter or Dot Diagram Method ' Karl Pearson's Coefficient of Correlation or Covariance Method Rank Correlation Method Lines of Regressions Regression Coefficients Properties of Regression Coefficients Properties of Lines of Regression Relation Between Regression Analysis and Correlation Analysis Curve Fitting Principle of Least Squares Fitting ofA Straight Line Fitting ofA Parabola (or Second Degree Curve) Fitting of A Curve of The Type^ = ab* Fitting of The Exponential Curve y = a(*>x

299 299 307 311 314 315 318 326 327 333 333 333 335 336 336 336 337 344 351 351 352 353 354 366 366 368 369 370 370

b Fitting ofA Curve of The Type y = ax2+~

370

Practice Questions

380

!

5.1 52 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 5.40 5.41 5.42 5.43 5.44 5.45

Concept of Probability Random Variable

U N IT -5 STATISTICS

Distribution Function Probability Density Function Theoretical Probability Distribution Binomial Distribution Hypothesis For Binomial Distribution Characteristic o f Binomial Distribution Mean, Variance and Standard Deviation o f Binomial Distribution Poisson Distribution Conditions Under which Poisson Distribution is Used Mean Variance and Standard Deviation o f The Poisson Distribution Normal Distribution Standard Normal Distribution Properties o f Normal Curve ' Method to Find The Probability when The Variate is Normally Distributed Rectangular Distribution or Uniform Distribution Mean, S.D. and Variance o f Rectangular Distribution Mean Deviation o f a Rectangular Distribution Gama Distribution Distribution Function o f Gama Distribution Mean, S.D. and Variance o f Gama Distribution Beta Distribution o f First Kind Commulative Distributive Function o f Beta Distribution o f First Kind Mean S.D. and Variance o f Beta Distribution o f First Kind Beta Distribution o f Second Kind Mena, S.D. and Variance o f Beta Distribution o f Second Kind Exponential Distribution Distributive Function o f Exponential Distribution Mean, S.D. and Variance o f Exponential Distribution Mean Deviation about Mean o f Expobebtial Distribution Test o f Significance , Null and Alternative Hypothesis Level o f Significance Chi-square Test or x 2 - Statistic ’ Properties o f x2 Distribution Conditions for The Validity o f x^-Test Uses or Application o f x ^ T e s t /-statistic and /-distribution Computation or Working Rule o f /-distribution Fisher’s F-test F-statistic Assumptions in /--test Computation o f F-test Fisher’s z-distribution Practice Questions Previous Year Solved Papers

389 J89*: 390 392 406 406 406 407 407 422 423 424 435 436 437 437 452 453 454 456 456 456 458 458 459 460 461 463 464 465 467 472 472 472 472 473 474 474 492 493 505 505 505 505 509 512 O

' F u n q t j o n s o f C o m b w x V a r d w i .e s — ..■■■"■■" — ■...................... ■■■■.................................. w f o M -wfw i m ^iilit . >■— — in t r o d u c t io n ^

!

I 'w - - .- - ; * * * * ■ -

We havife 'fafafl that there exist one-one correspondences between the set of real numbers and me points on the real line. The concept of ^ ( - l j is meaningless, since there is no real number whose square is a negative real ^70t*l783) was the first mathemati­ cian who ii^oellticed the symbol i with the property that ? = - l l i was Gauss (1777-1855) who first studied thatan algebraic equation with re&f A Kf ?M\i^ : * ftp

ii:.; i"

&]^n p M v ia b le . Ateb i a r i t f y ^ respectively called real and I

imaginary patfsrafa SonKtimes we express sr as z ~ x + i y * (*,31).

P ° ^ 'a l » £?•'

M -v

!

r

t ; ' ' :"r ’

The complex conjugate, ar briefly conjugate o f z=x+iy is I =x-iy.

For exgmp&4ao)iigitic of tx m - 3 - 5 i is z * r 3ft# •i i ‘ It is easy to verify th at: R(z)=x =Z- ? * , and l(z)= y = ^ ~ . ! 'v ; i •- i

U

,

■

.

POLARFORMS OF A COMPLEX NUMBER

’'

2,1

'..'it--

Consider a point /* in the complex pl{(he corresponding to a complexnumber z=x+iy. From the k ;\ jO ' i'i i ■f { t .••< ' f- ■ y adjoining figure, putting x = rco s# , a n d y = fg in ^ .f.*i

P(x,y)

-

;

Then

r = ^ ( x s + >*)

;

-

,j r

. ' w

'

■

k

If follows that r = x + iy ,

X

jr

•' ' '■

K

1 V **

sy S

/ ■

y ■j

\

'*

D

? iA (^W +il8m ff)=rei& ...(1) L-: x M X A It is called/?o/<^ybrw dfthe coirt^T®i
2 | Enginehung MAmEMAncs^il Remarks :

1. |*,*2H*,||*2|

* A

'*>2£*. , '|4 ^ %MI * i m * $ & C ■

3r

- --D , -

"

- '

■ •■■ -

7-

4. Equation of circle |r - a j = r where a is center and r is a radius, 5. Re (z)=x and Im (z)=y 6. . Jzj2= zz ^ -ij.<•. 7. The equation ‘T ~ ~ represents * “ *2 ^ (i) actoje if A 0 and (ii)» straight line If A»i..

1.4 FUNCTIONS O F ACOMPLEX VARIABLE To each value of complex variable z, if there correspond*4tatfMr more than one value of a coppkat vari*We H^then we say that w is a function sof a ccjpjplex variable z. Also we write w - f( z ) = u + iv ■, ..... . w is said to be single valued or many valued functions of z according as for a given value of z, there corresponds one or more than one valye of w, and u and v are ftmctiom of x aftd> For examples: 1. f(z)= z2

2. f(z)»&D.z

3. /(^ « lo g *

4. f ( z ) * z 2

1.5 LIMIT OF AFUNCTION A number / is die limit of a function f(z) as z tends to a and write lim f{ z )-l.

1.6 CONTINUITY O F /ft) A function f{z) is called continuous at z = a if

1

lim /(a)= /(o ). A function f(z) is said to be continuous in a domainD ifit is continuous at every point o f D.

1.7

m ffe re n h a b ility o f /^ A function

fin) in domain D, is called differentiable at a poiptz =f a, if tim —
z*0 1 4 : I f f(z)= "4,

Prove that iM

X w

->0 as g -> o ekmtgltirf m lto 4ettM kdnat 4» jt ■-»%xj»**y

> tf manner. Suppose *>-*0, along the radios vectef y » « * , then we have

x 2y (y -ix ) m M z M *-to z - 0

'

. i J i i - u !2 *-*o (x+*y)-0

[Since y ~ m x th en « -►0=>x+iy -* 0 =».*+*rox -►0 => x -+ 0] . IjJ -

= lim[ ,

----- 1

= lim - g f c & : * 0. bl!6 n {.< f.-v. **C(x7+M2)$ + im ) ■ ■ • i 'i.

' ■*

Ans.

■ -!• I.

. .•

.

Further, if g -* 0 along the eurve y = m x2 (ie., in any manner) then we have x 2y (y ~ ix)

< £± £ 1 *-*«

* -0

(* + y r)-0

[Since y = mx2, then z ->,0 => x -* 0] _ Hni (!

1J

x 2mx2(mxl -ix y fj. ljm mxi (m x -i) (x4 +m2x*)(x+imx2) z~*4>x5(1+nt2Xl+iinx)

-m i = :----- 2 ~ does not exist, [because limit depends on m i.e. not unique]. 1+ rn •• t ' :u : < \ r '

Proved.

ANALYTIC FUNCTION . _ A single valued functionffz) in a domain P is said to be analytic at.a point z = a if there exists a neighbourhood \ z - a \ < 6 at all points of which /"(zjeadsts. If f ( x ) exiits at every point o f a d&main D exceptar a flnite^tanber of exceptional points then f(z) is said to be analytic in Z > .T h e « exceptional points are called singula- points or singularities of the functionjjjft). If f { z ) exists at every point of D, then we say thatf(t) is regular in D.

The terms regular and holomorphic are also sometimes used as synonyms fo r analytic. For examples : 1. f ( z ) = z 7 2. f ( z ) - e * t* R w irl : '■ ■ H V' : £!-> : Singular p o in t: A point z =zn is stid to be a singutotgoint o f a fraction f(z) if f ( z Q) does flat exist. •; A necessary condition Ifcat w =f(z)=u(x,y)+i v(x,y) be analytic in a donwin'OiMljat du dv , d u dv •*— - **-—and ——=— (■<■ - jv j 1 dx dy dy dx

y^- - vv o p tio n s, ^

««*.-.?<;■

iiiui - - • • -u., at- - i t or ux =vy and uy ~ - v x. , i Theorem [1.1]; Necessary condition fo r f(z ) to be analytic: t f fi(z) = u+ iv is analytic in adom ain D, then u, v satisfy the C-R equations •

, - I .- J

dx

/W /J f*

8$

„«***; _ r’s->-

= «■+ iv JfMQanalytic functionIn S f f i g i g ? d f e n S _ mvr.. where « and v are the function of * and y. Let, S u oftL the H)c*enients of wand v

f(z + Se) - (u +Su) + i(v +Su) f(z +s j ) - f ( z ) (u +Sy)+*iv~±$$^(u +iv) We have

JZ

,

s * * ''> ■ ....

........... "'r^~ ' *'“■■* “ . ti. i -

1 1 • '

_ Su + iSv _ 8 u’ ' .Sy Sz 5z Sz ^

or

^

Sz-*0

j.,

i: ‘

or

fj«

^ 2* ,

. S'9.p

•? fito * >«,.N

. f(z )-

Sz)

I'jJ; : no «SR3Q9b”mnu>. ■

v.i’.-

tV V -.i’

*

p?/• ■ ? • ■■ .

Sino*, & M » ajjftrwtch zrw&aiongany path. .fes/o' w Along real axis : On real axis y=0 so that z =x,

“ -& U * 1*0

'M- -f-o

i>norr

».

!te tR

'•

an^ d'y=^, ( t r s f

,-dri: • ,.V

FUN =(foln

n:L&i

C

|f

Su

J 1 dy dy , k f(z) is dilYertmiaWe^ei* two values o f f! (i) must be the same i.e., -3 * .• vi

■• 'tf ,

r-■

\r

du equating (2) and (3) we get, ■

; ;S

. dv _ . d u d v . +l dx~~* dy*dy ■v: * - ;

^ -',■

.•

— - idu dv Equating real and imaginary pats : ^ - ^ a n d — = - — ■

j do \ -du

J-Jence the necessary condition fbrj^T’t^'be analytic is thatthe C-R equations must be satisfied. v Proved. .

t Thefunction w = f (z)= u+ iv is analytic in a domain D, if |f> U, v are dtfferentiaple in D and

=t>,, u r =-,%*

iti.f i’ .. V\ -.I!, ■

i

y,U-

v, . ;

(it) The partial derivatives ux, vx, uy, vy all are continuous in D. Let,f(z) x=u + iv be a single-valued function possnG&iqgpaftial derivatives

du du dv dv Qy

at each point ofa region Z) and satisfyirtgOR equations. ’

du dx

dv dy

, du dy

dv , dx

We shall prove that f(z) is analytic, i.e., f \ z ) exists at every (idiot of the region/). : By Taylor's thtorem 4or functions of two variables,; we have f(z + Sz) = u(x +Sx, y + S y ) f i v(x + Sx, y + Sy)

[on omitting seoon^ and higher degree terms of Sc and Sy]

'

=' ( s + is ) & + ( " i +‘i ) ' Sj'

[V /< « )« » + »wj

[Using C-R equations]

6 | Enginebrmo MftnENAncs-Ili

f(z + S z ) - f( z ) Ss

(du .5 u V du i— <5cc+ — + i— \i$y dx) ta * dx) dx dv du dx dx du | . dv dx dx

<

7w

&-*> & du dv Thus f'(z) exists, because — >— exist.

dx

dx' Proved,

Hence f(z) is analytic.

1.10 CAUCHY-RIEMANN EQUATIONS IN POLAR COORDINATES Let

w = f(z) => u + iv=f(x + iy)

Put

x=rcos0, y =rsin0,'then u+iv=f(rcosd+ir&wff)=f(rel9)

u+iv= f(re10)Differentiating (1) partially w.r.t r, we have + *—

= / '( r e " ) . * "

dr dr Differentiating (1) partially w.r.t 6, we have

. ( du t .dv' du . dv + i — = f'(r ew).ire* = irl V5r dr, d Q d d 1

)

[RGPV June 2609]\

[ve* = cos#+isin 0) •••(I)

...(2)

[Using (2)]

dv . du + ir — . dr dr Equating real and imaginary parts, we get = -r—

du dv , dv du = - r — and — = r dQ dr d6 dr du _ 1 dv , du _ 1 du dr ~ rdO ° dr ~ r d9 Which is the polar form o f C-R equations. Example 1.1 ; Test the analytic behaviour o f log z. Solution. Let /(z) = legz => u+iv = \og(x+iy) zz>

u +iv = log(rcos.f? + irsinff)

=>

u+ iv = log(re*)

=>

v + iv = lo%r + id

[•.•elB ~ co8#+isin#]

F unctions of C omplex Variables | 7

Equating real and imaginary parts, we get u ~ logr and v ~ 0 Differentiating partially w.r.t. r and 6. du 1 , dv ^ — = — and — = 0, or r or and — = 1. 39

— =0 dO

du

It is clear th at -

Id v

=

. dv

and -

=

1du

=> The function log z is analytic except at origin.

Ans.

Ill HARMONIC FUNCTION d2u d2u A real valued function u = u(x,y) is called harmonicJunction, if 7 T +7 T »' ox oy 2

i.e., u satisfy the Laplace’s equation : V u =

= 0*

d2u d2u

qgftmark : Harmonic Conjugate Function : I f the function f(z) - u(x, y) + iv (x> y), is analytic in a domain u " D, then the Junctions u and v are said to conjugate Junctions. J& Alpit 1.2 : Prove that, u + tvis analyticfunction in domain D, then « and v are harmonic. M ention : Since f(z) = u + iv is analytic => Cauchy-Reimann equations are satisfied du Le-

dv

~ te~ dy

d u __dv m d

- (1)

d2u d2v d2u Differentiate: 7 7 - . ~ and dx dxdy dy~ Adding:

d2u d2u

~

d2v a dydx

_ / ^ u *s harmonic.

Again from (1), differentiate w.r.t. y and x respectively, we get d2u dydx

d2v d2u and dy dxdy

d?v ,
d2v d2v On adding: T T + T T => *>is harmonic. ox oy i.e., this shows that u and v satisfy the Laplace's equation.

Proved.

ORTHOGONAL SYSTEM Twofamilies o f curves u(x,y)=c] and v(x,y)=c2are said to form an orthogonal system if they intersect at right angles at each o f their points o f intersection.

F unctions o f C omplex Variables | 7

Equating real and imaginary parts, we get u = logr and v = 0 Differentiating partially w.r.t. r and Q. du 1 , 5d _ TT = " «nd T = 0, or r or du — =0 80

dv , and — =1. d9

du Id v It is clear that - = —

, dv 1 du and - = - -

=> The function log z is analytic except at origin.

Ans.

HARMONIC FUNCTION d2u d2u A real valued function u = u(x,y) is called harmonicfunction, if T T +T T ox oy i.e., u satisfy the Laplace's equation : V u =

d*u
rk : Harmonic Conjugate Function : I f the function f(z) - u(x, y) + iv (x, y), is analytic in a domain D, then the functions u and v are said to conjugate functions. 1.2: Prove that, ifflz)* u + iv is analyticfunction in domain D, then u and v are harmonic, n : Since f(z) = u + iv is analytic => Cauchy-Reimann equations are satisfied ie "

du _ du ~dx ~~dy

du dv ~dy~~~dx'

d2u d2v d2u d2v Differentiate : T T ’ T r and r r =" -“T" dx2 dxdy dy~ dydx Adding:

d2u d*u

=

,n

^ u is harmonic.

Again from (1), differentiate w.r.t. y and x respectively, we get d2u dydx

d2v c?u and dy2 ™ dxdyi

d?v dx*

d2v d2v On adding: T T +T T =* v *s harmonic. ox oy i.e., this shows that u and v satisfy the Laplacefs equation.

Proved.

ORTHOGONAL SYSTEM Twofam ilies o f curves u (x ,y )-c i and v(x,y)=c2are said to form an orthogonal system if they intersect at right angles at each o f their points o f intersection.

8 1 ENOiNaaaNG M n m u n c s 4 li

Theorem fl.3 |; Iff(z) = u + ivisan analyticfunction in a domain D, prove that the curves u = const and v = const form two orthogonalfam ilies. Proof Let f(z) = u + iv be an analytic functions of z, so that Cauchy Riemann equations u r ~vx

uy = -v x are satisfied.

To prove that the curves u(x,y) =c,, and L’(.T,.y) - c2 are orthogonal. Let, m, = slop of the tangent to the curve

m= c,

and

m2 = slop of the tangent to the curve v =c2. Taking differential of or

du .

du ,

dy , dx u.

or

and v(xty ) - c 2 _

m 1 (say) and

r u. V

mtm2—

V

dv j dv . and T x d * ^ d y - «

~ vx

'

dy _ -vx _ "h (say) dx

- Ux Vx _

u xv x

_

|

[v by C - R equations]

J

Proved. u =c, and v =c2 form an orthogonal system. Example \J ; I f u(x, y) and vfx, y) ure harmonic functions in a region R, prove that the function r du 51;") . f du ( dvy v dy rk» d*x) + l {d x dy. is an analytic function o f z = x + iy. fRGPV DEC 2004J Solution.

du dv . ^ du dv Let $ = ----------and t =-— + — . dy dx dx dy Then we shall to prove that s +it is analytic, i.e., s and t satisfy the C-R equations. ds _ dt ds _ dt The C-R equations are : ^ ^ anti ^ ^ ■ ds We have,

dt =

d2u dxdy

d_ f du XT dx Kdy d2v ^ d2u dx2 dydx

dvs dx;

d fdu dy Kdx

d2v dy2

d2v d2v _ n r T + 7 J “ 0 => v is harmonic. oy ox

dv> dy j Since

dxdy

du dydx ...d)

Function* of C omplex Variables | 9 da Sim ilarly,^

^

v i

dt d (d u ^ ==>

fa )

____d (d u cbe^d*

dv dy

d2u d2v _ d2u d2v dy2 dydx dx2 dxdy

d2u d2u n -.(2 ) ~TT +~ rT = 0 => k is harmonic oy ox It is clear that Cauchy-Riemann equations arc satisfied, because from equation (1) and (2), u and v satisfy Laplace's equation. Hence s + it is analytic. Proved.

Example f .4 ; Show that the function f(z ) ~ >/| x y | is not analytic at the origin, although the Cauchy-Riemann equations are satisfied at that point Solution.

Given, f (z) = -^1xy | => u + iv =»yj\ x y\ + i.O => u(x >y ) - a/1 *y t and At the origin: du - lim dx (0.0) x- ° du [d y (0,0)

= lim '-*0

=0

x^O x

y-+o

y

Similarly, i = lim dx (0,0) *-*

x-*0

x

'd v

and

_dy\ [0.0)

^

Hence C-R equations are satisfied at the origin. Further,

A O ).]im /fr)- f i 9 . 1 i m ^ T~ ° - Um *-»o z *-+Q z *+*y-*o (x +iy) Along y=mx, then we have

Vlmx21 J\m \

,

f (0)= lun ----- ;— - =-—:— = depends on m he., not unique. *->0 (X + fWMC) 1+tflt => AO) does not exist. Hence function is not analytic at origin.

Proved.

10 I ENCai«BEJtiNGMATHEMAnC$411 Example 1.5: Show that an analytic function with constant modulus is constant [RGPVDec. 20021 Solution. Let f(x ) *=u+iv be an analytic function with constant modulus. Then, |

u + iv I=constant

=> Vu2 + V2 = constant * k (say) Squaring both sides, we get u 2 + v2 * k 2 Differentiating equation (1) partially w.r.t x, we get - du ^ dv du dv _ 2u— +2v— = 0 => u —-+v— =0 CQC^

OX

OX

...(2 )

ox

Again, differentiating equation (1) partially w.r.t y, we get du ~ dv du dv _ 2u—~ + 2 u --= 0 u — +u—-=0 dy dy ^ dy dy t du ■ G

f M

dv

, dv du ...(3)

S "

Squaring and adding equations (2) and (3), we get (u 2 +

[v »

+v —c & 0]

|A * )|2= o =*

|/ W h o

f(z)is constant

Proved.

Example 1*6 ; Prove that the function f(z)= \ z f is continuous every where but no where differen­ tiable except at the origin. Solution.

Given f{z)~\ z \ 2 = |s+£;y|2 - x 2 + y 2 •»

=> u = and u =0 => u and v being polynomials in x and >>are continuous evecy where. Hence flz) is continuous every where ~«du

dv

•••(I)

T

F unctions o f C omplex V ariables I 11

Cauchy-Riemann equations arc: du dx

dv , du dy*n $y

dx

=> 2x = 0 => x » 0 and 2y = 0 => y= 0 . Hence Cauchy-Riemann conditions are satisfied only at origin. Thus the given function is now here differentiable except at the origin.

Proved.

1.7 : Show that the function f(z) * e '* * , andffO) - 0 is not analytic at z « 0, although Cauchy-Riemann equations are satisfied at this point [RGPV 2001] Here,

f(z) = e ’ 4 =«-<**>'4 a- e — (* ♦U J<*z -M«y)‘

...(1)

j h ^ ixUyt~ 6xV) ~ e l(** ♦/>* J _ e " 7<**+/>* 6xV ]

U) = P *

IL

'

‘byOic2-y 2) . . 4ay(**-/)

J 006— -;----- r-r---- K S in -* tf+ f)*

( ^ + y s>4

Equating real and imaginary part, we get «4+y*- (kV1

u -e

00B

‘fap|(*2-y 2) (x 8+ / ) 4

and At* = 0,

v = e z^ w d ]8ini ^ V

)

f o g |im «*(*»<>)- « ( 0 , 0 )

dx

*-*<>

[Using (1)]

x

r e-*~4 - 0 v 1 = lim ----- — = h m — “ —3lim-

ux4

*-*o

= lim *->o

and Now, and

= lim i1 H-----r 1 + ----=1 + x 4 2 jc

- lim dy y-*0

dx dv

4y

= 0

~-j~4

du

dv

1

1 1 *->o X + — + ---- - + JT

=lim I f ----*-*°

y

=lim x-+0

lim —= 0

=lim y-fO

lim —= 0 .

x->0*

y-O y

[Using (l)]

12 I ENGINEERINGMMHEMAnoMtf Hence Cauchy-Riemann equations are satisfied at 2 = 0. Further,

~ lim ---z

/ '( 0 ) = lim e-

-

; if z -> 0 along z ~ rei7t,A

e~r \ e = lim --- r*77- = «>

r_*® re^

.*. f \ z ) does not exist at z = 0 and hence f(z) is not analytic at z - 0. Example 1.8: P rovelkat: Solution.

a*

+

a*

Proved.

. a*

-* } £ £ ■

Since z - x + i y and z = x - i y , then x = ~ ( z +z) and y = ~ ( z ~ z ) 2

2i

dx _ 1 dx _ 1 dy __ 1 ffy _ J_ dz ~ 2 ' d z ~ 2' dz ~ 2i and d z ~ ~ 2 i ' Since,

f = f(x ,y ) so that / = f(z, z ) .

ax Sz and

or

Sf 5?

$y’dz

2 dx

df dx | df dy _ i
dzd? dzdz

SsvcSjJ

d 2‘ *f dzdz

d^ I f4 { dx2 * d y 2

d2 t d2 dx2 dy2

4 \d *

2*

2^5x

1 a/ 2idy

iY a/ ( .a/~j 2^a* 3yJ

Y— s + 1— .eO

dy. {d x

d2 dzdz

dyt

...(2 )

[by (1) and (2)]

Proved.

Example 1.9 ; Ifffz) is a regularfunction o f z, prove that

<£_

a

la * 2 + d y ')

i/r*>i*=4i f w r .

[RGPV June 2004J

F u n c tio n s

Solution.

Recall that :

L.H.S. =

of

C om plex V ariables ) 13

- - 2 ~*Szdz' dx•2 + dyJ

' a2

a2

cbc2

dy*

I /(*>!*

= 4-^i-1 /(*) |J= 4-£_(/(*)/
Example 1.10 :

[v|2p=2z]

dzdz

dzdS

[As differentiate w.r.t. £ ]

oz

[v \ z \2= zz] Proved.

= 4f'(z)f'(z) = 4\fXz)\2 = R.H.S. if a regular function o f z» prove that

IL

JLY

3*, + V ;

Solution.

Recall that:

dx2 d2 i, dx

L.H.S.

dy2

= 4dzdz

dy J

logl/'(*)!

—i = 4 - ^ - I -Mog|/'<*) dzdz [2

[logtfW X *)}]

- 2 -^~=:[log/'(2) + log /'(*)] = 2 A

A *)

[ v | z | 2= :r * ]

= 0.

[Since f \ z ) and / #(z) treated as constants w.r.t. to z] ~ R.H.S. Proved. Example 1.11 : Show that a harm onic fu n c tio n sa tisfies th e form aI differential equation

Solution.

.(>*<

d2u =u. ozoz

Recall that:

‘ a2

dx2

dy2

dzdz

d2u d2u _ Since u is harmonic function => T T + T T “ • ox oy

...( I )

14 | Engs*«?wg H*thc>mjbc*4II s2 A 4 dzSs dx2+ dy2

[Using (1)]

&u = 0. dzdz Example 1.12 .*Ifffz) is an analytic function ofz, prove that

Proved.

|i H / w i * - 2 i / r * ; i *. Solution,

Let /(z ) = u+iv. Then R f { z ) ~ u => i/=s du2 _ 5u Now, differentiate, we get -r— = 2u— 6 cte dbc

av

Again differentiating, we get — j- = 2 dxr 2 ..S

Similarly,

= 2

(du*

3ya


(d u )3

0 *u

[ a j +lV

a2u + u3y2

...(2) f 32u

Adding (1) and (2), we get

*r2

By2

52u

[W

Since /(z) is analytic => u is harm onic =>

or

aV a2u 3 i ? - +^ =2

ox)

or

...(3)

\o y )

v 0u; aw an . dt> Also, Since / ( * ) * “aT aT+ <32? * TC Kt * G X l OX

d2u ^ d2u Q a *2 + dy2 ”

'

x,

dll t dx dx

irw f - ( £ ) ’ * 0

Hence, from (3) and (4), we get

aV

aV

dx2 + '*~> or

= 2| /'(z) I

s2 "N ! U/Tz) |2= 2 1/'(*) I2 {dx2 * f t 2)

a2

Proved.

F unctions o f C a a n A x V ariables | 15

t»13: Prove that the function ffz)= x y + iy is everywhere continuous but not analytic. Given: Hence

f(z) = xy+iy => u + iv = xy+iy u = xy and v = y. du dv . du dv . Differentiate: ^ - y - ^ - 1 . 0. du dv , du dv cleafly- * _ a n d - * - => Cauchy-Riemann equations are not satisfied. => ffz) is not analytic. Evidently, u and v both are polynomials in x and >>and so they are continuous everywhere. Aas. 1.14; Determine whether ~ is analytic or n o t? X

Let

[RGPV June 2003]

f(z ) = u +iv => u + iu = — x

U +W ~ lT ily = x 2+ y 2 Equating real and imaginary parts, we get x x + y*

U = — ------f>

du dx

U=

-y 2 x l +y 2

(x2 + y 2) .l- jc . 2x (x2+ y 2)2

y 2- x 2 (x2+ y 2)2 ^

dv _ 2xy dv dx (x2 + y 2)2 8,1(1 dy


[Rationalize]

du _ - 2xy dy (x 2 + y2)2

y 2- x 2 {x2+ y 2)2

du dv du _ dv Clearly' a T * Thus C-R equations are satisfied. Also partial derivatives are continuous except at (0,0). Therefore, 1/z is analytic everywhere except at z = 0. Proved.

M3 METHODS FOR CONSTRUCTING Al* ANALYTIC FUNCTION M ethod!; If u is given function then to find : ,

dv j

d v * ,''

dx

dy

dv » — d x + — dy du .. mH ?

du , ~di y

= M d x +N d y

[By C-R equations] ...(I)

16 I EhGMBBUNGJtaMBAIttKS-111 where

M_ ™

du oy

du and ^ ~ ~ r ~ dx

dM d2u dN d2u Differentiate : - r “ = ” T T and — = zTT oy dy* dx dx2 U U, U a7u dhtU A Since u is harmonic r^> r y + y 580 d x 2 + dy‘

dM _ dN =>

[By C-R equalions]

dy ~ dx

Hence equation (1) is exact differential equation. \ dv can be integrated to get v. Method t i : MILNE'S THOMSON METHOD: Type (i) : To construct analytic function J{z) in terms of z, when real part is given by the following formula: i

/(* )» | iM z’ 0 )-i^ (2 , 0))dz+C

...(1)

. , . du i / x du $ (*»?) - — and ri (*»y) - ~r~ dx oy Type (ii) : To construct analytic function f(z) in terms of z, when imaginary part v is given by the following formula,

where

where

$ (*»y) “

and (*>y) ~ ~ oy dx Type (iii) : To construct analytic function f(z) in terms of z, where u - v is given : Let U ~ u - v then (1 + i)f(z)= J [&(z, where

.,

,

dXJ

0)]dz+ C t

.

...(3)

v dU and f i (*»?)55

Type (iv) : To construct analytic fu n c tio n ^ in terms
j r . dV ., . dV fS (x,y) - — and (*.30 = — . oy ox

...(4)

t

F unctions b n n o n n c * s ¥ a — im g s | 1 7

1.15 .* Use Cauchy-Riemann equation to fin d v, where u = flo?y-y?. [RGPV Dec. 2001] Here u = 3x*y - y* Differentiating (1) partially w.r.t x and ^ respectively. We have

—= = vx

...(2)

4 (x >y)>say

^ = l x 2 - 3 y 2 =t2(x,y), say

Also

...(3)

Cfu &U a / n \ ==*>y+v-oy; = u=> u is a harmonic function.

Now putting x » x, y * jCUft (2) and (3), we get (z, 0 ) = 0 , and# 1(ar, 0 ) = 3** i

Hence by Milne-Thomson method, we have /( * ) = J [ ^ ( z ,0 ) - i ^ 2(zt0)]dz+c or or

[.*. Real part u is given]

f(z)= J -i3 z 2dz+c= - 12!*+c u+ iv = -4(x+iy)3+ c i - ~*(x8 + 3ixuy - 3x0? - 1i f ) +ci

[v c=ic]

or u+ iv = ( 3 x * y - f ) + # 3 x ^ -# * +c) Equating real and imaginary parts, we have u=3x2y - y 3, u=3j^2 - x 3+ c .

Ans.

Example 1.16 .*Show that u = ~ lo g (x* +y*) is harmonic and Jtod Us harmonic conjugate. Soiuthn.

Here du dx and Clearly

d*u d x2

, du and — x 2 + y2 ~ dy

x 2+ y 2

y* - x 1 } d%u _ x* -y * (x2 + y2)2 an dy2 (r* + y a)* S?u

n

= u ^ « is harmonic.

Let the hgqnonic conjugate of u be v. Then , dv , dv , du j dv = — dx +— ay = dx dy dy

du

[By C - R equations]

1 8 1 Enqb« b M M 4 M ! maiio»4II

or

dv = —

Where

Af = — **+/ aw dy

Clearly

x*+y*

/-* * (x* + / ) *

and

dN dx

y * -x * (x* + y2)*

dM _ diV _ ~ljy " I d x ^ * ®Quat*on (1) is an exact

Its solution is

J - 1jao*at ? + ? <3^X+{krwmof N uhkhan fm fivmx)

+C

= -y ~ t a n ''( ^ ) + 0 + or

i; = -ta n ~ ,(^)+c.

Ans.

Example \M : Show that u - 2 x - x l +3xy* fr harmonic and determine its conjugate. - [XGTV June 2002J Solution.

Given

u -2 x~ x* |^ = 2-3**+3/ and dr dx ~ = 6 ^y

Then

d*u d*u d ? + dy*~

and

^

- 6x

^ ~ = 6x

t t l s ™ur,nonic*

Let the harmonic conjugate of w be v, then we know that , dv * du , dv = - - d x + —-d y dx dy du . du , =~— ax+—-a y dy dx or

d v ~ - 6 x y cb+(2-3x*+3;y2)dSy

or

du = Mdx+ AWy,(say) dM

dN

$

[v B y C - R equation] ...d )

R w c m n tg t C o w w y m m » 1 19 Clearly

dy

* — => equation (1) is an exact. dx

/. Its solution v -

| M dx+ J (terms of N which are free from x)dy + c * * •-< jreouM

u = - 6 y j xdx+J (2+3,y:i)cj(y + c x* 0

_

or

3y3

v=s"6^,Y +2;y+ 3 +c

or i/ = - 3 b c * ,y + ^ y + d Ans. f 1.11 ^x) ; Prove that u • x2 - y 2 - 2ap - 2x + 3y is harmonic. Find a function v such that ffz) = u + tv is analytic. Also expressf(z) in terms o f & fRGPV June 2005J

»

Given : u = x 2- y 2-2ay-2x+ 3y

^DUaferetrtiate: G N*V-

^ = ^ 2^ _2x+ 3

i ? =2’ ^

-d )

-' 2

52u d2u —r- = 0 => w is harmonic. dx2 dy Further, let v be harmonic conjugate of u, then we know that , dv , dv , d v ~ — dx +— ay dx dy Adding,

du ,

Here,

[v By C-R equations]

= -{ - 2y-2 x+3)d x+(2x-2y- 2)dy - Mdx + Ndy, (sty) Af = 2 y + 2 x -3 and N - 2 x - 2 y - 2 dM dy

So clearly,

du ,

dy

and =^ dx

dN

[Using (1)] ...(2)

.

I T 2 equation (2) is an exact

Its solution,

- J (2y+2x-3)dx + J ( - 2y-2)d[y+c = 2xy + x? -3x-*at* - 2y+ c •

Ans.

201

Finally, by Milne-Thomson method, analytic function /(*) “

J

0 ) - i^ (z , 0 )]dz + c

...(3)

A

where,

M x >?) = -r- = 2 x - 2 y - 2 => ^ (z , 0 ) = 2z ~ 2 ox

and

M x>

=

- "

2y-2x+S=> ^(z, 0) = - 2z + 3.

Hence (3) becomes: f(z) ~ J [2 z - 2 - i ( - 2z + 3)]2[2,0]]flfe+iC Here

... ( 1)

du dv t (x, y) = ^ and ^ (*, y) *

We have ♦i (*..>') = | j “ = 3X2 - 3y2 Replacing x -> r and >>-> 0, we get *, (z, 0) = 322. Again, we have ••

dv

# *•' i

t

FuNC|W .^ .q 9 W ^ J f |}S?|W5s, | 21 du

=

du

dv

[■:■ by C-R equation ^ = ~ ^ ]

” - [0 - 6xy] = 6xy Replacing x -* z and y 0, we get ♦i <*, 0) « 0. Hence (1) becomes / ( z ) * J(3z 2 +i.0)dz + iC = z* + #C.

Ans.

jip/e 1.19 .* Finrf /A* im aginary p a rt o f the analytic fu n c tio n whose real p a rt is x? - 3 x f +3x* - 3 y f . tion.

Given :

fRGPV Dec. 2005, 2008(N) and June 2011]

u - x * - 3 xy2 + 3x2 -3 y 2

=>

— = 3x2 - 3 y 2 +6x and ~ = ~ 6 x y -S y . dx dy Let v be the imaginary part of the function u. t

,

dv '^c

dv , 'dy

du , 'dy

du . 'dx

[By C-R equations]

= - ( - 6xy - 6 y ) d x + (3x* - 3y* + 6 x)dy « ( 6j ^ + 6y)
is an exact differential equation because

...(1)

dM _ d N ^T

.*. Its'solution : l>=

J (6ry + 6y)<2x + J (terms in N which do not contain x)dy + c y const.

( X2 > 3y 3 -£ - +c - 6 y J (x + l)d* + J - 3 y 2dy+c = 6y ---- + X <2 J 3 = 3x2y + 6jor~
Ans* •J

|

1.20 .* I f u ~ x ? - y * b o t h lVi

u + tv is not an analytic function o f z> Or

[•<•< :

Prove that

y

u and v sa tisfy L aplace's equation but

-y * and v = ^ +y t ^

torm onkfunctions o fx a n d y but are not

harmonic conjugates.

\

22

Solution,

Let

u - x 2- y 1

du _ d2u * du ~ d*u n Differentiate: ^ - = 2 * ^ — = 2 m i — ‘ - Z y => ^ = - 2 d 2u

3 2w

Sbr

3>r

,,

T T + T T = z - z = u => ttjs harmonic /.e.„ u(x, y) satisfies Laplace's equation. Again let,

* * x 2 +y*

dv -2xy Differentiate: ^ ( ^ 7 7 7

=>

5u

x 2- y 2

d2v

d2v

6x*y-2y3

- 6 x 2y+2y*

8 y ~ ( x 2 + y2)* ^ dy2= (*’ V ) 3 a Ju

a 2u

j + T 7 = ® =* vis harmonic

i ? +^

i.e., v satisfies Laplace's equation. du

dv

dv

du

But* Imd Therefore « and v are not harmonic conjugates. Henoe ar+iv is not an analytic function. Proved. Example \2 \ : Show that the function e*(coa y + i t i n y ) is holomorphic (Le. analytic) andfin d its derivative. fltiGPV, June 2008] Solution.

Let,

f(z) = ex c o sy + ie x sia y => u + it; * e* c o sy + iex sin y

Then, u « ex cosy, v ~ e x sin y Differentiate: du dv , . du x . dv x — = excoa.y, —- = e erny and *£r~~e “ “ y* t ~ = e cosy dx dx oy oy du _ d v du _ dv Clearly, and * = " * • Hence, C - R equations are satisfied. Therefore f(z) is analytic. Now, we know th at, / #<*) * ~ + ^ * « * c o s y + i V 8in y Sbe dx * eJf(cosy + iBiny) * e*.e* = ex+^ = e*.

Ans.

Example 1.22 .*Sfcw tta* the function f( z ) =z\ *\ is not analytic anywhere. Solution.

We have,

f(z ) * (x +ty) \x + iy \

[v z = x + iy]

F u n c tio n s o r Co»fe*£X V&sablEs | 2 3

f i s t i n g real and imaginary parts, then ;

■'

u = xyjx2 + y2, and v * y ^ x 2 + y 2. du

r~~i +J x ‘ +y‘ ’ a n d ^ "

x2

Differentiate:

+

du _

.*y

.

a;

xy

y2

and

dx =>

dy

J x 2+y

yjx +y2

x -y .

du dv Which is satisfied only for x * y and ^ * ~"^T* Bk

f;

Hence C - R equations are not satisfied. /fa) is not analytic everywhere. Proved. : Shaw that the Junction v m sin kx cosy is harmonic andfin d its harmonic conjugate.

jfc

Gives:

^

dv , tfu . , Differentiate: — = coehx cosy => ^ - j = sinhx cosy

...(2)

and

...(3)

v = ainhx.cosy

...(1)

~ = s-sin h x sin y =>

- sinhx cosy

d2v & v dx2 + dy: . du . du , dv . We know that ®u ~ — a x + — ay -

dv .

[v B y C -R

or

du = (-sin h x sin y)dif-(coshx cosy)dfy = Mdx + Ndy, say

Here,

M = - (sinhx siny) and N - - (coshx cosy) dM . . ... r—- = -s in h x cosy ^ **»

So clearly

dM _ d N

dN cfe

- sinhx cosy

=> equation (4) is an exact.

...(4)

M

| Cu n m—

i- w

J U

.*. Its solution: u=

J M dx + J (terms of 2V which are free from x)dx +c jC&Wt.

= ~ s in y j s in h x d r + J Odx+c

= -ainy coshx+c.

Ans.

Example 1.24 ; Show that ex(x c o a y - y sin y ) is harmonic function. Find the anaiytic'function

Solution.

fo r which ex(x cosy - y sin y) is imaginary part [RGPV June, 2004] Or Show* that ike function e* (xcosy —y sin y) is harm&nlc and fin d its conjugate. [RGPV, Feb. 2010! Given imaginary part, v ^ e x(x o o s y - y siny). . ■•(1) Differentiate: “

- ax(xcosy - y sin y )+ e x(cosy)

...(2)

i

and

— = e*(- xsiny-einy-ycosy)

(3>

Again differentiate (2) and (3), we get ^ 4 r * e*(zcosy - y sin y )+ e* cosy+ ex cosy dx2 = e*(jccos
...(4)

a- ex( - x cosy - cosy - cosy + y siny)

- ex ( - x cosy - 2 cosy + y siny) Adding equations (4) and (5), we get

...(5)

d2v V V &v UV A T T +TT =u v is harmonic function. dx2 dy2 Now, MilneThomson. method, If vis given: A * ) 5* /

W ( * ,0 ) + i ^ ( a r ,0 ) } d e + c

where

M x »y) “ t - “ ex(* o o s y -y s in y )+ e JC(cosy)1 ox

Then

^(* ,0 ) = e*(z - 0)+«*(!) * ex(z +1).

and

dv M**y) m-r r = e * (-x sin y -sin y - ycosy), dy

...(6 )

! F w c H ft—r t» I t i w W f i H M i i i M

jfe

then

| a t

4(*,0) = e * (0 -0 -0 ) * 0 .

Hence (6) becomes: /(*) 35J *(* + Oe*dz+c = i£(* + l)e* - J e*darj+c = i [(2 + l)e* -e * ] + c - i z e 9 +c.

Ans.

0 5 : Determine the analyticfunction f(x) = u + iv , If v ^ lo g b f + y*)+ x-fy. Given imaginary part, v = log(*2 + jy2) + x -2 y . dv 1 s . Differentiate: ^ = ”i"” T ^ + 1

and

--(0

1 /n V „ ~T~— f7 T '(2 y ) -2 , qy

...(2)

r+ y

By M :foe Than son method, if v is given:

/ ( * ) = / W(^,0) + i^(z,0)]d2r + c . z

where

v

2y

...(3)

0

s ^

~

^

^ ( z»0) * -2.

and

J

Hence (3) becomes : /(* )*

d z+ c

= - 2* + i (2 log 2 + z) + c =±*21ogs£+(*-2)js+c.

Ans.

Example 1.26; T/' u = ----------------------, /bwf the corresponding analytic function.

coah2y+ coa2x

■* Solution.

fRGPV Dec, 2003/ Here, given real part of/fo) is , , sin2x u~c o s/i2 y + cos2jc Differentiate.

du _ (coe/i2y 4-co82 x)>2 co62 x - sin 2 x (-2 sin 2 s) (cosA2^+ooe2x)* 3

2cos2jcco8/i2y+2(co83 2jc+ sin 2 2x) ■

—

.

■ -

-

-

—

- - J - — — ----------------

(cosft2y+ cos2x)

26 I FrmngTnwini MinM~iiivnrfii IH 2 cos 2xcosft 2 y +2 _ x ( \ (coah2y +cos 2x )2 1 X*^ ’ ^

dtt_ (co8 ft2 y+coa 2 x)( 0 ) - 8in 2 x( 2 gin/t 2 y) c>y (cos^ 2y+cos 2 x)2

31x1

28in2xsin/i2y

~ 7I7Z ^~ ^77^2-r2\x >y)> say (cos^7^r~ 2 y+cos 2 x) Now putting x = z, y '» 0 in (!) and (2), we get

...(2)

2 cos 2 sr+2 2 -----------------------------2 -—=sec2,

(l+ co s 2 z)

l + cos2if 2 cos 2

and fa (z, 0)= 0 . Hence by Milne-Thomson method, if u is given : f(z)= J [<}>j (z, 0 )-«|>j (z,Q)]dz+ci= J sec 2 zdz+ic=t8inz+ic.

Ans.

y Example 1.27: In a two-dim ensionalfluid flow , the stream function is V = ~^ j ~ y » finding the velocity potential f . Solution.

Given:

y/ = — ——x 2 + y2

...( 1)

_ Differentiate:

^

2 rv

We know that,

cty = — d x + ~ d [y = — dx - — d|y dr §y dy dx

cb

(x + y )

y2- x 2 "aT=7 1 ---- ITT & <*+ y)

by C - R equations— d* Of

< b - J ^ r f j J dy= M dx+Ndy, say

Clearly

6M _8N

Its solution

=> equation (2) is an exact

/

O'8-**)

^

+ j taj2 °^e+| Oc^y+c

$y

an d — = - — dy cbc ...(2)

| 27

,.28 s i f w ~ f + i\k represents the com plex potential fo r an electric fie ld and T* x*

fin d the function o f +

Here, imaginaiy part of w, Differentiate: ^

§y

,x},k:

ag' 6bc

“ >* +^ T ~ T

« - 2 y - ;-

..0 )

frr *rt <*»*). say

(x2+y2)2

(s 2 + y * )fl)-* ( 2 s) (x 2 + y 2)2

Now putting x - z , y = 0 we get A (*»0)«0, and 4%(z,0)-2z ~ Z~ 1

Hence by Milne - Thomson m ethodjf yf is imaginary, then

/(*)= J

W
= J [0+ i(2 z— ~ )] d z + c = ij ^2z — yjc£z+c=^z2+ — + c „

(x+ iy)2+ —- ^ - |+ c = i x 2 - y 2 + 2i ^ + - ^ y \+c L x + iy ] I x +y J

Equating real parts, we get # = - 2*y+>

v

Ans. +c‘

^ . I 29 ; M ow th a t th e fu n c tio n u = e~**8in (x* - y * ) is harm onic. F ind the ■ ~ conjugate function v and express u + iv as an analytic function o f z, [RGFV Dec. 2006 and June 2007] to u ,v

Here, u = e “2ao'sin (x*-y*)

; Differentiate: ~ * - 2y e '1* ain(x 2 - y 2)+2xe ~2* coe(x 2 - y2) ox -~-r = 4 ya e~** sin(x* - y2) - 4 xye~** cos(x* - y i )+2e~**f cos(x 2 - y 2) oar -4jg'e~2v co 8 (x * -y * )-4 x , e~**ain(x2 - y 2)

...(1)

= - 2 x « '2j5f sin(x* - y2) - 2ye~i*y cos(x* - y 2). 3y

\

and ~ rr = 4*2e '2* sin(x8 - f y + i x y e * * cos(x* - y*)-2€ 2^ cos(x2 - ;y2) dy

+ Axy e 239 cos(x 2 - y 2)- 4 y 2e~2*ysin (x 2 - y 2)

•■•(2).

Adding (I) and (2), we get

NOW,

$ (2 ,0 ) = 22C082*

V $ (x,y) * ~ a n d ^ ( x ,y ) = ~

(2,0) * -2zsinz*

and

By Milne’s Thomson method, if u is real part, then /(* )* J l#t(*»0 ) - i # t ( ^ 0)]
Put iz2 - t => 2 z d z -~ i

or

/(z) = i f e?d t + c ^ - i e * +c i J

==> u +iv = -ie * 2 +c =:-*i€l<*+t3,^ +c -

[vz=x + iy]

_ ieitf-y2^ ) + c * -ie ~ 2ao'.e <(*1~y2> +c

_ _ ^ “2^ [Co8(x2 - y 2)+ism(j?2 - ^ 2)]+c =e~2*ysin(x 2 -,y 2)+£[~ e*** cos (x 2 -.y 2)]+ a+ id v ~ -e ixy cos(x 2 - y*)+b.

fvc = a + ib] Ans.

Example 1.30: I f u - v = ( x - y ) (of + 4 x y + y* ) andf(z) • u + iv isMin analytic function o f z = x + ty, fhutffz) is terms e f t. ffUSPV June 2006] Solution.

Let U = u - v = (x - y )(x 2 + 4xy+y*). By Milne - Thomson method, if U * u - v is given, then

♦♦.(!)

| 29

^ (* ,y > » ^ -» ( ^ + 4 a g i+ /) + ( * - y ) ( 2 * + 4 y )

[by (I)]

=>

^ ( z yO)-z +2z =3z

[':x-*zandy-*Q\

and

*2 (*>?)=— = - ( * 2 + 4ji?y+ jy2)+(x - y)(4x + 2y)

dU

y s => 4 ( s ,0 ) = - z 2 +4z 2 = 3z2. Hence (2) becomes:

[vx-»zatt0]

(1 + i)f(z)= j [^iZfOy-ifaiz, 0)]dz+c= J (3*2 -
' ■_r''

‘nifi ;

, f(z) = -isr +C,, where c*

or

c

, 1- i . and ^ - r = -».

Ans. e ^ -c o s jc + s m x

& an analyticfunction o f z - x + iy, and **-v -

is 1 3 1 : / / W “ « +

flndf(z) subject to the condition

“ *

coah y - c o a x '*

# ¥ •

y7 e ^ -c o sx + sin x ^ “ ~ U' coshy-coax ’ By Milne’s - Thomson method, if i/ ~ - v is given, then

" (l)

(l + i)/(* )= J

...(2)

, /

where ^

0,0)] dz+c

^

SU

x

.,

.3 1 /

Now,

,

/

v

517

(sinx+© 08x)(oosh)y -c o 8 x )-(e y -cosx+ sinxX sinx) ------------------------ (coBhyleos*? ------------------------ '

Put x -frz a n d y ^ O , then (sinz+cos z )(l- c o s z ) -(l- c o s 2r+sinz) 8in 2r (1-COS2T)2

s in z - s in z cosz + co sz-co s 2 z -s in z + s in z c o sz -sin 2z (1 - cos *)2 coez-(cos2m2m+ sin 2 z)

22 ■ ...............

(1-COBZ)2

c ofsz -l -1 — =?'■» ■■— ■— (l-c o s z )2 1-C 0S2

9 0 1riwn—iMM iUnaiiimrniiil and

^ / „ _\ dU ey(ooBhy-Q98x)-(ey -cosx+ainaOsinhy. ti v^y/*”^*”0* 11 ‘/ v \2 • 1 1 - ’ dy (coshy-cosx) Put *->2randy->0, then (1-cos*)

1

(I-cosa?)

l-c o s*

Hence (2) becomes:

(1 + i)/w ' /

[ o

i - ' a ^ j ] ^ ^

=-(1+i)J a ^ ) +c""Pr)/co*

/(* )= c o tf|J+ < i> where

...(3)

But given ^ f y ] =~ 2^ ’ **len 0 ) becomes at * = -j.

3 -i .ff 3- i t — - e o t ^ c , ^ — -l+ e ,

3~ i - 1 - i = > c ,= — - 1 = — .

Hence (3) becomes: /(* )* co tf—j+ —j - .

Ans.

Example 132 : iff(z) m u + tv is an analytic function o f z, findf(z) (f u - v - e * (c o sy -sln y ). Solution.

Let £/=u-i>=ex(cosy~ 8iny) By Milne's - Thomson method, \ f U ~ u - v is given, then

—(1)

< i+ i)/(*)= J U M y i M z M d z + c ,

...(2)

where

****

Now,

#i <*,*)»—“ ««* (coey-siny). ox

Put * -> z and y-»0, then $(z,0) *e*(l-0)= e*. and

dU

h (x >y) =— * e * (-« n y -c o e y )

‘

FunctjJMsW C6WH& VW Sfitts I 31 and ;y-»0, then ^ ( 2,0) = e* (0 - 1) = -e*.

Put

Hence (2) becomes : (l+i)f(z)= J [V -i(-e*j^dz+c k

or

(1 + i) f{z) = (l + t)e*+c

f(z) = e* +-^-r - e +cr

-

Ans.

1+t

^jfiit 133 .*Ifffz) m u + iv is an analytic function o f z, flndffz) i f u+ v =

2a&n2x (e* + e ^ -2cos2x)

T/

2 sin 2 jif

Letimaginaiypart, v - “ + t' - (ej , + e -2, _ 2cog2x)

or

V -, .

8in2*

e2' + e"3'

'

-C082JC

...(1)

cosh 2y - c o s 2x

V 2 By Milne - Thomson method, \f V = u + v is given, then (1 + i)f(z) = J [^(z.O) + ifc(*, 0)Jdz +c 5V

...(2 )

5V

where

A (x >y) ~

Now,

,, v dV -2 8 w 2 x siiih 2y M x .y ) - a y - (c08h 2j,_eos2x )2 => * ( 2' 0>= °..

.

dV =—

., „

and

[v 8111110 = °1

2 co82 x(co8h 2y - c o s 2x )“ 28 in 2 2* = ----------- --

(cosh 2y -c o s 2x)

-------------

2 cos 22(1 - cos 2 ^ )-2 si» 8 22

A ( 2 '0 ) = ------------ ( I - c m W

--------“

_ 2 oo82z - 2 ^ in >2g+coe* 2 s) (1-cos22)* 2 (cos 2 2 - 1) (l-c o s 2 z )2

-2 (l-*cos2z)

1 1 -2 2s in 22

00860 **

Hence (2) becomes r 0 + 0 /(*) = J [0+ i(-cosec2z)Jck + c = i cotz + c or

i c t /(*) = -— rCotz+-— 7 * -( l-£ )c o tz + c l. 1+ t

l+ i

2

Ans.

m

Example 134 / Let f ( z ) ~ u ( r , 0) + iu (r, $) be am analytic function, whete u ~ -r fism 38. Con­ struct the corresponding analytic function ') in terms o f z. Solution.

Here

u = - r 8 sin 30 => — = - 3r* sin 30, and ~ = -3 r s cos 3# dr 58 By total differentiation formula:

dv = ~ d r +~ d e = ( “

+ ( r f : ) cW

[By C-R equations]

or

di> = 3r2cos30 d r - 3 r 8sin 30 dS

or

d v - (3r2dr)cos30 + r 3{-sin30.(3d0)} = d ( r 3cos 30)

On integrating, v = r 3 cos30+c. f(z ) = u + it; = - r 3 sin 3 0 + ir3cos 3 0 +ic = ir3(cos30+ i sin 3 0 )+ ic ^ ir8^8** + ic = iz 3 + ic. [v z = re**] Ans. Exam ple 1.35 : i f n is real, M«f u>= r* (c o s n B + ts in n O ) is analytic except possibly when r —0. Solution. Given: w = r*[cosn0 + isin n0]« u + iv =>

u(r,^) = r* cosnl, and v(r,0) = r* sinn0.

du «_i ^ dv . j . Differentiate: — = n rR ‘costttf;-— = nr* sin n 0 ,
tt

. . Ctariy

du 1 „ _iau ¥ = 7 nr «osn6 = ~ .

a#

- " n r^ s in n ^ ;-rr* w**cosn^. a* . 1 » • /i 1 du and F = ; nr « . n f i . — .

Thus Cauchy-Riemann equations (polar form) are satisfied. Hence w = rn (cos 0+ i sin nQ is analytic except r = 0.

Proved.

1.14 BILINEAR TRANSFORMATION 02 + 6 The transformation : w = f(z) ------- —, with ad - 6c # 0 ...( 1) cz+ a is called bilinear transformation or M obiuS transformation, where a, b, c, d are complex constants. Remarks : (i) The transformation (1) is said to be normalized if ad -b e = I. (ii) In bilinear transformation (1), the (ad - be) is called determination of the transformation.

aar + 6 -dw +b (iii) If w = ----- then w (cz+ d) = az+b =0- z ~ -------------- * / cz + a cw —a verse transformation.

i

v . say is known as in-

F u n c tio n s

u~-r^sh%SSL Cor

op

C om kex V a r ia b les | 3 3

rfiu a d - b e ]p o in t: Differentiating (1) with respect to 2, we get “ j“ * ^ + ^ 2 ■

U : . “Sr* cos 3#

d . dw du) . — >then ~r~ = 00 and if z * <», then t - = 0. c dz dz ■?=?— and z~ are the critical points. C u or Fixedpoints: The invariant points or fixed points of a transformation (1) potting w - z .

[By C-R equations]

00830)

: M>= — ,..(l) is a transformation, fixed points, put w * z in (1), we get 'j* —^ £

V a -1 => z * ± i

Fixed points are z « -* and /.

rotation, magnification and inversion are special type o fbilinear transformations. -Ratio : If there are four points z ]tz 2iz 3 and z 4 taken in order, then the ratio Wr J—,

” *^4) -

II J

,

-

cfsss-ratios are invariant under a bilinear transformation; i.e. ■WK'W.j:’

- u ; 4)

(g, - * 2)(2 3 - Z 4)

(w2 ~w3)(w4 ^tu,) (*2 - z 3)(z4 - z x) We may use (1) to find a particular bilinear transformation which transforms one given tilfcle into another given circle or straight line. Thus to obtain the required transformation, We fake any three distinct points z i,z 2,z3 on the first given circle and any three distinct inti wu w2,w3 on the second given circle or straight line and substitute them in (1). .*Find the bilinear transformation which transform the pbitHs z m 2,1, 0 into w « /, A I respectively. [RGPV 2001/ Ue< required bilinear transformation be :

*rd0' Proved. 1*

(wL~w 2)(w3 - w 4)h (zl - z 2)(zi - z 4) ^ t o f 2^ w 3)(w4 -m ,) (z2 - z 3)(z4 -* ,)

...(1)

*Nrting m/j = 1, tt?2 = 0, w3 - i , w 4 - iv, z, = 2,z2 = 1, z 3 - 0, z 4 =z, we get I#*™

c,

are ( 0 - i ) ( w - \ ) ~ () -< h ( z -2 ) ' => => =>

**y is known as in-

w -i i ( w - 1)

~z z-2

z u > -iz-2 w + 2i = -ziw +zi to)z~$w + w z**i*-2i+ zi w (iz - 2 + z) = 2 iz -2 i w=

2t ( * - l ) , which is a bilinear transformation. ( l+ 0 * ^ 2

Ans.

3 4 1Fii—nmfliM MwrakAnca-lB Example 137 ; Find the bilinear transformation which maps the points z mO , - i , - l into w ^ i, 1,0. fRGPV Dec. 20071 Solution. Let the required bilinear transformation be (2t - 22)(g3 ~ZA) (Z2 ~Z3)(Z< - 2,)

(wx - w 2)(w3 - w 4) (W2 - W3)(W4 -W ,)

Putting, zl = 0, zx = - i, zs » - 1, z4 = 2, and

- i, u;, = 1, u>8 * 0, w4 * u>, we ge&

(i - 1)(0 - w) _ (0 + i ) ( - l - 2 ) ( l- Q w ^ - t ( 2 + l) (1 - 0 )(u>- i) (- i + l)(z —0) ^ ( w - i ) 2(1 - i) =>

(I - i)2wz - (-iz - i)(w - i) (1 - 1 - 2i)i^z = - izw + i2z - i w + i2

=>

-2 iw z a - m u - 2 - iu > - l => izw + i w - 2 i w z - - 2 - 1

iu ;(l-2 ) = - 2 - 1

16) =

-(2 + 1) > -i(2 -l>

[+ -•]

- i(z + 1) w =■ , which is the required bilinear transformation. 2-1

Afts.

Example 1 3 8 : Find the bilinear transformation which maps the points z m- 1, i , 1 into, w m- 2 , i , 2. Also fin d the fix e d points. fRGPV June 2003 and 2005] Solution. Let the required bilinear transformation be

(^1 - ^ 2) ^

_ (*t ~*2>(*3 ~ Z4>

(w2 - w 3)(w4 - w t) Putting,

...(1)

(z2 - 23X 24 - 2,)

2 , = -1, z 2 » j, 23 - 1 , 24 = 2 , and

we get (-2 - i)( 2 - w) _ (-1 - Q(1 - 2 ) (i - 2)(u> + 2) (i - 1)(2+1)

i -1

i * - 1*

jM Y E z l)J i£ Y * z L ) U -iA ^ + 2 j U -U U +U -2

-{ 3 + 4 i)(u ,-2 )= _ .(2 -1 ) (2+1) 5 (u>+2) w“

2-i

*u;-2 u>+ 2

2 -i

4+1

5

J

(2-1) 5i (2+1) (3 + 4i*)

- 62 + 2i ■, which is the required bilinear transformation.

Fixerf points : Put w = 2 in (2), we get —6z+2i . 7 . * 2 = —:------- => 12 - 3 2 - 2i + 62 = 0 => i22 + 3 2 -2 i = 0 12-3

-.(2)

r

F u n c tio n s o p C o w iifcx V a B a ia a * jf 5 5

t

' * =>

z=

-3 ± ^ /9 ^ H 2i

-3 ± 1 2i

-2 -1 z = -T- and 9 ~~J~ i-e >z ~ 2<# and / are fixed points.

Ans.

139 ; Find the bilinear transformation which maps the points » = <*>, z= i, z - 0 on to the points w^O, ip * 4 a n d ic=oa Let tiie required bilinear transformation b e : ( t ^ - w 2)(w3 ~ w 4) (w2 - w i )(w4 -it?,) Putting,

(RGPV June 2004 am i June 2007]

(zl - z 2)(z3- z A) (z2 - * 3) ^ 4 - z {)

zl =<*>, z t = if zs = 0, z4 = z, and u \ - 0 , k>* « i, (0-t)(qo-i6>) (i-oo)(u>-0)

» 00, w;4 = u;, we get (oo-i)(0-;g) (i-O )^ -o o )

ao [v Remove the factors, which containing » i.e., in —form] 00

-

(H )-(B ) -£ w

=>

-z -2 => - i = - z w i

n> = - —.which is the required bilinear transformation.

Ans.

1.40; Find the bitintar tnwsformatioft which maps z - 1 , 1,-1 re^tectivefy o n to w ~ i,0 ,~ L [RGPV Dec. 2005, Feb. 2010 and Dec. 2011] Also fin d : (i) the image o f |z| <1, (U) the invatfant points o f this transformation. Let the required bilinear transformation b e : (Wj ~W2)(W3 ~W4) (Zy ~ z 2)(zy - z A) (fv2 - w 3)(w4 -u>,) ~ (z2 ~ z 3)(z4 - z x) Putting,

z } - 1, z 2 = i, z 3 = -1, \z4 - z, and wx ~i, wt = 0 , wn = - it w4 = wt we get (i-O K -i-w ) _ (l~ i)(-l-z) (0+ i)(w -i) ( i+ l) ( z - l)

^

-(w + i) ^ ( t - 1) (z + 1) (w ~ i) (»+1) (z - i )

36 | SMQMKMWQItoB M IIG lJI ; . ( t - D _ ( t - i ) > -& i + 1 i* - i 2 -2

~(w + i) i(z +1) (w -i) (z-i)

.1 J

~ (w + i)(z -1) = (i z+0(w - 0 -(w z + i $ - w - i ) = iz w ~ i2z + iw ~ i2 - w z - i z + w + i = wtv+ 2 + iu; 1 1. -iu2-i2u>+u;-iu;*=*+l+i3”ri - w z + w - i z w - i w = z( 1+ 0+ (l - 0 u ? ( - 2 + l- « - 0 = ^ ( i + l) + ( l- 0 => u>=

or -2

+

2 (“ i -!)+ (!-» )

1- i ( 1 - 0 2 ~2i ‘ 1+ i ” 1 - i 2 2

8)

Z -l

lt>= -z-i or

1 -2 w - - — , which is the required bilinear transformation. i+z

...(I)

(i) Image qf\z\< \: From (1), solve for z, we get - w z ~ i w = z - i iw - i ~«wz - z ^ i w - i => z *• ~w - 1

i -iw MJ+1

...(2)

Since given image in Z - plane is jzjcl i(l - w ) <1 1+ w

[Using (2)]

|i(l-t^)|< |l + t^ => (l-u -w j-c jl+ u + u ^ 7(1 ~u)2 +u2 < ^(l + ii)2 +u2' *

[•i*MJ=u+iu and |if= l] jv |x + iy|=
l + w8 - 2u + v2 < l+ u a + 2u+va 0<4u

u>0.

This shows that the interior of the circle x 2 + y 2 =1 in the 2-plane is mapped onto the entire half of the w-plane to the right of the imaginary axis. Ans.

!

J 37

Functions g*

(ii) Tofin d the invariant points : Put w = z in (1), we get z = -—- => 2 = — ^ i +z 1 -iz

z - i z 2~ l- i z = 0 => iz* + ( i - l ) z + l= 0

2*

2t

2

4

These are the required invariant points or fixed points. Ans. 2z+ 3 f t 1.41: Show that the transformation to - — — maps the circle i t +3? -* X = 0 onto the straight Une 4u+ 3= 0, and explain why the curve obtained is not a circle, (RGPV, June 2002, Dec. 2003, June 2005 and June 2011J Given

...(1)

z -4

Solve above for z, we get w z-4 w ^2 z+ 3

z

w -2

-

...(2)

Now the given equation of the circle x* +y* - Ax * 0 in 2-plane can be written as 2 2 —2(2 4-2 )=0

...(3)

j^einces - x + iy ,z = x ~ iy , x =—* Z and z z = x 2 + y 2j Substituting for z and z from equation (2) in equation (3), we get 4u>+3^ v + 3 ) ( 4w +3A V .~{4w+3 4 ^ + 3 ^ =0 2 ----------+■ w - 2 ) \\ w --22 J 4 ^ w - 2 w -2 j =>

\6ww + I2w+I21v + 9-2(4wuJ + 3w -% w -6 + 4ww+3w-%w-6) =0

=> 22(u> + «;)+33=*0 => 44u +33 =0 [vw=u + iuand w=u~iv] => 4u + 3=0, which is equation of straight line in w-plane i.e., not a circle. Ans. 1.42 .* Find the bilinear transformation which maps the points z m0, - / , I onto w=i, 4
(W] - w 2)(w3 ~ wa) (U>2 -W{)(WA-W X) Putting,

(*1 - ^ 2X^3 ~ z i) (z2 - 23X*4 - * |)

2 j = 0, 2 2 = -I, 23 = i, z4 as 2, and

u>, ss j, (i-0)(oo-u>) (0-oo)(w ;-t)

=0, u>3 = oo, w4 = u;, we get

(0 + l)(i“*2 ) (-1 - i ) ( 2 - 0 )

(co-u>)

*1

(O-oo)

J

s in c e ------- => -1

7— (w -i)

=>

- 2 ( i + l)

=> (z i + z ) i = w ( i - z ) - i ( i - z )

- z + zi~w i~ w z+ l+ iz => u;z - u;i = z +1

2 + 1 = -------z - I which is the required bilinear transformation. ii£/+1 Solve above (1) for z, we get z -■ w- 1 Now, since in z-pJane, |z |= l

...(2)

iw + 1 =1 w- 1 =>

[Using (2)J

I + iw H w - 1 1 II + i(u + iv) j=| u+ i v - 11

=>

[yu; =u+ty=>u7= u-ii;]

|l - u + i t t |« |u - l + i i ; |

[vl a + ib \~ ^a 2 +b2]

(1-u)2 + u 2= ( u - l) 2+u2 :=> =>

Example

1+ u2 - 2 y+w2 =»u2+1-2u+i;2 u - v - 0 or i.e.» straight line in vv-plane.

Hence the image of unit circle |z |= l in z~p!ane is a straight line making an angle */A to u - axis and passing through origin in w-plane. Aas. S-4x Show that the transformation w transform the circle \z =1 into a circle of 4z-2

radius unity In w-plane andfind the centre o f the circle. Solution*

Given:

w-

5 -4 z 4 2 -2

=> Solve above for z, then z =

2w +5 4o>+4

Since inz-plane, |z |= 1=> zz = 1 ^

f 2w + 5 Y 2ElT+ 1,410 + 4 yv +4J

=>

4ww + 25 + \0(w + ttF) = 16(ww +1 +

=>

4(u 2 + u2) + 25 + 20u = \6[u2 + u2 +1 + 2tt] 2

2

3

.

'' [Using (1)] + 57)

u +v + « - —= 0 , which is equation of circle in w-plane.

[v w = u + iv) ...(2)

F unctions o p C o * p t£ x V a w w u ® | 3 9

* Comparing this with equation of circle: u 2 +u 2 + 2£u'+ 2/v+ c = 0 =>

g = 1/2, / a 0 and c = - 3/4

Centre = ( - £ ,- / ) = (-1 /2 ,0 ), and Radius = V#2 + f 2 ~c = ^ “ + 0 + ~‘ = 1Hence (2) represents circle with its centre at (-1/2,0) and of radius unity in w-plane. Ans. a z+ b \ M : Find the condition that the transformation w - ------— transforms a straight Une o f cz+ d the z-plane into 0te unit circle o f the w-plane► Or Find the condition that the transformation w = gm *

make the circle |ic|= 1 cow -

to a straight line in the z-plane, We know that the equation of a straight line in the z-plane is z-z2

=1

-.(I)

The equation of the unit circle in w-plane is | w |= 1

...(2)

a z +b a z + (bJa) w>——,---- ----cz +d c z+ (d lc ) Now, the transformation (3) transforms the unit circle (2) into the curve The given transformation is w =

l-|u > i=

a z + (bta) c z +{d1c)

-

—s

z+(jbla) z +(dfc)

...(3)

c a [v zx = -bla, 2j =d/a]

1= Which is the required condition. Example 1.45 .*Find bilinear transformation whose fixed points are 1 and 2. Solution. Let the required bilinear transformation be a z +b 1 wcz+ d

[Using (1)] Ans.

...(1)

\

Since,

z - 1 and z - 2 are fixed points, then

( z - \ ) ( z - 2 ) = 0 => z 2 - } z +2 = 0 Comparing (2) and (3), we get d _d

...(3 )

—b

------ --3 and — - - 2 => b - -2c and d = a - 3 c c c

az-2c ' cz + a - 3 c In particular, taking a = 1, c - - 1, then bilinear transformation

Hence the transformation (1) becomes : w - - ------- r~-

w=

z +2 4- z

Ans.

1.15 CONFORMAL TRANSFORMATION Let Ci and Ci be two curves in the z-plane intersect at the point P and the corresponding curves C, , C2 in the w-plane intersect at point P' under transformation w - f ( z ). If the angle of intersection of the curves at point P is the same as the angle of intersection of the curves at point P ' , both in magnitude and sense, then the transformation is said to be conformeda t point P.

Definition. A transformation, which preserves angles both in magnitude and sense between every pair o f curves through a point, is said to be conformed at the point.

1.16 SOME STANDARD TRANSFORMATIONS (1) Translation : w ^ z + c , where c is a complex constant, i.e., c =a+ib Let

z - x + iy and U) —U + , then w —z + c becomes z u + iu = (x + iy) + (a +jb) - ( x + a) + i(y + 6)

=>

u —x+ a and v —y + b This transformation represents a simple translation of the axes. Accordingly, the transforma­ tion maps a curve in die z-plane into a curve in the w-plane of the same shape and size. (2) Rotation and Magnification : w = cz, where c is complex constant. Let x = rco s0 , y = rs in 0 , u = R cqb#, v - R s i n ^ then

z —X +iy - r e l0 and w = u + iv

F unctions o f C om kjex V ariables ( 4 1

let, c = p e ia, then the transformation w ~cz becomes : R 8 ^ * /7 e w.re , Re* « pre*0+a) => R = p r and ^ = ^ + a . i Thus the transformation maps a point, />(r,0) in the z-plane into a point p ' (pr,0+ a) in the w«, plane. Hence the transformation consists of magnification of the radius vector of P by p~\ c | and ^ + rotation through and angle a - amp (c ). 1

0 ) Inversion : w = — v' z

Let

z =re10 and w =Re‘*

then the transformation w =— becomes : Re* * - e * => R * — z r r g

1

Thus under the transformation w;

z

^--9 .

, a point P (r, 0) in 2-plane is mapped into the point

plane.

in w~]

Example 1.46: Under the transformation w=—t find the image o f \z-2i\= 2. [RGPVDec. 2006]

Solution. The given transformation is 1 w * - or z

1 u;

1 u - iv or x + i y - -----—= -^ — fu +v

u + iu

[v z - x +iy and w = u +iv J Equating real and imaginary parts, we get u -v x s s ~ — 2 and y = ~~2— 2 u +v ir+ ir

- 0)

The given curve is \ z - 2 i \ ~ 2 or |x + i(y-2)|=*2 ri': 1

or x * + 0 '- 2 ) 2 = 4 or x 2 + y 2 - 4 y - 0 which is a circle in the jr-plane with center (0, 2) and radius 2 . Substituting the values of x and y from (1) ki equation (2), we get (u2 +V2)* =>

ua +u*

(u2 + 1^)*

4t>

A

■♦•(2)

u*+v

1

7- 3----- — r * 0 = > - y — j - + —

4t»

A

( ir + ir ) a +i> u +u u +u l + 4t>= 0 , a straight line which is the required image of the given curve in w-plane. Ans.

Example 1.47 ; Find the image o f the infinite strip Solution.

4

Also show the regions graphically. The given transformation is : 1 1 w or * = — z w

X

under One transformation e c - ~ . z

[RGPV Dec. 2002, Dec. 2008(N)J

1 u + iv u2 +v2 Equating real and imaginary parts, we get or

u x = 2 2 U +V

[v z - x + iy and w - u +u>]

-v y “ W2 +IT2

1 -v 1 When J ' * - , then ---- r = T 4 u + ir 4

=> u 2 +i>2 +4v = 0 => i*2 +(t;+2 ) 2 =4, which is the equation of circle in w-plane whose center is (0 , - 2) and radius 2. 1 -v 1 When ^ then —— 7 = ” 2 ir+ ir 2 => u 2 +i>2 +2v = 0 => u 2 +(i>+l)2 =1, which is the equation of circle in w-plane whose center is (0 , - 1) and radius 1. Graphs: u

w-plane Hence the infinite strip i £ y $, i

is transformed into the region between the two circles

: u 2 + (u + 2)2 = 4 . center( 0, - 2) radius 2 and

u 2 + (y + 1)2 ■ 1, center (0, -1) radius 1.

Ans.

x~ i Example 1.48^ Show that unfter the transformation w - — the real axis in z-plane is mapped 4H*I into the circle J 1. What portion o f the z~p!ane corresponds to the interior o f the circle?

Functions op g b # & r V A a ta u s t 4 3

0M.

Given:

“' = 7* 7T7l

•••(«)

Since, given the real axis in z-plane (Le., y - 0) i.e., z - x transforms into (1), we get

x - i (x - i )2 x 1 - l - 2ix w * -------------------------- =------x + 1 x2 +1 x +1

x 2 - l .( - 2x) 5—r + l“ 2— Tx +1 x +1 1 => | W |**1.

IW

Hence the real axis in z-plane transforms into the circle I w |= 1. Proved. Further, put z * / in (1), we get w * 0. Hence the half z-plane above the real axis corre­ sponds to the interior of the circle I u; 1*1. Ans. 1

Vj&mmple 1.49 ; Show that under the transformation » ■ -» circle ^ ([ i...

into a straight line in w-plane.

Solution.

Given: w = — /.e., z = —

-6 x = 0 Is transformed

...(1)

w

Again, given in z-plane: x 2 + y2 - 6 x = 0 or

g - * f e ± I > ,0

or

zz - 3 ( z + z ) ~ 0

z * = x 2 + y 2andx = ^ y

=>

1 1 ,[1 tl _ — - 3 —+ r r = 0 W W lw w j

=>

1 - 3 (w + i ^ ) = 0

=>

1- 3(u+ iu + u - iv) = 0

=>

1 - 6 u = 0.

Which is equation of straight line in w-plane.

[from (1)]

[v u; = u - iu] Proved.

Example 1.50; Show that the transformation pj = x* maps the circle \ z - l \ = l into the cardioids

p = 2(1+ cou
Solution.

pet* in the w-plane.

Given: put

w -z2 x = rcos0, y ^ r s i n Q in z ~ x + iy = > z ~ r e w

and

u = pcos^, y » p s in ^ in w ~ u +iv = pe* .

...( 1)

Hence (1) becomes : pe* ■ rV * =>

p * r 2 and ^ =

•♦•(2)

441 Now the given equation of circle in the z-plane is | z - l | = l =5. Ix + O' - 1 1= 1 =>

|( x - l) + iy | = l => ( x - l ) 2 + j 2 = l

=>

x 2 + y 2 - 2x = 0 => r 2 - 2rcoe 0 *O

[v p u t x = rc o sl, y = rsinfl]

r = 2 cos# => r2 —4oos2 0 =>

r 2 = 2(1+cos20) => /? = 2(l+coa^).

[using (2)]

The circle | x - l | * l in z-plane transforms into the cardioids p - 2(l + cos^) in w-plane. Proved i-z Example 1.51 .•Show that w = j —~ maps the real axis o f the z-plane into the circle |ic |= l and the

ha(f-ptaney > 0 into the interior o f tke unit circle |ic |= l in the w-pfana. Solution.

Given:

w= - *——. l~Z ■ i +z

*

.....

...( 1)

i- z =1 i +z

Since

|u;|ss 1 =>

=>

\i-z\= \i+ z\

=>

|i - x - i ,y i = U + x+ ’iy|

[from ( 1)]

[v z - x + iy]

x 2 +(1 - y ) 2 = x 2 -n(l+y )2 =>

\ + y 2 - 2 y = l + y 2 +2y => 4^ = 0

,y = 0 .

This shows that transformation maps the real axis of z-plane into the unit circle |ia>|= 1. Proved. Further, since | m-'| < 1=^> =>

i-z <1 i +z

[from ( 1)]

|i- x - i,y |< li+ x + < y | => x 2 + ( l- y ) 2 < x 2 + ( l + j )2

=> x 2 + l+ y 2 - 2 y < x 2 + I+ y2 +2y => 4 ^ > 0 = ^ y > 0 . Hence the given transformation maps the halffllanej^ 0 of Die z-plane into Die interior of the unit circle Iw {= 1 in the w-plane. Proved. Example 1.52 : Discuss the conformal transformation w m Solution.

Given:

fRGPV Dec. 2005]

w =J z

Squaring both sides, we get w2 ■ z =>

(u+iv)2 = x + iy

[v w = u + iv and z - x +iy]

F u tC TQ W S Q ^^P IjE X ,V 4 ^A e jU E a |* S

=> u 2 - v 2 + 2 iu v = x + iy Equating real and imaginary parts, we get u 2 - v2 = x —(2) and 2u v-y Now following two cases arise : Case I. When x = constant say, (a and b). i.e.y p u t x = a a n d x ~ 6 (where a> b > 0 and b < a) in ( 1), we get corresponds to the hyperbolas u 2 - v 2 - a and u 2 - v 2 * b respectively in the w-plane.

#*i!

w = Jz

w - plane

It).

Case IL When y = constant, say (a and b ) ' i.e., Put^ = a and>> = b (where a, b > 0 , and b > a), (2), we get the area included between the rectangular hyperbolas 2uv = a and 2uv ~ b. v

w = 4z Ans. •*> u O

w -plane

Sxample\.S$: Under the traatfpm atbn n? = —, show that the image o fthe hyperbola 'Ex. * .tn is the lemniscate p* * cos 2 4 • Solution.

\

Given:

1

=J

1

w = — => u + w - ----- — z x + iy

p c o sj+ ip sm j

1

rcoefl + ir s in tf [v x « r c o s l, y « rs in # an d u * p coaf, v ~ p e m f ]

46 ) EiiwhwwiwI I w m kihci M *>C* !S—W relv

{v e“ = c o sr 4-isinjj^

Pj* Given hyperbola in 2-plane: x 2 - y 2 - 1 =>

(rcoatf)* - ( r o n ^)2 =*1 =3 r a{ooe20 - 8in 20 ] = 1

=>

r 2ooe20 = I => W

co«2(-#) «1

[Using (1)]

=> cos 2^ m p 2. Which is a tenmiaeafe in w-plane. Proved Example IM ; Determine Ufa region in the w-plane In which the rectangle bounded by the lines x™ Q ,ym 0 , x * 2 , y - 1 Is mapped under the transformation w = >[2ei(K/4) z Solution.

Given, Since

e** sc o s tf+ ia in # u + iv = ^2 ^coe-j+ isin-jj(ac+ iy) = (!+ i) ( x + gr) * (x - y ) + i( x +y)

=> u = x - y , and v**x+y Given : (i) Line x ■ 0 is mapped into u = ->\ v = y => u = -v (ii) Line>> ~ 0 is mapped into u ~x, v = x => u = v (iii)L m ex*2 is mapped into u = 2 - y , v = 2 + y => u + u = 4 (iv)Line,y*l is mapped into u = * -1, v = x + l => v - u = 2 Y u+v=4 v—u *= 0

Ans. w-plane

Example 1.55; Determine the region o f the w-plane Into which the region is moped by the transformation u>~z*.

and

F u n ctio n s o

Given:

w =z 2

=>

u + iv = (x+iy)2 = x 2- y 2 +2ixy

=>

u = x 2- y 2 and v - 2xy

r

V a riab les j ^ 7

(i)When x = ~ , then (1) becomes: u = ~ - y 2 and v = y

(ii) Whenx = I then (1 ) becomes: u - \ - y 2 and v = 2y => u2 = - 4 ( w - l )

(iii) When y ~ —> then(1) becomes : u = x —~ and v = x ^ 4

2

4

1

v -u + —

(iv) When y - 1, then (1) becomes i u = x 2- 1 and y = 2* => v2 = 4 (u + 1)

y

i x =*-

2 x=\

v* I

y„ 1

•■Plane

W-Pt*DC

Thus, the rectangular region bounded by the lines x = i , x = l and y = Z£* y = 1 maps into the region bounded by the parabolas : if

^ ~~

an d » 2 = cU + ‘J» *** = ^ ( « + l).

Ans.

Example 1.56 : Show that the transformation w = z+ — maps the circle |z |= c into the ellipse X-

u c [ c + ty c o s ta n d v = Solution.

sin 6. Discuss the case when c mL

Given in z-plane the circle | z |= c, => z = ce“* Also given w = z + - = > » + w = ce* + -e~*

[Using (1)]

=>

tt + u> = c(co80 + i8in0) + i(co80-icos0) c

fv Euler theorem® 1

U+i„ = (c + i ) costf+ - i jein0 a = fc + ijco 8 0

Then

x

—(1)|

)

*

and

U= ( C~ c ) 8ine which are the parametric equations of an ellipse. When c - 1 then (1) and (2) becomes: u = 2cosd and v - 0.

"•(2)|

Since |coa#|
fv -l£ c o s 0 £ l] Ans.

1.17 SOME DEFINITIONS Zero o f an analytic function : A zero o f cat analytic function f(z) is the value of z for which /(* ) = 0 . //

z + *

\

For example : / W - z (z _ 2)' For zero off(z), we put z +1 *=0 =>z = -1.

i

Order o f zero: If, f(z ) ~ ( z - a)mf(z), Such that /(a ) = 0, f*(a) = 0, /'( a ) = 0, ..... /
/(*) -

-7 2 +1

For zero off(z), put (z - i)3 = 0 => z - i of order m = 3.

Singularity o f an anafytic function : A singular point of function f(z) is the point at which the function not analytic In etfMr words a point at which function/^ is not defined. For example : f( z ) 35

has a singularity at z * -yti.

Remark: For finding the Singular point of analytic functions fiinz

put deaom&itor

= 0.

F .r e t* .p k : «*>= (z_1)(* _ )r/2) For singular points, we put (z - /) (z - n/2) = 0 => z = /, n il i.e., i and n il are singular points of . f(z).

1 F unctions o f C o n R£X VM BttLes [ 4 9

0d*1td and Non-isolatd singularity: i * a is said to be an isolated singularity of the funetiofrf(z) rff (t) is analytic at each in some neighbourhood of the point a defined by | z - a |< 6 , except at the ptfteit a ftstff, ise it is called non-isolated singularity. fc if 2 ~ a is a. singularity of the function f(z) such that there exists no other gularity within a small circle of circumference with centre at th e^ o in t a, tftfn * a is said to be an isolated singularity. . example: ' \_

Z+ 1

Consider the function / W - z (z + 2 ) ' is analytic everywhere except at z ~ 0 and z - - 2. Thus z - 0 and z = - 2 are the only ttngfS&ftfe* of this ftmcfion. There are no other singularities ofyfcj toili#neigftbdurtood of t *= Z Hence 2 * 0 and 2 ~ 2 are the isolated singularities of this f(z) = cot (#7 2) ------\ —Let

tan . — I

It is not analytic at the points where tan! — = 0 = tan n it i.e., =? — = fin \z ) z =>

2 = —, where (w = 1.2, 3«..) n

Thus, z = 1,

........ are the isolated singutarties of f(z) excejjf 2 = 0 because in the z. 3 J neighbourhood of z ~ 0, there are infinite number of other singwlartfes. f 1 , J.e„ z = — if n is very large => 2 = 0 is the non-isolated singularity of f(i). f

; * f1 rfc

types o f singularity : ^lu p p o se f(z) is analytic within a region R except at i * it, whtett to in isolated singularity. Let C be a circle, whose centre is *a’ sueh that 0 <] !< R. 00 eo Then by Laurent's series, f(z ) * V aK( z* a)n + 6* (2 - o ) " \ !!«« »«i

doc; . •

The part 2h (z ~ <*) *...(2) is called principal part off(z), at z - a . n*\ There afe three distinct possibilities: (i) Removable singularity: All are zero i.e., fto firth in principal part offfz).

•40

BA | CMmMaarida U m u cIim um JII B J R J N W n M I W lV B V 8 J M I I C W * t ll

■OPf j

00 .

i.e.,

'« .(* -« )* .

|* - o |< f i

jM

Thai z * a is said to be removable singularity o f f(z). Alternatively, jjj} /(*) exist finitely, then z « a is a removable singularity, v sinz f\*> --------has removable singularity at z = 0 since

For example:

£

sin z

1^

z—

z3 z5 3! 5!

,

t z2 3!

z4 5!

= 1----- - H-----— ...

It has no term containing negative powers of z Le.t principal part offfz) has no term. However^ the singularity z - 0 can be removed and the function be made analytic by defin^l

sinz . . . ------ = 1a t z ss 0. z r (ii) Isolated essential singularity: If the principal part off(z) atz = a contains an infinite number of terms, then z - a is called) isolated essential singularity of f(z). i.e., if there exists no finite value of m such that lim ( z - a )m/’(z) * (finite non-zero x-*a

then z = a is called an isolated essentiaI singulatity of f(z). For example: The function eu* has essential singularity atz = 0, since i/g t 1 1 elZ = 1+ —+ ----1 - + ---r- + ... * 2\t 3lz

has infinite number of terms in negative powers of z. (HQ P ole: If the principal part of/^z) atz * a consists of a finite number of terms, say m, then the singularity at z - a is called a pole o f order m off(z). A pole of onler 1 is called a simple pble. Alternatively, If lim /(z) * oo, then z - a is a pole of fiz). I-W

Remark: For finding the pole of f(z), we put denominator tjp zero. \

For example: Let f(z) - -— ------- - . Then z = I is pole of order m = 3 and z * 3 is a pole (2 - l ) 3(* -3 ) of order m = 6. Rem ark: If lim /(z) does not exist, then z = a is an essential singularity.

FucnOW^CONM£XVA8>ABLE8 |# 1 ilte -

cat** v IS 7 : Find Ike kind c f singularity o f the fknctUm -— at z - a and * = » . [RGPV Dec. 2002] ft \

LP

c0* * *

o o s jc

G‘vcn: * ( * - a ) 2 rim teO r-a)2 For the pole of f(z), taking denominator to zero. i.e.,

sin/nr. ( z - a ) 2 =0

=> ( z - a ) 2 = 0 *®d ain*z=0 => z = a is a pole of order 2. and ain*z=8inrwr t*n

=> z =»**(where » = 0, ± I,±2.............. .) => z=0, ±1, ±2, are simple poles. ,\ If z *ao is a limit of these poles ^hen z=*oo is a non-isolated essential singularity. Ans. 138 : Find the kind o f singmlnrity o f the function f(x ) =

^ - jj

[RGPVDec. 20041

For Poles o f f(z) denominator o ff(z) = 0 i.e., ** = 0 order 2. k N*(a .

i.e., z * 0 is a />
For zeros of/fzj, the numerator of /fz) = 0 te., (z - 2) sin| z = 2 and ~^— =nn\ te.y z - 2 , z = — +1. (w heren»±1, ±2,.... ) z —1 nx Thus z = 2 is a simple zero. The limit point of the zeroes given by

z =— +1, X» = ± 1, ±2,...) is z • 1. [v lf n is large] rur Hence z * 1 is an isolated essential singularity. Ans. 1 3 9 : Show that the fitnction e* kns nn M a ted essential singntdrity at *s ®. jfeh*[RGPVDec. 2003] L „• ttV0 ‘

Given f ( z ) - e M *>,

Putting

4

.;

. .(1)

to ( l) b e c o r a e s :

;bn

..... 8 + ......00

aZ -J EIWSMBStfNG nXMMVKMH

Which contains infinite number of terms.

• * • •• v ' . . • ». - • • ■ Hence / = 0 is an isolated essential singularity of

=> z -oo is an isolated essentia

singularity of ez.

Ans

Example 1.60 : Find the kind o f the singularity o f f(» ) * s in

z~l.

f i >s Solution.

Let /(*) = sin

\l-z ) For zeros of f(z), put numerator of f(z) - 0 i.e., sin 7:— r = 0 (1 - z )

or

—^ 1 -2

[

V

8

i

n

^ L

= 0] J

or 2 =1 — !—(where n -± I, ± X .... )♦ Ifw is huge, Clearly 2 - 1 is a limit point of these n it zeroes: Hence 2 - 1 is isolated essential* singularity. • Ans. iJLV Example 1.61 ; Show that e '■**' has no singularities. Solution.

(i)

■ f(z)= e

For zeros

e ^ ^ = 0 = e~*y =>—- = »

V7 0

-

[v N um erator of f(z) =0]

=> z 2 =0 => z = 0 isa zero of order 2. Since there is no limit point of zeros, then/ft) has no singularity. ,

[v e^° =0]

(ii) For poles e~iU*2) *0 [ v Denominator off(z) = 0] Which is impossible for any value of 2, r^al and complex. Hence no singularity. Proved.

Ivl8 C O M P L E X IN T O G H A m W ? h Historically, Euler was the first to obtain the value of a definite integral by replacing a real variable by a complex variable, whereas P.S. Laplace (1749 - 1827) investiga^ (J J82 the validity of such a process. S.D. Poisson (1781 —1840) is believed to have been the first to use line integral in the complex plane. , ... It is interesting to note that the concept of indefinite as the process of inverse differentiation in case of a function of a real variable is extended to a function of a complex; variable if the complex functionf(z) is analytic. It means that \ff[z) is an analytic function of a complex variable 2, and if f f(z)dz=F(,is » > J */

then differential of F(z) is equal to f[z);i.e’.9

'• V '

f

I I#

\

rver, the cofK^pt of defmrte integral of a fimcntkm, of a real variaWfdpesnot^xteo^cd out, rightly domain of complex variables. For example, in the case of real variaMe, the path of integration of 6 J f(*)dx a

Iways along the real axis from x = a to x —b. But in the case of a complex function f(z), the path of the definite integral b J f(z )d z | may be along any curve joining the points z - a and z * b; so that its value depends upon the path v (curve) of integration. However, we shall see that this variation in the value of definite integral will F'Uisappear in some special circumstances. For instance, the variation in values can be made to disappear if the different paths (curves) joining z = tftoz = 6are regular path? (corves).

} SOME DEFINITIONS Domain (Region): A set S of points in the Aigand plane is said to be connected set if any two of ' its points can be joined by a continuous curve, till of whose joints belong to S. -*»An open connected set is called an open domain. If the boundary points of S are also added to an open domain, then it Ts called cloied domain, Jordan A rc: Let x(t) and y(t) be continuous function of a real variable / in the interval a s t <,fi. Then the set of points x in the Argand plane given by die equation is called a continuous arc if, corresponding to one value of t, there exists more then one value for z, then z is said to be a multiple point A continuous arc with no multiple point on it is called a Jordan curve. $ Contour : By contour, we mean a continuous chain of a finite number of regular arcs. If the contour is closed and does not intersect itself, then it is called closed contour. h Example: Boundaries of triangles and rectangles. * Sfrrtpiy and Multiply Connected Domains : f/ A domain in which every closed curve can be shrunk to a point without passing out of Die region is called a simply connected domain. If a domain is not simply connected, then it is called multiply connected domain. f

I I | &wg«mi»3 Mxmounc»4H

IMCOMPLEX LINE INTEGRAL Suppose,/?# is continuous at every point of a closed curve C having a finite length, Le.; C is a rectifiable curve. Divide C into n parts by means of (it + 1) points z 9 Let

a = z Q,b * z H,

We choose a point £* on each arc joining z*_j to z k.

Form the sum

Sn=t. /(*•)(*,~*r,l). \

rml Suppose maximum value of (zr - z r_,)

0 as n -* « .

Then the sum S„ tends to a fixed limit which does not depend upon the mode of subdivision and i denote this limit by hi : . £ f(z )d z or I f{z)d z which is called thecomplex line integral or line integral offfz) along C. An evaluation of integral by such method is alsocalled ab-initio method i

L

r* :

By the symbol

f(z )d z we mean the

along a boundary C intfce positive sense.

In case of closed paths, the positive direction in anticlockwise. The integral along C is often called a contour integral.

F u N rtiO H iw C n w H flf V m b w u s | 5 5

1.62 ; Evaluate

](3x* + 4xy + 3y* ) d x +2 (x* + 3xy + 4y* )d y ]

(I) a lo n g y = x* (U) a lo n g y = x . Does the value o f the Integral depend upon the path ? (i) Along curve, y = x 2 => d y - 2x d x and limit jc -►0 to 1. I * £ [(3x2 + Ax3 + 3x 4)d x +(:2x2 + 6x3 + 8x4)2xdx] = £ (3x2 + 4x3 + 3x4 + 4x3 + 12x4 + 16xs)dx = (ii) Along curve, y * x x> dy “ dx and limit x -> 0 to 1. I » £ (3x2 + 4x2 + 3x2)dx + (2x2 + 6x2 + %x2)d x = Clearly, both values are the same, hepce the value of integration does not depend upon the path o f integration. Am. pRppfc 1.63 ; Evaluate

J[

(z)2dz, along

t,

(I) the real axisfrom z ~ 0 to z ~ 2 and then along a line parallel to y-axisfrom z m 2 to Z m2 + L (ii) along the line 2 y - x . IRGPV June 2905 and June 2007]

M o n.

Since (z)2 = (x - iy)2 - (x2 - y 2)~ 2ixy (i) Along the path OMP where M is (2,0) and P is (2, 1). 2+i . Qc) d t =

fat

i

(x2 - y 2 -2 ixy)d z+

(x2 - y 2 -2 ix y )d z

...(0

M | Now, along OM, y**Q => dz = dx and x varies from 0 to 2. Jo*

x 3

~ y2 ~ 2“ 0f)cfe =

Also* along MP, x ~ 2 ,d z ~ id y and>- varies from 0 to 1 Imp

“ y 2 " 2ixy^d z = I (4 “ y2 ~ Aiy ^ y

4 iy ~ i^ Y + 2y7

~4

i i + 2 = 2 +iif 3 3

Hence 0 ) becomes, r2+t 8 +2_ + — II .1 = — 14 + —i. 11. /r = 1r * ( z ) d z = — * 3 3 3 3

Ans.

Example 1.64: Evaluate f * (2x+/;y + t)d z along the two paths: (i) x - t + 1, y - I t 2- - I (ii) The straight line joining (1 - i) and (2 + i) Solution* Let

''

fftGPV Dec. 2009 (N)f

£** (2x+iy+ l)dz

(i) Alongpath: x = / + 1 andy - 2t2 - 1 => dx ~ d t and dy = At dt. Then dz = dx + idy. Also from path x ~ t + 1 : when x = 0, then t ~ 0 when x = 2, then t = 1 i.e.. limit = / -> 0 to I. Hence (1) becomes : i / = j[2(t +1) + 1(2*8 -1) + l]( ^ +1 = 2jc- 2 2^1 => y+ 1 = 2 x - 2 => y = 2x - 3 => dy ~ 2dx

.-(1)

Fuwcttowi I to 2. Hence (I) becomes : / = J2[2x + i(2x - 3)+1 )(d z+ i2dx) = j^[2x +1 + i(2x - 3](2i + l>ijc » 4 + 8i.

Ans.

mxample 1.65 : Evaluate £ (** + 3z+ 2)dzt where C is the arc o f the cycloid x - a ( 0 + sin 0), f ■ y = a (l- c o s 6) between the points (0, 0) to (no, 2a). 'Solution. Let us consider the path of integration along a curve C consisting o f : I (i) the part ofthe real axis from the point (0,0) to the point (*0,0). On this line: z = x , d z = dx and x goes from 0 to na. (ii) followed by a line parallel to the imaginary axis from the point (m , 0) to the point (na, 2a). On this line: z ~ m + iy, d z =■idy an^y goes from 0 to 2a. a Let, / = £ (z 1 + lz + 2)dz ' “ F

+ 3x+ 2)c£x + £ ° {(>ro + iy)2 + 3 (fro+iy) +2]idy

or

I =f"i x*+-x2 +2x1 +i|"-(?ra +iy)s+^(> ra+iy)2+21ylJ0 L4 2 J0 [_3 2

or

/ =|j ( W

+ |( ^ a ) 2 +2*aJ + i[i(^ a + i2 o )s +^(>ra+i2a)2 + 4 o - i( x a ) 8 |_3

1

2

3

-^(/ra)2 l 2 J

3

I - 2 ;ra + - ( * a + i2a)4 + —(tra + i2a) + 4ia . Ans. 3 2 Example 1.66: Integrate z2 along the straight line OA and also along the path OBA consisting o f two straight line segments OB and BA where O is the origin, B is the point z = 3 and A the point z * 3 + L fRGPV Dec, 2004] Solution. Since, z = x + i y => d z - d x + i d y . Along Die curve C, we have * or

£ z 2d z = j c (x + iy)2(dx+idy)= £ (x2 - y 2 +2ixy)(dx+idy) (O'ThOpoint Afe z ** 3 + /, i.e:%A fa (3,1). The equation of the line OA is : y - 0 = y ^ ( x - 0 ) le .%x = 3y

...(I)

5 S | FmrnOTi^Kdi rattATjca-MI Hence, x*=3y=>dx « 3dy and .y varies from 0 to 1.

Now from (1), we have

JL *2<3k = £ ( ^ 2 - ^ 2 + 2i3y.y).{3dy+icfr) = £ (&+6i)0+i)y2dy = 0 8 + 2 6 0 ^ j = i(l8 + 2 6 .-)= 6 + j ^ .

00 Again,

z 'd z =

z 'd z = J ^ ^ +

...(2 )

*2
(x2 - y 2 +2wy)(dx + M f y ) + ( x 2 - y 2 + 2ixy)(dx+ idy)

...(3)

On the line OB, y = 0=>dy = 0 and x varies from 0 to 3. On the line BA, x = 3=>dx = 0 and >>varies from 0 to 1. Hence, (3) becomes: 3 x3 z 2dz * x 2dx+ £ ( 9 - y 2 + 6iy).idy * + i 9^--^-+3i,y2 3 0 3 = 9 + ir9 -I+ 3 ij= 6 + i^ .

...(4)

Thus equations (2) and (4X show th at:

z 2dz = I

z 2dz.

L U l CAUCHY’S INTEGRAL THEOREMOR CAUCHY’STHEOREM JOBA

Statement:

Am .

i

I f f ® & m anotytic function and f( x ) k continuous at each point wtthim an# on a simple closed curve C, then

£ f(z)d z=0.

[RGPV Dec. 2002 and Dec. 2011]

F u N C T fO N S ^ rC o tfftex V a ria bles |* W

Let R be the region bounded by the curve C. Let f(z) - u ( x , y ) + iv(x,y) = u + iv and z - x + iy => dz = dx+idy, then £ f( z )d z =

(u + iv)(dx + idy)

= £ ( u d x - v d y ) + i £ (vdx+udy) du du dv dv

Since f ( z ) is continuous, the partial derivatives So using Green's Theorem

M dx +Ndy »

are also continuous in region R.

JJ

1) becomes: I f(z )d z = JJ

//

...(2)

Since f(z) is analytic so using Cauchy-Rientann equations, we get du_dv d u __ du 8 x ~ By and S y ~ 3x fc*• 6) Thus, (2) becomes : jj, /(*)<& = 0. rw nit *• Extortion o f Cauchy's Theorem: Ifffz) is analytic in the region R between two simple closed curve C\ and C* then

/(*)<** = k f( z )d z .

CAUCHY'SINTEGRALFORMULA Statement: Iff® is anotytic within and an a etaaed curve C and *a*is any point within C, then .
^ " 5 5 1: Je -a )dX' °T i

oeC

[RGPVDec. 2007, June 08/

f(z) Proof. Consider the function------ > which is analytic at every point within C except at z 2 -0

■

cl

Draw a circle C\ with a as centre and radius r such that C\ lies entirely inside C. ie., C\ :\ z -a \ = r. » f Thus function ——- is analytic in the region between C and Ci. z-a By using extension of Cauchy's theorem, we have f Jc ( z - a ) Since,

C, : | * - a | = r

f JV Ldz Jc, ( z - a )

m *(1)

or * - a = rew or z = a + new

6 0 1 Encmhebwhq

=> dz ss i r e ^ d S and 0 -►o to 2 n . Hence (1) becomes: r k

j w _
J ^-d z (z-a) f ^ ’^

- )'.irei9d& * i

f t a + re ^ )d 0

...(2)

In Die limiting form, as the circle C\ shrinks to the point a, i.e., r -* 0, then (2) becomes: Jc Hence

<** * * f / W * - * » r 60 a ^ < a >

/(° ) = ^ < J e

...(3) froved.

Deduction : Differentiating both sides of (3) w.r.t. a, we get

/ ’(a) = — — f ' 1 2 x i d a * ( z - a) Similarly,

r(a) =^

l

2«

f ..ftfcK
(4 | ~ (4)

J ^ f dz'

-(5) \

In general,

/<' >(<,) = ^

t

(I-a p * * 1

-< 6)

Results (4) to (6), are Cauchy's integral formula fo r the derivative qff(z). Remarks: (i) If point a

©/< then fc

(ii) Converse of CMefcy'i Theorem (Morera's Theorem): Let f(z) be continuous in a simply connected region R and let for every simple closed curve C, such that £ f( z ) d z = 0. Theft fl(z) is analytic, (iii) Cauchy Inequality : Iff(z) is analytic within a circle C given by |s -a |s = jR a n d ifl/(* } |£ M on C, then

Proof. Since

\ / W =

f f(z)dz Jc ^ , qy»*i

P*y Cauchy integral formula]

FuNciidJferWGbiAfcEX V ariables |(fcl abv.

|/»(o)l» J l L f J W ± , ' ' 2xi 2in. Jc (z -—anY** )*+l [Since, z ~ a = Re* =5 dz - iRa'8dOaxi&\dz\-\iRewd6\= R d o \ n\ e !/(*)! \dz\ 2jtjc K ^ - o)**1I

[vU I-1]

fix n! M r2* 2* f lB+l

I

**J£-2*r 2n R n+i Mn\ (iv) 0 -y

Proved.

Liouvllle's Theorem : If a ftinctionjfc) is analytic for all finite values of z and is bounded, thenf(z) is a constant 1.67: Using Cauchy's integral formula, evaluate

e**

JLr * + i / d z, where C is the circle [RGPV, June 2011J

r

|* |- 3 .

jUiwftm.

By derivative of Cauchy's integral formula: -.(1) Here a = - I, n = 3, and f( z ) ~ e 2* ,2 i

2>riJc (z + 1)4 Since at pole z = a = - t, so that | z |=| - 1 1= 1 < 3 i . e point z = -1 He in circle C. /(*) = e2z => A * ) * 2e2x => f \ z ) = 22e2* =1

=>

f m(z) = 23e2* => / '( - I ) = 8«“2. ‘• i ) ‘

Hence (2) becomes : 8e

_2

3 f e2* , ~ ~ r L 7— TT®* M ** (Z + l)

*♦•(2)

62 1

M il Mill—IIIiTlB III

Example 1.68 : Using Cauchy's integral formula, evaluate I Solution.

dx * where C is the circle * (s+ * i)

|* + 3 I|= 1 . For poles (or singular points), put z ( z + m ) = 0 => z = 0, « * -» * At pole z = 0 |* + 3 i|= |0 + 3 i 1=3*1, te ., pole a tz ~ 0 out side of C. At pole z - -tri : |z + 3 i|= |-* * + 3« M (3 -;r )i |= 3 - ;r <1, /.e., pole at z = -jd, insideofC. " & z ( z + m) ■

L

m

= ± f m k

* Iz

-1]* z +1 nJ*

[Using partial fractions]

± _ ± f L -dz z m *c (z + » )

1 = -~ [0 3 -A l2 * ./( - » ) ] sa m

Because f ■ Jc { z - a )

m—~ .2 * X l m = -2.

3 ^ +7»+l z-\ f ( 3 ) , f ' ( l - l ) and f X l + i ) .

Example 1.69 ; I f /(S)® £

Solution.

° eC a«C

\ 0 ,

[v /(z ) = 1

/(-m ) = 1) Abs.

fMiww C Iv the circle x? +y* =4, fin d the values of

The given circle C is x 2 + y 2 * 4 or C :|« |= 2. The point z » 3 lies outside the circle \ z \ - 2 while z = 1 - / and z - 1 + / lie inside the circle.

[B ecause!*M31*34:2, \ z |* |l - i |- V 2 < 2 , \ z \ - \ \ + i \ - j 2 <2]

3z2 + 72 +1 j j 3 * 2 +72 + 1 . j ^ dz a n 4 ---- -—-— in analytic within and on C. ( 0 /( 3 ) = X *~3 2 -3 By Cauchy's integral theorem,

~ \ C = |z | = (ii) By Cauchy's Integral formula:

f(z)

Ik -(z^ -rgd) z

\

0

z=i-l)

CO II

+ •M II N

3z2 +72 + 1 dz = 0, ic 2 -3

Ftacfflotf &CowtoVAMAMJir I 69 £ ^LIJJLLLcIz * 2xi[3z* +7z +1]^ / ( # ) - fcri(3£*+7#+l) f \ 0 - 2**(6£+7) Hence

/ ,a - 0 = 2 jri{ 6 a -0 + 7> Ans.

* 2*1 (13-60 = 2* (6 + 13i). f

(iii)Also,

[From (2)]

A » - 2 « ( 6 ) « 12 *i

Ans.

ra + o * i2 « . m fii 1 .7 0 :1/F(t) - £

z -%

&***’ where* is the ellipse:

[RGPV June 2009]

Fi*rf the values o f ft) F(3.5), M F®, F f - /) ami F 'f - i). 1

fution : The given ellipse c

Here the point z - 3.5 lines out side of the ellipse c, while z * j, - 1 and - i lie inside the ellipse c as shown in figure. t (4z* + z + 6) j ^ 3 .5 ) = 1 (z _ 3 5) d*

0)

•••(»

since z = 3.5 e c ie .t outside, by using Cauchy integral formula

i

(4z2 + z + 5) z - 3.5

dz * 0

Hence (1) becomes F{3.5) = 0. (ii) By Cauchy integral formula:

f(z)

“ In i/lt), if / € c

2

(4z* + z + 5) I

,y*>

z —t

and Putting * = /, in (2), we get

•dz as2«/[4z2 + z + 5Jx- ,

FX/) *2wi(4^ + / + 5) F'(f) ~ 2fti (8/ + 1) F " (0 -2 * * (8 ) F( 0 * 2*/ (- 4 + i + 5) = 2*/ (1 + 0 * 2*(/ - 1).

fc j |

Putting / - - 1 in (3), we get F \ - 1) « 2*/ (- 8 + 1) * - 14*/. Putting t = - / in (4), we get

F"(-0=2itf(8) - 167«.

Ans.1

1J3 RESIDUES OF/jy ATPOLE Since analytic function fiz) can be expanded in a Laurent's series /(z) = y a „ ( * - a ) » + T 6 . ( » - o ) n-0 n-1 _ _ I , Then the coefficient of (z _ a y i.e., bx is called the residues off(z) atpole z = a. It is denoted by

[Res f ( z ) \ ^ or Res. f(a). fa68 f(z )\-a = fr| - ~ r 2m For example:

x/ v

sin* Z*

f( z )d z , where C is the closed curve around the point 2 - a. *3 *5 z 7 1 2 ——H ---- ------+ . 3! 5! 7S *

1 1 1 1 1 3 =— j— ——+ —2 “ —* + .... »3 3! ' 5!^ 7! ZS

... [R e s ^ z ) ^ 4-0 = Coefficient of

=~

=

A«s.

1.24 METHODS OF FINDINGOUT RESIDUES OF/fcJ ATAPOLE (i)

Residue off(z) .at simple pole z - a (i.e., pole o f order I) :

(ii)

[Res /(z )L « = lim (z -a )/(z ). *-KJ Residue off(z) has pole o f order m at z = a :

(iii)

Residue o f f(z) at infinity : Le., Residue oifiz) at z = a> = lim {-z./(z)} = - [Coefficient of — in the expansion of fiz) .for values of z in the neighbourhood z j by z = oo ] -3 < lt where C is the closed contour enclosing alt the singularities offiz) except at infinity.

.....-Him

^ i r u r i j a w r " 16I

CAUCHY'S RESIDUE THEOREM

r

Statement :

r Iff(z) is analytic function, except at a finite number of poles a,, a 2, a3,....an within a | > dosed contour C and continuous on the boundary C, then

J

f(fz)dz

= 2/ri.E R es/faJ nt th e pole a,

at , .... ,o„ in C

£ = 2xi (sum o f re sid u es n t th e poles w ith in C). >41

Let c,, c2,.... ,cH be the circles widi centres at 04, 02*03.......o»* respectively and radii so small that they lie entirely within the closed curve C and do not overlap. Thenf(z) is ^alytic within the multiply connected region enclosed by the curve C and circles Cu C2,.... C ,. So using Cauchy's theorem i Q j,. - for multi connected regions, we have / :f

£ f ( z ) d z =* £

fi z )d z + £

f{ z ) d z +....+ £

Q

f{z)tdz

* A = 2m[Re8f(z)]gma) + 2m[R£*f(z)}^2 +.... + 2»IRes f ( z ) \ ^

,

O'

[By definition of residues]

= 2j»fsum of residues 0# /(z ) a t poles alta2, .......a* i*kC]. ^ •V

Q

Proved.

i-2 z 1.71: Find the outer o f each pole and restore at it i f z ( z - l ) ( z - 2 ) ' IR**PV Dec. 2001] Let

1- 2z f/_\_ A * /* ", rr* z(z-l)(z-2)

For the poles of/Ts$> : put * (z ~ l)(a? - 2) « 0 => We know that

(i)

1,2 aH are simple poles. [ R e s ./f a ) ^ « lim (» -o )/(z )

- *£<*-«»«*> = lim

=i

(ii) (Bi) [lte .rt« > |0 - lm < * - 2 ) r t2) = lim

Ans. Ans-

=

Abs.

now

Rumple 1.72 .* Determine-the pots o f the function /fe>= f y - x f f c + 2 ) and the residue at each pole.

[RGPV June 2002]

661

Solution.

Let f ( z )~

*2

(z - 1)2 (z + 2)

For the poles off(z) : put (z -1 )2(a + 2 )-0 =>

2 = -2 is simple pole and * = 1 is pole of order 2,

We know that

[Res. f { z ) \ a = lim (z - o) f(z) M-HI

(i)

+

Abs.

We know that [R m / W L . . = («)

~ o)'" /(z )! '

[R«s ;r« ] , . , m. ! = ^ ^ 7 lim £ « z - l ) 7 M ) . ;lim i L * L U lim «-■ d* [* + 2j “

(z + 2)(2z) -z * (l)| (z2 + 4z) 5 (z + 2) j W *-*» (z + 2) “ 9«•

Ans.

** Example 1.73 ; Evalute the residues o f (z _ j ) ( z _ 2 )( * -3 ) <**2 = 1* 2, 3 and infinity and show that their sum is zero. Solution.

Let f(z) =

M

z2 (z-l)(z-2)(z-3)'

For the poles off(z) : put (z =>

[RGPV Dec. 2002]

1) (z - 2) (2 - 3)=0

z =1,2, and 3 all are simple poles i.e., each pole of order m = 1.

We know that [Ree./Xz)]^ * lim (z -a )/(z ) z8

(i)

[Res. A*)],., * Km (z - l)/(z) = lim

(io

[ * « . / ( C - 5 9 c»- » / » = Hg ( 7 : ^ - 3 ) '° a l i l j — 4-

* -* 1

,

«-»i ( z - 2 ) ( z - 3 )

2

(BO [ R « . / W L = S (*-3)/(*>“ ten (zi 4 - 2 ) = ^ (iv) Residue off(z) at z = 00= lim {-2 f{z)} -z =lim l “MO

z2 (z -1 ) (z -2 ) (z

) 4 [by definition]

F u n c tio n s

flim

of

C

o m p ie x

V a m a m je s | 6 7

-z 8

z \ l-X ) (i-^ ) (l-^ ) -1

a-o>a-o)d-o)

=-i.

j 9 i _g + 9 _ 2 Thus, sum of residues of /(*)■ —- 4 + —-1 ------- -------=0.

Proved.

1.74; Find the poles, order o f the poles and residue at Itfo r the function

1

fRGPV June 2003] Let /(*) = ~T—r z A+1 For the poles off[z) : put z 4 +1 * 0 =» z 4 * -1 i.e.,

z* = eoe?r + £8in*, = cos(2n + l);r + i8in(2n + l);r z = [oos<2n +un +i sin(2n+ l)jr]l/#

^

z 4 *[cos(2^ + l ) ^ + i8in(27i + l ) ^ ]

=>

z - et<2n+v/ i , where n - 0,1,2,3.

or

z - ei*f4le3i*,4fe5i*,*fen*u are four simple poles.

[y e* = coed + tsintf]

We know that [Ree./X*)]^ =Um (z - a ) f ( z ) (i) [Res./X*)] ^

= lim ( z - e ix' *).-}— , w "4 * +1

[which is — form, so using L*hospital rule]

=5lim

I -— 1 e-3^/4 = — 1r cos----3x ism . . — —r Az >A 4[ 4 4 .

-I . 1 v r '- ^ j (ii) [Re8./0z)l

i( i+ » ) *~1T'

w . * lim ( z - e 3m!4).—^ — , now using V hospital rule W "4 * +1

Ans.

68 I ENGlJte£Hl^MArrtEMATlC&-llI (iii) [Res./(2:)je_eJ«/4 = lim

4

( z - e 5jn/4).—p— , using L’hospital rule z4+ 1

Ans.

4

(iv) [R es./^l^T .M = lim

( z - e 7j”/4) . - - — , using L'hospitairuJe Z +1

= lim —- = lim z-H?*'* 4Z Z^e1*'*

3

z

I

V z =— -z

1 1 1

1

UJ

An*.

v s in z . . Example 1.75; Find the sum o f residues o f the function t(z) = —— — in the circle \z = 3. - X COS Z

Solution.

1 1

For the poles of f(z), : put zco sz = 0 => z **0 and cos r = 0

** '■ 1

=> r = 0 (simple pole) and z = ± ^/2, ± 3 ^ /2 ,± 5 ^ /2 .... are simple pole.

At the pole, z ~ 0, then | z |=| 0 1- 0 < 2 inside the circle |

2.

Atthepoles , z = ±a72, then |z |= i ±»72)=/r/2 = 1.57 < 2, inside the circle |* |= 2, At the poles z = ±3jt/2, then |z |= | ± 3 ;r/2 |= 3 * /2 = 4.71 <% Which is outside the circle |z j= 2 Hence poles z = 0, z = /r/2 and z - - t / 2 lie within the given circle C. (j) ^'

[Res./(z)] 1

'V\v

V

- l i m ( ^ - 0 ) - ^ ^ - = l i m 8*n ? ~0. «-*<> ZCOSZ
(ii) [^®s•f <*)Lrf - Hjg (* ~ M) zc~ ” •

[which is “

= Hm (sin z +(z - ^ycos z) 1 *-*'A cosz-zsinz (0 -f)

from, using I *Hospital rule]

2 ;r*

(iii) [Res./(«)]*»-*/ = lim (z + ^ ) - -------, so using L ’ Hospital rute zcosz sinz+(z + ^ )c 08z -1* 2 = am --------------------— —-r---- r-——. *—% co sz-zsin z j *) n r* j 2 2 Sum of all residues of f ( z ) =0 — +—=0. 7T K

Ans,

F u n c tio n s o f C omdex V /W a k -e s j 69! i,

M f# 1*76: Determine the po/e and residue at each pole. I f f(z) -

1-e*2

m m - .. Let ‘ (2z) i- I+ I!

(2 z)2 (2z f +.... 2! 3!

V 4 •> 2 -» (2 + 2z + - z + 3 .... ^

v

-.(I)

V*

Hence f(z) has pole at z = 0 of order 3. /. Residue off(z) at z = 0 of order m = 3 i i»•r

I lim (3-1)! *-x> dz

[c -o f . S z f l

ton:1 2-2z-~z2 3

tin*

[Using (1)]

8 2 * lim --------.6z~ 3 3

Ans.

2 z-*Q

1*77 .*Determine the poles and residue o f the function sin z tathm,

Let f(z)=— sm z

Ii*vIvihKU'A^

poles : put sin z=0 ■=? sin z - sin tin

=>z = nn (where, n »0,±1, ± 2..... ) Hence z = «/rare simple poles. [By definition]

[Res./Xz)]^, = lim ( z - a)f(z) [R e s.ftz )]^ , = lim { z - n n ) z-*mr

- lim

z-*n/r

(2 Z - T17Z) COS Z

sinz nn coann

[Since — form, using L Hospital rule] nn = (- l)n nn. (-1)*

Ans.

70 | E N G i c ^ l i i t ^ t i c s 4 I I Example 1.78 : Find the residue at z ~ O of the following function : J +e (i) f(*) - ~ a m z +z c o s z Solution.

[R«* •/(*)],_ - tim ( z - a ) /(*)

[By definition]

(i) [Res./Xz)! =lim ■r-^—— — u 0 sinz+zoosz 1+ 1

1 + e*

- lim r-»fl

1.

1+ 1

+ C0S2

sin z v lim ------= *-*o z

x ^ x* (ii) Now, since c o s x * l- — +— ....... . then /(z )= 2 C o a fi]= z [l-—^ - + T -i-T - ...... \ z ) |_ 2! z* 4! z J This is the Laurent series about z 530.

2z 24 z8

Residue of /(z )a tz » 0 =coefficient of — ^ ( z - 0)

...d)

[By definition] Ans.

Example 1.79; Find residue o f f(z)=

Solution.

Given

a/ Me origin.

zain(mz)

z 1 z* 1+* + — +—-+. 2! 3!

/(*)

5!

3!

H +jrjr j mz* 1 -

m’z* m424 3! 5!

[Since (1 - x)~l = 1+ x + x 2 + .......] 1 j»2

22

23

1 + * + — r + - T

2!

3!

+

.

1+

[Because for all sufficiently small value of 2]

3!

I 71

// )= X— 1 .-Y 1 +— 1 . -1+ .......... ... /(* m z m z which is Laurent series about z = 0

...(1);

Hence, Residue of /(*) a t z =0 = Coefficient of

1 ( z - 0)

J_ Ans. m r ((4 * -3 - 4z )d )a z 1.80; Evaluate Jp z (z - i ) ( z ~2)* w*tere ^ k the circle | 2 1=3/2. [RGPVJune 2008J Let

f(z)=

[

" lZ z ( z - l)(* -2 ) Using residue theorem:

^

jc f(z)dz = 2ni [S Res.of f(z) at each pole with in C.]

...(1)

For the poles off(z) : put z (z - 1)(2 - 2)=0 => 2 = 0, 1, 2 , are simple poles. At pole z = 0, then 12r|~| 0 1—0< 3/2, which lie inside the circle C. r'

Atpole 2 = 1, then |ar J =| 1 1—1< 3/2, which lie inside the circle C. At pole z - 2, then | * | =121= 2< 3/2, which is not lie inside the circle C.

"

We know that : [Res f(z))gma = lim ( z - a ) f(z). Then (i) [R* f ( z ) U

0)

=1=2.

[Res/( 2)1, i =lim (2 —I ) — ——

— =—-—=—1.

2(2 —1) (2 - 2)

1

(-1)

Hence (I) becomes : | c z ( J l i ) ( z !_2) ^ = 2*-i (2 + (-1)] = 2ni.

Ans.

z* tommple 1.81 .*Determine the pole o f the function f(z)=—~—rrrr — -7 and the residue at each r (z-iy(z+ 2) f

Z*

pole. Hence evaluate I -------- =-------- d z where C :\z |=3. * (z - 1) (z + 2) WnHon.

For the poles off(z) : put (z - 1)2 (2 + 2)=0. f(z) has a pole of order m ~ 2 d X z - 1 and a simple pole at z = -2. d

f / __\i

d

72 I = *-*11

(z + 2)

J

= Uml i l Hi l =£ 1(«+2) J 9

*2 4 /(z )U 2 * Um [(2:+ 2) /(*)]= lim -— - j = - . z~*“2 *-*-2 (2; —1) 9 Using residue theorem:

(ii)

Jk f (z )dz ~2 m [I Res. of f(z) at each pole with in C],

...(1)

At pole « = l,th e n lr |= |l|= l< 3 , inside in C. At pole 2 = -2 ,th e n |a r|= |-2 != 2 < 3 , inside in C. Hence (1) becomes : f 7----- %-------- d z ± 2 x i \ —+^-\ =2ni ( * - l) 2(* + 2) [9 9] iia

Ans Ans*

Example 1.82: Evaluate f ----- - f ------ where C Is the circle Iz + *l=>/3. Jc (Z - t ) * & Z + 3) . Solution.

12z-7 ' ( z - \ ) 2 (2 z +3)' Using residue theorem: /(*)=

f(z)dz=2m [IRes.of f{z) at each pole with in C]. For poles off(z) : put (z - 1)2 (2z +3) -0 3 =>z = 1 (order 2) a n d z =— (simple pole) 2

At pole z - 1, then |« + i|= |l + i |« -J\2 +12 =
S S ^ . T ^ _a)m /(2)]

=> [R®«/(a)],.i. « . i =7f l*m 1! *-»i a z [

— { W 7 .1 ( z —l) (2 z+ 3) J

=Bin ± ( U * z l ) = K * r ( 2 £ ± 3 m z Q 2 jz Z H

*-* cfe\ 22+3 J

[

(2z+3)2

...(1)

6 ° - 10= 5 0 =2 25 25 Hence (1) becomes: 122-7 [ -----i g - L — cb =2m[2]=4m. k (2 --1) (22 + 3) I ) 2 (2* pxttmpie 1.83 : Evaluate

Ans.

z-3 - j —- — —dz, where C Is the circle |2 + l+ i|= 2 ,

U t f(z)=

z 3 22 + 22 + 5 Using residue theorem: £ /(z )d z = 2;n[IRes. of /(z) at each pole with in CJ.

...(1)

-2 ± > /4 -2 0 - 2 ± 4i For the poles of/fe) : put z +2z + 5 = 0 => z -------- ----------— -— => z ~1 ±2i => simple poles are z =-1 + 2i and - l - 2 t . At pole z = - l+ 2 i, then 12+1+ t|= |--l + 2i + t + £ H 3 i 1=3*2, outside in C. At the pole z - - 1- 2i, then 12 +1 + i M -1 - 2i +1 + i H - i 1=1<2, inside in C. Thus only pole z =-1 - 2i lie inside in C. • [Res /(*))*—1- 2. = ^ -

<* + *+ 2i)f(z)

te + l + 2 Q (g-3)_ Km (z + 1 + 2i)(z-3) 2Z+ 22 + 5 (z+ l + 2i)(2 + l - 2 i )

2 -3 —1 —2£ —3 2 + 1-2 t - l - 2 i + l - 2 i Hence (1) becomes : =

..li m

Jc 7 ^

4 + 2i 4i

1 i. 2

----------------------------------------------- s=----------- ---—

E

dz = 2m [ v ih

Ans-

( 2 +i ) Z*

Example 1.84 .* Find the residue o f f(z)= -------- ------------------ at its pole and hence ( x - l ) ( z - 2) ( z - 3 ) evaluate Jj, f(z )d z , whereC: |* |= 6 /2 . Solution.

For poles of ffz) : put (2 - 1)4 (z -2 )(z -3 )= 0 => 2 =1 (order 4) and z =2, 3 (simple poles) Residue off(z) at z “ 0 “ jj®}

~

[By definition]

...... T

741

(i) [Res f(z)\z-i = lim (z -2 )-— — - —-— w *-»2 (z - \ ) a(z - 2 ) ( z -3 ) =lim

8 ~-8 (-1)

(2 -I)4(2 -3)

(ii) [Res/(2)]z_3=lim (2-3).

*-3

(Hi) (Res « * ) U . ■=—

... [Res A * )U i,«-4 =

(* - 1)4(z - 2 ) ( z - 3) «-5(*-1)4 (*-2) 16'

Um

{(z-a.)" f(z)\

(4-0! tfe3

(* -D 4.

1 ' d3 5— I 6 dz3' (z-2)(z-3)J

[By definition]

(* -1 )4(z -2 ) ( 2 - 3 )

Z*l

*-i

(19« - 30)

1

2 +5+ 6 —
d3 f , 27

*=»

[v as dividon algorithm]

81

J_ — r
6

27

■= lim-

[v resolve partial fractions]

f 27 8 1 «fe2l (z-3)J+(*-2)JJ Z*1

J_ _ d [ 54 6 d z\(z-Z f

16

]]

(z-2)3jJi.1

J_ -162 48 •+ 6 J 2 - 3 ) 4 (2 - 2)4

A

-162 +48U J l = i ° i . 16 J 16 16

Further, using residue theorem : f(z)dz-2ni[ZUes.oi f(z) at each pole with in C]. At pole z= 1, then 12 |— 11 |=l<5/2, inside in C. At pole z - 2, then | z |^| 2 1=2<5/2, inside in C.

At pole z - 3, then 12 |-j 3 l=3*5/2, outside in C.

~.(!)

i

F u n c tio n s O F ^O N rtJix V a r ia b les I 75

Clearly, only the poles z = 1 and 2 2 2 lie inside the circle C. Hence (1) becomes :

I

1

Wfamtple 1.85 : Evaluate Jc (Natation.

f(z) =

where C it the circle |z - l |= 2 .

1 z 2 sin hz

v sin hO =

* -e°

z 2(e* - e"*) 2

*2 *3 1+ * + 2T+ 3 !+'

f,

*2 *3

T ‘ *+ 2T‘ 3f+1

*2 *4 1 + 3T+ I T +- •

2zJ 2Xs ^ i r i T -

,

( z 2 z* = z 3 1+ ---+-----+.

=z -3

,6

120

, z 2_z* 6 120

+

‘ "

-I

z _ J ___ 1___ *_+ 120*.... z 3 6* 120 which is the Laurent's series. By Residue Theorem: ...(2)

£ f(z)dz=2m[2.Res / ( z ) at each pole in C]. From series (1 \f(z) has pole at z - 0. At pole z = 0, then |z - l |= ! 0 - l 1=1 <2, lie inside the circle C. Resf(z) at z = 0, * coefficient of

1

in (senes)

1

•

Hence (2) becomes : f ___ l ___ d z = Jc (z2s ta te )

I 6)

3

.

Ans.

Example I M : Evaluate f -------- d z, where C is the unit circle | * |= L Jc COS71z Solution.

Let /(*)=-

COS K Z

Using residue theorem: ..(I)

£ /(z)dz=2/»[£Re8.of /(z) at each pole with in C).

\

76 |

For the poles offtz) : put cos n t * 0 => Jtz = ± —, ± — ,± — ........ 2 2 2 1 3 5 => z = ± - , ± - , are all simple pole?. 2 2 2

At poles 2= ± “ , then | z|=I ±-~ < 1, inside the circle C. 2 Z Z 3 3 3 At poles z ~ ± —, then |* H ±—1=—* 1» outside the circle C. 2 6 / Hence only two poles * = ~ and ~ ~ He inside the circle C. • [Res/(z)] | =lim ( z -1/2).*

which is — from, using V Hospital rule

=limi -a in x z .n Z~*2 Similarly, [Res f (z)]

= lim -i

tf.(-sin ;r/2 )

[v sin ^ /2 = 1]

it

, = lim (z + 1/2).____ usingV Hospital rule C08XZ

(-s in nz)n

(-;r)sin(-* /2 ) (-> r)(-l)

erX n

Hence (1) becomes : r

ez ■dz = 2ni cos/rz

e^ K

e^ 7t

( e% V

- - A i s i a h —. 2 mple 1.87 («) ; Evaluate

J^ _ j^ z _ ^ _2r

fon.

j f

L

Let

■**>“ 771

Ans,

4

C is the circle {z \ * ■?.

F e f 20101

FuN crtbxg

V dw »ai a s | 7 7

We know that

1

| f(z )d z * 2ni * [Z Res. off(z) at each pole in C]

... (1)

c

ft.-

For the poles of J(z) : put ( z - 1) (2 - 2 ) = 0 => z = l and z = 2 are simple poles. At pole z * I, then |z | = | 1 | = 1 < 3, inside the circle C. At pole z - 2, then 12 1* | 2 }* 2 < 3, inside the circle C. Hence both simple poles are He inside of C. 2x

e2x e2 = lim------- = ----- = —e2 z-*i (z-2 ) (-1)

V | J)-

e2* (ii) [R es-/(z)]...2 = J i» (^ -2 ).( i _ 1)(2_ 2)

[

-

js

^

o

-

s

I;

Hence (1) becomes :

[•

1 ( z - I ) ( z - 2 ) A = 2tu [e* - e2].

r t iMxsmple 1.87 (A) : Evaluate Jc

-*

cos k z * , l)(z-2)

Ans. , , where C is the circle \ z \ - 3. [RGPV June 2003 and Dec. 2011J

n.

cos x z 2 ^ z) = U - l ) ( z - 2 ) '

Let £

f(z)dz=2m[ZJte8.of f(z) at each pole with in CJ.

•■•0)

t■

M \‘

For the poles off(z) : put (z -1 )(z -2 ) = 0 => z ~ 1 and z = 2 are simple poles. At pole z - 1, then |z |= | 1h i < 3, inside the circle C. At pole z = 2, then | z |=121=2 < 3, inside the circle C. Hence both poles are He inside of C. ro

VI

v

/

(0 [ R - /W L , =lun <*

COSJTZ2

- Im

CQ8/TZ2

COB7T

-1

,

78 | EwnnmwiQ>fcBW|Mnc»m (ii) .[Res/’(«)]„, *Um (z -2 )

cos**3 (* -1 X 2 -2 )

cos4* (2-1)

1 I U

Hence (1) becomes: d*« 2*i (1 + 1) *4*i.

Solution.

Let By residue theorem: Jc /■(*)<&=2*i[E Res. of/(*) at each pole with in C]. For the pole of f(z) : put (* - * /6 ) s * 0 =>

z * * / 6 is a pole of order m = 3

At the pole *=*/6, then |* |s |* / 6 | * 0.523 <1, inside the circle C. We know that:

(m -1)! *-** dz"1'* ^

[Re8 f (*)I,«,/8(M«3

{(* -o )m/(*)} sin** '( * - * / 6)a

lim [30 sin4 * cos* * - 6 sin6 z) 2 «-**/•

Hence (1) becomes:

t

FuNC^W olaH*VW dAB^s I 79 Example 1.89 : Evaluate f t a n z d z , where C is the circle |* | = 2. n.

„ sin z Let /(z) = tanz = - — . COS 2

By residue theorem:

f /(z)dz=2*»[ERe8.of /(z) at each pole with in C]. I* l^t;y

...(1)

For the poles o f/fe): put coaz = 0 => z * ±*72, ± 3*72, ±5*72.......... all are simple poles. At the poles z=± */2, then | e |=| ±*721= jt/2 = 1.57 < 2, inside the circle C.

\

At the poles z= ±3*/2, then | z |=l ±3*/21=3*72 = 4.71 < 2 , outside the circle C. Hence only two poles z * *72 and z * -*72 lie within circle C. [Res/Xz)],.*,, = lim (z-> r / 2 ) ^ ^ , z-wr/l

COS 2

0 [which is ~ form, using V hospital mle] Jim Qinz + (z-nl2)ca9>z _ 1 ^ ^ -sin 2 -1

-

. ; ^ y(-

and

/(«)],._,/* “ ,55/, (z +*f2)

r+v'

= lim *MZ + (z +ffl2)cQ&z _ - 1 _ L -sin 2 1 Hence (1) becomes:

using L’hospital mle.

f tanzdz = 2*t[-l + (-1)] = -Am.

jc

Ans.

APPLICATION OFRESIDUES TO EVALUATEREALINTEGRALS f2*

Wn

\ Type (I) : Integrals of the type I i?(cos0,8in0)<20, where F (cos#, sin0) is a &89V rational function of cos 9 and sin£, such integrals can be reduced to complex line integrals by the substitution, dz put z = e i0>so that dz=ieie dO, => “ ^ “ 7 7 U"

Also,

o o . , = ^ l = i ( , + i ) and Sin*

As 0 varies from Oto 2?r, then z moves one round the unit circle in the anti clockwise direction, i.e., \z\ = \ f

F(cos 0, sin B)d8~

= £ « * )& ,

where C is the unit circle Iz 1=1. Then above integral on the rigit can be evaluated by using the residue theorem. f(x) —— tfcc, where f(x) and F(x) are polynomials in x such

C

that xf(x) >0 as x

oo, such that F(x) has no zeros on the real axis.

Consider the integrals / =

~ ^ d z , over the closed contour C consisting of the real axis

from -R to R and the semi circle C, of radius R in the upper half plane. We take R large «*mgh,

f(z)

so that all the poles of F(z) in the upper half plane lie with in C. By Residue theorem we have m

= r

f k

[ sum of the residue of f (z )d z cR fix ) , ' F(z) "+ 1/? 'F(x)

/(*)

.

*n ^

pi®0*]

u P P er

„ . f(z) m ^ sum of ^ residue of f H ) in

plane]

...(1)

If we put z ~ Re*e in the first integral on the left side then R is constant on C\ and a» z moves along C| and 8 varies front 0 to n. •

H

c F(z)

For large R,

r

•» F ( R e ie) W e' F(Rel

Re

R e ‘’ i d 0

ia&\ is of the order of real part of

f

f(R)

1

j" f

.

t

F unctions o t C ompuex V ariables | 0 1

" r

F (fle T ^ R e ‘) t d 0 -*° when « - » «

Hence from (I), we have a (tv -» f {X) , ^ f(z) E -^ ~ ^ d z -2 m [ sum of the residue of in the upper half plane] 'e 1.90: Using contour integration evaluate the integral

f2*

d8

2 + cose* [RGPV June 2002, Dec. 2003 and June 2011/

Let

2* H

dO 2+cos£

Put z ~ e iB => dz= et9.id $ = > d 0 ^ -^ - =» d 0 ~ ^ el9.i iz «** +e~ie 1 Since eos£*— —-----=> co&0=^ z + — \, then we have

3-

/ * f2# *# _

2 f

or I - j )c

r 2 + co80

az dz t-

1

dz

g + 1 ( z + 1 '\ iz

IR )

f

2z

dz

$ 4 z + z ^ * l iz

where C is unit circle |z|*1

...(1)

I Let /(*)=■ z 2 +4* + l -4 ± V l6 ~ 4

For poJes : put z + 4 2 -f 1 =0 =>z «------ ------- o z 2

- 4 ± 2 > /§

2

=> z =-2±V ?=> z = - ? + V ia n d z - - 2 - V 3 are simple poles. ^4/ pole z = - 2+ V?, then | ? |=1 -2 + VJ 1-0.267< 1, inside the circle C. ifc/>e/e z a - 2 - V5, then U N - 2 - V II*2.73 ft 1 i.e., out side of G Thus, only the pole z =. -2 + V3 fws inside the unit circle |z |= l but z = -2 -V 3 * s outside the circle C. Hence by Residue theorem :

Jc f(z) dz =2*x [Ree /(-2 +V3)] =2** f lim [ z - (-2 + V§) /(«)]]

$2 | En6m zkm u M x n n w ic s ^ fl =2xi

=2m

r (»+ 2 -Vi>) 1 r ~<-«^>(*+2-,/3)(*+2+v/3)J L

*

Hence (1) becomes: fi* 2z f ■T f** adO v 2 m 2x /= I 2^ 0 ’ 1 1 m d * =m * n 2+CO80 eU dQ Example 1.91 •*Evaluate 5+ 3sia0 Let

pin H

dO 5 + 3 sin 0

e i0 -e~i9 if O Since, s in 0 = — —:— => sin 0 = —• z — 2i 2i\, zj (2* Jo 5 + 3sin0

r i:

dz 5+3 [ S

dz

-lie -H


or

/ = 2 ^ /(z)dz

R

« (z + 2 -V 3 )(z + 2 + ^/3)]

4 fRGPV Dec. 20011

Put z =ei6 => dz=i ei0d&=> ~ = d 0 vz

-

+

Am,

r

Solution.

*

~

I "

F u n c tio n s

where 'C is the unit circle, Le., C : |zj = 1 and

op

C o s W lex V a r ia b le s

—t:— r 3z2 + 1012-3

For poles: put 3z2 + 10iz-3= 0 -10*±>M 00+36 6

-6i±4i

-

-10f±/8

-K+4i

. 0 — 4i

*=— 3— • ^ en a " — 3— ond

— 3— (say)

ar=-~ and p * -3 i are simple pole 3

At pole

z = a = -i/3 ,th e n 1z |=| —i/ 31=1/3 < 1, inside the circle C.

At pole

z * £ -- 3 i,th e n |z H - 3 £ | = 3 < 1 outside the circle C.

'• f ^Z)=3 ? + lOi* - 3 ~ 3(* - a )(* - £) We know that:

" (2)

1 [Ree/(^)=La = Km J-*«f (z - ar)/(*)-lim !«*« 3 { z - P ) i

i

_2.

By using residue theorem: Jc f(z)dz * 2xi (IR es/(z)at pole z =« in C). 0 . 1 2* =2bnx—* — =—. 8t

8

4

Hence (1) becomes : /■ 2 Jc f(z)dz=2. *r/4=*72 Example 1.92.*Evaluate £* (g + ^ « e y »,a > 6 > 9 ' Solution.

Let 7= I** ------— ---- r ■*> (O + &CO80)2 Put z^e* => dz = e* idO ..

e +e 2

z + l/z 2

= ~~ iz 2*+l 2z

CO80 -------------- -------- ----- ----------- .

Ans. [RG PVJune2004/

8 4 -I

and limit 0-+O to 2/r reduces to C : \ z \ - \ j - f2*

^ f (a + fccos)2 *c

■ it

1 IZ

a + b (z2 +1) 2Z

zdz ( 62; + 2az+b)z *6* Jc ( ,

\

2a

V

dz

I2 +T 2+,J

£ f(z)dz, where C :|« |= l 11 Here f(z)~

...d)

2o 2 + — *+1 2

For the poles o f/fe j: put *2 + ~ * + I = 0

or

I Z~2

~P

- a ± J ( a 2 - b2) 6

4

— a+-v/(o* - 62) —a —J(a2 -b*) Let <*=---- — --------- and /?=------ ------------ are two poles of order m = 2. 0 o Since, a > 6 > 0 ,l/? |< l and that |a/?|= l, it follows |a |< l . Hence 2 = a is the only pole of order two with in circle C. Hence pole 2 = $ outside o f C. f(z) can be written as : ^

*

Now, Residue ^t a pole 2 = 0 of order m ; d"

-

(m -l)t dz*

[(2 -a r/(2 )}

.<*)•(* - f f \ =lim ± \ J L -,} -Km *-** d z \ ( z - f l ) J *-** = 1-m z - P - 2 z (a + P) *-* (2 -^ )3 “ ( a - /? ) 3

(z-P)

I F u n c tio n s o p C o m p lex V a ria b le * | 85*

-2 a ab2

*(-i)x-

By residue theorem: ** Jc

(Sum of the residues of the poles which lit: with in C) ~2j A

rab2 1 \ 4 (a2 - b 2)3/2 J

Hence (I) becomes : =—~x2m ib

d0

2jta (a%- b 2),3/2

2xa (a* - 62)3'* •

f*« cos 26 Example 1.93 .*Evaluate J, 5 + 4 00*0 Solution. •

[Using (2)]

f ( z )dz

1 ab2 4 (a 2 -& 2)3/2

*•• f 1,0 — (a + fccoe0)2

-..(2)

Ans.

Put z = e ‘s => dz= ie'e => ~ = d S iz Since

„ e ^ + e '" if O l |V + l \ o o rf-------— ‘ 2 {2 + z h { — ;

and

cos 20=

/ = f2' 00826 i c Jo 5 + 4cos0

1' 2

=

f Jc

1 'j

-h ie S &

1

I

+? j =I V

*

J

2 v *2 J
i

[RGPV June 200SJ

Av contour Integration.

l L i * mh l c *«>*■

J ...(0

8 6 1E m w m p m M

z A+ 1 where /(*)=- 2/_ 2 - —„v z (2z +5z +2)

For the poles of f(z) : put z 2(2z2 +5s+2)=0 -5±V25-16 i.e., 2 = 0, pole of order 2, and z - --------------4 Le.t z=--5 + 3 - 5 - 3 4 z ~~ \ anc* -2, ^ simple poles. At pole 2 =»0, then | z |- | 0 |=0<1, inside the given circle C. At pole 2=-1/2, then |z |= |-1/21=1/2 <1, inside the given circle C. At pole 2 = 2, then Iz 1=1-21=2 <1, outside the given circle C. ... [R e»rt2)ko,m_2, =—

- i nn—

z^odz[

Um — {(*-0)2./(*)}

■i 1+ ■ 1) 1■ (*4

z2(2z +5ar+2)

[B*>/(*)Jz-x = J j k + Y d 'f W

= lim (z+1/2)-^---- & t.fj------

L --------i-------------------- *--------- 1

* .2(*+ l/2 )(z +2)

-M m

.JZ .

L **(2**+fo + 2)

2«*(*+l/2)(« + 2)J

.

‘

2z*(s + 2) 12 By using residue theorem;

Jc f(z)dz=2xi* (Sum of residue at each pole in C)

I Hence (1) becomes: r2*

Jo

cob 2 #

5 + 4cos0

4 12J jq

1 m _ 2j X 3

12 3

x

~6

Am*

FuNcnbNroF CommJexVahables | 87 t

r T ‘[(L +2CQ80) j (5 + 4co8g)

***” ' Sincc

1 2^°

(l + 2co«0) , (S+4©m8)

l + 2coe0

[RGPV Dec, 2004/ r [v fl» r- « - / < « ]

= co80+ iein0

/ = Real part (R.P) of i f Put

z - p*6 => ~r- = d$t and cos 0 = *~e t2

7 = R.P of i f

1+2*

4*

2z

—

where, C is the unit circle Iz |= 1.

(1 + 2z)dz = R .p .o f ijr -^ p 2i Jc ^2 Z+ 52 + 2 f = R .P.of — f ± 2i Jc (2 + 2)(l+ 22) - R ^ o f ^ r f r-irrc fe 2i Jc (2 + 2) 7 = RP*of ^ For the pole o

Jc f(*)dz

where, f(z) = ^

-

...(1)

f : put 2 + 2 = 0 =>2 = -2.

z = -2, is a pole but it does not lie within C le. 12 1= 1, because (2 1=4—2 1= 2*1. Hence by Cauchy's residue theorem £ f{z)dz = 2m(0) * 0. Hence (1) becomes: J = R.P. of ~ x 0 = 0 2i l + 2CO80 (5 + 4cos0)

r

A A" 5-

Example f .95 .*Evaluate by contour integration, £ Solution,

Put Z = Since sin 9 -

dd

TTJ

(5-3«in0) 2

dz d z - i e ^ d O => -r- = d 0 iz -is -I e -e z-z 2i

-r

*

1 2i

2i

dz iz

(5 -3 sin ^ )i 5 -3

where C is the unit circle | z\= 1. 1 - I 5-

3<*2 -1) 2iz J -L

where

az dz iz

r 4iz "T" Jc JC (10w -3z2 + 3)2

4 izdz (3 « + l)J(w+3)*

4 izdz ( 3 z - i ) 2( z -3 i)

4iz f(z) = ( 3 z - i ) 2( z - 3 i ) 2

For the poles off(z) : put ( 3 z - i ) 2( z - 3 i ? = 0 => z - if 3 and z - 3 i are poles, each of order 2. At pole z - i/3, then | z | ={i / 3 1= 113 < 1, inside the circle C. At pole z - 3/, then |z|=J 3i |*= 3 < 1, outside the circle C. Thus, pole z ® i/3 of order m = 2, inside the circle C. [R a o /M U s..., =

<*&

lim .£ { . ( ^ - ‘/3)24 «

4i d — lim -------

1

~ l im 9

(z -3 i)2 ( \ ) - 2 z ( z - 3 i ) (z - 3£)4

Functions or By residue theorem:

ft* i.

f(z)dz = 2 » (sum of residue at each pole in C.)

SHJ = — 5r . =* 2 «. x f^ - — r Hence (1) becomes .

f2» dO _ 5jt 28 J, (5 -3 sin 0 )2 * 32*

An8‘

Example 1.96; Apply calculus o f residue to prove that co*2Q ( l- 2 a c o a 0 + a * )

r Solution.

2u a 1 ( 1 - a 2)* <1^* [RGPV June 2003, June 2009 and Feb. 2010]

Put, z = et& => d z = et0.idO => dO - —

Since,

-

e*+e*

1 (

«»* =— — - j

v il

?J

I-S

,

2»

"

1)

r+-J

T—TTTl+O” T

dz

22(z-o )(l~ aar)

1f

(*4+l)
2i t

z ( z - a z - a + a z)

2i

*4 + l where C is the unit circle, C:| z i= 1 and f(z) - — w, .• 2 ( z —a )(L-az ) For the poles offfz) : put z 2( * - a ) ( l- a z ) = 0 =£>z = 0 is a sole of order 2 and z = a, z = — are two simple poles. At pole z - 0, then | 2 |H 0 1= 0 < 1, inside the circle C. At pole z = a, then | z |=| a 1= a < 1 (v a 2 < 1), inside the circle C.

At pole z - \fa, then | z |-1 1/a |< 1 (v a 2 < I), outside the circle C.

...(1)

90 1

I Thus only pole z = 0 (order 2) and z = a (simple pole) are inside the circle C. [Ree* f ( z )\z-a = lim [(z-o )]/(z)

(£ l± lL _ ]=liin (£l±iL„ toliu = lim '(Z-„) *-KI 2: (2 - a ) (1 - cup) J *-** z*(l-az) a2 ( l - o s) ‘ and

1!

dz [ Z ‘J i z - o H l - a z ) ) ^

{ d z [ z - a f - a + a 4z JJ^

f ( * - o z 2- a + a 2;g)(4z3)-» 3) - ( z 4 + l) ( l- 2 az + a 2) ] | (z -a z 2- o +o V By Residue theorem: = 2m

(1 + a 2)

“

o2

/(z)dz = 2*x[ERes/(z)at each pole in C.]

o 4+l o 2( l - o 2)

l +o2 o2

4wro3 2\

*

Hence (1) becomes: r*» cos 2# 1 4i^o* / « f * ____ —"■■■ d0 = — x (l-ao co sd + o * ) 2i ( l - o 2) *

Solution.

Put

f

f «

Proved.

(id 4k (2 + cos 8)2 ~ 3^3 *

z = e*, then cosO = —^z + i j =

• J _ f2* rifl ■*° (2 + cos0)2

2na% l- o *

1

j

dz 7- , where, C is circle |z | = 1.

2+ i £ ± l 2 s

...(i)

where,

/(*) “

(z2 + 4z + l) 2

*

F unctions o f C o m m a Variables | 9 1

For the poles of f(z) : put,

/ .

. „ i , , „ - 4 ± x /l6 - 4 -4±2>/3 + 4 z + l) = 0 =>2r + 4ar+ l= 0= > z = ---------------* ----- ------

i.e.

z - -2±V 3 are poles of order two.

Here, pole

* = -2 + ^ 3 Hes inside the ciicle |*|= 1,

because [2 1=|-2 + >/3|=-2 + ^3 <1. But pole * = -2 - V3 outside of C. Residue at (2 * -2 + i/3 ) of order m = 2 is

(2+2->/§)*.« j * ,-11 k f <& (2+2-Va^(2+2+V3)*J i#»-s+wS dz 1 (2 + 2 + $ )* J] (z + 2+V3)2.l-g .2 ( g + 2 + V3) (z + 2 + ^ 3 )4

-2+V3

2 + V5 + 2->/3 = 4 1 ~ (-2 + ^ + 2!+«$)* * 8.3%/3 “ 2.3^3 “ i j s By Residue theorem:

1

£ f(z)dz » 2 » [£R et/(z) at pole in C ] s ^

x^

1

r r** 4 *i 4* Hence (1) becomes: ^ = J, (2Ve0itf)r = T Xi ^ = i ^ ' #st BMVlr 1.M r AmfeMir j,

Mutton.

Put z = e*8 =»

^

resUuethMfem'

c o a 6 a ^ ( * + ;z) anc* S’n ^ = 2 i(Z - z J

/ = (*' --------- d* £$ ------------f t «» (3 -22co80+ein0) c o s0 + sin 0 ) ” ^Jc

I *(*2 +l) 3 ” 2 2z

dz (*2 - l ) ' w 2*2

ffi

*

A“ -

9 2 1EngiM ^ |I # I 9 I M | qbA -ic

*1(l-20+6*e-(l + 20

2 r L ( l- 2 i) *

dz

1 £b= (rH )Jc (1-2©

...d)

(1 - 2f) j

3 6iz (1 + 2*) For pole of/fe>: put * + j ^ — f T ^

=»

«2(l-2i)+ 6i*-(l + 2i) = 0 - 6 i ± J - 36+4(5) - 6i ± 4t -2 i , -10* z ~ -------—----- -------- --------------------------a n d -------2(1-2i) 2(1-2*) 2 0 -2 * ) 2 (1 -2 0 “ j~^ 2 ^,ap l , #

i e *z ~ a

—I At pole z = a = — , then |* |-

y^ arc S*®P^® P°l®s*

-i 1-2*

which is inside the circle C. -5 i At pole z - p = ——— outside the circle C. (l-2 t)

[Resf(z)]Zma = lim (* - a).f( z) = lim (z - a). *-K* 9-UX v /(*) = 1

( -4 , 5i } U -2 i

1

6w (1+20 F^aT T ^T

(z-a)(z-0)

(1-20 4i

1-2* J

By Residue theorem: f( z )d z ~ 2m (£R es/(z)at each pole in C] . (1 -2 0 = 2jbx X— i = - ( l - 2 0 . 4t 2 Hence (1) becomes: Ans.

C<&lf*fcVARIABLE8 | 93) mmpie 1.99: Prove that J8

sin*6 JO %n. p iT T E * (a +b coa 9) = jJT»a ~ v 'a )> wAcrr a > b > 0. [RGPV June 2008/

m dn

“www-

u t/. p

- J S S * * ,
Jo o + 6coe0

.g rP g g L ^

2(a+6cos0)

f*' 1 -V * ^ = Real part of J0 2a + 2*ooe* Putting z = e10 =>

and cos# =*

/ = Real part of f - —

-2 we get

rr»T“ > where C is the unit circle \z\= 1

Jc 2o+b(z+afl) iz

-R ealp arto f j £

1 1

<** = Real part of £ /(*)<&,

*t \ _ (I"**) where f M m w ? 7 £ i 7 i ) For the poles o f/fc): put i(bz2 +2az+b) = 0 => bz1 +2az+b = 0 -2a±nJ{4a2 -4 6 2} - o ± V(o2 -&2) Z = . . . . . - . - . - - - . s . . - ------------26 6 I)

Let u

S'

or ------- Jti--------- and P * ------— ------- are simple poles. o o Since a > 6 > 0, so that Ip |> 1, hence pole z * p is outside the circle C.

Also I ay? I = 1, so we have I or |< 1. Hence z *= a is the only pole which lies within

C :|* |- l. [Res f(z))tma = Urn ( z ~ a ) f ( z )

[By definition]

(I-**)

ten-

r

ft \ 1” ^ /(*)= T i(bz* + 2oz + b)

I -2 * ib(z - or)(z - P)

.H m = , g , - . l (g I - . . l£ . *-•« i6(z->£) i b ( a - p ) ,ib (a - fi ) i b ( a - p ) ib

[ v ^ = l]

94 | Engbo&ung M/mtEMAncs-IIl Hence by Cauchy's residue theorem (1) becomes : / = Real part of £ f(z )dz - 2m [sum of residues at poles within C]

Proved. Example 1.100: Using the calculus o f residue, prove that r xain(ax)

J.

Solution.

Let

2

/(*) =

y 2.***

y dz,

where C is the contour consisting of I 1 (i) Real axis from -R to R R (ii) Semi-circle Cr of radius A:! z \ = R pR Hence £ f(z)dz = J ^ f(x)dx + J c#/(*)<& To prove that, as R -+<*>,

Ssv\ ^ 11

y S

' 0

+R ...d)

f(z)dz -►0.

On the real axis z = x and on the semi-circle z ^ R . e * and limit 0 varies from 0 to n. Now consider £ f{R»J0) . B . e P i d 0 * £

-r -r

f( z ) d z

—(2)

R e ie eia(Rco*e+iR'toe) Reie M v « * )R .e ^ .e *

ze z*+k2

MB R 2e2Uf +k2

[Now Divide N r and Dr by R 2e**] ~ r Since

. e l« lit I 11+ lr./2 (cos20 - i sin 20) |

g-sfirin0.1 I J l + k 2j r 2(co e2 ?-im n 2 0 ? I

[.• e“2W = cos20-isin20]

F unctions o f C ompI ex V ariables | 9 § -afteinff

' e

faking real pamoniy]

-aRnnB ...(3)

(l-fc ’i r 1) Therefore from (2) and (3) we have e* .id 6

(J lV W ) ...(4)

[v

r

a - f c ’/ r V Jo0

io r0 £ 0 iirl2 ,^~ r-'z.‘l l x e

^ ^ •d e

2 r*/2 -alt— e * d6 £ --------T---- 5-1 a - f c , R s) »

[-

...(5)

By Jordan inequality — £ It

&

^ 1or sin# £ ~

ft J

T#/* ,-alHUVir (1- k lR * )

~aR .~

ft J

,-aR(2*)/jr

-] *72

l-Jfe2i r J [ ' a R

- * ___(e*^ -1 )= * 0 “ «““*) o fl(l-fe 2i r 2) a f l ( l - * 2j r 2)

...(6)

/.A s ft -> oo, we get Jc
Jc f(z)dz = j* f(x)dx =

f(x)dx + 0

f ( z ) d z - 2m (Sum of Residues)

[vBy residue theorem]

The pole off(z) are »**±ik o f which z = +ik ties m upper tetfptane 1*1=1. £

f(x)dx - £

'2 4

:# 4 .j^ j

2m

-z~ e 2

* ^ \ 2ia j

*»*

...(7)

»«** Henee-

£

x ’^ h k r d x =

Equaling (he imaginary parts x sin a x , * 2 +**

£

or

r

or

Example / ,101: Solution.

x sin ax , 7 T F

jr _<* T‘ '

iy contour integration

f* cosmx . |

fRGPV Dec. 2*06]

Let J * |* BE?*1* dx. ,0

X* + 1

.mix Taking* /(*)==

(* + 0

/= f /(«)<&- f *4— ^

* * (*2 +i) -* Where C is the con**sr consisting 6f the (i) Real axis from -R to R

o ».

(ii) Semi-circle CR of radius R in the upper half plane. Then £ /(*)<£* - £* / ( x ) d x + / ( * ) d *

i n: .

To prove that as £ -►*>, : f " r- •/.'• 5

f( z ) d z -> 0

m

F u n c tio n s o f QomnJExVABUWGS | 97

z |2 - I |

N

[.-. U 2 + l |- U 2 - ( - t ) |a |z |2 -l] On semi-circle z * R ^ 6 and Q varies from 0 to it.

... ..

T

\t\=R=>\dz\-\R.t?°.ide\=RdO and

I

M

j_ j

g-foJfsInPgfatRcosff | _ e~mRnn6

to

2fl

r /J

-s

€-m R 2 0 lx

•

40

j By Jordon’s inequality — n

0

or — ^sintf^O Jt

Ktl 2R (« * -1 )

-2 R x x r (tf-D M l

m itt ir .

,,

xR m (fi2 - I )fl Hence

/(z)<2z|-»0 ae fl -> °o

Hetce ;i) becomes: £ f(z)dz - £

f(x) dx => £

f( x ) d x « 2m [$um ofresidue]

...(2)

Nowjfe) has simple pole at z * ±* of which only z = +i lies in the upper half plane; 1*1-1. imt

(z -Q e (z-i)(t+iy

.-it

2i

9 # 1BkmmmHwo Mxnmucnc*ms Hence (2) becomes: —r— dx =■2m x — —- m x 2 +l 2i Equating Real parts on both side, we get

£

cosmx ,

r° cos m x ,

C

Example 1.102 : Evaluate by contour integration £ y ^ Solution.

k

■m

Ans.

» i/o > 0 .

Let /(«) = — — 4 and L /(*)<** = f ; - 4 dz, z*+a* Jc z +
/(* )d * +

f(z)dz

...(1)

For fee poles f[z) put z 4 + a 4 * 0 => z 4 * - a 4 => 2 = o (-I) (- i) !/4 = (eos(2n + l)/r + i s i n ( 2 * + i » l/4] /.e. z4 « o 4. e ^ l)' => 2 = a w h e r e w = 0, 1, 2, 3 le. The poles are at z = oe*>/4,aei3' /4, ae*5*M, aeilx!4 of which first two lies in the upper half plane 1*1=1. Let a denote any one of these poles then a * =

[v z4 +a* - o]

Now [Res. f(z)]gma = h m ( z - a ) f ( z ) > l-W

formusingL’hospitalruleJ

= lim (z-or)—r—^—r ar F— z4+ a 4 LO t - = —*=V lun—-7 *-** 4z 4a'

Of 4a4

a

4o4

»

...(2)

i* /4

(i)

[By (2)1

W**- /<*>]«**" ae,»3jr/4

4a

[By (2)3

F u n c tio n s

- 1 r . iwU .

.•. Sum of the Residues = ^ r t 0 ®*** +ae - U 2 i \ e i" * - t i "*'\

IXmUj __

|jc. A * H “ | k =

r*

i?d<9

— 3--------r - » 0

Jo J f - a 4

C om plex V a ria bles ( J l i

I |c**^4 t C*3* ^ ]

i

iJ2

j- i r

Now,

op

...(3)

7 ^ ^ . D

as If -to o

Hence (1) becomes:

B;

£ f ( z ) d z = f t f(x)dx+0

=» f° —T— -T- f /(*)
4a

x l2o> '

dx a - f 7r Jo x 4 + a 4 * a3

r

dx x jc4 + a 4 i j l a 1

Ans.

4a 3

14? TAYLOR'S SERIES If a fu n c tio n ^ is analytic at all the points with in a circle C with its centre at the points 'd ;r* ’ and radius r, then / ( * ) = £ a* (* -« )* ; where £*,

/W -£ *-0 or

n'

with 12 -

n!

0 1< #

f(z)=f(a)+^— ^(z - o ) + ^ p (2 - d f +...... + 1! Z!

A!

(z - a)* +

LAURENTS SERIES jr h

If/(^ is analytic in a region A, bounded by two concentric circles C\ and C2 of radius Ri (jR, >R2) with center V then for all z in &

/(z ) =

2

*-0

a ,(* -a )" + 2 ] 7 -^ x r , with |z-o|
R\ and

1 0 0 1Enokvesmo NUnffMAncfrJH _

where

i f /(O * ~ 2 * Ar, f t - o r '

and

Example 1.103; Expand log z in a Taylor's series about z ■ /. Solution. Let /(z)=logz. By Taylor's theorem : « * > « £ (2 - o r ^ ) B-0 *• =f(a) + ( z - a ) f ' ( a ) + —~

...(I)

f ’ (o) + ....

N o w ,r (* )=z-. m = - 4z. r w = - ^z . r c * ) = -z4 ..... z /a )= io g i= o ,A i)= i. r a ) = - i . r a ) = 2 !, / ,”ia ) = - 3 ! ,../" ia ) = ( - i) " '(n -i)!e tc Therefore ( I) becomes : /(z) = log* = /(l) + (2 - 1) A l) + ^ p - / ' ( l ) + ^ p / ' " ( l ) +... + or

/(2)=(2 - l ) - ^ i z l ) l + ^ ^ ~ . . . . + ( - l ) - , f c ^ + .... 2 3 n

+ .. Ads.

. sin z

Example 1.104; Find Laurent series about * =ji fo r the function f ( z ) Solution.

sinz (* -* )’ we can write sinz = [s in ( z - * + * ) ] - - sin (z-;r) Expanding the series as ein(z - /r), we get

z -n

Since /(*)-■

f, v (z - jr)3 ( z - t ) 5 « * * = - < * - * ) ----- 5TL + ^ j —

v sinz • A*)=-(z-Jf)

[.’.sin(;r+0)=-6in0]

(z -* )7 , | 7 ? - + ......

1 (z-ff)

, , (z-n)1 (z-n)K. ( z - x f * - 1 + — 51--------- u ~ +~ v . —

...(I) [By (1)]

....

Reamrk ; Binomial Expansion: (1—jc)-1 =\+x + x 2 + x 3 +......qo, if |x | <1.

Ans'

FbNcnQNS o p C omplex V ariables | 1 0 1

1 in the regions V - 3 * + 2) (U) 1 < \ z \ < 2 ( W ) \ z \ > 2 (tv) 0 < \ z - l \ < l [RGPV Dec. 2005] or +

1.105 ; Expand in the series the/unction f (z ) (i) 0 < \ z \ < l

Expand A * ) *

[RGPV June, 2009]

( *- 2) ** *** reS^>n \z\> 2.

Resolving into partial fractions, we get /(*)=

1 I *2 -3 z + 2 ( z - 2 ) ( z - \ )

z

1 1 -2 z- 1

(i) When 0
F ro m (l):/W = — - 2^ - -

• < -3

— - ih

i

tv a - * r , = £

= lt-0 1 x 4 ta*0 i f ’" S

j 1 2**1}**• if 0<|*|<1.

(ii) When 1<|*|<2: i.e, \z\>\ and |z |< 2 => | j | <1 and |j < 1

Using binomial expansion, we get

Ans.

1 2 1^ 1 <1 and z 2 2 z

(Hi) W henUf>2 : Since \ z |>2:=> From (1): /(*) =

1 2r —2

I

- K '- r - K '- i r Using binomial expansion, we get Ans.

m

=- t ( i f - t ( 2 " - l ) - i r . if |z |> 2. z t 0 \Z ) *£* W ft 3? 1 1 (iv) When 0 < |* - 1 |< I : 1 ar-2

1 ar-1

1 -fl-(* -l)]

t 2~l

From (1): /(*) *

z - 1-1

z -1

, -i

= - Z (2 -1 )* -< * -!)-*

[‘.•(I - * ) " 1 =S**J. Ans.

J*»#

(a?- 2 ) ( z + 8 ) Example 1.106; Obtain the expressionfo r ( , + !)(* + 4 ) which m e voUd when a) u i < / Solution.

Let,

m

i< \z\< 4

m

u i> *

z 2- 4 , 5*+8 , 1 / ( * ) - —---------- ==>1 - - ----- —— s - - l — z 2 + 5*+4 (*+ 4)(*+ l) 1+* ,

(i) For |* | <1: we have

4 4+ 2

{v Reaolving partial fractions}

Functions cwC ompiex Variables j 103 =l- l- £

(-!)»*• - 1 - j ; (-1)” ^ " . H»1

= - l + £ (~1)*+1[l + 4~*]z*; when |ar|
1 z 1 \z I (ii) For 1 <| z |< 4: we have 1—: < 1 and — < 1 i.e., z <1 and 4 <1. \z | 4

,-> -K 4 r-H F -4 [ ,- w - -

z fz ‘- 4 +U ' -

= S (-o ’ Ml which is valid in 1<\z |<4. 4 1 1 (iii) For I z |> 4: we have 7—7 < 1 and 7—7 < —< 1 IzI \z \ 4 I)"

,

•4 [4 A+l

„*+i

= i + i ^ a + 4 * * 1) •-0 z

= 1 + £ (-1)*(1 + fmmpie 1.107; Expand f(z ) «

z +3 — -——

(Z —Z —Z /Z

wk®n \ z\>4.

Abs.

in powers o f s.

(!) within the unit circle about the origin, (ii) within the onnutus region between the concentric circles about the origin having radii 1 and 2 respectively. (Ill) the exterior to the circle with centra as origin and radius 2 Le. fo r |z|>2.

1 0 4 | E N 6 ffe £ R ra M M H m n c » in

Solution.

z46+ 3*' (* t/ \ _ _ > +* j3)) Let « « * t f _ z _ 2) z ° * (* + l) (z -2 )‘ Hence resolving into partial fractions. „ .

3 2z

2 5 3(2+1) 6 ( * - 2)

(i) For 0 <1 z |< 1, we have -i

[v |z |<1]

- t t

Ans.

[§<-»■-$)■]*•I

(ii) For 1 < | 2 1<2 =>

and M < 2 = > |||< 1 . we have

,

3 2 ( , 1Y‘ 5 (. zY ‘ / w " - * + 5 l 1+; J - l a ^ - a j = _ i.+ 2 f , . i 4 2z &z I z s?

4 £

J

\

J

_ ± f i +£ + 12|_ 2 2

J +J 2

**■

~ i — 2 -f (-i)*-L—l y r * r **

3*a

**

i i (iii) For \ z |>2, => j T p J .

=

3 2z

2 r.

1

3z L

*

“ S 'W

121 , |z| we have

X 1 *2 z3

I 5 r, 2 l 2 23 1 + -^jK + -*+ “ 5-+“ 5'+... J &*[_ * * z3 J

Ans.

F unctions o f C o m pi £ x V ariables 1 1 0 5

>1.108; Obtain the Taylor's or Laurent's series which represents the function. f= lf/+ V > 7 * + i> ' (l) | r | < / m / < | z | < Z Resolving/fo) into partial fractions, we get 5|_* + 2 (i) For | z | < 1, we have

z2+ lj

/ w 4 i [ !+f ) ' 4 <2- 2)0” ! r ' 10

-[(1 -z2 + z 4 - z 6 + ... + (-I)**2" +...]

2"

Ans.

10

This series being in the positive powers of z represents ’Itoylor’s expansion for ffz). (ii) For 1 <) z |<2, we have

!-it “ r & t » CIRCLE ANDRADIUS OFCONVERGENCE • Let power series about z = a. w: fr m

'■

/(*> = £

...(1) m* The interior circle |z - a\ < R which include all the values of z fotf(z), is called the circle o f convergence of series (1). The R is called radius of convergence of series (1). bar*: r„ ,

l Let z = a be centre of circle andf{z) be analytic function in the circle j s - o j e i ? , except the poles a ,,d a ,a ,

soon.

Then, R = mm. distance of poles a ,, o 2>._from the centre z = a.

1 0 6 | E n g in eerin g M athem atics -H I

1 . Expand the function in Taylor's series about z ** 2 ani (z-l)(z3 ) indicate the circle o f convergence. 1 1 Let f{z) = ( * - 1 ) 0 - 3 ) z 2 - 4a:+ 3 t = -11 - ( z - 2)2]'1 = - [1 + (2 - 2)s + (z - 2)4 +...] ( z - 2)z ~ l

Example 1.109 .* f(z) =

Solution.

or

/(z )-

(~ 2) , whjch is Taylor’s series about z = 2.

Circle o f convergence: Here centre is z = 2 i.e., C enter* (2,0) The poles f(z) are z = 1 and 3 i.e., z = (1, 0) and (3, 0). distance of pole z = (1,0) from the centre z = (2, 0) d, =V( 2 - l ) 2 +0J =1 Also distance of pole z = (3, 0) from the centre z - (2, 0) di = yj(3 - 2 ) z +02 =1 Radius of convergence; /? = min {d,, Thus circle of convei^ence = |z - a |<

} = I. z - 2 J<1.

Ans.

Exercise-l(A) 1.

Prove that the function | z \2 is continuous every where but no where differentiable except at the origin.

2.

Determine analytic function f(z) = u + iv, if u = log(x2 + y 2) + x ~ 2y.

3.

If iv = A log

'z - o ' \ z +a )

, where a is real, show the lines u ~ constant, v - constant, form two

orthogonal families of coaxial circles. Prove that this function may be used to represent a source and sink combination. Show that f(z) = z + 2z is not analytic anywhere in the complex plane. Show that the function f(z) defined by x +y =0

z = 0.

is not analytic at the origin even though it satisfies C-R equations at the origin.

F u n c tio n s o f C o w i£ X V a r ia b le s |

L

107

If f(z) = u + iv is a regular function of z in any domain, prove that (0

(ii)

! t + t t ] i /<*> !p= p 2-1a * ) r 2 -i m dy ) i

L

+ i L j iM

V.5x

r=

p (p _ o . i u r

M

i2 •

f\z) i2 .

dy

I

Determine a, b, c , d so that the function

' i

/(z) = (*2 +
|i

Detennine the analytic function f(z) in terms of z whose real part is (0

cosxcoshy

(ii)

e_Jt(x 8 iny-yco8y)

(iv)

e~x {xcosy + y&\ny)\f{0) = 1

(v)

sin 2* ^ j ^ y + cos2*‘

(iii) e* cosy

10. Find the analytic function f(z) in terms of z whose imaginary part is i

x -y

it- (0

(fii)

x 2+y2

0*)

sinhxeos^

&ry-5x + 3

(iv) ^ p - + coshlcos^*

Prove that u = x 2 - y 2 - 2 x y - 2 x + 3 y is harmonic. Find a function v such that f(z) = u +iv ^ . is analytic. Also find f(z) in tenns of z. An electrostatic field in the jy-plane is given by the potential function f - x 1~ y 7, find the stream function. cos* + simc — L' If f ( z ) ~ u + iv is an analytic function of z and u - v - ----------------- -— .prove that 2coax - 2co8hy f*

f(z ) =

-

when

= 0-

(4. Uffz) ~ v + iv is an analytic function find ffz) if u + v [5. Find the analytic function f(z) = u + iv, given t?

(i) u = a(l+ co s0 )

(ii)

r* 0 .

x

—•, when /(I) = 1. +12

1 0 8 | E n g in eerin g M athematics-H I

16.

Prove that y = lo g [(* -I)2 + ( y - 2 ) 2] is harmonic in every region which does not include the point (1, 2). Find a function ^ such that # +it// is an analytic function of the complex variable z - x + iy. Also find f(z) in terms of z.-

2. f(z) = -2z + 2ilog(iz) + c. 9. (i) c o s z + c

1+ i z

10. ( i ) ------+ c

Answers-1(A) 7. a = 2, 6 = -1, c = -I, d = 2.

(ii) i z e 'z +c

(iii) e* +c

(ii) isin h z + c

(iii) 3z2 - 5iz + c

11. v - x 2- y 2 +2 x y - 2 y ~ l x and f(z) = (]+i)z2 ~(2 + 3i)z.

8. k * - l

(iv) cot z +c (iv) —+ icoehz + c. 12. t// = 2xy +c.

14. i z i +c. 15. (i) a(1+co80 + isin01ogr) 16. ^ - - 2 tan

0 cos <9+ ( r —0 sin0 + c. (ii) ( r +— I - rj I r)

and /(z) = 2 ilo g ( z -l-2 i) + c.

»•«* «* K * '* (• C ; Exercise-1 (B) 1. 2.

Prove that bilinear transformation maps circles or straight lines into circles or straight lines. nnm te /? /van Ka mrt In fir Prove that every bi Itnikar tronrfnrm otirtn 11/itk hi/A (in ita r t

'v)-a' \w ~ P )

z - a '|

U -P p J) Prove that every bilinear transformation which has only one fixed point a can be put in the form 1 = ----1--■f A.5 ------w -z z-a Find the fixed points of the following transformations : z (’) W ~ T2 ~ - Z7

3 z -4 (") w = TZ —T\

3i’z + l W W = - Z7 +7lT

(2 + i ) z - 2 (iv) w = -------- :— . v 7 z +i

iz+2 Show that the transformation w = ~^Z+J maps the real axis in the z-plane onto a circle in the »plane. Find the centre and the radius of the circle and the point in the z-plane which is mapped on the centre of the circle.

F u n c tio n s o f C o m p lex V a r ia b le s | 10 9

6.

p ds Show that J ~ in an inverse with respect to the transfonnation az +b w --------j . ad -6 c = 1, where ds = J d x 2 + d y . cz +a

7. 8. 9. 10.

11. 12.

5 -4 z Show that the relation : w = ——- transforms the circle 1z 1= 1 into a circle of radius unity in the w-plane and find the centre of this circle. Find the Mobius transformation which makes the set of points in the z-plane (i) a, b, c (ii) 2, l+2i, 0 correspond to the points 0, 1, oo of the w-plane. 1+ iz Show that w = maps the part of the real axis between z - 1 and z = - I on a semi-circle in the w-plane. Find the Mobius transformation which maps. (i) 1, —f, 2 onto 0, 2, respectively. (ii) I, - I, oo onto 1+ i, 1- i, 1 respectively. • 1+ z Show that the bilinear transformation w = ----- maps the region I z |S 1 onto the half R(w) > 0. I 'z z-i Find what regions of the w-plane correspond by the transformation w to

(i) the interior of a circle of centre z = -j. (ii) the region y > 0, * > 0, |z + i|< 2. 13. Show that inverse of point a with respect to a unit circle is i/o . i f z + 2^ Show that the relation u>= - . - - - j transforms the real axis in z-plane to a circle in w-plane. V Find the centre and radius of circle and the point in z-plane which is mapped onto the centre of the ‘ circle. i$. For the transformation w = z 2, show that the circles I z - a 1= c (where a, c being real) in the zplane correspond to the limacons in the w-plane. \6.

Show that the mapping z = 4w transforms the family of circles | tt>-l |= X into the family of leminscates. 1z - 1 1| z + 11= A, with focal points at z = ±1.

(7. Show that the transformations w.(z +i)2 =1 maps the interior of the circle Iz 1= 1 in the z-plane *' 1 on the exterior of the parabola — = 2(1 - cos^) where w - pe'*, in the w-plane. P 1$. Show that the transformation w = - ^ z + — 'j maps the upper half on the circle Iz |< 1 on the upper half of the w-plane.

1 1 0 | E n g b e e r m o M a th e m a tic s-!!!

j 4, ft yr \ ft ----- I represents I ar |< 1 on the strip ~ y ( u ; > < -

(

of the w-plane. 20.

If w = 2z + z 2, prove that the circle | z |= 1 corresponds to the cardioid in die w-plane. Answers-l(B)

4. (i) z = 0 and 1

(iii)z = i

(ii)z = 2

(iv) z = 1± t.

7 1 ( 7 9 5. In vv-plane u1 +v7+—v — = 0, centre 10* ~ —I i.e. w = -~ -i and radius ~4 2 V *>/ 8 ® In z - plane, z ~ —. 4 7. Centre is 10. (i) w =

8. (i) ut

• 2(z-» (1 + 0 ( 2 - 2 )

m m

'Jr M

-

b~c ( b-a{z-cj

.(2- z (ii) w - i z j

.... z +1 (it) w ------z

W KiT

' W Exercise-1(C) 1+2z 1 1 , 2 J ---- T ” - T + — t + z - z +z .... where 0 < |z |< 1. z ‘ +z z z Find Laurent’s series expansions in powers of z of the function^,) given by Prove that

Z(\ + Zl )

3.

by series of positive and negative powers of (z -1) Represent a function f(z) = <*-!)<*-3) which converges to f(z) when 0 <1 z - 1 1< 2.

4.

Obtain the expansions for

(i) UI<1

(z -2 )(z + 2) which are valid. When (z +1)(z + 4)

(ii) !< ]z |< 4

( iii) |z |> l.

F u n c tio n s o f C o m p lex V a r ia b le s | 111

2* 25 Prove that: ta n -1 z = z - — +

when |z |< 1.

(z — (z —|)^ Show that: logz - ( z - l ) ---- —— + — ------ when I z - 1 1< 1. Expand f{z) ~ ( * - 1)2 K

I Find the expansion of ^ 2 + 5)^2 + 2) 'n ^ (i) !*!<•

9.

Find expansion of

(ii) l< |z|< > /2 z 2 +3

(i) |a r-l|< 2 10.

il.

about 2 = 1 , as a Laurent’s series. P°wers of 2, when

(iii) )z|> 42.

valid for:

( ii) j* + l|< 2

By considering the Laurent expansions of —^ -r about the points z - ±1, show that the function 1- z , 0 <11 -JT t< 2 . ; /(2) = 7 4 Discuss the nature of singularities of the following functions and also calculate the residues at the singularities. 1 sing

® z ( i - z 2)

^

(z-x)

pQ*' Find the zeros and discuss the nature of the singularities of ( z - l)

Also calculate the residue at the poles.

1 71 — at z ~ —. (sin z -c o sz ) 4 Discuss is the nature of the singularity at z = 00 of the function f(z) = cosz - sinz.

13. Discuss singularity of 14.

15. Discuss the singularity of the function f(z) - —

at z = 0.

“ *1 *

16.

n Discuss singularity of the function f(z) = tan — at z - 0.

\zj

17. Find the nature of singularities of the following functions (i) -—— at z - 00 (ii) coaec] —) at z = 0. l + e* \z j

1 1 2 I E n g in e e rin g M a th e m a tic s-!!!

Answers-! (Q 2. f{z) = — z + z —z +..., when 0 <| z |< 1 z t( \ ~ * 3‘ /(Z)~ 2 ( z - l )

3^

f z -lV I 2 .

e2 2e2 2e2 4e2 2e2 . ------- r-+------- T + ----- + ----- + -----(z —1) +.. ( z - l ) 2 (z -1 )2 z —I 3 3 ,2n « n-0

j»»0

pt-1

»2»+2

M ) n( l- 2 " ) n-0

_2 » + 2

I 1 11. (i) The points 0, 1,-1 are three simple poles with residues 1. - — — respectively. (ii) ;ris a simple pole with residue -1. 12. 2,1 + — are zeros; 0 is a pole of order 2 and 1 is an essential singularity. The residue at z = 0 ii nn | 2 cos 1 - sin 1. 13. Simple pole. 14. Isolated essential. 15. Non-isola!ed essential. 16. Non-isolated essential. 17. Non-isolated essential, Non-isolated essential.

»

p Exercise-! (D) Use Cauchy integral formula to evaluate : sin to2 + cos® 2 d z , where C is the circle | z | = 3. L ( z - l) ( z - 2 ) Evaluate the following integral ic “ Cosz d z t where C is the ellipse 9x2 + 4y2 = 1. Use Cauchy integral formula to evaluate :

4.

3z + z £ (z2 -1) dz, where C is the circle (z - 1| = 1. Evaluate by using Cauchy's integral formula z -l L
f. F u n c tio n s o f C om pi e x V a r ia b le s | 113

i: R

Evaluate f k

— dz, z 2+1

Jr t

r sin3 z Evaluate by using Cauchy integral formula : I --------dz, where C:| z |= 5.

w h e re

C is Iz + i 1=!. 1 1

2

r |

Evaluate by using Cauchy integral formula

■,

f * +Z+ dz, where C is the ellipse 4x2 + 9y 2 = 1. * z 2 - 3 z +2 Evaluate by using Cauchy integral formula

Z + —

L

I,

j*

COS 7TZ

M

QQg jgp

{ —----- dz, around a rectangle with vertices 2 ±i and - 2 ±i. )c z 2 - \ Evaluate by using Cauchy integral formula } ------ dz, where C is the circle [ z - l [= 3 . Z~~7T

,i

*

16. Evaluate f \ - dz, where C is the circle | z \ ~ 4. K sm h z II.

Evaluate f --------- dz, where C is the triangle with vertices (0, 1) (2, -2) (7, 1). ■fc z" sins: [Using residue theorem to solve the following integrals :

J2.

where C : |2 - l |* l .

b-

£ w f * ? ' where C : |z “ i,=2

j I '

f 3z 2 +2 dz, where C : \ z - 2 \ ~ 2 . Jc (z ~\} (z 2 +9)

I5-

Zz + l1 ( 22 Jc (2z - ] ) 2

*

I

Ii.

I
I*.

£ ~ T ^ Z‘

(

?

, C C :|2 ,= L T

“ h e re

C : U , = I-5 '

1 1 4 | E n g in e e rin g M a th e m a tic s-III

19.

£ z 2eUxd z ;C :\z \= \.

20.

Prove that

sin20 ,a I n . / I TT. I r ---- ~ d d ~ —y { a - s a - b ), where 0 < b < a. Jo a +bc080 b Using residue theorem to solve the following integrals :

f r z s -

Jo

add a 2 + s i n 'd ' ^ a>

22‘

_dt> _ 17 - 8cos#

25.

Prove that £

26.

Prove that

27.

C

f ' 1+ 2CO80 de " 5 + 4cos0 [“

1 ■ dx +^

Jo

S -4 co s0

“ 4^ ' ° > ° '

f” —^ ^ —5— -r-d * = — (a > 0, 6 > 0) J-« (* + a )(x +6 ) a +6

ctr , 3■ » (x +1)

28.

r *sinax , . L , —, » <**. (o > 0 ). x*+fc*

Answer*-1(D) 1.4ai

2. 2«i

3. 4jb

2m 4 'T

5. m

7. 0

8.0

9. - 2 m

10. -2*»

11.

12. - 2«i

13. «716

14. m

15. »

16. 0

17.0

18. - 2 m

8m 19 3 ?

7t 2L VI+ i r ^0 f

22.0

23. /r/15

24. * /4

27. 3*/8

28. | e - “ .

2i{-\y 2 n «•

6. 2»

N um erical A n a ly sis - I LI INTRODUCTION Numerical Analysis play an important role in solving many real life mathematical, physical and engineering problems. In the eighteen and nineteenth centuries, great mathematicians like Gauss Newton, Lagrange, and many others developed the numerical techniques which are still widely used. The digital computer, however, has enhanced the speed and accuracy of the numerical computations. Numerical computations is an approach for solving complex mathematical problems using only simple arithmetic operations. The approach involves formulation of mathematical models of physical situations that can be solved with arithmetic operations. One of the most useful tool ofNumerical Analysis is interpolation. According to their ''Interpolation is the art o f reading between the lines o f the table Interpolation is the technique to obtain the value of a function for any given functional information for it* Let y - f ( x ) be a functional relation between the independent variable x (i.e., argument) and the y be dependent variable (i.e., entry). Until otherwise stated we shall consider/(or) to be a polynomial of all discussions.

L2 APPROXIMATION AND ERRORS

[RGPV Dec. 2003]

In any numerical computations : Error = (True value) - (Approximate value) I'M* * Approximate Numbers : 9 The numbers such as 6, 9, —> 15, 36 are exact numbers. But these are numbers such as 4 ii r — ,n/2, it, e which can not be expressed by a finite numbers of digits. We may approximate them by numbers — **3.6667, >/2»1.4142, n * 3.1416 and e * 2.7183 respectively. Such 3 numbers are called approximate numbers. Significant Figures : The digits that are used to express a numbers are called significant digits or figures. The numbers 89653,2.8659,0.33857,119.36 each contains five significant figures. But the numbers 0.00869, 0.00000108, 0.098500 have only, three significant digits (869, 108, 985), since the zeros serve ***»’ only to fix the position of the decimal point.

1 1 6 | Engineering Mathematics-111

Rounding O f f :

20 We know that — = 2.8571428.....with is never ending. In practice, we have to cut it down# 7 ; ’ a manageable form such as 2.86,2.857,2.85714 and so on. This process ol nttiing off superfluous digits and retaining as many as desired is called rounding off. Truncation E rro r; They are caused, when an infinite process is replaced by a finite one. jg3 ^5 For example: a im = x~ — +

jp7

..too° = T (say).

If it is replaced by * - — + — = Tx (say). Then, the truncation errors- is |r~ 7|(. Absolute, Relative and Percentage Errors : Ldt *i be the approximate value of the exact number x, then : The absolute error Ea is defined as E„ = \x - x \j The relative error is define as : Er ~

X-Xi

- — » and

Ea The percentage error is define as : E p - —- x 100. X

Example 2.01 : lf 0.333 is the approximate value o f 1/3, fin d the absolute, rehtivc and percem# iR d P V J m e m errors. Solution.HtK x = 1/3 and xa = 0.333. (i) Absolute error : The absolute error Ea is given by ga _ | x

1 3

xa I - - - 0 .3 3 3 3

1000-999 3000

3000

333 1000

= 0.00033.

(ii) Relative error ; The relative error En is given by £, = — = ■ * 1/3

= 0.00099.

(iii) Percentage error : The percentage error Ep is given by Ep = Er x 100 = 0.00099 * 100 = 0.099.

At

i* •. »> Numerical Anaiysm-I | 117 &&82 :Find the relative error o f 2/3 is approximate to 0.667. Here x -- 2/3 and xa = 0.667

fRGPV June 2006/

Relative error E, x 2 = 3

x 667 1000 2/3

2/3

12000-20011 2/3x3000 Ans.

2,9# :tVhat are different types o f errors and safeguards against them T fRGPV Julie 2003J of e rro rs: The procedures are adopted to eliminate all mistakes whether these are human Le„ due to computer of technical. To the calculating device, when all such mistakes have been eliminated from a computation, the solution is not generally exact oh account o f inherent errors of various types. They are classified as follows : (t) ftttflcation errors : They are caused, when an infinite process is replaced by a finite one. For example, sin jc

=

Xs x & x 1 X ~ 3! + 5! ~ 7 l

x3 *5 If it is replaced by x - — + — = T, (say). o! 5! Then, the truncation errors is |T - Ti|. (J) ftmindin'; off error The errors are unavoidable in most of die calculations will be non-terminating decimals and for practical reason only certain numbers of figures will be carried in a calculation. If a number is rounded of with decimal place a change in the number of 7 * 10~(" + may increase <>i decrease.

Ea Perceirt;»>;0 error E P = — *100. x

A ns.

1 1 8 f E wb— ew wb M athem atics -111

2.3 SOME IMPORTANT RULES FOR ABSOLUTEANDRELATIVE ERROR! If x be any number expressed as jc = O .^i di di ... x 10", (where d\, di, d j , ... are decimal digits). Then, we have the following rules regarding the absolute and relative errors : Rule I : Absolute error because of truncation. Ifxa is the approximate value of x after trunca to m digits, then jir-*d < i0 " -m Rule II : Relative error because of truncation. If xa is the approximate value of jc afl truncation to m digits, then

Rule III : Absolute error because of rounding-off. If xa the approximate value of x aft rounding-off to m digits, then [ r - x a | <0.5 x io»~* Rule IV : Relative error because of rounding-off. If x„ is the approximate value of x aft rounding-off to m digits, then <0.5 x 1 0 -"+I. The above type of analysis is useful when we want to keep track of the errors that occur if tii numbers in a calculation are truncated or rounded-off to a certain number of decimal digits. Example 2.04 :Round~offthe number 537.261 to four significant digits and then calculate absohn error, relative error and percentage error. Solution. Given number is 537.261 (= x say). After rounded-off to four significant figures, the given number would be 537.3 (= X\ say) .\ (a) Absolute error, Ea - ii -Xi| - 1537.261 - 537.300j = 0.039. Am (b) Relative error,

x-xx .039 _ oc _ X Er = — ----- 5 3 7 2 6 I _

Am,

(c) Percentage error,

Ep = Er * 100 = 7.25 x 10“i.

Ans.

Example 2.05: Evaluate the sum S = -Js + ^ 5 + J l to fo u r significant digits andfind its absoiuu and relative errors. Solution. J s - 1.732, Vs = 2.236, 77 = 2.646 (rounded-off four significant figures) Hence sum, S = 6.614 and Ea = 0.0005 + 0.0005 + 0.0005 = 0.0015. The total absolute error shows that the sum is correct to three significant figures only. We take, 5 = 6.61 then

Ans.

N um erical A nalysis -! 1 1 1 9

Overflow and Underflow: An overflow is said to have occurred when the sum of two n digits occupies n + 1 digits. The number is larger than the largest number that can be stored in word. This condition is called an overflow. On the other hand the result is smaller than the smallest number which couid be stored in a word. This condition is called underflow. For example : 0.9998E1 + 0.1000E - 99 = 9.99980E100 ~0.9998E101 This is known as overflow. 0.9998E (-5) + 0.1000E98 = 0.9998E (- 104) This is known as underflow.

ALGEBRAICANDTRANSCENDENTALEQUATIONS An equation of the form/fr) = 0 is called algebraic and transcendental equations depending upon whether flx) is purely a polynomial of jc or contains exponential, trigonometric or logarithmic functions. For example; 2Jc3 + 3jc2+ 17 = 0 is an algebraic equation white Sx3 + 4 log x + 2sin x = 0 is transcendental equation. The process of finding die roots of an equation (or zeros of an equation) is called the solution o f j(x) = 0. We shall discuss some numerical methods for the solutions of algebraic and transcendental equations, rla : 0) lf/jc) is divisible by jr - a, then jc = a is a exact root of the equation/*) = 0. 00 An equation of the n* degree has only n roots. (iii) Every equation of odd degree has atieast one real root

BISECTION ORBOLZANO METHOD: Bisection method based on die repeated application of the intermediate value theorem. This method one of die simplest iterative methods. Let/jc) = 0 algebraic or transcendental equation and let/a) a n d /i) are opposite sign i.e.,fid^b) < 0 then root of equation/jc) = 0 lie between a and b. Then a +b the first approximation to the root is x\ = ^ •

Now ifX ri)= 0. Then x\ is a root offlx) = 0. otherwise the root lie between a and xi or x\ and b according asflx) is positive or negative. Then we bisect the interval as before i.e., second o + jc, x, +b . approximation to the not root x2 - —^— °r *2 = —g— and continue the process until the root is

found to desired accuracy.

IS O | E n g o s e r in g M /u ic o y m c s - IlI

a +b

From the figure root lie betwen a and b i.e., first approximation to the root *1 = —- — . a + Xi Then the second approximation to the root lie between a and X[ i.e., *2 = — „— • Similarly, the third approximation to the root He between X| and

i.e., *3 - ~

and so

Exmmpie 2.06: Find a root o f the equation xi ~ x - 4 = 0 between I and 2, tofour places q f decim bisection method. Solution. Let f{x) = jc3 - x - 4 = 0 Initialisation: M x - Ij{1) = (l)3- I - 4 = -4 and atx=2,/2)=(2) 3-2-4=2.C!early1j = - 8 < 0. Therefore, the root o f/x ) = 0 lies between I and 2. First iteration :

a+ b 1+ 2 xj = —- — = —- — = 1.5,

Since X O = - 4,^2) = 2, f a ) = (1.5)3 - 1.5 - 4 = -2.125. Since/1.5) is negative andX2) is positive ».
1.5 + 2

X2 = — ~— = 1.75,

Now, X 1.75) = (1.75)3 _ 1.75 _ 4 = _ 0.39062. Clearly/1.75) is negative and/2) is positive. /. The root lies in interval (1.75, 2). Third iteration :

1.75 + 2 x$ = ---- ----- = 1.875,

Now, / I -875) = (1.875)3 - 1.875 - 4 = 0.71679. S ince/1.75) is negative and/1.875) is positive. /. The root lies in interval (1.75, 1.875). Fourth iteration :

1.75 +1.875 x\ = ------- -------- = 1.8125,

Now, /1.8125) = (1.8125)3- 1.8125-4 = 0.14184. S in c e /1.75) is negative a n d / 1.8125) is positive. The root lies in interval (1.75, 1.8125). Fifth iteration :

1.75 + 1.8125 x5 = ----------------- = 1.78125.

Now, / 1 . 78125) = (I.78I25)3 - 1.78125 - 4 = - 0.12960. SinceXL78125) is negative andXl-8125) is positive. /. The root lies in interval (1.78125 1.8125).

N lm b u c a l A naly§ » 4 | J 2 1

Sixth iteration :

1.7126 + 1.8125

-------------------- - = 1.79687,

Now, / l .79687) - (1.79687)* - 1.79687 - 4 = 0.00477. S ince/1.78125) is negative a n d /l.79687) is positive. .*. The root lies in interval (1.78125, 1.79687). Seventh iteration :

1.78125 + 1.79687 , xy - ----------- ------------ = 1.78906,

Now, X I.789.6) =(1.78906)3- 1.78906-4 =-0.0672. S in c e /1.78906) is negative and/1.79687) is positive. The root lies in interval (1.78906, 1.79687). Eighth iteration :

1.78906 + 1.79687 xg = ----------- ------------ = 1.79296,

Now, / 1 . 79296) = (1.79296)3 - 1.79296 - 4 = - 0.02913. Since/1.79296) is negative a n d /l.79(687) is positive. The root lies in interval (1.79296,1.79687). Nineth iteration :

1.79296 +1.79687 xg ~ ----------- ------------ = 1.79491,

/ l . 79491) =(1.9491)3- 1.79491 - 4 =-0.02513. S ince/l.79491) is negative and/1.79687) is positive. /. The root lies in interval (1.79491, 1.79687). Tenth iteration :

1.79491 +1.79687 xio ------------- £----------- = 1.79589.

Now, / 1 . 79589) = (1.79687)3 - 1.79589 - 4 = -0.00375, S in c e /1.79589) is negative and/1.79687) is positive. Eleventh iteration :

1.79589 + 1.79687 xn = ----------- ------------ --- 1.79638

Now, / 1 . 79638) =0.0005 S in ce/1.79589) is negative a n d /1.79638) is positive. The root lies in interval (1.79589, 1.79638). Here, we see that the digits in the first three places of decimal are the same in the interval (1.79589, 1.79638). Therefore, the value of the root to three places of decimal is t.796. Ans. Example 2.07 : Find the real root o f the equation x logto x “ 1.2 by Bisection method correct to three decimal places. Solution : Let / x ) = x logjo x - 1.2 = 0 ...(1) Here / l ) = -1 .2 X2) = - 0.59 and/3) = 0.23, clearly/2)/3) < 0. Hence a root lies between 2 and 3.

1X 2 | E n o m e e h n g M athematics-H I

First approximation to the root is, x, = 1 ( 2 + 3) =2.5 Now, fo x ) = /2,5) = - 0.205 (-ve) Hence root lies between 2.5 and 3. Second aproximation to the root is, *2 = ± (2 .5 + 3) =2.75 Now, ftx 2) =/2.75) = 0.008(+ve) Hence root lies between 2.5 and 2.75 Third approximation to the root is, *3 = ^ (2.6 + 2.75) = 2.625 Now, f a ) =/2.625) = - 0.099 (-ve) Hence root lies between 2.625 and 2.75. Fourth approximation to the root is, * = | (2.625 + 2.75) = 2.6875 Now, ftxA) =/2.6875) = - 0.046 (-ve) Hence root lies between 2.6875 and 2.75. Fifth approximation to the root is, X5 = | (2.6875 + 2.75) = 2.71875 Now, ftx s) 2.71875) = - 0JI9(-ve) Hence root lies between 2.71875 and 2.75. Sixth approximation to the root is, | (2.71875 + 2.75) = 2.734375 Now, ftx 6) =.*2.734375 = - 0.00546 (-ve) Hence root lies between 2.734375 and 2.75. Seventh approximation to the root is, x, = | (2.734375 + 2.75) = 2.7421875 Now, fix-,) =/2.7421875) = 0.0013 (+ve) Hence root lies between 2.734375 and 2.7421875. Eighth approximation to the froot is = |(2 .7 3 4 3 7 5 + 2.7421875) « 2.73828125

NUMERICAL ANALYSIS'! | 12 3

Now, j{xg) =/2.73828125) = - 0.002 (-ve) Hence root lies between 2.73828125 and 2.7421875. .\ Nineth approximation to the root is ** = i (2.73828125 + 2.7421876) = 2.740234 Now, / x 9) = /2 .740234) = - 0.00035 (-ve) Hence root lies between 2.7402 and 2.7421. /. Tenth approximation to the root is, x,0 « \ (2.7402 + 2.7421) = 2.74115 mt

Now, jix io) =/2.74115) = 0.00043 (+ve) Hen^e root lies between 2.7402 and 2.74115 .’. Eleventh approximation to the root is, x„ = 2.740675 Now, / * „ ) - /2.740675) = 0.00025 (+ve) Hence root lies between 2.7402 and 2.740675 .\ Twelth approximation to the root is x X2 = - (2.7402 + 2.740675) = 2.7404 2 Now, Xx,2) =/2.7404) = - 0.00021 (-ve) Hence root lies between 2.7402 and 2.740675 .’. Thirteenth approximation to the root is xo = \ (2.7404 + 2.740675) =2.7405 Now, fi? n ) =X2.7405) - - 0.00012 (-ve) Hence the root lies between 2.7405 and 2.740675. Since X12 = xI3 = 2.741 upto three decimal places, hence the root of given equation is 2.741. Ans.

ITERATION METHOD OR SUCCESSIVE APPROXIMATION METHOD To find the real root of at equation/x) = 0...(1) Which can be expressed in the form x - j ( x ) ...(2) Firet we find an initial approximate value xo for (1). A better approximation xi for the root is obtained by replacing x by xq in R.H.S. of (2) I.e.. xj = (*o) A still better approximation x2 for the root is obtained by putting x = xj in the R.H.S. of (2). Thus x2 = <*(*,)■

1 2 4 j EnGINEQONG MAnftMATTCS-UI

This procedure is continued and we get x3 ~ 4 (*2> M = 4 fo) Xn If the sequence xq, x \, x^ ,......... x„..........of approximate roots converges to a limit a, thefi a is taken as the root of the equation/*) = 0. Remarks: The sufficient condition for convergence of iterations : (i) If I is the interval in which the root a of the equation x = f (x) lies, then
xH?+1) =100=>x= where

10

=>x=${x)

10 j(x) = J x + [ '

'(x) = 10 x Y * (* + I)3' 2 = (x + lp/z =>

5 |^'(*)| ** j(x + 1)3/* j < 1 in flw interval (4, 5).

So that the iteration method can be applied. Since the root is nearer to 4. 4-v Taking xq = 4.2. Then approximations are : • «*1*f ' 10 .. - .,- t (*o) = J 4 2 + i = 43852 x2 = *(*i) =

10 3852 + i = 4 3092

10 *3 ~ # fo ) * ^4,3092 + 1 " 4 33990

N umerical A nalysis -1 1 1 2 5

10 « - * < * > - A 3 3 9 9 0 + 1 ’ 4'32750 10 V4.32760 + 1 " 4 33256

*s * JC6 = S(x5) =

10 33266 + 1 = 4-33050 10

*7 *=

= ^3306

o

+ T ~ 433106

10 xg

(x7) - ^

“ 4:33,06

Since xy-x% = 4.3311 correct up to four decimal places. /. root of given eqution is x = 4.3311. Ans. frawyiV 3J& : Find the positive root o f the equation x 3 - Sx + 8 - 0, other than 2, by the method o f simple iteration, correct to 3 places o f decimals. Let f{x) - x3 - 8x + 8. Then/0) - 8 ; / l ) = l;/2 ) = 0 i.e., J(0 ) > 0 ; / l ) > 0 ;/2 ) = 0; (i.e., exact root at x = 2), so that / l . 5) =-0.625 < 0. A positive root other than 2, lies between 1 and 1.5. We rewnte the equation asx = 1 + I — I ,

...(I) 1 Vs3

i.e., in the form x = $ (x) where ^ (x) = 1 +

=*

(fl

3x2 3 2 4>'(x) * “i " =>1*>'(*)1= —x 2

3 {Because j*'(125)| = - (1-25)2 = 0.5859 < 1.] O So that iteration method can be applied. Let us take xo = 1 {which is near to root)

<

1, V * e (1, 1.5)

1 2 6 | E n g in eerin g M athematics- III

= 1.2043

1+

=

I

2

)

- 1.2183

i f 1-2183 f = 1.2261 *5 = 1 + { —

2

0%= l + ( *7

J "

2261 f 2 ~ -J = 12304

= 1 + ^1 .2 3 0 4 ? j

= 1.2328

(Vj , f 1-2342 'f X9 = 1 + [ — — - 1.2350

J

. ( 1.2350 )* x .0 - 1 ^ — J - 1 . 2355

,

.

( 1.23651

+

J = 1.2357

f 1.2357 f j = 1 2359

x'2 = 1 + { — —

.

( 1.2359 f { 2 ~ I = 1,2360

.

( 1 .2 8 6 0 ?

*l3 =

Xl4=

2 ~ J

= 12360

Since *13 = X]4 = 1.2360 correct up to four decimal places, root of given equation is x - 1.2360

Ans.

N umerical Analys» 4 | 1 1 7

Example 2.10 .*By iteration method find the value o f (48)I/3, correct to three decimal places. IRGPV Dec. 2003/ Solution. Let, X = (48)l/3=>x3 = 48 Suppose A x) = x3 - 48 = 0 ••■(I) Put x = l,th e n /l) = -4 7 = -ve Put x = 2, then/2) = - 40 = -ve Put x = 3, then/3) = - 21 = -ve Put x = 4, then/4) = 64 - 47 = + 16 = +ve Rewrite the equation x3 = 48 48 48 jc = - 5- => x = ^(x), where ^(x) - - j x1 t\x)

-96 x3

-96 = < l,Vx € (3,4) x3 Xs Hence adding 50 x both sides of equation (I), we get 5Qx = - x3 + 48 + 5Qx !#'(*)!

Because

=>

x ~ TR (48 + 50x - x3)

=>

x = ^ (x), where $ (x) = — (48 + 50x - x9)

00

...(3)

Clearly |^'(x)f < 1 for all x e (3, 4). Hence iteration method can be applied. Taking initial rootx© = 3.6, then iterations are: ,_ 48 + 50xo - x* 48 + 50(3.6) - (3.6)3 * ! - # « > - -------- go------ ^ = ------------50--------------- 1627 48 + 50x! - x? 48 + 50(3.627) - (3.627y> X2 - 4 M ---------- j g --------= ---------------- 50------------------3'633 48 + 50x2 - x | 48 + 50(3.633) - (3.633)8 „ _ *3 = *(* 2) = -------- ------------------------------------------------ =3-634 48 + 50x3 - xf 48 + 50(3.634) - (3.634)8 * * = # (x 3) = -------- g0-------- -- ---------------- 50--------------- = 3 *634 Since x3 = X4 - 3.634 correct up to three decimal places. Hence root of given equation is 3.634.

Ant.

1 2 8 | E n g in eerin g M xruS N A ncs-iH

2.7

METHOD OFFALSE POSITION ORREGULAFALSI METHOD To find the root Ax) = 0 SinceAxo)Ax\) < 0, then root of equation (1) lie between xq and x>. Let xi =

...[ —- .

In this method x-±will be taken as the point where a straight line or chord AB through the point fixo) a n d /x i,/x i)) intersect the x-axis i.e.. intersect at point fo, 0). .\ Equation of straight line AB is

y~Ax o)

=

/( * i ) - /( * o ) *i

( jc — JCo)

Since straight line intersects at fo, 0) so that 0 -Axo) =

/(* i) - (*o) *, - x 0

(* 2 - X o)

*>X

(*1 - X o )/ (X q )

*2

/ ( * i ) - A*o) B(xt MX

or may be used the following formula : X2 =Xj -

(*i - X o ) f(x, ) - / ( x

* f ( x ,) 0 )

Which is an approximation to the root of equation (I). If/xo) and/x 2) are opposite signs then the root lie between x q and x2. So that replacing X| by x i in (2), we obtain next approximation x }. However, ifAx \)Ax2) < 0 i.e., root lie between X| and X2 then we obtain next approximate*x$ accordingly. We continue the iterations till the root is found to the desired accuracy. Remark: This method is also known as method o f linear interpolation. Example 2.11 : Find a real o f the equation x? - 9x + 1 * 0 by the method offalse position. fRGPV June 20O7j Solution. Let, Xx) = x3 - 9x + 1 here A 2 ) - - 9 ,/3 ) - 1 = > /2 )/3 ) < 0 root lies between 2 and 3. Taking xo = 2, x\ - 3 so that/xo) = - 9, and/xj) = 1. By method of false position, First approximation : (*I - *o)

x

N um erical A nalysis -! | 129

Now, fo r ) = /2.9) = (2.9>» - 9(2.9) + 1 = - 0.711 Clearly/xi).X*2) < 0, therefore, the root lies between 2.9 and 3. T ak in g x0 = 2.9, xt = 3,/xo) = - 0.71 l.jfo ) = 1 in (1), then second approximation: root

xi - 2.9 - ~—- —— (-0.711) = 2.9416 1 + 0.711 Now, / x 3) = /2 .9 4 16) = (2.9416)3 - 9(2.9416) + 1 = - 0.0207 Clearly/xO/xj) < 0, so the root lies between 2.9416 and 3. Taking x0 = 2.9416, xj = 3 ,/x 0) « - 0.0207,/x O = 1 in (1), we get third approximation root

x4 = 2.9416 - 2 ^ 1 (-0.0207) = 2.9428

Now, y(x4) - - 0.0003 so that root lies between 2.9428 and 3 Taking xo = 2.9428, xi = 3, we get x} - 2.9428 -

(-0.0003) = 2.9428,

Clearly x4 = x5 = 2.9428. Hence the root is 2.942 correct to three decimal places. Ans. 2.12 : Find a real root o f the equation xi - 2 x - 5 ’* 0 b y Regula-Falsi method correct to three decimal places. fRGPV, Dec. 2011f Let /( x ) = x3-2 x —5 = 0 ... (1) Here / (0) = - 5, / ( I ) = - 6, /( 2 )= -l, /( 3 ) = 16, Clearly root lies between 2 and 3. First approximation :

Taking x0 = 2, Xj = 3 ,/(x 0) = - l,/(x i) = 16. By Regula-Flasi method an approximate root: (X) Z lo }

x = X* ~ ~77Z f ( x \ )~- f777T ( x 0) x f ( xo) - 2

—— x ( -1 ) = 2 + — =2.059 16 + 1 17

Now, from (1), we have / (2.059) - - 0.389 and/( 3 ) - + 16. => next approxim^root lies between (2.059, 3). Second approximation : Taking x„ = 2.059, xj = 3 ,/( x 0) = - 0.389,/ ( x ,) = 16.

...( 2 )

1 3 0 1 E n g in eerin g M athem atics -111 (3 -2 fl5 9 )x H ^

16 + 0389

1 y ' j|

i.e., From (1),

x = 2.081 / (2.081) = - 0.150 and/(3) = +16. => next approxim ate root lies betw een (2.081,3) Third approximation:

,\

Taking x0 = 2.081, xt = 3,/( x 0) = - 0.150,/xt) = 16 , , 2 .q8 i - <3 - 2 0 8 I > - ^ ! 5,)> 16 + 0.150

[b s f f l 1 y ( n

= 2.090 Now, from ( I ), w e have / ( 2 . 0 9 0 ) = - 0 . 0 5 ! a n d / ( 3 ) = + 16 => next approxim ate root lies betw een (2.090, 3).

Fourth approximation: T aking x0 - 2.090, x, = 3 , / ( x 0) = - 0 .5 1 ,/ ( x , ) = 16. o w _ f l z 2 j9 0 )x (-a 0 5 0 16 + 0.051 = 2.093 H ence root o f the given equation is 2.093.

: iywi Ans.

Example 2.13 : Find the real root o f the equation x log/g x - 1.2 = 0 by the method o f false position coned to three decimal places. {RGPV Dec. 2002, Dec. 2005 & Dec. 2001) Solution. Let f ix ) = x logio x - 1.2 Here

f i 2 ) - 2 log,0 2 - 1.2 = - 0.59794 f i 3) = 3 log,0 3 - 1.2 = 0.23136

Clearly root lies betw een 2 and 3.

For the first approximation, Taking xq = 2 , jq = 3 , / x o ) = - 0.59794 a n d / x j ) = 0.23136. By the m ethod o f Regula Falsi o r false position method

0 N ow ,

(3 - 2) x (-0.59794) (.23136 + .59794) ” 2-72102

f i x 2) = y (2 .7 2 102) = - 0.01709 and / 3 ) = +ve

i.e., The root lies betw een 2.72102 and 3. Taking xo = 2.72102, X| = 3,y(xo) = 0.01709 a n d / x j ) = 0.23136 in equation (1) we get.

Second approximation,

x = 2.72102 + -------------------------- x 0.01709 *3 (0.23136 + 0.01709) = 2.74021

N um erical A nalvsis -I ( 131

Now,

/ x 3) = X2.74021) = - 0.00038

clearly A * t)./ta ) < 0 root lies between 2.74021 and 3. Taking xo = 2.74021, X| = 3,/*o) = - 0.00038, fix\) = 0.23136, we get .■. Third approximation x^ =

2.74021 -

0 25979 ■

(-0.00038) = 2.74064

Now, Xx4) =-0.00001 Thenfourth approximation is , 5 . 2.74064 -

(-0.0001) - 2.74065

Since X4 = xs = 2.7406 correct up to four decimal places. /. Hence root of given equation is 2.7406. Ans. XmpU 2.14 : Find the root o f equation xe* * cos x, using Reguia Falsi method, correct to four decimal places. fRGPV June 2009J Given c o sx -x e * = 0 Let / x ) = cos X- xe* ...(1) /(O) = cos 0 - 0 = 1 [taking mode of calculator is Radian] X I) =-2.1779 HenceX0) a n d / 1) are opposite sign. Root of equation (I) lie between (0,1). Taking xo = 0 , x\ = l,X*fc) = UX*i) = - 2.1779 By Reguia falsi method: First approxination :

(Xj - x„) ' - ' - u M

v

- f a r ™

'(2 )

= ° “ (- 2 .1 7 7 9 -1 ) X l = 0 -3147 Now, from (1), X0:3147) = 0.5198 ClearlyX0.3147) andXl) are opposite sign, so that next approximate root lie between (0.3147, I)Taking xo = 0.3147, x, = l,X*>) - 0.5198,/x i) - - 2.1779 Second approximation : from (2), we get x « 0.3147 -

(1 - 0.3147) x .5198 ---- J - - = 0.4467 (-2 .1 7 7 9 -.5 1 9 8 )

From (1) XO-4467) = 0.2036 Clearly next approximate root lie between (0.4467,1). Taking xq = 0.4467, x, = 1,/xo) = 02036,/ x ) = - 2.1779

132 |£«eB«i*saNaM*rHEMAiTcs-UI Third approxunate : from (2), w e get (1 - 0 .4 4 6 7 ) x 0 .2 0 3 6 r = U .4 4 b 7 ------------------------------------- —a 4Q40 x ( - 2 .1 7 7 9 - 0 .2 0 3 6 ) U4y4U From (1), w e have / 0 . 4 9 4 0 ) = 0.0709 Clearly root lie betw een (0.494, I) Taking x0 = 0.4940, x, = 1 ,/x o ) - 0.0709, f a , ) = -

2.1779

Fourth approximation • Prom (2 ) becomes a AriAn (1 - 0 -4 9 4 0 ) x 0 .0 7 0 9 = i).4y4U-----------------------------= n cfiQi ( - 2 .1 7 7 9 - 0 .0 7 0 9 )

UM ,yi

N ow from ( I ) , we have / 0 . 5 0 9 I ) = 0.0261 H ence n e x t approxim ate root lie between (0 .5 0 9 1, 1)

Taking

= 0.5091, x, - l.yfo) = 0.0261 an d /x t) = - 2.1779

Fifth approximation : From (2), becomes _

0.509i _ £ 1 ^ 0 0 ^ * 0 0 2 6 1 ( - 2 .1 7 7 9 - 0 .0 2 6 1 )

u 'M 4y

From (I), we have ■

../ 0 .5 1 4 9 ) = 0 .0 0 8 7

Clearly root lie b e ty e e n (0.5149, J). Sixth approximation : From (2), becomes (1 - 0 .5 1 4 9 ) x (0 .0 0 8 7 )

A From

- u .5 i4 y -------------------------------- = a si as ( - 2 .1 7 7 9 - 0 .0 0 8 7 )

(1), we have

X 0 .5 1 6 8 ) = 0 .0 0 2 9 Clearly root lie between (0.5168, 1). Seventh approximation : From (2) becomes (1 - 0 . 5 1 6 8 ) x 0 .0 0 2 9

r = U .O lbO -------------------------- :-------- = n 517A.

x

(-2.1779 - 0.0029)

101/4

We have / 0 . 5 174) = 0.0010 H ence root of given equation x = 0.5174.

Am

Example 2.15 ; Find the root o f the equation cos x ~ 3 x - l f using the Regula falsi method correcti three decimal places* . Solution. le t , f[x) = c o s x - 3x + 1 J(0) - 2 [taking m ode o f calculator in radian 7 (1 ) = - 1.45969. H e n c e /0 ) a n d / l ) are opposite signs. /.T h e root lies between 0 and 1.

Hence

N umerical A nalysis-! | 133

/. Taking x0 = 0, x, = l , / x 0) = 2-X*i) = - I-45969. we get First approximation:

(1 - 0)

or

x2 = 0 “ f[-1.46969 r.M fln --n7 * 2 = 0.5781 - 2]

N ow ,

J(x2) = / 0 . 5 7 8 I ) = 0.1032098 = +ve

root lies betw een 0.5781 and I. Taking xo = 0.5781 and X| = 1 ,/x o ) = . 1032098,/ x i ) = - 1.45969, then

Second approximation:

Xx = 0.5781 + ------- ( - ~ - 5 — ------ - x 0.1032098 [1.458969 + 0.1032098] = 0.606 N ow ,

•

(Using (1)]

/ x j ) = / . 6 0 6 ) = 0.003955 = +ve

root lies betw een 0.606 and !. Taking xo = 0.606, X| = 1 ,/x q ) = 0.003955,/ x j ) = - 1.45969, then Third approximation :

x*

= 0 .6 0 6 + -------- (1 ~ ° - 0 6 )------- - x 0 .0 0 3 9 5 5

{1.45960 + 0.003955]

= 0.6071 N ow ,

/ x 4) = /0 .6 0 7 1 ) = 0.0000059 = +ve

/. root lies betw een X| = 0.6071 and I . T akingx0 = 0.6071, x , = l , / x 0) = 0.000059, fix) = - 1.45969, then

Fourth approximation: x = 0 .6 0 7 1 + ------- -1-

x

—

-------- x 0 .0 0 0 0 0 5 9

[1.45969 + 0.000059]

= 0.607 Since x4 = X5 = 0.607 correct up to 3 decim al places. Hence, the root o f given equation is 0.607.

A ns.

Example 2 .1 6 : Find a root o f the equation 3x + sin x - e * * 0 by Reguia falsi method correct to four decimal places. Solution. Let j(x) = 3x + sin x - e* ...(1) T hen

/ ( 0 ) = - 1, / 0 . 1) = - 0.705. / 0 . 2 ) 0.423 [taking m ode o f calculator in Radian] /0 .3 ) = - 0 .1 5 4 ./0 .4 ) = 0.0975 C le a r ly /0 .3 ) a n d / 0 .4 ) are opposite signs. T he root o f / x ) - 0 lies betw een 0.3 and 0.4. . 1■

1 3 4 | E ng in eerin g M athematics-111

Taking xo = 0.3, x, = 0.4, / ( xq ) = - 0.154,/x[) = 0.0975 By Regula falsi method : (*i ~ *o) *2 = X0 - [ f i x , ) - f ( x o)] f(*o) First approximation: (0.4) - (0.3) (-0.154) (0.0975) - (-0.154) 0.1 x 0.154

0.3612

Now, flx2) = X0.3612) = 0.0019 = (+ve) The root lies between 0.3 and 0.3612 Taking xq « 0.3, jc, = 0.3612,/ x q ) = - 0.154,/x,) = 0.0019, then Second approximation:

‘ (0-3 , + O

< 0 i5 4 ) " o 36
Now, fix )) =/0.3604) = - 0.00005 = (-ve) root lies between 0.3604 and 0.3612. Taking x0 = 0.3604, x { = 0.3612,/xo) = - 0.00005,/x j) = 0.0019, then Third approximation: (0.3612 - 0.3604) (-0.00005) (0 .0 0 1 9 )-(-0 .0 0 0 0 5 ). [Using (2)] = 0.3604 +

0.0008 (0.00005) = 0.36042 0.00195

Since X3 = X4 = 0.3604 correct to four decimal places, root of given equation is 0.3604.

2,8

Ans.

SECANT METHOD OR CHORD METHOD The secant method is an improvement over the method of false position, such that condition/jc0) J(x 1) < 0 are not required. The general formula for successive approximations is given by (*n ~ %a-1 )/(*«)

N u m erical A nalysis -1 I 1 3 5 r

-

(•*!

x 0)

f/Y \

i.e. for n - 1

« - *>

for n = 2,

(x2 - x,) n x3 = x 2 ~ , , , _ 77" A * * J and so on. [ / ( * 2) - /(* k )l

i r

2.17 : Find the root o f c osx -x e* —0 by using Secant or Chord method. Let, A x ) ~ cosx-xe* /(0) - cos 0 - 0 = !(+ ve) / I ) = -2.177979 (-ve) root of equation (I), lies between 0 and 1. Hence/0) a n d /l) are opposite sign. Using Secant method : r

~ *i»-l) f ( Y \ i/d .) -/(*„)]f i , )

• "-(2)

First approximation :

(■^i ~ *o) \ f0Tnml' x^ ^ X l " 7 ( x ly - - f ( x 7 ) fiXl)Taking xq = 0, x x = 1, jfa>) ~ l,X*i) = - 2.177979, we get , X2 =

(1 - 0)(-2.177979) [2.177979 + 1] = 03147

Now, fix2) = /0 .3 !47) = 0.51797 Second approximation : for n - 2, in (2), we get ■ ** ' ( / ( S - / ( i , ) i /(X2> * 04467 .•. Now,

/ f o ) = /0 .4 4 6 7 ) = 0.2036 Third approximation : for n - 3, we get

(x3 - x2)

,/ , f e W f e » / ( l a ) ' 0'53' 8 Now, /[xA) = / 0 . 5 3 ) 8 ) = - 0.0432. Fourth approximation : for n = 4, we get

(x4 - x3)

..

. "

Now, f [xs) = /0 .5 1 7 3 ) = 0.0014. Fifth approximation : for n = 5, we get

/. Now,

f[xb) = / 0 . 5 l 7 7 ) = 0.0001.

...(1)

° '5173'

1 3 6 | Ewoiw ttH H to M athem atics -!!!

Sixth approximation : for n = 6, we get v

-

(-r 6 _ * 5 )

\

ft

*? = 6 _ [/(*6) - / ( * 5 ) ] 6 a5177‘ Since xg = *7 = 0.5177 correct to four decimal places. Hence root of given equation is 0.5177. Example 2.18 : Find a real root o f x* - 5x + 1 = 0, by using Secant Method. S o l u t i o n Let / * ) = x 1 - Sx + 1 / (0) = + I /( ! ) - 1-5 + 1= -3 Clearly/( 0 ) and/(I) are opposite sign. Hence root of equation (1) lie between (0,1). Using Secnat M ethod :

for « = 1, 2, 3..... .. F/ntf iteration : Put n - ! in (2), we get T

Here x„ = 0, jc,

<*»-■*<>>

■; f ( r >

*2 ' ' 1 /(* > )-/(* 0) = 1 ,/x 0) * /(0 ) * ! ,/( x t) = / ( ! ) = - 3 * ■ ' ' - { ^ x<- 3) ■ ' “

-0.25.

Thus x2 = 0.25 and / (jr2) = /(0.25) = - 0.2344. Second iteration : Put n = 2 in (2), we get xf(x)

= 0 25 "

(0.15 - I) x (-0 2 3 4 4 ) ----------£— ----------- (-0 2 3 4 4 + 3)

= 0.1864. Hence x, = 0.1864 a n d /(r3) = / (0.1864) = 0.0743. Third iteration : Put« = 3 in equation (2), we get (Xj - x 2)

( / ( * a ) - / ( * 2)) = 01S61

(Q-1864- 0-25) * ^ 0743) (0.0743 + 0.2344)

= 0.2017. Hence x4 = 0.2017 and/(.r4) = /(0.20I7) = - 0.0005

N umericalA nam s » -I (>137

Fourth iteration : Put n = 4 in (2), we get '< /(* .)-/< * » / M . 0 2 0 1 7 - (^ 0 1 7 - ° l8 6 4 ) .( - 0 .0 0 0 5 )

(-0.0005 - 0.0743) = 0.2016. Hence = 0.2016 and/(x5) =/(0.2016) = 0.0002. Fifth iteration : Put n = 5 in (2), we get (x5- * 4) _

0 2 0 1 6 _ (0 2 0 2 6 Z 02017)

0.0002 + 0.0005 = 0.2016 Since xj = X6 = 0.2016 Hence root of given equation is 0.2016.

Ans.

NEWTON-RAPHSONMETHOD Let xo be an approximation value of a root of the equation/x) = 0. Let X| be the exact root closer to xo, so that x\ = xo + h, where h is small. Since xi is the exact root o f/x ) = 0, we have/xj) = 0, i.e., J[x0 + h) = 0 h

h2

/(x 0) + — f'(xo) + — f ' ( x o) + ... = 0

i.e.,

[By Taylor’s theorem]

Since h is small, h2 and higher powers of h may be omitted. Hence

/(*o) / x 0) + hf\xQ) = 0 => h = _ ^ j /(x 0) ^ j

x i = x 0 + /r=*o

...(I) [Using (1)]

Taking xt as an approximation value of the root, a still better approximation x 2 can be obtained by ... * *i -

*2 ‘

f(* i) A x ,)

T he iterative process is continued until un we get the required accuracy. The iterative form ula is

x„+1 = x * ~

/(*„)

^

,

(for n = 0, 1 , 2 , .....)

is called th e Newton-Raphson formula o r N ew ton’s iteration formula.

Remarks ; (i) (ii)

W hen the derivative o f / x ) can be easily found and is a sim ple expression, then the real root o f the e q u a tio n /x ) = 0 can be com puted rapidly by the N ew ton-R aphson method. N ew ton’s form ula converges provided the initial approxim ation xo is chosen sufficiently to close to the root.

138) Engineer^ Mathematics-HI (iii) Condition fo r Convergence o f Newton-Raphson Method: fix*) f'(xn)

The Newton-Raphson formula is

*»+i = ixn)

The iteration formula is

fi x ) fix) We know that the condition for convergence of the iteration method is |<#'(x)! < 1. The corresponding condition for the Newton-Raphson method is From (1) and (2), we get

,

=x -

...(1)

...(2) [Using (2)]

f / ,2( * ) - /( * ) • /'( * ) 1 f i x ) . f ‘(x) , f'Hx) I f ' 2(x)

=> \f( x).f\x)\ < {fix)}2, for all x in the interval in which the root lies. 2.10 GEOMETRICAL INTERPRETATION OF NEWTON-RAPHSON METHOD Let x0 be a point near the root a of equation fix) = 0, then tangent at /*0 {(x0, _/fa0)} is y - fi x o) = /{x0X* -

Xo).

f ( x 0) It cuts x-axis at (xu 0) then x i ~ *o “ Y’(x ^) *

Which is first approximation to root a. If P\ corresponds to X| on the curve, then tangent at P\ will cut x-axis at X2, nearer to a and is therefore second approximation to root a. Repeating this process, we approach the root a quite rapidly. Hence the method consists in replacing the part of the curve between P Q and x-axis by the means of the tangent of the curve at P q . Example 2.19: Solve by Newtons-Raphson method : x * - 3 x + / * ft /RGPV Dec. 200/ & Dec. 2007j Solution.Let, fix ) - x3 - 3* + 1 ...(I) fiO) - + 1 = + ve / ! ) = I - 3 + 1 = 2 - 3 = - 1 = - ve. Clearly/0) and fi\) are opposite signs.

I

N um erical A halysis -I | 13 9

root of equation (1) lies between 0 and 1. Since |/(0)| = 0 + 1 = 0.5 n* i.e.,*© = —-—

Taking initial approximation xq = 0.5 Using NewtorvRaphson method.

Since => Hence (!) becomes:

/(*») x" * ' ~ *" " T o o fix ) = x3 - 3x + 1 / t x ) = 3*2-3

11

- (2)

- 3x, +1) 3xf - 3xn - xg + 3x, - 1 3x2-3 ~ 3(4-D

2xg - 1 x" +1 ~ 3(xf - 1 ) Putting n = 0, in (3), then first approximation :

" <3)

2xp3 - 1 _ 2(.5)3 - 1 _ JC| “ 3(x$ - 1) ~ 3(,52 - X) _ 0 3333 Putting n - 1 in (3), then second approximation: 2V - 1 X2 ~ 3(x,2 - 1 ) = 0,3472

Putting n - 2 in (3), then third approximation:

Hence x2 = X3 = 0.347 correct to 3 decimal places. Thus root of given equation is 0.347. Example 2.20 : Solve by Newton-Raphson method : x* - 3 x - 4 - 0. Solution. Let, / x ) = x 3- 3 x - 4 Then fiO) = - 4 = - ve; / l ) = l - 3 - 4 = - 6 = -ve f i l ) = 8 —6 - 4 = —2 = —ve, fi} ) = 27 - 9 - 4 = 14 = + ve .'. fil) and/3) are opposite signs. Hence root lies between 2 and 3. Also since /2)( < jfl3)| i.e., root near to x = 2, so taking xo = 2. By Newton-Raphson formula:

Since

* A x .) fix ) = x3 - 3x - 4

Ans. fRGPV June 2002/ ...(1)

~ (2)

1 4 0 | E n g in eerin g M athematics-111

=>

/'(* ) = 3x2-3

Hence (2) becomes :

x„ + , =

or

x „-»+1 + , ~ 3(X2 _ w

(xf - 3x„ - 4) 3x% - 3xn - x* + 3x„ + 4 - ... i --------------= ------- 3 ( x f - l ) 2x2+4 ...(31

Putting u = 0 in (3), w e get

2xg + 4 First approximation:

2(2)3 + 4

20

j ) “ ^(22 - 1) ~ I T = ^-222

xt =

Putting n = I, in (3), w e get (2 x f + 4 ) 2 (2 .2 2 2 )3 + 4 2 5 .9 4 7 2 *2 = x2 = ^ _ J} -= ^ . 2 2 2 ) * ' - 1] ' T l s T T b = 2,197

Second approximation : Putting n = 2, in (3), w e get

2xf + 4 2 5 .2 0 9 x3 = 3^ 2 ^ ) “ 1 1 4 8 0 4 = 2,19 6

Thin!approximation: Putting >? = 3, in (3), w e get

2xf + 4

Fourth approximation :

X4 =

2 5 .1 8 0

^ ~ 21467 ~ 2.196

H ence x 3 = x 4 = 2.196 correct to 3 decim al places. .'. root o f given equation is 2.196.

Ans. Example 2.21 : Find the real root o f x 4 - x - 10 ** 0 by Newton-Raphson method. [RGPV Dec. 2003 & Dec. 20/// Solution. Let f(x) = x 4 - x - 1 0 = 0 ...(I) T hen

7(0} = - 1 0 = - v e I ) = I - 1 - 10 = - 10 = ( - ve) 7 (2 ) = 1 6 - 2 - l0 = 4 = (+ v e ). Clearly fi 11 and 7(2) are opposite signs* root ot equation ( 1 ) lies betw een 1 and 2 . Taking initial approxim ationxo = 1.5 By N ew ton-R ephson m e th o d :

. •*»+1 = *»

Since => H ence (2) b e c o m e s :

/(*«> f'(xn) ’ " = 0‘ ’’ 2....... f i x ) - x * - x - 10 f'(x) = 4x3 - 1. (x* —xn —10)

4x* - x n - x< + xn + 10

N um erical A nalysis -! 1 1 4 1

Putting, n = 0 in (3), w e get

First approximation:

x, =

3(1.5 )4 + 10 4 (1 .5 )3 - 1

3**+10 4x^-1

2 5 .1 8 7 5 1 2.5 _2015

Putting, n = 1 in (3), w e get

3*j4 + 10 Second approximation :

x2 -

4 i3

“

3(2.015)4 + 10 4(2.015)3 - 1 = 1,874

Putting, n = 2 in (3), w e get

3x4 + 10

Third approximation:

x3 == 4jcg _“j xj

3(1.874)* + 10 "

4(1 .8 7 4 )3 - I

= 18 5 6

Putting, n = 3, in (3), w e get

3x$ + 10 Fourth approximation :

=

4^3 _ j

3(1.856)4 + 10 ~

4 ^ g g g )3 - 1

= 1

Since, X3 = X4 = 1.856 correct to 3 decimal places. root of given equation is 1.856. Ans. txsmple 2.22 : Solve the equation 3x = cosx+ 1, by using Newton-Raphson method. fRGPV Dec. 2002, June 2004, Dec. 2006 and June f Solution. Let X*) = 3x - cos x - I ...(H X0) = 0 - cos 0 - I = - 2 = (- ve) / I ) = 3 -cos (I) - I = 1.4596 = (+ ve) ClearlyX0) andXO axe opposite signs root of equation ( 1) lies between 0 and 1. Taking initial approximation xo=0.5 By Newton-Raphson formula: ) x „ ^ ^ x n ~ J ~ , (« = 0, 1,2,3...... ) ...(2)

1

2011

SinceXx) = 3x - cos x - 1 =>/'(x) - 3 + sin x [3x„ - cos(.tn) - 1] Hence (2) becomes: x„+ 1 = g+ ^

...(3)

Putting n = 0, in (3), we get First approximation :

x, = *0 " = 0 .5 -

(3x0 - cos Xq - 1) 3 + 8m(Xo) [3(0.5) - cos(.5) - 1} 3 + sin(.5)

n _ (-0.37758) ' a 6 - T J 7 9 5 T ' 0'608 Putting n ~ 1, in (3), w e get .'. Second approximation:

Xl

1 = 0.607

[3^, 3 + sin x,

(0.0032079) 3.57123

1 4 2 | E ngjne£RING M athem atics-III

Putting n = 2, in (3), we get Third approximation :

_ [3*2 - COS *2 - 1 ] _ 0 g07 _ (- 003629) 2 3 + sin *2 ' (3.57041) = 0.607 Since x2 = = 0.607 correct to 3 decimal places, root of given equation is 0.607. Ans. Example 2.23 ; Find by Newton’s method correct to five places o f decimals the root o f the equation x loge x = 4.772393. [RGPV Dec. 20041 Solution. Let J{x) - x logx - 4.772393 ...(I) A O = - 4.772393 = (-v e) f{2) = - 3.386099 = (- ve) / ( 3 ) = - 1.476556 = (-ve) f{ 4) = 0.772784 = (+ ve). Clearly/3) and/4) are opposite signs. .'. root of equation (1) lies between 3 and 4. Taking initial approximation x$ = 3.5 By Newton-Raphson method: Xi

f ( x n) *« + 1 = x» “ "f'(xn) ’

Since => .. Hence (2) becomes :

(" = °» *» 2......... )

f[x) - x log x - 4.772393 /'(x ) = togx +1 _ I*„ log(xn) - 4.772393] x„+ j = * „ ---------- (i T io g a ")--------

Putting n = 0, in (3), we get ... . .. .. First approximation:

„ [x o lo g x o -4.772393] x\ = x 0 --------- ——------- -------(1 + log x0) 0 _ [-0.3877226] - 3 8 - 1 2.282763) - 3 6721098'

Putting tt = 1, in (3), we get Second approximation : X2

X*

[x\ log X\ -4.772393] (1 + log*!)

3 67-1098 - 3.6721098 Putting n - 2, in (3), we get , Third approximation :

( ° 04163956) , r71n,nr ^ 3 0 0 7 6 6 3 7 4 ) = 3'6739196

[X2 log X2 - 4.772393] xx = x 2 --------------— --------- ---------(1 + log x j

(.008328323) ■ 3.6739196 - i ■ 3.6703006

- ( 2)

N um erical A n Aly sis -I I 143

Putting n = 3, in (3), we get Fourth approximation: xa =

3.6703006 - - ° Q- 0{^ 4— = 3.67029979 (2.300273566)

Putting n = 4, in (3) we get Fifth approximation :

x$ - x *

[*< log x 4 - 4.772393] (1 + log *4) = 3.67029

Hencex* = x$ = 3.67029 correct to 5 decimal places, root of given equation is 3.67029. Ans. 2.24 : Find the real root x 4 - x - 9 ~ 0 by Newton-Raphson method correct to 3 decimal places. IRGPV June 2005 & June 2008/ Let, /( x ) = x 4 - x - 9 ...(1) 7(0) = - 9 = (- ve) 7{l) = ] - l - 9 = - 9 = - v e 7(2)= l 6 - 2 - 9 = 5 = + ve. Clearly/1) and/2) are opposite signi. root of equation (1), lies between 1 and 2. Taking initial approximation *o= 1.5 By Newton-Raphson formula: /(* .) *« + 1 = x* ~ f ’( x j *

(» = o, 1, 2 , .......)

(*2 (4*3 - 1 ) I

v fix) = X* - X - 9 then/(x) = 4x3 - 1

4*« - ** - *2 + ** + 9 4x2-1 3** + 9 x„+ j — ^ 3 _ j ‘

or

•••(2)

Putting n = 0, in (2), we get First approximation :

3x^ + 9 3<1.5)4 + 9 24.1875 *> = 4xg - 1 ~ 4(1.5)3 - 1 = 12.5 = 1,935

Putting n ~ 1, in (2), we get Second approximation:

3*? + 9 3(1.935)4 + 9 51.0577 x2 ~ = ir3 “ 4(1.935)3 4/1 _- 11 ~ 27.9803 97 QftDS ” 18248 4*? -_ 1i" "

Putting n = 2, in (2 \ we get Third approximation:

Xj -

3sj+ 9 42.2646 _ j - 23 3056 = 1,8135

1 4 4 |i£N q*ifflppa#l«H EM A ncs-IH

Putting n - 3, in (2), we get Fourth approximation :

3x^ + 9 41.4483 X4 = 4 j,| _ j = 228568 = I -8 *34

Putting n = 4, in (2) we get Fifth approximation :

xs =

3x< + 9 ^ - 1.8134.

Hence x4 = X5 = 1.8134 correct to 4 decimal places. root of given equation is 1.8134. Ans. Example 2.25 : Find the root o f 4 x - e * ’m0 that lies between 2 and 3, by Newton-Raphson method. Solution. Let, /{x)= 4x-ex ...(I) Then / ( l ) - 4 - e = + ve /(2) * 8 - e2 = 0.6109 = + ve J{3) = 12 - e3 = - 8.0855 = - ve. Clearly root lies between 2 and 3 and root closer to 2 because ]/[2)\ < J/(3)|, Taking initial approximation x q - 2 . By Newton-Raphson formula: 1

11

fix*) f*(X ) * (" _ (4x„ - ex»)

or Putting n = 0 in (2), we get First approximation :

.......) r [ v /'W = 4-e»]

4x„ - x.e1" - 4x. + 4—e*" ('i _ _ e*n ~lt +l f I = x" (4 ^ T j‘

-(2»

e * * { \ - x 0) «2(1 - 2) Xj = (4 __ gxo) ~ (4 _ e 2) ~ 2.18026963

Putting n = 1 in (2), we get e1’ (1 - Xj) Second approximation :

x2 =

(1_ ^ )

~

10.443822 4.87019074 = 2 ' 14444213

Putting n = 2 in (2), we get Third approximation :

x3

e^d-X j) 9.77041972 = (4 _ g n ) - 4 53727722 = 2.15336627

Putting n = 3 in (2) we get e*3(l - xs) Fourth approximation :

x* =

9.93487335

^ _ gX3^ ~ 4 61280604 = 2-15329237.

Hence x 3 = X4 = 2.1533 correct to 4 decimal places. root of given equation is 2.1533.

Ans.

N um erical Attu,V8(&4 | 145

2.26 : Find the smaller root to the equation e~x - sin x = 0 by Newton-Raphson method. Lei; f(x) =e~*~ sinx ...(1) /(O) = I —0 = 1 — ve) / (I ) = e~1- sin (I) = -0.4736 = (- ve). /. Clearly/O) a n d / 1) are opposite sign. Hence root lies between 0 and 1. I.' Taking initial approximation xq - 0.5. By Newton-Raphson formula: /(*«) f'M

xn + 1 ~ « ~ xn —

(e~Xn -■sin x„) ( - e~xn - cos xn)

(e-*n _ sinx*) *„+i = *» + {er** + cos x n)

..-(2)

Putting n = 0 in (2), we get .\ First approximation:

x\ = xo +

(e-*# - sin x 0) (e~*o + cosx0)

(e~5 + cos(0.5))

n e 0.12711 = 0.5 + _ = 0.58565. 1.48411 Putting n = I in (2), we get Second approximation:

(0.00400) (e xi - sin Xj) + (e x> + co b x x) = 0.58565 + a.39010) = 0.58853.

x2 =

Putting n = 2 in (2), we get /. Third approximation:

058853 +

Xi

(0.0000) 1.38690

‘ (c‘x* + cos**) = 0.58853. Since x%= X3 = 0.5885 correct to four decimal places. /. root of given equation is 0.5885. Ans. 227 ; Find the real root o f the equation x sin x + cos x * 0 near to x - it correct to three decimal places by Newton-Raphson method. Let ^ x ) = xsinx + cosx ...(1) Given that initial root xo= *=3.14159 By Newton-Raphson formula: /(* ,) fXxJ [xn sin x„ 4 c o a x j [ *•’ f ( x ) = x cos x + sin x - sin x]

1 46 | E n g ineers ^ M athematics -! 11

(x„ sinx* + c o s x j or

x„ + i = x *

..Oil

Putting n = 0 in (2), we get ____ __ .. (*osinx0 + cos*o) x, = * 0 --------------------------- = 3.14159 -

/. First approximation:

l

COSXq

= 2.82327 Putting n = 1 in (2), we get (jtj sin xr + coajq) Second approximation:

xi = x\ ~

AT] COS

. 2 . 8 2 M 7 - (- 08? 15.49>. . 2 79859 (-2.681433) 279859 Putting n = 2 in (2), we get Third approximation:

x3 - x 2 ~

(x% sin x 2 + cosx2) x z cos x 2

= 2.79859 — (^000537)_ (-2.635569) Hence x 2 ~ x3 = 2.798 correct to three decimal places. root of given equation is 2.798. Atis, Example 2.28 •• Using Newton's iteration method, find the real root o f x logio x = 1.2 correct to fix decimal places. [RGPV, Feb. 20/0/ Solution. Let, fix) = x logio x - 1.2 ...(I) J { \ ) = - 1.2 = (-ve) J{2) = 2 logio 2 - 1.2 = - 0.59794 = (- ve) A3) = 3 logio 3 - 1.2 = 0.23136 = (+ ve). Clearly/2) and/3) are opposite signs. root of equation (1) lies between 2 and 3. Since f[x) ~ x logio x - 1.2 v logio x = log* x log* 10

log* 10 = x logio

e-

x - 1.2 = .43429 x loge x - 1.2

[v log^o e = -43429] f ( x ) = 0.43429 (1 + log, x) = 0.43429 + log10x

By Newton-Raphson formula: Xn+ 1 = x» or

fjxj f \ x n)

n

■43429xw + 1.2 Xn+ 1 " .43429 + logio xn

(s , logl0 xn - 1.2) (.43429 + logio *n) ...(2

N um erical A nalysis -! | 1 4 7

Taking initial approximation :xq = 2.5 Putting n = 0 in (2), we get First approximation:

.43429x0 + 1.2 2.28575 x\ = .43429 AOAOn +, log i_~l0 x“0 QOooo .83223= 2.74651

Putting n - 1 in (2), we get .'. Second approximation:

.43429xL +1.2 xi = .43429 AOAn + log _~10 x,=

2.39278 ononn .87307~ 2.7407

Putting n - 2 in (2), we get Mird approximation:

.43429x2 + 1.2 x , - .43429 + ',ogl0 ^

2.39026

- 2.7406

Since x2 = *3 = 2.741 correct to 3 decimal places. .*. root of given equation is 2.741. Ans. 2.29 ; Using Newton-Raphson method, evaluate

to four decimal places.

Let, and let

x = 728 =>x2 = 28=*Jta - 2 8 = 0 J(x) = x2 - 28 AO) = - 28,/H ) = - 27,/ 2 ) = - 24,/ 3 ) = - 19, M ) = - 1 2 ,/5 ) = - 3 , / 6 ) = 8. Clearly/5) a n d / 6) are opposite signs. Hence root lies between 5 and 6 . Taking initial root: xq = 5.5 By Newton-Raphson formula: fi*n)

or

+i “

Xn ~

(xj ~ 28) 2x„

iv m

x? + 28 » (« = 0, 1, 2 , ....) 2x.

...( 1)

= 2x] ...(2)

Putting n = 0 in (2), we get Firj/ approximation:

xg + 28 Xl " 2xo

(5.5)2 + 28 2(5.5)

58.25 11 ~ 5-29545

Putting n = 1 in (2), we get .*. Second approximation:

x2 =

x? + 28 _ (5.29S45)2 + 28 = 5.2915. 2Xj 2(5.29545)

Putting h = 2 in (2), we get Third approximation:

xf + 28 _ (5.2915)2 + 28 = 5.2915 2x2 2(5.2915)

Clearly JjcC2 =X = x 3 = 5.2915 correct to four decimal places. root of given equation is 5.2915.

Ans.

.1 4 8 j E ngineering M athematics -111

Example 2.30 : Find the cube root o f 2 approximately by Newton-Raphson method correct to ft decimal places. Solution. Let x - (2),/3 => x3 = 2 => x3 - 2 = 0. Let f{x) = x3 - 2, then/’tr) = 3X2. Now/1) = - 1 and/2) = 6 => root of equation (1) lies between 1 and 2. By Newton-Raphson method: (4-2) ■*n+ I

f'(xn)

34

2*3+2

2 (

34

= 3 P

1 1 + 4>

Taking initial approximation :jco= 1.5 Putting n = 0 in (2), we get First approximation:

1.5 +

(1.5)2 J = 1.29630

Putting n = 1 in (2), we get Second approximation:

2 1 x2 = 3 1.29630 + (1.29630)2 = 1.26093.

Putting n - 2 in (2), we get Third approximation :

1 2 xj = 3 1.26093 + (1.26093)2

= 1.25992.

1 (1.25992)2

= 1.25992.

Putting n ~ 3 in (2), we get Fourth approximation:

1.25992 +

Since jc3 = x4 = 1.2599 correct to four decimal places. root of given equation is 1.2599. A Example 2.31 .*Find the negative root o f the equation x 3 - 21 x + 3500 * 0 correct to three decimal places using Newton-Raphson method. Solution. Let, j(x) - x3 - 2 Ix + 3500 ...(1) / - 5) - + 3485 = (+ ve) A - 10) = + 2710 = (+ ve) A - 15) = + 440 = (+ ve) A - 16) = - 260 = (- ve). C learly/- 15) a n d /- 16) are opposite signs. root lies between (-15) and (-16). Taking initial root: xq = - 15.5

N um erical A nalysis -1 | 1 4 9

?fcBy Newton-Raphson formula: fix*) *« + t

=

x *

~

(* | - 21x„ + 3500) (3x2 _ 21)

2x3 _ 35QQ (3x2 _ 21)

- ( 2)

Putting n = 0 in (2), we get .'. First approximation:

2xg - 3500 _ -10947.75 = - 15.645. xi = (3x§ - 21) 699.75

Putting n = 1 in (2), we get Second approximation:

x2 =

(2x? - 3500) _ -11158.729 = - 15.644 713.298 3xf - 2 1

xj =

(2 *2-35 0 0 ) -11157.260 — 7T* = 3xf - 21 713.204 = - 15.644

Putting n - 2 in (2), we get Third approximation:

Since x2 = x3 *=- 15.644 correct to three decimal places. .‘. Negative root of given equation is - 15.644. Ans. 2J2 ; Find the Newton-Raphson iterativeformula to fin d the pth root o f a positive number /Vand hencefin d the cube-root o f 17. . Let N up = x. Then xP^ N or fix) = xP - N = 0. Newton-Raphson iterative formula:

xn+i = *»

(*£ - N ) A xJ

pxF

(p - l)x£ + N or

pxV

.••(I)

The cube root of 17 lies between 2 and 3. Taking W= 17, xo = 2 and p = 3 in (1), we get 2 x 23 + 17 = 2.75, 3 x 22 2 x (2.75)3 + 17 = 2.58264. X2~ 3 x (2.75)2 Similarity x3 = 2.57133; x4 = 2.57128; xs = 2.57128. Since x* = X5 = 2.5713. /. The cube root of 17, correct to 4 places of decimals = 2.5713.

A ns.

1 5 0 j E n q n s o b n g M athematics-111

Example 2-33 .*Find the root o f the equation x* ** 100, coned to 4 places o f decimals, using Nem Raphson method. Solution. Let, f[x) ~ x* - 100 = 0 =>J\x) ~ x* (1 + log x) Since / 3 ) = - 73 and/4) = 156. The root of the equation lies between 3 and 4. Newton-Raphson iteration formula : x„ +\ =

f (xH) “ ~fXx~)

Let us take xq ~ 3 ,/x q ) = - 73; /'(*&) = 56.66253 *' = 3 + 8 6 ^ 5 3 * l 2 m i * <3' 4> Since the value of the root has to lie with in (3,4), our choice of x$ is wrong. Now taking) = 4»/*o) = 156 andf 'f o ) = 610.89136, we get /. First approximation :

x\ = 4 - ^

156

39136 = 3.74464

Now,

/ * , ) = 40.3506 l ; / ’(x1) = 325.65909 Second approximation : jc2 “ 3.62074. Now, f t x 2) = 5.50! 80; f '( x 2) - 241.24870 .*. Third approximation : xj = 3.59793. Now, / x 3) = 0.14718;/y(jcJ) = 228.37149 Fourth approximation : X4 = 3.59728. Now, / x 4) = - 0.00115;/X*4) = 228.05119 ,\ Fifth approximation : x$ = 3.59729. Since *4 = jc5 = 3.597 correct to 3 decimal places. Required root of the equation is 3.597.

Aw

2.11 ORDER OF CONVERGENCE OF AN ITERATIVE METHOD Let,/*) = 0 be given equation, and let xq, *i, x i,...... x„ be the successive approximations of tht root a o f/x ) = 0. Let e„ be the error in the root n = 0, 1,2,........Since a is the exact root fix ) = 0, then e„ =x„ - a and en+ j = *„ + 1 - a. For any iterative method p is said to be order o f convergence, if \e„ + 1! S k\etf, where * is a positive constant. If p - 1 then the convergence is linear and \fp = 2, then it is quadratic.

2.12 CONVERGENCE OF BISECTION METHOD According to bisection method, we have seen that, the original interval is divided into half interval in each iteration and bracketing the root in new interval. If we take the mid points of the successive intervals to be the approximations of the root, then the one half of the current intervals is the upper bound to the error. Thus, ife„ is the error in nAiteration, then e„+1 is the error in (n + 1)* iteration, which is given by en+1 ~ 0.5 en

N u m er ic a l A nalysis -! J 1 5 1

'Now comparing (1) with order o f convergence p :

M

£ ty X

Hence p = 1 and k = 0.5 The shows that, the bisection method is of 1“ order convergent or linear order convergent.

CONVERGENCE OF SECANT METHOD ®

X|, X2.......are successive approximation obtained by SECANT method :

(*n ~ *»-t) f(x n) /(*») ~ /(*„_,) tod a be the exact root of equation/(.t) = 0, then Let x„ = a + e wandx„+i = a + e„+\ where en,e„+] being errors in n* and (n + 0 th approximation respectively. Hence (1) becomes: a + e „ + i = a + en

or

...d)

(e„ — ■/(« + O f ( a + e„) - f ( a + en_r)

en_i f ( a + en) - ew/( g + e».i) e" + 1 _ /(a + e j - /(a +

f(a) +e j ' ( a ) +£ f ' ( a ) +....

~«n /(«)+«,-! A«)+-|f-/•(«)+....

/(a)+€kA«)+^/'(«)+....

-

/*(«) + .... (e„ -
/ '( a ) +....

/'( a ) +

or

eB+i

2!

/ '( a ) +

tv /«) =0] -.(2)

/ '( a )

[Neglecting high powers of em e„ _ i due to very small] Since by order of convergence p is en+j = k e% where k is constant =*

en = k e£_, [v « - > « - ! ]

-..(3)

1 5 2 | E ngo ^ A

ng

M athem atics-II)

e)lp ~ k vf e„_ i => e„. t = k~l 'P e^/p

=>

p ( a) e,V r k -Vp e„ +\ = ----- 5----- ei * / ’(«)

(2) becomes :

ff-np

'n f ( a ) -+i

ke£ = ~ z - <

or

~

t

f'(a)

[Using (3)1

Comparing both sides, we get 1 , , ^ = p

=>

p2 - p

..

k - " P f" (a ) 2 f \ a ) = constant

- 1=0 1± ^ T 4 1 + y/E /n p = ------ -------= — - — (Remove -ve sign because p > 0).

=>

/. Order p - 1.618. Thus order of convergence p = 1.618.

2.14. CONVERGENCE OF REGULA FALSI METHOD Let f(x) - 0 be given equation. If the ftuiction/(x) in the given equation/(r) = 0 is convex in the interval (*o> * 1) that contains the root, then one of the points x0 or *i is always fixed and other points varies with n. If the point Xo is fixed, then the function / (jc) is approximated by the straight line passing through the points (x0, / (x0)) and (xmf (x„)), n - 1, 2, 3,.... Then the error equation (2) see in above Secant method, becomes : - 1 /"(<*> - „ 2 / ' ( a ) 0’ " e0 = x0 + a is independent of n. 6„+j —k e„

Here

... (A)

1 /"(a)

where k = ^ f ' ( a ) e° iS tl,e asymptotic error constant. Now by the definition of order of convergence p is e„ +1 <; k\e„\p Comparing (A) and (B), we get p - 1 => Regula - Falsi method has linear rate of convergence.

...(B)

2.15 ORDER OF CONVERGENCE OF NEWTON-RAPHSON METHOD Let a be the exact root of the equation fix) = 0. Let x„ and x„ + 1 be the approximations for the root after and (n + I)**1iterations respectively. Let e„ and e„+ 1 be the corresponding errors. Then e„ = x „ - a and e„+ ( = x„ + 1 - a /.e., x„ = + a andx„+ 1 ~e„+J + a. By Newton-Raphson formula, x„ +1 = xn

f(*n)

f ’M

N tM B ubM 'iA m nrsfr'I | l5 S

>• 9 ....

/( g + <*n) f \ a + en)

<* + e» + I = a + c «

_ f( g + e j e" + > “ * A * + 0 [Since expanding X<* + e*) and/ '( « + e„) by Taylor’s series, we have fi a + «.) - /(« ) + ^ /'( « ) +

= e*A «) + f | /'< «) + ....

-•(I)

(since/or) = 0]

and f \ a + e„) -/Xflf) + e /" ( a ) + .... Putting these values in (I), we get % A « ) + g /■<«> + . . . e*+i - e» -

or

en+ i

A «) + «,/*(«) + /'( a )

2[

2f'(a) [

A « ) + e . m i

e* T («) 2A

« )

1-e .

f\a ) f\cc)

-]

f\a ) v e. 1 ---- -

^ A«)._

[v (1 + x )'1 = 1 - X + X2 - ....]

el f \ a ) e„+1 = *^*y7^y [Neglecting small quantities]

or

...(2)

Comparing by definition of order of convergence e„+\ < k \e„\P. Hence p = 2 and

h _ 1 /'( « ) “ *j f'(a) *

This show that order of convergence of Newton-Raphson method is 2, i.e., Newton-Raphson method is quadratic convergent Example 234 : What do you mean by convergence o f the method and its important in numerical analysis. fRGPV June 2003] Solution. A few method which are applicable to both algebraic and transcendental equation/*) = 0to finding the real root of an equation, we first find an approximate value of the root of the gfven equation and then successively improve it to some desired accuracy. The general technique to (. finding the root of equationX*) - 0 is that we start with an initial approximate value, say xq then find the better approximations successively xj, x2, ......x„. By repeating the same method the root more and more closely, i. e., is called method converges. LetXx) = 0 be given equation, and let x& xj, x-l......x„ be the successive approximations to the root aofXx) **0. Let e„ be the error in tire root xn ; n ~ 0,1 ,2 , ......Since a is the exact root o f/x ) = 0, then e„=x„- a and e„+ i =x„ + 1- a For any iterative method p is said to be order o f convergence, if |e„+l| £ k [ e jf , where k is a positive constant Ifp = 1 then the convergence is linear and i f p ~ 2 then it is quadratic.

1 5 4 | E ngineej^ n s M athematics-III

2.16 COMPARISON STUDY OF ITERATIVE METHOD Nameof Method

1. Bisection

Convergence Condition

Iterative formula

Order of Convergence P

*" +! 2 where x„ and x„. t are enclose roots

Gain of one bit /(d) and/(a)

2 Regula Falsi xn*i

per iteration

Guaranteed

must be opposite convergence sign i.e., Ad)Ab)< o. m

xn_ j

Reliability of Convergence

m


Guraranteed convergence

Or (x„ X f( x ) X"+,~ " f(x„) —f ( x n~\) / U ) Xa and i are enclosed roots 2. Secant

4. Newton-

(xn ~ *«-l) u f(x ) p - i+Vs -----—or *"+1 " / ( * > - / < * - ■ > n n) y 2 x„ and x „ _ , are enclosed roots p~ 1.618

***' *"

f(x„) f'(x„)

P= 2

f(Xn)*f(x„- |)

No guarantee of convergence if not near the root.

f(x„)
Convergent fast, if starting value near die root.

Raphson

2.17 GRAEFFE’SR G O I’SQUARING METHOD In all the previous methods we require prior information about die root, but in this method we do not required any previous information and it is capable of giving alt the roots of algebraic equation fix) ~ 0 at a time. Consider the polynomial: fix) =xfl + + aix” ~ 2 + .... + a„-\x + a„ = 0 ...(1) Now, separate even and odd degree powers of x and then square of it, we get (xn + 02*" “ 2 + —¥ = (<*!■*"- 1 + Now, substitute x2 =y and simplifying, we get y + b ^ y - 1+.... + bn —o

“ 3 + —-J2—(2)

N unerjcaI^Analysis-1 1 15 5

where,

b\ - - o f + 2o2 bz - o f - 20]0a + 2a,

A* - (-D»a». If x,, X2, ....x* be the roots of equation ( I X then the roots of (2) are xf , x f....... jc*. Now, after squaring m times, let the new transformed equation be «" + ifcja*- 1 + kit/1" 2 + .... + kitH2 + iii-1« + ^ = 0 ...(3) whose roots are c$|, <5j,....4 a « such that 4 = *?".(» = 1,2,...si.) Now, from (3), we get 14=-*, £
=(-1)"^ J and from these equations we can obtain the result for <5as ty•'

«?i «

«!

and since Sl = xfm

* ~ T - » ............... . «2 *i = (^ )1/2j" -

f

k Y /aM ~T~ Hence, we can determine the actual ^ *»-i /

roots xi, X2, ....x* of the equationX*) * 0. Example 2 3 5 : Apply Greffe's root squaring method to solve the equation x i - t x 2 + 1 7 x -1 0 - 0 fRGPV Dee. 2003/ Solution. Given x3-8x 2 + ! 7 x - 10 = 0 ...(1) Rearranged given equation by separating even and odd degree powers of x, we get x* + 17x = 8x? + 10 Squaring both sides, we get (x3 + Mx')1 * (8X2 + 10)2 => x* + 34** + 289x2 = 64x4 + 160c2 +100. Now, let x2 - y, we get / + 3 4 / + 289y * 6 4 ^ + 160y+ 100. => / + 129v = 3 0 /+ 1 0 0 . [As rearranging by separating even and odd degree terms] Again squaring both sides, we get

'<■

=> =>

( / + 129y)2 - (30V2 + 100)2 / + 2 5 8 / + 16641/ = 9 0 0 / + 6 0 0 0 / + 10000 /+ 1 0 6 4 1 /= 6 4 2 /+ 1 0 0 0 0

Now, let / = z, then we get r3 + 1064 Iz =642^ + 10000

1 5 6 | E n g in h w n o M athematics-IH

Again, squaring both sides, we get (z* + 10641 z f = (642Z2 + 10000)2 =>

z6 + 21282z4 + I13230881Z2 = 412164z4 + 12840000z2 + 100000000

Put z2 = h, then we get - 3 9 0 8 8 2 m2 + 100390881 « - 10s = 0

Comparing (2 ) with m3 + k\u 2 + k2u +k%

= 0

...(2)

and hence

*, = - 390882, k2 - 100390881, *3 = - 108.

Let the roots of equation (2) are *i =

<%,

then we get

= ( - ki)m

(390882)1'8 = 5.004 « 5

*2

- f - S s s r )

*3

(^ 1 0 0 3 9 0 8 8 1 J

Thus, roots of given equation are 5 , 2 and 1 . Ans. Example 2 J6 : Solve the equation x 3 - 6x* + l l x - 6 “ 0 by Graeffe’s root squaring method. [RGPV June 2006J Solution. Given x3 - fix 2 + 1lx - 6 = 0 ...(1) Rearranged given equation by separating even and odd degree powers of x, we get x3 + I It = 6x2 + 6

=> x(x2 + U ) = 6(x2 + i ) Squaring on both sides, and putting x2 = y,we get y(y+ 1 1 ) 2 = 3 6 ^ + 1 ) 2 => Xy2 + 22y + 121) = 360^ + 2y+ 1) => y3 + 22y2 + 121y —3 6 )^ + 72y + 3 6 =>

y* + 49y - My2 + 36 yiy 2 + 4 9 ) —(My2 + 3 6 )

=> Again squaring and putting y 2 = z, we get

z(z + 49)2 —(1 4 ^ + 36)2 =>

r 3 + 98z2 + 2 4 0 1 z = 1 9 6 z 2 + 1 0 0 f e + 1296

=>

z3 + I3 9 3 z = 98Z2 + 1296

=>

z(z2 + 1393) - (98Z2 + 1296)

Further, squaring and putting z2 - u, we get => => =>

« (« + 1393)2 = (9 8 « + 1296)2 m3 +

2786m2 + 1940449m = 9604m2 + 2 5 4 0 1 6 a + 1679616

« 3 - 6 8 1 8 m2 + 1686433 m - 1679616 - 0.

Comparing (2 ) with m3 + k\u* + k$u + *3 = 0 , so that i ] —— 6 8 1 8 , k 2 = 1686433, k3 * - 1679616.

...(2)

N um erical A n a l y w *! 1 1 5 7

Let the roots of equation (2) are S\, <%, then we get x, = ($)>*==(-*i),/8 = (6818)l/8 = 3.01444 * 3

/rvw« f kz)Ua ( 1686433Y/8 - ( - j g i g - J

,• ^

x*

/Jtu /s-f ^

feaV/8

-1 .9 9 1 4 .2

f 1679616f ' 8 U 686433J

Thus, roots of given equation are 3,2 and 1. ^KACRSTOW’S M ETH O D O R M ETH O D FO R C O M PLEX R O O T

0.9999 « 1. Ans.

i 3Rbs method is useful to obtain complex roots of an algebraic equation/*) = 0. The complex roots -Ofsuch an equation occur in pairs a ± ib. Each such pair corresponds to a quadratic factor, which j®as follows: ‘’t

{ x - ( a + ib)} { x - ( a ~ ib)} =x2 ~2ax + a2 + b2 - x 2 + px + q,

where p and q are real. ’Let f i x ) - x” + a\x"~ 1 + .... + On- \x + a„ be given polynomial If we divide/*) by (jc2 + px + q), we obtain a quotient

Qn-2

...(I)

= X n ~ 2 + b ]Xn - i + „.. + b„-2

■ and a remainder R„ = Rx + S Hence (1) becomes: f ix ) = (x2 + px + q) (x"-2 + - 3 + .... + b„-j) + Rx + S •—(2) IfX2 +px + q divides/x) completely, the remainder Rx + 5 = 0 i.e., R - 0, S’= 0. Obviously R and S both depend upon p and q. So our problem is to find p and q such that Rip. q) = 0, S{p, q) = 0. ...(3) Let p + Ap, q + Aq be the actual values of p and q which satisfy (3) then, Rip + Ap, q + Aq) = 0; Sip + bp, q + Aq) = 0 To find the correction Ap, Aq, we get the following equations : Cn-2*P + Cn-S&I * K -l \ (Cn~l - V l W

+ * ,-2 ^ = K

J

" W

After finding the values of bp and c,s by synthetic division scheme, we obtain approximate values of Ap and Aq say, Apo and Aqo Ifpo, q0 be the initial approximations then their improved values are P\ ~ P 0 + Apo, q\ - qo + Aqo.

Now taking p\ and q\ as the initial values and repeating the process, we can get better values of p and q. This method works well only if the starting values of p and q are close to the correct value.

1 5 8 | E N O o saaN o M A n o u n cs-n i

Remark : The synthetic division procedure are asfollows: a* :

a0(=l)

<»1 -pb o

a2 -Pbi -flbo b2

-p c 0

“ PCt -
ho(=D

c0(—1)

Cl

- P ^ 2 .... - f * n-3 -< £ 1 .... -
On-1 -P*>n-2 -P
~pb„. i - p b n_2

*PCn-2 -<*n-3 C*-l

-p ~Q -P ~Q

Example 2.37 : Find a quadratic factor o f the polynomial X4 + ix* + 3x? - 5* - 9 ® ft Starting with p g ~ 3 , q o ~ ~ 5 by using Bairstow^s method. fRGPV June 2003] Solution. Synthetic division scheme is as follows :

c*_ j —b„_ j = —36 + 1 ——35. The corrections Apo and Aqo are given by Cn-2&P0 + C » -3 ^ 0 = V ! => 10 Apo =~ 1 and (c * .| - b „ - ]yipo + c„^ 2 ^!o = l>n => -3 5 Apo + 10 =4 On solving (1) and (2), we get, Apo = - 0.09, A?o = 0.08 Thus pi, q\ the fjgp approximations ofp and q are given by pi =Po + Apo = 2.91, = qo + ~ ~ 4-92 . Repeating same process Le., dividing/x) by x2 + 2.91x - 4.92, we get 1

5 -2 .9 1

1

2.09 -2 .9 1

1

-0 .8 2

3 -6 .0 8 4.92 1.84 2.37 4.92 9.13

-5 -5 .3 5 10.28 -0 .0 7 -2 6 .5 7 -4 .0 3 -3 0 .6 7

-9 0.20 9.05 0.25

—(1) ...(2)

-2 .9 1 4.92 -2 .9 1 4.92

N um erical A nalysis -! | 1 5 9

At this step, the corrections Ap, and A^i are given by, 9.13 Ap) - 0.82 Aqi = -0 .0 7 - 3 0 . 6 0 ^ + 9.13 Aqt =0.25 => Ap, = - 0.00745, Aq\ - 0.00241 Hence second approximation of p and q are given by, Pi + Api - 2.91 - 0.00745 = 2.90255 q2 = q\ + A?, = - 4.92 + 0.00241 = - 4.91759. Thus a quadratic factor is, x2 + 2.90255 x - 4.91759. Dividing the given equation by this factor, we can obtain the other quadratic factor.

Ans.

2.19 SOL UTION OFSIMULTANEOUS LINEARALGEBRAIC EQUATIONS We come across very often, simultaneous linear algebraic equations in various fields of science and engineering. We have solved such equations by mathematical methods such as Cramer’s rule (determinant method) and matrix method. These methods involve a great amount of labour, when the number of equations exceeds four. On die other hand, the numerical methods, which involve less amount of labour and are also best suited for computer operations. These numerical method are of two types namely ; (i) Direct method (ii) Iterative method. In the direct method the given system of equations will be transformed into an equivalent upper triangular system. In the iterative method, the solution of given system of equations is obtained by successive approximations.

2.20 GAUSS-ELIMINATION METHOD (DIRECT METHOD) Consider the n-linear equations in n unknowns, a u X i + O j2*2 +

+ ° \ n x n ~ ^1

021*1 + <*22*2 + •— + °2nxn = ^2 .-(I) + °»2*2 + ■■■*+ a«rt*n = K where dy and b, are known constants and x,’s are unknowns. The system (1) is reduce to matrix form : AX = B

where

A =

On <*12 •• °1n 021 °22 °2n _anl ®rt2 •• ann .

N ow our ob operations.

. x=

*1 *2 .Xn.

...(2)

V and 5 =

h

A.

1 6 0 | F itith a m m l U n g m n c s - m

On Now take the pivot an, so that multiply the first row of (3) by “ ~ and add to the ith row of ^11 [A : B], where i = 2 , 3 , n. So that all elements in the first column of [A : J?] except a\ y are made to zero, then (3) reduces to

[A : B] ~

°ii °iz *• °21 a22 ••

..(4)

a »l ar>2 "^iw Now taking 622 as the pivot, using elementary operations, we make all elements below bn as zero, then (4) reduces to Oil °12 0 ^2 [A : B]

~

a 13

*•

°ln

C2

^!3

0

0

C33

•>

0

0

C*3

••

...(5)


Cnn

Again taking C33 as the pivot, using elementary operations, we make all elements below C33 as zeros. Continuing the process, all element below the leading diagonal elements of A are made to zero. Hence

[A : B] ~

Oil

°12

°13

„

a ln

fc l‘

0

^22

kt3

..

^2n

C2

0

0

C33

c 34...

^ 3n


0

0

0

0

a«.

-(6)

From, (6), the given system of linear equations is equivalent to a„*! + 0 x2*2 + 0(3*3 + ™ + “i ^%2*2 + ^Sl3*3 + ••• + = C2 C33*3 + —CstXn = dj <*nnXH = By using back substitutions, we get values ofxm xn. 1, x„_ 2, .... *2, * 1. Example 2 .3 8 : Solve the following system o f equations by Gauss-eUmlnatktn method : 2 x - y + 3z » 9, x + y + z * 6, x - y +z = 2 fRGPV Dec 2001}

N um erical A nalysis *! j 1 6 1

elution.

The given system in matrix form : AX = B 2 - 1 3 A = 1 1 1 1 -1 1

where,

Augmented matrix:

2 - 1 3 [A : 5] = 1 1 1 1 -1 1

..(I) [Taking «n - 2 as pivot]

Operating

j

and -^3

**3 “ ( 2 ) 3 -1/2 -1/2

'2 -1 0 3 /2 0 -1/2

[A : B]

we get 9 3/2 -S/2

[Now taking 622 = 3/2 as the pivot] Operating: ^3

(1 ^ +I &2 , we get 3 3 /2 - 1 / 2 0 -2/3 -1

'2

[A:B\

0 0

9 3/2 -2 J

.(2)

Which is in upper triangular form. From (2), we get 2jc -

y + Sz = ®

( 3 /2 ) y - ( l/8 k = 3 /2 (-2 /3 )2 = - 2

Using back substitution, we get z = 3 , y = 2 , x = I. ,\ The requked solution is x = 1, >>* 2,2 = 3. Example 2*39: Solve by Gaim-elimiHOtkm method:

Solution.

Ans.

lO x + y + 2 z - 13 3 x + lO y + z = 14 2 x + 3 y + lO z = IS Given equations write in matrix form : AX~ B. e*

A =

...d) 1

0rH

where,

3 2

(R G lty June 2002]

X 10 1 , X= y » z_ 3 10 '

13* 14 15

1 6 2 | E ngin eerin g M athem atics-!!!

A ugm ented matrix :

[A : B\ =

‘ 10 3 2

O p e ra tin g : R 2 -* Rz

— |

1 10 3

and

[A:B]

2 1

13 14

10

15

^

- j “ j ^ i . we get

" 10 0

97/10

1

2 2/5

13 10 1/ 10

0

14/5

48/5

62/5

_ ^ - ^ 2* we get

O perating :

10 0 0

[A:S\

1

2 2/5

97/10 0

460

13 1 0 1 / 10 ' 460

Which is in triangular form. From (2), we get

IOjc + y + 2z = 13 (97 / 10)y + (2 / 5)z = 101/10 4602 = 460 Using back substitution, we get

2 = l,y = l,x » 1 /. The required solution is x = l, > ’= l >z = l . Example 2.40 : Solve by Gauss-eliminatioH method : 3x + 4y + 5i - 18 2 x - y + 8z = 13 S x -2 y + 7z =20.

Solution.

A ugm ented matrix

[A : B] =

Ans.

3

4

5

18"

2

-1

13

5

-2

8 7

20

fl,1 ■ we get 3

[A : B) ~

0 0

4 -11/3 -26/3

18 ' 5 14/3 1 - 4 / 3 -10

N um erical A nalysis -1 | 1 6 3

Operating: &2

R3

we get

2}

5 18' ~3 4 [A : B] ~ 0 -11 14 3 0 13 2 15 Operating:

+ — j ^2 • we get 18 5 ‘3 4 14 0 -11 3 [A : B}~ 0 0 204/11 204/11

Which is in triangular form. From (1), we get 3x + 4y + 5 z = 18 -lljy + 14z = 3

204 204 ■z 11 11 Using back substitution, we get 3 - 14z -1 1 The required solution is x = 3, .y = 1, z = 1. Z

=

1

, y

=

1 8 - 4 y - 5z =

ample 2.41 : Using Gauss-ellmination method, solve the system 3.15x - 1.96y + 3.85z = 12.95 2.13x + S.12y - 2.89z = - 8.61 S.92x + 3.0Sy + 2.1Sz » €.88

Solution.

3.15 -1.96 3.85 [A : B]= 2.13 5.12 -2.89 5.92 3.05 2.15 O perating:

^ 2

12.95 -8.61 6.88

>Rz+(“fill) Ri ***3 [A : B] -

=3

] ' x

12.95 3.15 -1.96 3.85 0 6.4453 -5.4933 -17.3666 0 6.7335 -5.0855 -17.4578

Ans.

1 64 | E n g in eerin g M athem atics-III

f 6 .7 3 3 5 )

Operating : -“ 3

n 3 ~ I 6 4453 I™2’ we get

12.95 3.85 3.15 -1.96 0 6.4453 -5.4933 -17.3666 0 0.6534 0.6853 0

[ A: B] ~

...(1

From (1). we get 3.15* - 1.96y + 3.85z = 12.95 6.4453y - 5.4933* = -17.3666 0.6534* = 0.6853 Using back substitution, we get 0.6853 0.6534 ‘ , 048Sy =

5.4933 x 1.0488 - 17.3666 6.4453

l '80° 5,

1.96 x (-1.8005) - 3.85(1.0488) + 12.95 = 1.7089 3.15 Ans. /. The required solution is x = 1.7089, y - - 1.8005, z = 1.0488. Example 2.42 : Solve by Gauss-elimination method: lO x- 7y + 3z + 5u = 6, - 6x + 8 y ~ z ~ 4 u = J, 3x + y + 4z + llu = 2, S x - 9 y - 2 z + 4u = 7. x =

‘ 1

Solution.

[A: B] =

-6 3 5

Operating :

0.7 0.3 0.5 0 .6' 8 -1 -4 5 4 1 11 2 -9 -2 4 7 R2 + 6R\> R3

Rj - 3/?i, R4

[Since dividing first equation by 10

R* - 5R\, we get

"l 0.7 0.3 0.5 0.6 0.8 -1 8.6 0 3.8 [A : B ) ~ 0 3.1 3.1 9.5 0.2 0 -5.5 -3.5 1.5 4 O perating: Ri

.l )/?2 and R4 -> (3.8)&i + (5.5 1 0.7 0.3 0 3.8 0.8 [A : B] ~ 0 0 9.3 0 0 -8.9

0.5 0.6 ‘ 8.6 -1 39.2 -25.9 0.2 62.5

N um erical A nalysis -! | 165

Operating :

/? 4

—►(9.3) R4 + (8.9) # 3 , we get

0.6 1 0.7 0.3 0.5 8.6 -1 0 3.8 0.8 0 0 9.3 39.2 -25.9 0 350.74 350.74 0 0

[A:B]

.-(I)

From (1), we get x + 0.1 y + 0.3z + 0.5 u 3.8,y + 0.8z - u 9.3 z + 39.2u 350.74u

= 0.6 ~ 8.6 = -25.9 = 350.74

Using back substitution, we get m= 1, z = - 7,>> = 4, x = 5 /. The Solution of given equations is x = 5, y - 4, z = - 7, u - 1. xampie 2.43 ; Solve the following system o f equations by Gauss-ellmination method:

’elution.

Sxj - x 2 + x 3 ~ 1 0 2xt + 4 x 2 = 12 x , + x 2 + Sx 3 - -1 The given system in matrix form is AX - B Augmented matrix: '5 -1 1 10' [A : B) = 2 4 0 12 1 1 5 -1

Operating : R\

R}, we get [A : B] ~

1 1 2 4 5 -1

5 -l‘ 0 12 1 10

Operating : R2 -* Rz~ 2R\, R-} -> R3 - 5R,, we get 5 -1 1 1 14 -10 0 2 [A: B] ~ 0 -6 -24 15 O perating: ^2

f1 U J]

5

1 1

1 1 0 1 0 -6

7 3

[A: B ] ~

we get -l' 7 15

Ans.

1 6 6 | E n g in eerin g M athem atics-111

O perating : Ry -> Ry + 6/fr, w e get

[A : B) -

1

1

5

-1

0 0

1 0

-5 -5 4

7 57

Using back substitution, w e get From the last row, we have - 54*j = 57 57

-1 9

= “ 54 = From the second row, we have *2

18

5xi = 7 19 = 31

x-> = 7 - 5 x — = —

18 “ 18

From the first row, the have x\ + X2 + X\

=- 1 95 = 46

= -1 - — 18

18

23

.*. The required solution is jcj =

~^~
18

31

23 9

19 and *3 - - — 18

Ans.

2.21 GAUSS-JORDAN ELIMINATION METHOD (DIRECT METHOD) This method is a modification o f G auss elimination m ethod, in this method, the coefficient matrix

A o f the system A X - B is brought to a diagonal matrix o r unit matrix by making the matrix A tra only upper triangular but also low er triangular by making all elem ents above the leading diagonal of

A also as zeros. By this way, the system AX = B will reduce to the form.

Oil 0 0

0 0 0 b22 0 0

0 0
0 0 ^nn V

0

Using back substitution, w e get the solution : x n _ ......... . x2, X\. Example 2.44 ; Solve the following system by Gauss-Jordan m ethod: 5xf + x 2 + x} + x 4 = 4; Xi + 7x2 + X3 + X4 = 12 x ] + X 2 + 6X 3 + X 4 - - 5;

xi +X2 +Xs + 4xi = - 6 Solution.

Interchange the first and the last equation, we get

Augmented matrix,

[A : B] -

1

4 ~6

1

1 6

1

1

1 12 1 -5 1 4

1 1

1 7

1 5

N um erical A nalysis -! | 167

Operating : Ri -» Ri - R\, Ry -» Ry~ Ru Ra

[A : B)

(V O p e ra tin g : ^2

6j

"l 1 0 6 0 0 0 -4

1 1 0 1 0 0 0 -4

Operating : R\ -> R\ - Ri, Ra

1 0 5 -4

4 -0.5 -3 -19

-6 3 1 34

1 0 0 0

0 1 1 0 0 5 0 -4

4.5 -0.5 -3 -21

-9 3 1 46

1 0 0 0

0 1 1 0 0 1 0 -4

4.5 -0.5 -0.6 -21

-9 3 0.2 46

R\ + 4R}, w e get

1 0 [A:B] ~ 0 0 -^ 4

-6 ' 18 1 34

j - ^ 3 > w e get

[A:B]

O perating:

4 -3 -3 -19

- R2, then Ra -* Ra + 4/?2, w e get

[,4:2?]

Operating : Ra

1 0 5 -4

we get

[A : 5] -

O perating : / ? [ - »

Ra ~ 5/?i, we get

0 1 0 0

0 5.1 -9 .2 ’ 0 -0.5 3 1 -0.6 0.2 0 -23.4 46.8

1 1 0 0

1 5.1 -9.2 ‘ 0 -0.5 3 1 -0.6 0.2 2 0 -1

' 1 N R* >w e get { 23.4

[A : B]

“l 0 0 0

16 8 | E n g in eerin g M athem atics-H I

Operating:

—►JR3 - (0.6)

-*■ R? - ( 0 . 5 ) R i , R l -* R\ + (5.1)/L», we get

1 0 {A : B] ~ 0 0

0 1 0 0

0 0 1 0

0 0 0 -1

1 2 -1 2

Using back substitution, we get xi = l, x 2 - 2, X3 = - 1, jt4 = - 2 . Example 2.45 ; Apply Gauss-Jordan method to find the solution o f the following system: I 8x +y + 2 ™12 ; 2x+ I0y + z a 13; x + y + Sza 7. Solution. Interchange the first and the last equation, we get 1 5 7' 2 10 1 13 [A : B) = 10 1 1 12 "1

Operating : R2 -» R2 - 2R]> R)-+ R3 - 10/?i, we get 1 1 [A : B] ~ 0 8 0 -9 O perating : Ri

[g

5 7 ‘ -9 -1 -49 -58

w e 8et

1 1 [A : B) ~ 0 1 0 -9

7 5 -9 /8 -1/8 -49 -58

Operating : R^ -»• R$ + 9R2, we get 7 5 rl 1 0 9 / 8 1 /8 1 [A : B) ~ 0 0 -473/8 -473/8 Operating: R 9

1 ^ ’ weget 5 7 ‘l 1 1 /8 [A : B] - 0 1 - 9 / 8 1 0 0 1

Am

N um erical A nalysis -1 1 1 6 9

R} - R?. we >yi

Operating

I u 49/8 57/8' [A : B] -

o

0 Operating: ^2

1 -9/8 -1/8 1 0 1

•fta + [ g ] -^3* -^1

^

8 I

we get

'l 0 0 I [A : B) ~ 0 1 0 1 0 0 1 1 Using back substitution, we get x = l,y = I, z = 1.

Ans.

CROUT’STRIANGULARIZATION METHOD OR CHOLESKEY’SMETHOD ORLUFACTORIZATIONMETHOD(DIRECTM E T H O D ) This method is based upon the fact that any square matrix can be expressed as the product of a lower and a upper triangular matrices (All the principle minors of the matrix must be non-singular). i.e., if A ~ [o,y], Then \A\ * 0. Let us consider the equations : 0)1*1 + a J2*2 + °13*3 ~ &! <*21*1 +

022*2

+ °23*3 = &2

...d)

° 3 l x l + ° 3 2 x 2 + °33*3 = ^3

Which can be written in matrix form : A X - B

•••(2)

V

where, Since

X\ «1! °12 °13 A « 021 °22 023 > x= H and B = b2 _*s. bs .°3I °32 O33 _ A = LU ' 1

where

L=

0 1 hi

matrices. /. System (1) becomes : We assume UX = V, where V Then system (4) becomes :

O' 0 and t/ = 1 LUX - B

0 0

...(3)

Uj2 “13 °22 ^23 are lower and upper triangular 0 ^33 _ l()

»>2 -.(5) y3 LV = B

-(6)

1 7 0 | E ng in eerin g M athem atics-!!!

1 0 0' ^!1 i 1 l>2 4i ^2 1 3 .

■

=

V *>2

k h \V\ + V2 = &2 h i vi + httVi + ^3 = Solve these for Vi.e., vj, v2, v3 and putting in (5), we get U; i .T| + U,2*2 + « J^ X 3 = V,

U22X2 + 11*3X3+ = “ 33 X 3 = V3 j

x\, xi, xj. Example 2.46 : Apply Crout's method to solve the equations : 3x + 2y+7z = 4; 2x + 3y + z ~ 5; 3x + 4y + z —7. Using back substitutions, w e have obtain

> ’tution.

' fRGPV Dec. 2004 & Dec. 2007}

G iven system can be written in m atrix form : AX = B

' 1 Sincc A ~ LU w here L = hi

i

hi

M ' 1 hi hi

T^>

un ^!ult k l uU

o '1 0 and ( / = 1

0

0 1 hi

O' ui\

“ ii 0 0

Ui2 Wia

0

0

“ 22

1

0

0

U12 w13 41^13 + °2S 4 lu12 + **22 ^ lu12 + ^ “22 ^3l“l3 + ^S2°23 +

«23 “ 33.

-(I) Ui2 !i22 0

«% °23 “ 33 .

3

= 2 3

7' 1 4 1

2 3

3 2 7 2 3 1 3 4 1

Hence

u!3 - 7 u, 2 = 2 ^21u13 + ^23 = 1 ^21ui2 + u22 = ^ lio3 ~ ” 1 1 /3 ^21 — 2 / 3> => U-22 = 5 / 3, ^ /3]Uu = 3 h l Ul2 + ^32^23 + °33 ~ 1 ^!U32 + ^32°22 = 4 => U -33 = -8 / 5 => 42 = 6 /5 , 4n =

«n - 3, ^21^11 = ^

[v LU = A]

N um erical A n a ly sis -1 | 1 7 1

Thus,

1 2/3

0 1

1

6/5

0" 0 and {/ 1

3 2 0 5 /3 0

0

7 ] -1 1 /3 | -8 /5 j

UX = V

Suppose,

••(2)

V =

where

then A X - B => LUX = B=>LV= B

V3 1 2 /3

0

1

6 /5

0 " ii 0 u2

1

1

4 =5 5 7

_U3_

2 7 6 => V! = 4; - i'i + i'2 = 5 => v2 = - ; u, + - v2 + vs = 7 o

.i

5

L*3 ——. 3

5

Hence (2) i.e.%UX = V becom es : '3 2 0 5 /3 0 0

7 -11/3

=

> 2

-8 /5

4 7 /3 1/5

3% + 2 y + I z ~ 4 ^

(5/3)y - ( ll/3 ) z = 7 /3 - ( 8 / 5 )z = 1 / 5

Ans.

Using back substitution, we get z - - 1/8, y = 9/8, x - 7/8.

Example 2.47 : S'o/ve Ay Crout’s m ethod:

Solution.

10x + y + z = 12; 2x + lOy + z ~ 13; 2x + 2y + lOz = 14. Given system in matrix form : A X - B ' 1

0

0‘

h\

1

0

^3!

^32

1

Let, A = LU, w here L =

(I) un

u t2

UI3

0

«22 0

^23

and t / =

0

^32 _

LU = A, we get

Since,

«jt = h iuu ~ > ^21 = *3l“ ll >

10, 2 1/5, 2

*31 = 1 / 5 ,

u12= 1 *2IU12 + ^22 = :>

Uq2 = 4 9 / 5,

41^12 + ^32^22 = 2 ^

/g, = 9 / 49 ,

u13 = 1 + u 23 = 1

^

u 23 = 4 / 5

^31*% + ^ 2 U23 + u 33 ” 1 0

u 33 = 4 7 3 / 4 9

1 7 2 | E ngin eerin g M athematics -III

Suppose,

UX = V

-(2 )

f-’i where V =

. then AX = B => LUX = 5 => LK = B ^3

1 0 0‘ 0 U2 _ 1 /5 1 1 /5 9 / 4 9 1 -y3_ =>

=> =>

12

13 14

V[ = 12; /2 [V| + vj = 13 v2 = 53/5 /31 vj + /32V2 + V3 = 14 v3 = 473/49.

Hence (2) i.e., UX = V becomes : 1 ‘10 1 4/5 0 49/5 0 4 73 /49 0

x y z

12 5 3/ 5 473/49

lOx + y + z = 12 (49/5);y + (4/5)2 = 5 3 / 5 (473/49)2 = 473/49 By back substitution, we get z = l,.y = l,x = 1.

Ans.

2.23 ITERATIVE METHOD The iterative method is not always successful to all systems of equations. In order for the iteration procedure to give the solution of a system of equations, each equation of the system must contain one coefficient much larger in magnitude than the others in that equation and the large coefficient must be that of a different unknown in each equation. In other words, after rearranging the equations if necessary, the large coefficients must be along the leading diagonal of the coefficient matrix. i.e., the system of equations : will be solvable, if I Oj 1 I> I I + I <*13 I I O02 l>l @21 I + I ®23 I ’ I O33 I> I £*31 I + I O32 I

2.24 GAUSS-JACOBI METHOD Consider the system of equations of three variables in three unknowns : Oj* + l\y + cxz = dt 02* + f^y + CjZ = dz a^x + bsy + c3z = tf3

if,

.(1)

N um erical A nalysis -! | 173

11*! I> I bi 1 + I c, | I fife ! > | Oj I + I c2 I

| Cs |> | <*3 | + | 63 I then, iterative m ethod can be used for the system ( I ) Hence w e w rite x, y, z as follow s :

x

= ~(dt - b y - c i z ) >, <*1

y =

(dj - oax - c^z)

-.(2)

z = — (da - agx - b^y). If x<0>, >^0), z<°> are th e initial values o f x, y, z respectively, then

First iteration : ^ l ) = — (d[ - 6|y °> - c,z<°>) ^

«1 yl) = “ (A - O2X{0) - C^Z™) 0*

(3 J

■

2<1) = — (dg - <%x(m - b ^ y ^ ) ) c3

A gain using these values j

in (2), w e get

Second iteration: y(2) = — (d, <*1 1 y2) = — (<4 -

2(2) =

—

(di -

- C2Z<‘>)

0 3X<1> -

^ y « )

..(4)

J

C ontinue the procedure till the convergence is assured correct to required decim al places.

Remark : In the absence o f initial values o f x , y , r w e take, usually x = y - z ~ 0. Example 2.48 ; Solve the following system by Gauss-Jacobi method : lOx - 5 y - 2 z = 3; 4x - lOy + 3z = - 3; x + 6y + lOz - - 3 . fRGPV Dec. 2008(N)j

1 7 4 | E n g in eerin g M athem atics-!II

Solution.

Since |tO| > |- 5| + |~ 2|, |- I0| > |4| + |3| and |10| > |1| + \6\. Then Gauss-Jacobi method can be used for the given system of equations. Given system can be written as : x —

[3 + 5^ + 2 z]

v = — [3 + 4x + 3z] y 10

Let the initial values be x =y = z = 0, we get First iteration : * 1, = — [3 + 5(0) + 2(0)] = 0.3 yi) = - 1 [3+ 4(0)+ 3(0)] - 0.3 ^ [-3 - (0) - 6(0)] = - 0.3 Second iteration : Putting the values o f x ^ . y 1>, z*1*in (I), we get *2) = i - [3 + 5(0.3) + 2(-0.3)] = 0.39 y 2) = -L [3 + 4(0.3) + 3(—0.3)] = 0.33 = ^ [- 3 - (0-3 ) - 6(0-3)] ——0.51 Third iteration : Putting the values of xP \ y^2\ z<2) in ( I), we get & = ^ [ 3 + 5(0.33) + 2(—0.51)] = 0.363 ^

= ^ t 3 + 4<0 3 9 > + 3(~°-51)} = 0.303

aP> = ^ [ ~ 3 - (0.39) - 6(0.33)] = - 0.537 Fourth iteration : Putting the values of

/ 3),

in (1), we get

= j^ [ 3 + 5(0.303) + 2(-0.537)] = 0.3441 = ^ [ 3 + 4(0.363) + 3(— 0.537)) = 0.2841 ^4) = j L [ - 3 -(0.363) -6(0.303)] =-0.5181

N umerical A nai,ysis-I | 1 7 5

Fifth iteration :

10

y5)

3 + 5(0.284) + 2(-0.5181)] =0.33843

J_ 3 + 4(0.3441) + 3(-0.5181)] = 0.2822 10 JL_

10

-3 - (0.3441) - 6(0.2841)] « _ 0.50487

Sixth iteration :

x<(>)

J_ 3 + 5(0.2822) + 2(-0.50487)] = 0.340126 10

y(6)

J_ 3 + 4(0.33843) + 3(-0.50487)] = 0.282911 10

t<6)

J_ -3 - (0.33843) - 6(0.2822)] = _ 0.503163 10

X<7)

J_ 3 + 5(0.283911) + 2(-0.503163)| .. 0.3413229 10

y7)

J_ 3 + 4(0.340126) + 3(-0.503163)] = 0.2851015 10

Seventh iteration :

J_ -3 - (0.340126) - 6(0.283911)] = _ 0.5043592 10 Eighth iteration :

*8)

J_ 3 + 5(0.2851015) + 2(-0.5043592)J - 0.34167891 10

^8)

J_ 3 + 4(0.3413229) + 3(-0.5043592>] = 0.2852214 10

2<*>

J_ -3 - (0.3413229) - 6(0.2851015)] = - 0 . 5 0 5 1 9 3 1 9 10

x<*>

JL_ 3 + 5(0.2852214) + 2(-0.50519319)] = 0.34157 10

Nineth iteration :

J_ 10

3 +4(0.34167891)+ 3(-0.50519319)] =0. 28511

1 = TX -3 - (0.3416789) - 6(0.2852214)] = - 0.5053 10 Since eight and nineth iteration are sam e correct to 3 decim al places. Hence, x = 0.342, y » 0.285, z - - 0.505.

A ns.

1 7 6 [ ILnginleking M athem atics-111

Example 2.49 ; Solve the following system o f equations by using Gauss-Jacobi method correct to} decimal places: 8x - 3y + 2z * 20; 4x + H y - z -3 3 ; 6x + 3y + 12z ~ 35. Solution. Since |8| > {- 3| + |2|, (11| > |4| + j- !j and |i2| > |6| + |3|. Then G auss-Jacobi m ethod can be used for the given system o f equations. H ence, w e w rite jc, y, z as follow s

x = ±[20 + 3y - 2z] ' o J j [ 3 3 - 4 1 + *J i[3 5 -6 x -3 > U Let the initial values be x = 0, y -

0, z = 0, w e get

First iteration :

*0) = I [20 + 3(0) -2(0)] =2.5 O y n = _L [33 - 4(0) + 0] - 3.0 2(1) = A-[35 - 6(0) - 3(0)] = 2.916666 Xz Second iteration : Putting th e values o f

1\

>, r *1 >in (1), w e get

x<2) = I [20 + 3(3.0) - 2(2.916666)] = 2.895833 O yt2j = l - [ 3 3 - 4(2.5) + (2.916666)] = 2.356060 p . ) = ± - [ 3 5 - 6 ( 2 .5 ) - 3 (3 .0 )) = 0 .9 1 6 6 6 6 X^

Third iteration: * 0 ) = i [ 2 0 + 3 ( 2 .3 5 6 0 6 0 ) - 2 ( 0 .9 1 6 6 6 6 ) ] = 3 .1 5 4 3 5 6

O

y 3 ) = - ± - [ 3 3 - 4 ( 2 .8 9 5 8 3 3 ) + ( 0 .9 1 6 6 6 6 ) ] = 2 .0 3 0 3 0 3

2(3) = — [ 3 5 - 6 ( 2 .8 9 5 8 3 3 ) - 3 ( 2 .3 5 6 0 6 0 ) ] = 0 .8 7 9 7 3 5

12

"Fourth iteration : x<4) = ± [ 2 0 + 3 ( 2 .0 3 0 3 0 3 ) - 2 ( 0 .8 7 9 7 3 5 ) ] = 3 .0 4 )4 3 0

8

N um erical A n a i ysi% | | 1 7 7

y 4) = j y [ 3 3 - 4 (3 .1 5 4 3 5 6 ) + (0 .8 7 9 7 3 5 )] = 1.93293

2(4) = — [3 5 - 6 (3 .1 5 4 3 5 6 ) - 3 (2 .0 3 0 3 0 3 )J = 0.831913

12

Fifth iteration : x<5) = I [2 0 + 3 (1 .9 3 2 9 3 7 ) - 2 (0 .8 3 1 9 1 3 )] = 3.016873

y j ) = J L [ 3 3 _ 4 ( 3 .0 4 1 4 3 0 ) + (0 .8 3 1 9 1 3 )] = 1.96965^

i s ) = — [3 5 - 6 ( 3 .0 4 1 4 3 0 ) - 3 (1 .9 3 2 9 3 7 )] = 0 .9 1 2 7 1 7

12

Sixth iteration . = - [ 2 0 + 3 (1 .9 6 9 6 5 4 ) - 2 (0 .9 1 2 7 1 7 )] = 3 .0 1 0 4 4

8

y«> = - L [ 3 3 - 4 (3 .0 1 6 8 7 3 ) + (0 .9 1 2 7 1 7 )] = 1.985930 * 6) = — [35 - 6 (3 .0 1 6 8 7 3 ) - 3 (1 .9 6 9 6 5 4 )] = 0.915817

12

Seventh iteration jfrt = - [ 2 0 + 3 (1 .9 8 5 9 3 0 ) - 2 (0 .9 1 5 8 1 7 )] = 3.015770

8

yP) = j j [ 3 3 - 4 (3 .0 1 0 4 4 1 ) + (0 9 1 5 8 1 7 )] = 1.988550 2(7) = J - [3 5 - 6 ( 3 .0 1 0 4 4 1 ) - 3 (1 .9 8 5 9 3 0 )] = 0.914964

12

Eighth iteration : X») = - [ 2 0 + 3 (1 .9 8 8 5 5 0 ) - 2 (0 .9 1 4 9 6 4 )] = 3.016946

8

yP) = J L [ 3 3 - 4 (3 .0 1 5 7 7 0 ) + (0 .9 1 4 9 6 4 )] = 1.986535 2(8) = — [3 5 - 6 (3 .0 1 5 7 7 0 ) - 3 (1 .9 8 8 5 5 0 )] = 0 .9 1 164 I-

12

Nineth iteration: *9) = —[2 0 + 3 (1 .9 8 6 5 3 5 ) - 2 (0 .9 1 1 6 4 4 )] = 3.017039

8

^ 9) = I T 133 ” 4 <3 (> 1 6 9 4 6 ) + <° 9 1 1 6 4 4 )I = 1-985805

1 7 8 | B M » S B K > iU n e u n c s * IIl 4*) = i - { 3 5 - 6(3.016946) - 3(1.986535)] = 0.911560

Tenth iteration jfm = i ( 2 0 + 3(1.985805) - 2(0.911560)] = 3.016786 8 yiO) = _L[33 _ 4(3.017039) + (0.911560)] = 1.985764 2(10) = J - J 3 5 - 6(3.017039) - 3(1.985805)] = 0.911696

12 Since nineth and tenth iterations are same correct to 3 decimal places. Hence, x = 3.017,^= 1.986, z = 0.912.

Ass,

2.25 GAUSS-SEIDEL METHOD This is the modification of the Gauss-Jacobi method. Let us consider a system : ai* + 6|.y + cxz = d, 1

OiX + + C2Z - dz OgX + l^y + CgZ = dz Transform the above system in the following form : * = — [dy - 6)y-C izI )
- O3 X

Z =

J

C3 We start with the initial valuesy = yi°> and z = z<0\ we get x*1*from the first equation, i.e., *(1) = — (dy - c ^ 05) °i Now using the second equation, we use z = 2*°> and x = xin the third equation, we get C3

In finding the values of the unknowns, we use the latest available values on the right hand side. This process of iteration is continued until the convergence is assured. As the current values of the unknowns at each stage of iteration are used in getting the values of unknowns. Remark : The convergence in Gauss-Seidel method is very fast when compared to Gauss-Jacobi method.

N umbocal A nalys» I | 1 7 9

Example 230 .*Solve the following system by Gouss-Seidel m ethod: 27x + 6y - z = 85 6x + 15y + 2z - 72 x +y + S4i * 110 [RGPV Dec. 2002, Dec. 2006, June 2008, June 2011 and Dec. 2011] Solution. Since in each equation one of the coefficient is larger than the other, satisfying the condition for Gauss-Siedal method, we write the given equations in the following form : x - ~ j ( M - 6 y + z) y = — (72 - 6 x - 2z) y 15 2 = i-d lO -x -^ ) 54 we start with y = 0 and 2 = 0 and using the most recent values of x. y, z, then First iteration : XH> = ~ (85 - 0 - 0 ) =3.15 yi) = — (72 - 6 x 3.15 - 2 x 0 ) = 3.54 15 z(i) = — (110 - 3.15 - 3.54) = 1.91 54 Now using the most recent values of x, y. z, then Second iteration: *<2) = — (85 - 6 x 3.54 + 1.91) = 2.43 27 y2> = ± ( 7 2 - 6 x 2.43 - 2 x 1.91) = 3.57 e ± ( n o - 2.43 - 3.57) = 1.92 54 Third iteration: xf.1) = — (85 - 6 27 yf}) = — (72 - 6 15

X

3.57 + 1.92) = 2.426

X

2.426 - 2 x 1.92) = 3.572

20) = — (110 - 2.426 - 3.572) = 1.925. 54 Since second and third iterations are same correct to 3 decimal places. The required solution is : x = 2.426, y = 3.572, z = 1.925.

Ans,

T

1 8 0 | En g in eer in g M athfm atic S'111

Example 2.51 : Solve the equations

I Ox + 2y + z ~ 9, ~ 2x + 3y + lOz = 22; x + lOy - z = - 22 Solution.

by Gauss-Seldel method. {RGPV June, 2003] Since in each equation one of the coefficient is larger than the other, satisfying the condition for Gauss-Siedel method, we write the given equations in the following form : 1

y = — (-22 - x + z) 10 J_ z (22 + 2 x - 3y) J 10 We start with y ~ z - 0 and using the most recent values of x, y, z, we get First iteration: y n .

1

y> =

^

.

0.9

( - 2 2 - x < " ) = - 2.2 9

^

( 2 2 + 2.t “ > - 3 y f)) = 3 .0 6 7

Second iteration :

* 2)

= To*9 '

2yW ~ zW)

= 105,3

I - Yo (" 22 _ JC‘2’ + z<1>) = “ ' " 84 * 2> = ^

( 2 2 + 2 x < » - 3 y (2)) = 3

Third iteration jc<3> = — (9 - 2yW - z<2>)

= 0 .9 9 9 6 8

+ z(2))

=

-

y 3> =

( - 2 2 - * (3)

1 .9999

z<3) = ^

(2 2 + 2x<3> - 3 / 3)) = 2 .9 9 9

Fourth iteration :

*<4> - ^ (9 - 2 y 3> - 2(3)) = i y 4) =

(-2 2 - *<« + z<3>) = - 2

•••(I)

N um erical A nalysis -! | >181

2<4 ' = ^

( 2 2 + 2 * < 4 > - 3 vv<4 ) ) = 3

Fifth iteration : a<5> = ^ [ 9 - 2 < - 2 ) - 3 ] = 1

^

-

^ [ - 2 2 - 1 + 3 ]

~

= - 2

( 2 2 + 2 (1 ) - 3 ( - 2 ) ] - 3 .

The required solution is • x = l,y = - 2, z - 3. Ans. Example 2.52 .*Solve the following equations by Gauss-Seidel method: 83x + Hv -4 z =95 7x + 52y + /iz - 104 3x + 8y + 29z = 7/ Ztec. 200i a«rf 2006} Solution. Since in each equation one of the coefficient is larger than the other, satisfying the condition for Gauss-Siedel method, we write the given equations in the following form : x ^ J-(9 5

Oo

-

U y + 4*)

y - — (1 0 4 - IX - 1 3 2 )

52

r = ^ (71 - 3* ~ 8^ We start with y = z - 0 and using the most recent values of x, y, z, we get First iteration : ^1) = - L ( 9 5 - l l y « » +42<°>) 83

= iO O< 95 - 0 + °> = 1145 y j ) = J - ( 1 0 4 - 7x"> - 1 3 2 '° ’)

52

= — ( 1 0 4 - 7 x 1 . 1 5 - 1 3 x 0 ) = 1 .8 4 6

52

Ah = — ( 71 - 3 * t f > - 8 y > ) 29

= — (71 - 3x1.145 - 8 x 1.846) = j.821 ^9 Second iteration :

N umerical A nalysw-I | 181

z<4> = ^

(2 2 + 2 x (4) - 3 y 4)) = 3

Fifth iteration

V* = ^ [9 - 2(-2) - 3] =1 >*5) ’

{"22 - 1 + 3 ] ~ ~ 2

& = ^ [22 + 2(1) - 3(—2)] - 3. The required solution is

x = 1, v = •• 2, 2 = 3. Ans. Example 2.52 : Solve the following equations by Gauss-Seidel method: 83x + llv - 4 z = 9 S 7x + 52y + 13z - 104 3x + 8y + 29z - 71 fRGPV Dec. 2003 and June 2006} Solution. Since in each equation one of the coefficient is larger than the other, satisfying the condition for Gauss-Siedel method, we write the1given equations in the following form : x - -rr< 95 OO

+ 4z)

y , ± ( 1 0 4 - 7 x - 13z) Z= 2 ^ (71 ~ 3 x ~ 8y) We start with >>= 2 = 0 and using the most recent values of x, y. t. we get First iteration : *1) = ± ( 9 5 - liy°> + 4z<0’) 83 = ± ( 9 5 - 0 + 0) = 1.145 oa y(i) = ± ( 1 0 4 - 7x(n - 13z‘0’) 52 = ± ( 1 0 4 - 7 x 1 .1 5 - 1 3 x 0 ) = 1.846 52 4 i) » ± (71 - 3xlI) - 8 / » ) 29

= ± (71 - 3 x 1.145 - 8 x 1.846) = 1.821 29 Second iteration :

1 8 2 |f e i6 w a w > M « iB M iic s -lll

jp> = ^ - ( 1 0 4 - 7 * ^ ~13*W> = 1.412 2® =

(71 - 3x«> - s y 2>) = 1.956

Third iteration: *<3) = - ^ ( 9 5 - l i y 2> + 4z«>) = 1.051 OO

y3) = i - ( 1 0 4 - 7x<3> - 13*«>) = 1.344

52

* 3) = ^ ( 7 1 - 3 x < 3 > - 8 ^ 3 ) ) = 1.969.

Fourth iteration: * 4) = ^ ( 9 5 ~ H y 3> + 4*<3)) = 1.051 oo

y4)

= — (1 0 4 -

52

4&V The required solution is :

7x<«>

- 13*«>) =

1.34

~ 3l<4) " 8^ 4>> = 1-969

x = 1.051, y = 1.34, z = 1.969.

A ds .

Example 2.53 ; Solve the following equations by Gauss-Seidel method : 10X{-2X2~X3~X4 * 3

Solution.

- 2x] + 10 x 2 - x j - x j = 15 - x , - X2 + IOX3 - 2X4 = 27 —Xj —Xj —2x 3 + IOX4 ——9 fRGPV June 2004, Dec. 2006 A June 2009} Since in each equation one of the coefficient is larger than the other, i.e., satisfying the condition for Gauss-Siedel method, we write the given equations in the following form : 1 = 10 (3 + 2x2 + xs + x4) *2

JL (15 + 2xj + X3 + x4) 10

*3 = 10 *27 + x* + x* + 2X4 ^ 1 X4 = — ( - 9 + Xi + Xj + 2x3) we start with xi = x3 = x4 = 0 and using the most recent values of xj, X2, x3, x4, we get First iteration:

x,(l) = X o (3 + 2X2(01 + x3f0' + **(0)) = 0 3 *2’

10

(1 5 +

2x,;i> + x3<°> +

x ^ )

= 1.56

N umehcai. Analysis*! | 183

- 10 X4 = 10

27 -r *,<»' + x j " ^

2 x 4<°>)

= 2.886

-9 + *i<‘> + x2‘l> + 2r3‘l>) = -0.1368.

Second iteration

II

%

3 + 2x,<‘> + *3<»> + x4<” ) = 0.887 x,a> = 10 1 X2<2> = 15 + 2x,(2> + X3W + x4(1)) = 1.952 10 1 27 + x,‘2) + x*<2> + 2 x /» ) = 2.957 10 1 -9 + Xi<2> + x2<« + 2x3(2)) = - 0.025 X& = 10

Third iteration:

1 10 1 x,(3> = 10 1 10 1 Jt4<3) = 10 X,,3> =

3 + 2x2<2> + x3«> + x4‘2>) = 0.984 15 + 2x,<3) + x3(2) + x4(2)) = 1.99

II

27 + x,<3) + x2<3> + 2x4<2)) = 2.99 -9 + x,'3> + X2<3> + 2x3‘3') = -0.0046

Fourth iteration

1 10 1 10 1 10 1 x*<4> = 10

3 + 2x2(3) + X3<3) + x4(3)) ==0.997

II

27 + x,<4> + x2<4> + 2x4<3>) = 2.999

S'

15 + 2x,(4> + x3(3) + x4<3>) = 1.998

H U Jiu 11

X,W =

-9 + x,<4> + x2<« + 2x3(4)) = - 0.0007

Fifth iteration : ,(5) = —

1

10

3 + 2x.<4> + x3<4> + x4<4’) = 0.999

■*2,(5> = —10 15 + 2x,«» + x3«> + x4«>) = 1.999

1 6 4 | &NGlNE£JON6 MXTHfcMATICS-III

,r4^ - ~

( - 9 + x ,(8> + * 2<5' + 2 x 3<5)) = - 0.00044

1 he required solution is :

X] * 1, xi » 2, X3 « 3, x4 * 0. An Example 2.54 : Solve by Gauss-Seidel method, given the following system : 28x + 4 y - z »32 x +• Sy + lOz = 24 2x + 17y + 4z * 35 Solution. Since in each equation one of the coefficient is larger than the other, i.e., satisfying t! condition for Gauss-Siedel method, we write the given equations in the following form x = ^ (3 2 - 4 y + z) 1 j y (35 - 2x - 4z)

y

z = ~ (24 - x - 3y)

Start with y = 0, z / irst iteration :

= 0 and using the most recent values of x, y, z, we get =~

(32 - 4(0) + 0] = 1.1429

1 y*> = j f f35 - 2(1.1429) - 4(0)] = 1.9244 ^ (24 - (1.1429) - 3(1.9244)] = 1.8084 t'econd iteration :

*<2) = ~

(32 - 4(1.9244) + 1.8084] = 0.9325

^

1 = y j (35 - 2(0.9325) - 4(1.8084)] = 1.5236

^

1 * j0 124 " <0 -9325^ ' 3(1.5236)] = 1.8497

Third iteration :

J<3) = ~

(32 - 4(1.5236) + 1.8497] = 0.9913

y 3t =

(35 - 2(0.9913) - 4(1.8497)] = 1.5070

z<3) - ^ (24 - 0.9913 - 3(1.5070)] = 1.8488 I ourth iteration :

1 & “ ^ 3 (32 - 4(1.6070) + 1.8488] = 0.9936

N um erical A n a l y s e ! 1 1 8 5

y 4) = — [35 - 2(0.9936) - 4(1.8488)] = 1.5069 z<4> =

[24 - 0.9936 - 3(1.5069)] = 1.8486

Fifth iteration

* * = 28 132 ” 4 + (1-8486)] = 0.9936 y s> = 7 7 [3 5 - 2 (0 .9 9 3 6 ) - 4 (1 .8 4 8 6 )] = 1.5069 * 5) = “

[2 4 - 0 .9 9 3 6 - 3 (1 .5 0 6 9 )] = 1-8486.

Since 4th and 5,h iterations a re sam e, w e get x - 0.9936, y = 1.5069, z = 1.8486.

Ans. Example 2,55 : Solve by Gauss - Seldal method 5x + 2 y + 3 - 1 2 x + 4y + 2z ” 15 x + 2y + Sz =*20 [RGPV, Feb. 2010} ibttioit. Since in each equation one o f the coefficient is iaiger then the other, i.e., satisfying the condition fo r G auss-Seidat m ethod, w e w rite th e given equations in the follow ing form :

x = ±[12-2 y - z ] y = ± [ 1 5 -jr-2 ^ }

4

2 = ±[20-x~2 y] Start w ith y = 0. z = 0 and using the m ost recent values o f x, y, z w e get Fist ite ra tio n : jd » = “-[1 2 - 0 - 0] = 2.4

= ~[15- (2 4)-0] = 3.15 2<» a y [ 2 0 -2 ,4 -2 x 3.15] - 2.26

Second iteration x<« =• - [ 1 2 - 2 x 3 . 1 5 - 2 . 2 6 ] = 0.688 y(2> = - [ 15 - 0.688 - 2 x 226] = 2.448 4 *2) = j[ 2 0 -0 .6 8 8 -2 x 2.448] = 2.8832

186 j Engmshng Maihekatjcs-HI Third

iteration . x*'' = ^ [1 2 - 2 * 2 448-2.8832] =0.8442 v,.
Fourth ite ra tio n :

= i[I2 - 2 *2.09^4-2.9922] =0.9626 = -^[15-09626-2x29922] =20133 r J- = 4(20-0.9626-2 2.0133] = 3.0022 Fifth ite ra tio n :

*.<< = -{12-2x2.0333-3.0022] = 0.9942 5 vt5< = -(1 5 - 0.9942 - 2 x 3.0022] =2.0003 4

HM = -[20 - 0.9942 - 2 x 2 0003] =3 0010

5 Since 4* and 5* iterations are partially sam e, hence solution o f given sy stem is

x * 1.000, y a 2.000. z ~ 3.000 2.26 RELAXATION M E T H O D : Consider a system of equations : a\X * b\y - C;z - i/j OiX + £>2}' + c2- “ ^2 ay.x t bxv c p = d\. Let, Rx, Ry, R. be ‘residual’ defined by Rj = d\ ~ and

Ans.

- b - j - --\-

Ry = d2 - o^x - c;z R. = d} - GjX - b£' - cxz

...(1)

Now we start initially x = y = r = 0 and obtain the initial residuals, (i.e. the initial residuals are Rx = di, R> = . SRz = - ay i.e.. Rx, Rt, R. are reduced by aj. a2, ay respectively. Similarly, we can find the effects on the residuals when v and z are given increasements dy = 1 (keeping x and z constants) and S: = 1 (keeping x and y constantns). T h e operation table is :

N u c r ic a l A n a l y ss 4 | 1 8 7

Incrm ent in Variable Sx = 1 *y = 1 S z- 1

Change in R esiduals SRf

6RZ —®i

~ a2

-c.

-c2

8R t -* > 3 ~ C3

At each step then we reduce to almost zero the numerically largest residual. In order to reduce a particular residual, then the value of the corresponding variable is changed, for example if residual Rx is to be reduced by a quantity a, then a

Similarly if residual Ry is to be reduced by a quantity # then Sy =

JL

and so on. Continued such

processes till al! the residuals become zero or negligible, Hence the solution is obtained by adding the increments Sx, Sy, 8 z in x,y, z separately. (Nmi* ; The convergence conditions that were given for iterative methods must be satisfied for convergence , of relaxation method also. h n frff 156 : Solve the following equations by relaxation method : 1 Q x - 2 y - 3 z = 2Q5; | ~ 2 x + lOy ~ 2 z ~ 154; p tetafc

2 x - y + lOz = 120. Since, |10{ > h 21 + r 31; |10| > h 2| + f- 2| and |10) > h 2| + | - 1|. So we can use relaxation method. The residuals of given equations are : R t = 205 - IQr + 2y + 3r. R, = 154+ 2 r - I 0 y + 2r, R: = l20 + 2 * + y -t0 z . The operations table is :

SR ,

Sy = 1 Sz = 1

2 -1 0 2

2 1 -1 0

1

11

6R ,

«-i

X

6Rt o

Increm ent in variable j | j

2 3

18* | ftatN an teM m siM n cs-n i Relaxation table is : x =y=z =0 Sx =

205

R, 154

K. 120

0

195

161

20.5 195

1Q c a l9 S _ 180.5 1 0 , <>* = — — = 18 (approx.)

39

0

180.5

93

36

0.5

5x =9

3

54

18.5

6 y - 5.5

14

-1

24

St =2

20

3

4

<5x = 2

0

7

8

J* = l

3

9

-2

<5y = 1

6

-1

-1


0

0

0

' y=¥

Nowx = H6 x 3 32, y = Z 6 y = 26, z = E£z = 21 The required solution is : x - 32, ’ y = 26, 2 = 21. Example 2.57 ; So/we tikefollowing equations by relaxation m ethod: 10 x - 2y - 2 z - 6, —x + lOy —2z ** 7, - x - y + l O z - 8. Solution. Since, |I0| > (- 2| + h 2|; |I0| > |- lf+(~ 2| and |10| > (- i| + h 1|! So we can use relaxation method The residuals of given equations are : Rx = 6 - IGr + 2y + 2z /^ = 7 + J C -l0 y + 2z = 8 + x +y - IQz. The operations table is :

¥ II

Increment In variable 6x = 1 6z = l

5Rt

5*, -10 2 2

1 -10 2

1 1 -10

Ans.

N um erkn >A n a l y s e 1 1 8 9

tlaxation table is ;

x =y - z - 0

6.0

7.0

8.0

« , - J L 0.8 10 .6y =---8.6 . = 0.86 10

7.6

8.6

0.0

9.32

0.0

0.86

&c = £ 2 1 ^ 0 .9 3 2

0.0

0.932

1.792

& = 0.1792 8y = 0.129 8r = 0.062 & = 0.019 Sy = 0.0102 &r = 0.005 52 = 0.0016 By = 0.0008 &x = 0.0009

0.3584 0.6164 -0.0036 0.0344 0.0548 0.0048 0.0080 0.0096 0.0006

1.2904 0.0004 0.0624

0.129 0.1910

0.1020

0.0010

0 0.0050 0.0082

0.0112

10

0.0002

0.0011

0.0

0.0162 0.0002 0.0010 0.0019

H. Hence, x= I.Sy = 0.9999 « 1, 5 y = 'LSy m 1, x = Z 8 z = 0.9998 * 1 Ans. The required solution is : x - y ~ z = 1. 'jample 2.58 : Solve by relaxation method : i 3x + 9 y - 2 z - l l ; 4x + 2y+ I3z ° 24; fRGPV Dec. 2005] 4 x - 4 y + 3 i 'm- 8 Option. Since in each equation one of the coefficient is not larger than the other, i.e., not satisfying Ans. the condition for relaxation (iteration) method.

Exercise-2(A) 1. An approximate value of misgiven by 3.1428571 and its true value is 3.1415926. Find absolute and relative errors.

4jc2v3

2. If u = — — and errors in x, y, z be 0.001, compute the relative max. error in u when x =y = z-l.

1 9 0 1 ftg i e B n w r.lto M B M iiC 5 -lll

3. Let x = 0.5998 * 10 “ 2. Find the relative error, if x is truncation to three decimal digits. 4. Let x = 0.458529 * 10-2. Find the absolute error if x is rounded - off to three decimal digits. 5. Explain underflow condition and overflow condition of error in floating points addition and subtraction. Answer-2(A) 1. Ea - 0.0012645, 2. 0.009

Er = 0.000402, 3. 0.133 x 10-2,

4. 0.471 * I0-3

Exercisc-2(B) Use bisection method to find the roots o f the following equations as indicated, correct to three places o f decimals :

1. 2. 3. 4. 7, 10.

Find the smallest positive root of the equation x3 + x2 - 1 = 0. Find the root of the equation x3 - 4x - 9 = 0, that lies between 2 and 3. Find the root of the equation x3 + Zr2 + 2.2x + 0.4 = 0 that lies between 0 and - 1. x4 - x - 1 0 = 0. 5. x - cos x = 0. 6. 3 x -e I = 0. 9X2 - 1 —sinx = 0. 8. x6 - x 4 - x 5- 1 =0. 9. ta n x + x = 0. xe* = 1. tl. Compute cUbic root of 36. Use simple iteration method tofind the roots o f thefollowing equations, correct tofour places of decim al:

12. 13. 14. 15.

Find the root of the equation x3 + x2 - 100 - 0, that lies between 4 and 5. Find the root of the equation x? + x - 1 = 0, that lies between 0 and 1. Find the root of the equation 3x - logic x - 6 = 0, that lies between 2 and 3. Find the root of the equation 3x + sin x - e* = 0, that lies between 0 and I. 1

16. x =

(X + 1)2

17. x = ( 5 - x ) w

18. sin x - 10(x- 1)

21. x 1- 1 ^sin2* 19. x sin x = 1 20. e* ^ cotx 22. Compute square root of 30. Use Regula Falsi method to find the wots o f the following equations, correct to four places of 23. 24. 25. 26.

decim als: x2 + x 2 + x - 100 = 0 (Root lying between 0 and I).

x3- 8x +■40 = 0 (Root lying between - 5 and - 4). x tan x + 1 = 0 (Root lying between 2.5 and 3). x6 - x 4 - x 3 - 1 = 0 27. xeJt = 2. Use secant method to find the roots of the following equations, correct to three places of decimals: 28. x 3 + x2 + x + 7 = 0 29. x - e r x ~0.

NumacAL Axm.yws-1 | 191 Use Newton-Raphson method to fin d the roots o f (hefollowing equations, correct to fo i plac . o f decimals : !

X4 - x ~ 10 = 0 (Root lying between 1 and 2).

I!. xJ - x 2 + x + 100 = 0 (Root King between - 5 and - 4V x - e~x = 0 (Smallest positive root). xe* - cos x = 0 (Smallest positive root) [M. 2x - logio x = 7. 35. tog x = cos x x snx= —

38. tan * = x

36. x4 + 4 sin x = 0. 39. Mx - sin x) = 1.

e* = x3 + cos (25)x which is near x = 4.5. 41. 10* + x - 4 = 0.

42. Compute ^ 9 .

•43. Using N-R method, obtain formula for

and find ^20 correct to 2 decimal places.

44. Obtain the cube root of 120 using Newton-Raphson method starting with xo - 4.5, 45. Apply Newton-Raphson method to find an ap proximal ton solution of the equation e* = ¥ correct to three significant figures assuming x = 0.4 as an approximate root of the equation. 46. Find the root of the equation fi x ) = sin x -

^ = 0 near x = - 0.4..

Compute cubic root of 42, by Newtoo-Raphson Method Develop an algorithm using Newton-Raphson method to find the fourth root of a positive mcnber N and hence find ^ 32 . f, 49. Find cube root of 3 correct to three decimal places by Newton's iterative method. / .V 1_ 56. Prove the recurrence formula x* + j - 3 2xj. + —

\

51. 52.

53. 54. 55. 54.

for finding the cube root of N. Hence find cube root of 63. Solve the following equations by Graffe's root squaring method : x4- lflfcr3 + 35x*- 5Qx + 24 = 0. xJ -4 x 3 - 3 x + 18 = 0. Extract the quadratic factor o f the form x2 + px + q from each o f the following polynomials, using Bairstow i method and assuming indicated initial values o f p and q. X4~x3 + 6x* + 5x + 10;p= 1.14; = 1.42. x* + xi + 2x2 + x+ \',p = q = 0.5. x4 - 3x3 - 4 X 2 - 2x + %\p = 0.95; q = 1.05 x 4 - 2 x 3 + x 2 - 4 x + 4;/j = 0.5; q = 1.5.

I92r\ F fr y cnw tirH f h ih b w ics-H I Answers-2(B)

2.707. . 0.619. 10. 0.567. 14. 2.1080. 18. 1.088. 22. 5.4772. 26. 1.4036. 30. 1.8558. 34. 3.7892. 38. 4.4934. 42. 5.3852. 46. - 0.42036. 1 48. x"+1 4 3x„ +^-j, 2.3784. 51. 4,3.21. 52. 3, 3, - 2. 54. x2+x +•1 55. x2 +2x+2.

1. 0.755. S. 0.739. 9. 2 .020. 13. 0.6823. 17. 1.516. 21. 1.404. 25. 2.7980. 29. 0567. 33. 0.5178. 37. 1,896. 41. 4.4346. 45. 0.619.

2. 6

0.222. 0.392. 11. 3.3019 15. 0.3604. 19. 1.068. 23. 0.4264. 27. 0.853. 31. -4.2644. 35. 1.303. 39. 1.171. 43. 4.47. 47. 3.4760 3. 7.

1.813. 1.404. 12. 4.3311. 16. 0.4655. 20. 0.5314. 24. -4.1891. 28. -2.0625. 32. 0.5672. 36. -1.934. 40. 4.545. 44. 4.9324. 4. 8.

50. 3.979. 1.442. 53. 2 +1.1446*+1.4219. 56. x3+x+2. 49.

jc

Exercise-2(C)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Solve the follow ing systems by using (I) G auss-elim ination (ii) Gauss-Jordan (iii) Crout’s methods : t + 3y + 6z = 2, jc - 4y + 2z - 7, 3x- y + 4z = 9 2x + 2y + z = 12, 3x + 2y + 2z = 8, 5x + IOy - Sz - 10 2.1 Ox - 4.50y - 2.00z = 19.07. 3.0Qr + 2.50y + 4.30z - 3.21, - 6.00x + 3.50y + 2.50z = - 18.25 x - v + 2 = 1 ,- 3x + 2 y - 3z - - 6 , 2x - 5y + 4z = 5 x + 3y + lOz = 24, 2x + 17y + 4z = 35, 28x + 4 y - z - 32 x - 3y~z = -3 0 , 2x - y - 3z = 5, 5 x - y - 2 z ~ 142 5x - 9 y ~ 2z + 4w - 7, 3x +y + 4z + 1lw = 2, IQr - l y + 3z + 5w = 6, - 6x + %y - z - 4w = 5. 10x + y + z = 12,x + 10>' + z = 12, x + y+ 10z = 12. lOx +y + z - 18.141, x + 10y + z = 28.140, x + y + lOz = 38.139. 3x + y - z = 3,2* - 8y + z = -• 5, x - 2y + 9z - 8. 3x-.y+ 2z = 12,x + 2y + 3z = II, 2 x -2 ;y -z = 2. 2x - 3y + z = - 1, x + 4y + 5z = 25, 3x - 4y + z = 2. x + 2y + 3z = 6, 2x + 4y + z - 7 ,3x + 2y + 9z = 14.

N u m e ric a l A n a ly sis* ] | 193

tH. 2v Sv-

|5 LJA p7. 18.

' tv

Iy 1 ■ ' x ■\ 2> •" V x - :>•

' »•

9. vt r y - 4" t

vf -- - 2, Z.v + 3y — z + 2w - 7,

til 19. 4 . i 2 v

= 3,

- _ hw —2 y — rA - —2. U 5 ■ t-. >7 H. ?x + 3y + r - 7. 4. ' V - v 3.t - - 4. 2v - 3y - 5z = - 5. ■= - 12. 5v y r = M , Ax - y + 3z = 10.

y.his)

6. 2

< f 1 j + w - 2.

O lr *• 4 .9 3 , 1 ,88x - 4.62y + 5.50z = 3 . 1 1

1 . 1 0 v - 0 .9 6 v J - 2 . 7 2 r =- 4 .0 2

20. 6 y - v + z = 13. x + y + z - 9, lOx + y - z = 19 21. x + 2y - i 2- -f- 8m’ - 27. 5x + 4y + Iz - 2w - 4, 6 r - 12-v - 8; *• 3w = 49, 3x - l y - 9z - 5w = - 11 ,22. x + 0.5 v- + 0.33z = 1, 0.33x + 0.25y + 0.2z = 0, 0.5x +

0.33y + 0.25z = 0

2v + 4y + r “ 3, 3.t ^ 2y - 2z = - 2, x -y ° + r = 6 x +■y + : - h> = 2, 7.x ■+•y + 3z + 2w = 12, 8x —y + r - 3m- = 5, lOx + 5y + 3z + 2w = 20. 25. 2 x -r 4y - 8r - 4 1 . 4.v + 6y + lOz = 56, 6x + 8y + lOz = 64 23. 24.

26. 2v -r 2y - z + «• = 4, 4x +■ 3y - z + 2w = 6, S.v * 5;.' - or +- 4iv = 12, 3x + 3y - 2s + 2w = 6 .

A nsners-2(C ) 2. - 12.75, 14.375, 8.75

1. 4. - 2, 3, 6 > 7. I. 4, 5. - 7. 10. 1 , 1. i. 13. 1 , 1. 1 . J6. 1 ,- 1,2. 19. 4.2075. 1.3327, 0.2468 22. 55.56, - 277.78,. 255.56. 25. 1.5, 2.5, 3.5.

5. 0.99, 1.50, 1.84 8. 1, 1, 1 . 11. 1. 2, 3. 14. 1. 2 , 2, 2. 1.64, - 2.49, 0.32. 2, 3, 4.

17. 20. 23. 26.

1,3 1. 1,~ 1 ,- 1.

2. -

3. 1 .3 4 ,- 4.76, 2.58. 6. 39.2, 16.7, 19. 9. 1.234, 2.348, 3.455. 12. 8.7, 5.7, - 1.3. 15. 1, 1.2.

18. 1 ,0 ,- 1,2. 21. 3, - 2, 1, 5. 24. 1, 1, 1, 1.

Exercise-2(D)

1.

2. 3. 4.

,SV;/\v’ e/,‘c following system o f equations by (i) Gauss-Jacobi method 00 Gauss-Seidel method ( t i l ) and relaxation method. 5.v -■ 2y + r ~ - 4. x + 6y - 2z - - I, and 3x + y + Sz = 13 8.
1 9 4 | E n g in eerin g M athem atics-!!!

5. y ~ x + lOz = 35.61,* + z + 10y = 20.08,y - z + lOx = 11.19. 6. 3.I22* + 0.5756y-0.1565;-0.0067/ = 1.571. 0.5756* * 2.938v + 0.1103z - 0.0015/ - - 0.9275. - 0.1565x + 0.1l()3y + 4,127- + 0.2051/ = - 0.0652. - 0.0067*-0.0015y + 0.205 lr + 4.133/= -0.0178. 7. 10* - 2y + z = 12, * + 9v - z - 10, 2.r Hr = 20. 8. 1 0 * -2 y - z ~ / = 3, - 2* + \ 0 y - z - / = 15 - * - y + 10; - 2/ = 27, - * - y - 2z + 10/ = - 9. 9. S x - y + z = 18, 2x + 5 y - 2 z = 3 . x + y - 3 z - 16. 10. 2x +y + z = 4, .r + 2y + z = 4. * + y + 2z ~ 4. 11. 4* + 2y + ; = 8, * + 5y - z = 10, * +y + 8s = 20. 12. 8* + y + z = 8, 2* + 4y + z = 4, * + 3y + Sz - 5. 13. 14* - 5y= 5.5, 2r + 7v = 19.3. 14. * - 2y + JOr « 30.6, Ix + 5y - z = 10.5, 3* + y + z = 9.3. 15. 8* - 6y + z = 13.67. 3* + 11v - 2z = 17.59, 2x - 6y + 9z = 29.29 16. 7.6x - 2.4y + 1.3z = 20.396, 3.7* + 7.9y - 2.5z - 35.866, 1.9* - 4.3y + 8.2z = 32.514. 17. 10* - 2y + z = 12, * + 9 y - z = 10, 2x ~ y + 1Iz = 20 18. 5*- y - z = 3, - * + 10y- 2 z ~ l , - x - y + lOz = 8. 19. 2 r - y +z - 3, 2z + y - z - 1, ,r +y + z = 0. 20. 50* + 2y - 3z = 196, 3* + 65y + 2z = 8 1 ,- * + y + 33z = 63 21. 8 * + y - z = 8,*-7>> + 2z = - 4 , 2x+y + 9z=12. 22. 27* + 6y - r = 85, 6* + 15y + 2z = 72, * + y + 54z =110. 23. 9* - 2y + z = 50, * + 5y - 3z = 18, - 2* + 2y + Iz = 19. 24. 2 * - 3y + I Or - 3, - * + 4y + 2z = 20, 5* + 2y + z = - 12. 25. 6*( + *2 - *3 = 14, *i + 5*2 - *3 = - 18, 2xj + *2 + 9*3 = 6 8 .

26. 3* + y - z - w - 0. * + 3y - z + 2w + 3 = 0, - 2* + 2y + 3z - 2w - 4 = 0, * + 2y + z - 5w + 1 =0.

Answers-2(D) 1. 4. 6. 8. 11.

14. 16. 19. 22. 25.

2. 0.83,0.32, 1.07 3. 2.45, 1.62, 3.79 - 1.001,0.999, 3.0 2.5796, 2.7976, 1.0693 5. 1.321, 1.522, 3.541. 0.5835, - 0.4307, 0.0181, - 0.0044.7, 1.262, 1.159, 1.694. 9. 2,0.999,2.999 0, 1,2,3 10. 1, 1. 1. 2, 2, 2 13. 1.25, 2.40. 12. 0.876, 0.419, 0.574. 1.232, 2.262, 3.382, Actual 1.2, 2.3, 3.4. 15. 2.45, 1.62, 3.79. 3.23, 4.85, 5.76. 18. 1, 1, 1. 17. 1.3, 1.2, 1.7. 20. 4, 1,2. 21. 1, I, 1. I , -1 .0 . 23. 6.15,4.31,3.24. 2.473, 3.256, 1.931. 24. -4, 3, 2. 4, -3, 7. 26. 2 ,-1 ,4 , 1. a

N umerical A n a ly sis - I I ,-^jt

CALCULUS OF FINITE DIFFERENCES^* The calculus of finite differences, in contras^fo infinitesimal calculus, deals with the changes of the functional value, the dependent variitte'due to finite change in the independent variable (argument). The finite variation arguments may either be equal or unequal. Consequently, the finite variation (the differencing interval) is discrete in nature. On the other hand, in infinitesimal calculus, we study the variations which occur when the aigument changes continuously with a passage to the limit. Suppose that y - f i x ) has set of (n + 1) tabulated values with equally spaced x : xo, = *0 + ft, xi = .to + 2h, X}—xq + 3h, ,...x„= xo + nh and Corresponding value of y are y ■y(b y u y 2i ...-yn■To determine the values offi x), f(x ) or/ ”(*) for some intermediate value ofx, then the various types of different opereators are found useful.

DIFFERENT OPERATORS We will study the following operators : 1. The Shifting Operator (E ): Efix)=fix+h) =fix + 2h) EPfix) = fix + nh)

i.e., i.e., i.e.,

Here, n takes up integral or fractional, positive or negative values. For examples : E~ lf ix ) = f i x - h) i.e., E*f(x) = f(x +\ h )

i.e..

Eyo = y i Ezyo = y 2 £ nyo = yn

E~[yi =y\ E\ayx

Properties o f Operator E : (i) Operator E is distributive. (ii) Operator E is commutative with respect to constant. (iii) Operator E obeys laws of indices. 2. Forward Difference Operator (A): If x0, x \ , x i , x „ are equally spaced with interval of differencing h and ify =fix), then A/fr) =fix + h) ~A*) or Ay0 = y t ~yo i.e., Ay, =y,+ i -y , for i ~ 0, 1,2, 3..... n - I, The symbol A is called forward difference operator and Ay, is called first forward differences. Similarly, the second forward differences are A2yt - Ay, + 1 - Ay, For example : A2y 0 = Ayi - Aya ~ (y j ~ y i ) - (yi - y o ) = y i - 2y\ + y0.

Clearly any higher order difference can easily be expressed in terms of ordinates.

1 9 6 | E n g in eer in g M a th e m a tic s-1 11

Forward difference table: Arugments x

y~ f(x )

*0

yo

*1

y\

x2 Xi

y2

*«

y*

*5

y5

Entry

ys

First difference Ay

Second difference A*y

Ay0 Ay, Ay 2

A2y0

Ay3 Ay4

aV,

A2ya AV3

Third difference A3y

AVo aV i aV 2

Fourth difference

Fifth difference' A5,y

aV o

5 A4y, A *

Here the first entry yo is called leading term and Ayo, A2y& .... are called leading differences. Properties o f A : 1.A [/to± #x)] = A fa)±A #*) 2. A[c/x)] ~ c A/(x); c is constant. 3. AmA"fix) - Am+"fix), where m, n being positive integers. 4. Ac = 0 5. If given «-observations (x„ y,), then A" /= 0. Example 3.01 :Evaluate when A = /. Solution. A2x3= A[(x + 1}3 - x3] = A[x3 + 3x2 + 3x + 1 - x 3] “ Atfx2 + 3x + 1) = 3AX2 + 3Ax + A1 - [3(* + l)2~3x2] + [3{x + l)-3 x ] + 0 = 3*2 + 6* + 3 - 3x2 + 3* + 3 - 3x = 6x + 6 - 6(x + 1). Example 3.02 : Prove that n-\ = Ay* - Aye . *-o Solution, We have ji-l n-l +1 - y*>. now open the summation = A(yi ~yo) + A(yz-yi) + Afo -yz) +.... + A(y„ _ i - y*_ 2) + A(y„ ~y„~i) - Ay» - Ayo + Ays - Ayt + Ay3 - Ayz +.... + Ay«_ i - Ay„_2 + Ay„- Ay„_ t = Ay* - Ay0. Example 3.03 : Prove that A logf{x) « /o^ | J +

}.

[ v A1 = Ans.

Proved.

N um erical A n aly sis-1 1 | 1 9 7

Nation. We have

r

.

f f ( x +h )

log^x + h) - lo g /w = log| - y (x),„„j ' f ( x + h ) - f ( x ) + f ( x ) \ - log j [

( v A s add and subtract]

Proved. f ix ) J { fi x ) 3. Backward Difference Operator (V) : !f*o, X |.x 2, .... x„ are equally spaced w ith interval o f differencing h and y = fix) then Vfix) =fix) -fix-h) Vvi ~ y i ~ y i~ i> f o r »'= J, 2 , 3 ...... , n T he sym bol V is called th e backward difference operator and Vy, is called first backward differences . Also, the second backward differences are V2y/ = Vy, - Ay,_ 1? i = 2, 3, 4 ,..., n

For example :

- ^ i ) = Vy2 - Vy,

V2}^ = V(V>2) =

= iyi ~ y \) ~ Cvi -^o) = >'2-2>’j +/0 Similarly, the higher o rd er backw ard differences are defined as follow s :

Vky,

1-

Backward difference table: A rgum ent

E n try

X

* = /(*)

*0

Vo

*1 x2

Vi

*3

l>2 V3

*4

V4

*5

Vs

F irs t d iffe ren c e

Vy Vyj = y , - y 0 Vy2 Vy3 Vy4 Vys

Second differen ce

T h ird j F o u rth d ifferen ce i differen ce

V*y

V3y

i

V4y ..............

F ifth difference

Vsy

^ y ^ V y g -V y ! v 3y 3

v 2y 3

V3y4

v zy 4

v 3y5

v 2y s

V4y4 v 5y5 V4ys

— — ——„i

Example S. 04 :Find the first backward difference o f x? + 2x, Solution.v Vfix) -fix ) - fix - /i) => V(x2 + 2x) = [*2 + 2x] - [ ( jc-A )2 + 2(jc- / i)] = (x2 + 2x\ - [x2 +

- 2/tx + 2h - fi1.

- 2xh + 2 * - 2h]

Ans,

1 9 8 I E n g in eer in g M ath em atics -!!!

4. Central Difference (6): If a'o, JC|. X2........x„ are equally spaced with interval o f differencing h and y - fix), then the fia order central difference o f y is defined as :

SAx) - / ( * + £ ) - / ( * ■ - f )

Or

s f[x *f )

- f e * h) -f [x) .

The symbol 5 is called the central difference operator.

Central difference table : X

y - fix)

Sy

*0

Vo

S Vy2

*1 x2

Vi V2

6 ay

s 3y

5 zy l

6H

S 3Vsa

J 2y2

£ 4y 2

S\

*3 *4

# 4y

6H

5. Averaging Operator (ft)

A

h) + /J 1

f x +mM

=

I

I

h

- -

2

6. Differential Operator (D) Dfix) and Remarks :

A f(x) = f \ x ) dx

& A X) =f '(*) and so on.

(i) The operators E, A, V, <5. jj and D are all linear operators : (ii) We shall use the alternative notations.y* - fix)

i.e..

Ex

+ h) = y x +h.

3.3 RELATIONBETWEENTHE OPERATORS We can express each o f A, V, S, fi and D in term s o f shift operator E. These relations may be regarded and rem em bered as standard result. (i) A = £ - I. (ii) V = 1 1. (iii) 6= E* ~ E ~ 2 (v) E = ehD o r - I +A (vi) log ( I + A) = - log (1 - V) = hD

(iv)

+E~i)

fRGPV, June 2011] fRGPV, Dec. 2011]

N um erical A n aly sis-1 1 | 1 9 9

froof: (i) By definition o f A : H ence (ii) By definition o f V :

Afix) ^ fix + h) ~A x) - E fix) - fi x ) A = E - 1. Vfix) - fi x ) - f (x - ft)

[ v Efix) - fix + /;)]

-fix ) -E ~ ]fix) Vfix) = (l-£ -» )X x ) V=1

[v £ - • / * ) = /* -* ) ]

V /.r) Hence (iii) By definition o f S :

= f ^ / x ) - E - m fix) » ( £ '« - £ - 1/2)At) Hence (iv) By definition p :

or

= ^(S»^+£?-«'2)/(x) Hence (v) By definition o f E : Using Taylor's series, w e get

W

= /(* )+ hf'( x ) + — f \ x ) + ....... = f ( x ) + k D f ( x ) + l- ~ D 2f(x)-^..

urt + —^ h2D'- + . = I11, +hD

= e*Pfix) Hence, E = e*0. or eM) = I + A (vi) Taking iog both side o f ( I ), w e get log ( I + A) = h D A lso, V = I=> V = i - => - hD = log (1 - V) =^> hD - - log (I - V)

D- i

/(*) v ex = 1 + x +

2!

■+ .

... (1) Proved.

Proved.

2 0 0 | E n g in eer in g M ath em atics -IH

Remarks: 1. A = £ - 1 => A = ehD - 1 => A + 1 = ehD. 2. V = 1 1 V = 1 - e ~ hD. 3. Since V = I - e ~ hD => e~hD~ 1 - V => - hD = log (I - V) => hD = - log (1 - V)

|R G P V Dec. 20051

y2 V3 y4 1 V+ — +— +— +. 2 3 4 D =h 4. Since A + 1 = ehD hD = log (1 + A).

|R G P V Dec. 2005|

1 . A2 A3 A4 A----- + ---------- + . D =h 2 3 4

-(2)

N ote : Results (1) and (2) will be used in numerical differentiation.

3.4

DIFFERENCE OF A POLYNOMIAL Ifflx ) is a polynom ial o f degree n in jc,

i.e.,

fix ) = aoxn + a\x ? ~ 1 + azx"~2 + ..... + a„. <3() * 0. T hen &nfix ) = constant = ao n\hn (w here h = interval) and A" + 1 fix ) = 0. Theorem [3.1J : Show that the n'h difference o f a polynomial o f degree n will be constant and all (n - 11"1 and higher order difference are low. fR G P V June 20091 OR Iff(x) be a polynomial o f degree n in x, then the nth difference off(x) is constant and the (n + l)th and higher differences are zero. Solution : Let fix ) = a0 + a t.t + .... + a„x". ...(!) be a polynom ial o f degree n, w here a„ * 0. B y definition o f forw ard difference o p e ra to r: 4 / 1*) =f ix ^ h ) - f i x )

- [do + <3 |(x + h) + ...... + a„ (x + /j)n] - [oq + « |T + ........+ 0,,v"] [using (I)) = a\[(x + h ) - x \ + an [(* + h f - ~ x z] + ...... + a„ [(.v + h f - x"]

= a\h + aJpCixh + h2] + .......... + a„[nC\ .v" _ 1 /; + nCi .x"~ 2 h2 + ...... + "Cn h"\

- [a\h + a^h2 + ........ + a j f 1] + jrp c j crh + h 'l a^h2 ~ ..... + a j i " -1] ^ ...... + nc\ anh x n~ i

= b\ + b^x + ------ + b„_ ijc"-2 + najixn~ 1

...(2)

w here, b\, b i ............ 6„ _ | are constant coefficients. This show s that, first difference o f fix) i.e., Afix) is a polynom ial o f degree n - 1 .

N u m e ric a l A nalysis-11 | 2 0 1

w’ 1

Again, we have A~/(x) = A [A /x)] = A[/(x + h)-f ix) ] = Afix + h)~ Afix) = 0 i + b 2 (x + h) + ..... + b„_ | (x + f t y - 2 + na„ h (x + h y - * ] - [61 + bjx + ........+ 6„_ 1xn~2 + naj?xn~l] = h i [(x + h ) - x ] + by [( x + A ) 2 - x 2] + ........+ 1 \{x + h y * ' 2 - j c " - 2 ] + n a jt [(x + /»)*~ ’] = b2 + b y p C i xh + h2} + ...... + 6„_ 1 [n~2C\ x * - 1 h + = d2 + dyx + ..... + d„_ 1 x "-* + n ( n - 1)h2 a„x^~2

.... + h " ~ '] ...(3 )

where, d2, d^.........d„- \ are constant coefficients. T h is s h o w t h a t, s e c o n d d if f e r e n c e o f f{x) i.e., A 2 f { x ) is a p o ly n o m ia l o f degree n - 2 . T h e r e f o r e , c o n tin u in g a b o v e b ro c e s s w -tim e s, so w e g e t a p o ly n o m ia l o f degree n ~ n = 0 i.e, constant polynom ial is A"f{x) ~ n ( n - 1) ( w - 2 ) ....... 2.1 h"a„ = n \ h n cin ...(4)

= constant A lso A" ■*"1 f[x) = 0, An + 2f{x) = 0 , ..... so on. /. e A" + mJ(x) = 0, w hen m > I .

Proved.

tie 3.05 :Evaluate A2 (cos 2x) >n. We have [v A = E - \]

A2 (cos 2y) = ( £ - 1)2 (cos 2x) = (A*2 - 2E + 1 ) cos 2x = E 1 cos 2 x - 2E cos 2x + cos 2x = cos 2 ( r + 2 hi) - 2 cos 2 (x + h) + cos 2x = cos ( 2x + 4 h) - 2 cos (2x + 2ft) + cos 2x.

f v ET'ftx) = / x + nh)}

Ans. [RGPVDec. 2001}

1pie 3.06 :Evaluate A (e^.log bx) blution. We have A(c?°-'.log bx) = e°lx + ^U og b(x + h )~ e ^ .lo g bx = e*x +h) |0g fjfa + ga(x + h) ]0g i x + gafx + h) |0g b x -e ™ log bx

’

[v add and subtract]

e
j

. (^ax+ah _ ea*)l0g6x

_ eax.ah log^1 + — + €*“ (6°^ -I)l0g6x.

Ans.

3.07 :Show th at : hD - sin h~1 (fiS). blution.

We have

[.. £ = eAD]

2 0 2 I E n g in eer in g M ath em atics-1 1 1

v s in h 6 -

pS = sinh(hD)

Exampie 3.08 :Evaluate A Solution.

*

2 Proved.

hD = sinfr1

H ence,

e°

cos 2 x

We have

(x + h )2 cos 2 (x + h)

iI,— cos2x J1

cos 2 x

(~ + h)- cos 2.r - x 2 cos 2 (x + h) cos 2(x + h). cos 2 x

+ x2[cos 2 x - cos 2 {x + /t)] cos2
[(x -f /t)2 - x2] ct>8 2 x

(2 hx + h2)cos2x + 2x- sin(/i) sin(2x + h) cos 2(x + h). cos 2x a9[ Example 3.09 :Evaluate A

Solution.

5x + 12 x"s + 5 * + 6

]

Ans.

fRGPV June 2006]

’

We have

5x + 12 x2 + 5x + 6

5x + 12 [ (x + 2)(x + 3)

= A2 |

= A2

2 (x + 2)

3 (x + 3)

[ V

Resolving into partial fractions]

N u m e ric a l A n a ly s is - H | 2 0 3 6

(x + 2)(jc + 3)(x + 4)

(x + 3)(* + 4)(x + 5)

2(5x + 16) (x + 2)(x + 3)(x + 4)(x + 5)

Ans.

Example 3.10 : Show th a t: 8* (i)

Solution.

A=T Since (0

/ +Sr

8* +t S = £ l/2 -E ~ m

1

,H .S . ( £ 1 / 2 _ £ - 1 /2 )2

(g + g " 1 (E +

jE +

- 2)

y

r

l/2 x

jt

(£ 1 /2 _ £ -1 /2 )2

2) + ^ /2 _ jp_U2 ^J T + E + E 1

2 =

/p i/2

+ (E 1/2 - E~v 2)^1 + < ^ L J

+ (E1'2 - E - y 2 ) J - ( E ' /2 + E ~1'2)2

-----? ) +

(E + JS-1 - 2) 2

2

^£l/i _

£ - i / 2 ) i ( £ l / 2 + £ - 1 /2 )

(E - S - J) +

.2

£ + £ -> - 2 + £ 2E - 2 _ , A .......... ■■ ■ = --------- - iS —1 = A. 2 2 = R.H.S.

=

Proved.

(ii) We have L.H.S. =

- ~r(E1'2 + j^-v 2)(E i'2 - E ' y~)

{*,• By definition o fp and d'j

m

- - ( £ - £~ l) = i l l + A - (1 - V)J = I [A + V] = R.H.S. Example X I I : Evaluate (V+ A p (x2 + x), taking h “ I. Solution. We have (A + V)2 (x2 + ac) = [ £ - 1 + 1 - £ - I]2 (x2 +x)

[v A - E - I and V = 1 - £ 1j Proved.

2 0 4 | E n g in eer in g M athem atics-111

= ( £ - £ - 1)2 (x2 +x) ~ (E? ~2 + E~2) (x2 + x) = (£2- 2 + E - 2)x* + (E2- 2 + F 2)x = ^ (x 2) - 2X2 + E~ V ) + £*(x) - 2x + ^ 2(x) = (r + 2A)2 - 2X2 + (x - 2A)2 + (x + 2h) - 2 x + ( x - 2/») = ( x + 2) 3 - 2

x2

+ (x - 2 ) 2 + (x + 2 ) - 2

x

[ v £ £ - * “ lj

[ v E"fix)^fix + nh)}

+ (x - 2 )

[

- x2 + 4x + 4 - 2x2 + x2 -4 x + 4+ x + 2 - 2x + x - 2 = 8. Example 3.12: Evaluate (I + 4) (I - V). Solution. We have ( 1 + A ) ( l- V ) = ( ! + £ - l ) [ l - ( | - £ - i ) ] = £(1 - 1 + E~)] = EE~i = 1. Example 3.13: Evaluate Al 0f ( l - a x ) ( l - b x * ) ( l - c x 3)(1 - dx<)] Solution. We know that, if fix) = a^x” + a\x” ~ 1 +.... + a„. Then Anf ix )= a o n lh n. A10[(l - axX 1 ~ c*3Xl - <&*)] = A10 [abcdx' 0 + ......] = abed A10(x10) = abed 10!. [v h= 1]

v

A * 1]

Ans.

Ans.

...(1)

Ans.

f/iM Ee* Example3.14: Evaluate | - j f J « • {RGPVJune 2002, Dec. 2004; Dec. 2006, Dec. 2008(H)] { E Y '^ e * Or ( A 3 ) , Ee* ?* Provt that e* * k~p \e *7O ’ e y 'a V Solution.

{RGPV, Feb. 2010 and June 2011}

We have

aM £« x i?e* —r l e 1. — — = A a.£- le* £ J AV A2ex [Since *_/(x) = ^x - A) and £^x) - fix + h)] ox+h px+A = A V ' * . - ------ = e~hA2ez. AV A2€x = e~h.ex+h = e*. Example 3.15: Find the second difference o f x? - 5 x + 6 the Interval o f difference being /. Solution. Let - x 2 - 5x + 6 Afix) = A(x2 - 5x + 6)

Ans.

fix)

= Ax2 - 5Ax + A6

Now,

fix)

A?

= [(x + l)2- x 2] - 5 [ ( x + l) - x ] = 2x - 4. ~ A(2x - 4) = 2Ax - A4 = 2(x + 1 - x) - 0 = 2.

[v AC = 0] Ans. Ads.

- N um erical A n a ly sis -11 | 2 0 5

Example 3.16 : Prove that (i)EV= VE = A (Hi) 6= 4(1 + A>-,/2= V(1 - V rm x Solution,

m

(i) We have, and Thus (ii)

(iit) Also (iv) Also

(v)

A

(II) F= S E ~1/2 (iv) VA = A - V= 62

v £V = £(l - E~ ') - E - I = A S/E = ( l - £ - > ) £ = £ ~ 1 = A EV = AE ~ A. R.H.S. =
A V A2 - V 2 ( E - l ) 2 - ( l - f i 1)2 _ V~ A ~ VA “ (E - D(1 - E - 1) (E 2 - 2 E + 1) - (1 + E~2 - 2 E ~ l ) (e - d (i - e - ’ ) E 2 - 2 E - E ~2 + 2E~* £ - 1 - E E - 1 + E _1

_ ( E 2 - E ~ 2) -

2 ( E - E~»)

E + E -1 - 2 _ ( E - E -* ) ( E + E - 1 ) - 2 ( E - E - 1) ( E + E -* - 2 ) _ ( E - E - 1 ) (E + E~* - 2 ) (E + E - ‘ - 2 )

= =(} + A) - (1 - V) = A + V = L.H.S. Example 3.17: Prove that

fi =

r _ £* ++f*/* ~ \JJ + 2j l +A

Proved.

2 0 6 | E n g in e e r in g M ath em atics - 111

Solution.

We have

2+A 2^ +A

2 +E - l

l +E

2yfE 1 E J E + y[E

Also

f+ ~

.-(I)

+±(E**-E-"*y = ~yj4 + (E + E - 1 - 2 ) = - J E + E - 1 + 2 2 2

= I ^(El'2 + £-1/2)2 . A (£1/2 + £-1/2) = ^

2 From (1) and (2), L.H.S. « R.H.S. Example 3. I S : Prove that

...(2)

2

Proved

( E 1(* + E - " ' ) ( l + A )”» =2 +A

fRGPV Dec. 2005/

Solution.

L.H.S. » (£ ,/2 + E ~ l/2) (1 + E - 1)m = (£I/2+ £ -• * ) E lf2 = E + I = £ - l + 2 = A + 2 = R,H.S. Example 3.19 :Given u j » 13, u j - 28, u 4 - 49, us “ 76fin d A? u j and A? «> Solution, (i) A2h2 = ( £ - 1)2 ^ = (£? —2£ + l)u2 = £2u2 —2£«2 + «2 = a4 - 2 « j + »2 = 49 - 2 (28) + 1 3 = 6 . (ii) A3k2 = ( £ - I)3 «2

[v A = £ - 1J

Proved. [v A - E - 1] Ans.

= ( f 3 - 3 ^ + 3 £ - I )«2 = £ 3«2 - 3 £ 2»2 + 3 £ « 2 - «2 = «5 - 3«4 + 3«3 - «2

= 7 6 - 3(49) + 3(28) - 13 - 0.

A ns.

Solution.

o II £

1 **»

Example 3.20 .'Given y^ * 580, y i » 556, y 2 ~ 520 andy 4 - 385. fln d y > Solution. Given four observations, thus we get fourth difference will be zero, i.e., A^y = 0 => [As puty = yo] AVo ~ 0 => [v A = E - 1] => (1 -£ )* y o = 0 => (E* - 4£3 + 6E2 - 4£ + 1)yo =0 [Using binomial expansion] => y* ~ 4K3 + _ 4Vi +>5o = 0 [Enyx ~ yx+n\ 385 - 4 y ,+ 6(520) - 4(556) + 580 = 0 Ans. yj =465. Example 3.21: Prove that y 4 = y s + A y , + A*y, + Asy,. R.H.S. = y 3 + &y3 + (A2 + A3) ^ = E 2yi + AEyx + Az(l + A)^

['•’ E nyx = jy^+n]

N um erical A n a lysis - !! | 2 0 7

= 2? 2y, + =

AEyi + A 2E yj

[v A =

E - 1]

(E2 + A E + A 2E)yj = [E 2 + A # ( l + A)]y,

= [ E 2 + A S 2]*, = £ 2(1 + A )yt

=

- y 4 = L.H.S.

Proved.

Example 3.22 : Prove that Jo

Solution.

+ yi + yt + ....+ yn - "*JCiyf +

L.H.S. = y0 + = (1

+ E +E'1 +...... + E ") v0

(1 ~

[

(v E*vr = y ^ J

+ ^ 2y 0 + .... + E "y 0

1

En+l > - E J*>-(_

1 . ,.».i 1 A .

V 1 + r + r' +

J

= — [1 + n+1 C jA + rt+1 CoA 2 t .... + rt+1 C„+iA '1*5 - lb n A

Provwl.

= BtIC,y0 + "+,C2Ay0 + .... + *+,C^,A"y0 Example 3.23: Prove that y tx + y t x * + y 3 x* + .......... Solution.

+ ( J - x)* A y* + < T ^ * F * ' y ' + .....

L.H.S. = y,x + y2x2 + y3x3 + ..... = yj.x + EyiX 2 + E 2yxx 3 + .......= x[l + E x + E~x- 4 ........ ]yj [Herey is not function of x but it is a function o f som e other variable] = x(l - Ex)~ly 1

[ v As binnnval expansion]

= x[l - (1 + A)x]-Jyi - x[l - x - Ax]-1 y,

1 -x

l-x

Example 3.24 : Prove that

Ax 1 l-x yi +

(*.* A =

i yi = l - x Ayi +

Ax A2x2 1 + -------- + --------- - + ( 1 - JC) ( l - x )2 A2yt + .... = R.H.S.

E - 1]

yl

Proved.

2 0 8 | E n g in eerin g M * n sM A n c s> IU

Solution.

We have _ 1 ___ 1 x +h x

= A"'1

1)A* - 2

(

[l^X + ft

* (-l)A "-2

= (-DA*-2

x +h

XJ

ln-1

x(x + ft) A ll- « x x + ft

V* + 2ft

* + A JJ

1 jc + 2ft

x

?/ 2 x(x + hVr + 2ft)

* (-1)2A"-2

____ 2 !A*__ x(x + h)(x + 2ft)

= (-1)3A"'3

________ M W ________ x(x + A)(x + 2ft)(x + 3ft)

[As similar processes]

In general, A» f —1 = (-1 )" -------------------------- . Vx J x(x + ft).......(x + nft)

3.5

Proved.

FACTORIAL POLYNOMIAL The factorial polynomial of degree n denoted by the symbol

or aW. Thus

xM = x(x - h)(x - 2ft)......... (x - n ^ l ft). Here ft ?c difference and n is positive integer. In particular ft = 1, we have jrf‘1 = x J&l = x ( x - l ) xPJ =x( x- IX*- 2 ) and so on. If a is negative integer, then reciprocalfactorialfunction of order n and denoted b y ^ -") or.tl-,,i. Thus. 1 (x + ft)(x + 2ft)(x + 3ft).........(x + nh) In particular ft = 1 we have 1

WwEWCtu^tew rw s-II j 209 I

** 2J = (x + IX * + 2)

________1________ **~3} ~ (x +XKx +2Xx +3) Remarks: 1.

30(150on-

AxW = (x + /r)W -*MI - [(x + AX*X*- h).......( * - n - 2 ft)l- [x (x - h)....... ( x - n - 1 h)] ~ x ( x - h).......( x - n - 2 A)I(x + h ) - ( x - n - 1 A)] = nh x(x - A)........(x - n - 2 A) “ nh x**"lJ.

For example : Ax^i = ^

= 3x1*1.

Similarly

A2xM ”=n(n - I) A2xi" ~ 21 A«jrl»l = «!/i»

2.

A*xW = 0 for k > « and Ax1*1 = — x1*1 = ux^*1). dx

X f — x**l =* ----- t + C, where C is constant A J n. + 1 Example 3.25 (a) : Represent the function : f(x ) «=x* - 12x?+ 24x*-36x + 9mudIts successive differences in factorial notation. fRGPV June 2004} Solution. Given/x) = x* - !2xJ + 24c2 - 30* + 9. Letjfc) * + lftcPJ +Oc£l + Hafii + E in factorial notation =>x*~ 12x3 + 24x2-3Qx + 9 = Ax(x- IX *-2X x-3) + BxOc- IX*- 2 ) + C x (x -l) + Dx + E ...(1) Now, putting x = 0 on both sides of (1), we get E = 9 Again putting x - 1, on both sides of (1), we get 1 -1 2 + 2 4 - 3 0 + 9 =D + £=>/> + £ = - * = > £ - - 17 Further, puttingx = 2, on both sides of (1), we get 16 - 96 + 96 - 60 + 9 = 25 + 3C + Z) 2C + 2D + E = 35 => C = - 5 Similarly, putting x = 3, we get 6fl + 6C + 3i> + £ « - I 0 8 = » 3 = - 6 Finally, equating the coefficient of x4 on bolh sides of (IX we get A = 1. Thus, the required polynomial in factorial notation is f e ) - xW - 6 - SxPl - 17xt‘l + 9. Ass. 3.

'2 1 0 | Engbveebng H m e u n c s J II Hence,

4/1*) = 4xiJl - 6 x 3xl2>- 5 x 2xW - 17 x I + 0 = - 18brt2l - 10xW- 17, Also ^ A x) ~ 12xt2l —36xt11—10, A3/ x ) = 24*m - 36, A4/x ) = 24. AsJ{x) = 0. A bs. Example 3.25 (b) : Express y *=2xJ - 3x3 + 3x - 10 in factorial notation form. /RGPV, Dec. 20J1] Solution. Given f (x ) = 2x3- 3x2 + 3x - 10 Let f(x) - A xW + fltf2! + Cx 1*1 + D ... (1) in factorial notation. => 2x} - 3xJ + 3x - 10 -~Ax (x - I) (x - 2) + Bx (x - 1) + Cx + D ... (2) Now, putting x = 0 on both sides of (2), we get Z) — 10 Again, putting x = I in (2), we get C + D = - 8 =*C = 2 Finally, putting x = 2 in (2), we get B = ~ (-2 C - D ) ^ s - 3. a*

Also, equating the coefficient of x3on both sides, we get A =2 Thus, polynomial in factorial notation (2) becomes ; f i x ) = 2rP) + 3xt2J + 2xf J - 10. Example 3.26: Obtain the function whosefirst difference is Xs + 4x? + 9x + 12. Solution. Letfix) be the required function, we have A/tx) - x3 + 4x* + 9x + 12 or 4/1*) -AxW + + CxI'l +- D in factorial notation => x3 + 4x2 + 9x + 12 =Ax(x - lX x -2 ) + Bx(x- I) + Cx + D Now, putting, x = 0,1, 2 on both sides of (2), we get D * 12 C + D = I + 4 + 9+ 12 = 26 o C - 14 2fl + 2C + £> = 8 + 1 6 + 18 + 12= 102 D ^ 1 and equating the coefficients ofx3, we get A - I. Hence (1) becomes: Aflx) = xt3l + 7x^21 + 14xt’l + 12, on integrate, we get 2fHl xW mi fix) = —— + 7. —- +14. —— + 12x'1’ + k, 4

= *<*

o

Ans.

...(1) ...(2)

[k being constant]

a

- j ) +

+ 7 * < » - 1) + i 2 r + *

= —(x4 - 6 l 3 - 7 x 2 ~Gx) + -~(x 3 - ix ‘ *2x) +1(x '2 - x ) +12x +k 4 3

N u m e r a l Analysis-!! | 2 1 1

*4 ( 6 7^1 3 ( 7 _ , f 6 14 _ . A . — + — ■+— x3 + ------ 7 + 7 x 2 + — + ------7 + 12 x + fc 4 I 4 3J i 4 J I 4 3 J 49 = —x* + —x ? —r *2 + — x + ft. Ans. 4 6 4 6 Example 3.27 : Obtain the function whosefirst difference is 2x* + 3x* - Sx + 4 fRGPV,.Feb, 20l0f Solution. Proceed as above example, we get the function y - ( = i x ‘4' + 3x,J1 + 4x111+ k.

Ans.

Example 3.28 : Find the function whosefirst difference is Px2 + //x + 5. /RGPVDec. 2004/ Solution.

Lety(x) be the required function, we have A/{x) = 9x2 + llx + 5 or A/(x) = A x ^ + fltM + C in factorial notation 9x2 + 11x + 5 = A x (x -‘ 1) + B x +

=>

c

-( I) [ V * l !J = X, x*2l *=x ( x - 1)]

Puttingx = 0 we get C = 5. Putting x - 1, we get 9+11 + 5 = 27 + C = ? iJ;=20. Equating die coefficient of xJ, we get A - 9 Hence (1) becomes: A/W = 9 x ^ + 20xni + 5 v(3] y\X\ Ax) - 9 ~ + 2 0 ~ + 5 x W + f e

=> =>

/ x ) = 3 x (x -lX x -2 )+ 1 0 x (x -l) + 5x + /k Ax) =3x3 + x2 + x + *, where k is constant. Ans. Example 3.29 : A second degree polynomial passes through (0,1) (I, 3), (2, 7) and (3,13). Find the polynomial. Solutbn. Let the polynomial beX*) = + + c. ...{1) The difference table for the given values of x and/x) is as follows :

0 1 2 3

f(x)

Af(x)

As f(x)

1 3 7 13

2 4 6

2 2

A*f(x)

From (1)

Ax) + +c &Ax) = 2ax + b =s> A/(0) = b=> b ~ 2 and A2/(x) =2a=> A2jt0) = 2a=s>2a = 2=>a = 1 Again putx = 0,^0) = c ^ c = 0 Hence (1) becomes: Ax) =x<2>+ 2x<»+ 1 = x (x - l) + 2x+ I = x2 + x + 1.

[v Mx) = 2] [v A2/ 0 ) = 2] [ v / 0 ) = 0]

Ans.

2 1 2 { E ngwejmwg M A W tom cs-lH

F xample 3.30: Find A*fix ), if f(x ) * (2x + 1) (2x + 3) (2x + 5) _ (2x + 19). blution. Given J[x) = (2x + 1X2* + 3X2x + 5)....(2x + 19)

[ v There are 10 factors]

- - R ) K ) K ) ..(-4) ,r 19 f o1 Ax) - 210 I + » [where h «= 1 and by definition of factorial polynomial.] (

19 V95

Hence &ftx) = 2tox lO lx + — I

tffix) =2^x90^ +^ !

[81

A^*) =2*°x72o(x+y j \f«1

=210x6040^+y j

, 2>«*504o(*+ ^ } ( * + ^ ] ( * + H } ........

=80640(2*+9)(2x+ll)(2x+13)....(2x+ 19).

Ans,

Frampfe 3,31 .'Prove that

* y x « j w - HC iyx+n-t + * c ,y x. n t — + ( - i y iy x. R.H.S. = y s„ - nClE ^ y x^ + "C ,£-2y « » -.. + (-1) *C»-E-*yx+»

Solution.

= {1- "C,£-J+ KC2E~i - .... +(-l)» * (1 - £?_l )*>*♦„ [As binomial expansion] = { (I-# * 1)* .£*}>* * ((X-E~l ) E y y x = [ E - \ Y y x =&*yx -L.H.S. L
Sy

y

5 6 7 . .

8

9

4 3 4 .

10

24

Proved.

— I1 1 I

6 14

From the difference table AVs = 0.

2

5 8

Q

O

oo

0

Ans.

N umbkcal A nalysis-11 | 2 1 3 (ii) We have

[v A = £ - I)

A*ys = ( £ - l)*ys = (£ < -* C ,£ 3 + 4C j£ 2_4C 8£ + 4C4)yb

Iv E*yx -

= > 9 -4 ^ 8 + 6y7 - 4 y 8 +y8

= 24 - 4* 10 + 6 x 4 - 4 * 3 + 4 -2 4 - 40 + 24-12 + 4 *0

Ans.

Example 3.33 : Given x: 1 2 3 y: 2 5 10 Find the value ofV*y$ Solution, (i) By using the difference table to find V2y5 : >

i

y» * 5

Vy

3 5 7 9

2 2 2

5 26 [RGPV June 20(hf

V3y

2

3 4 5

J? Ii 1&

u

i

X

3k m >

1 1 I i

4 17

yt = & >1 = 17

0 0

y, = 26 From the difference table, we get V2^ = 2. (ii) Without using the difference table to find V2^ : We have V2^ = (1 - E~ 'J2ys = (l + E - i - 2 E~')ys

Ans. [v V = 1-&'■')

[v E~*yx = yx.„ j =ys +y3~2y 4 Anv = 26 + 4 0 - 2 x 17 = 3 6 - 3 4 - 2 . Example 3.34 : Estimate the missing terms in the following table, 4 3 x: 0 1 2 81 ? y=f(x): I 3 9 Explain why vmtue differs from 3* or 27. (RGPVDec. 2002, Dec. 2008(N)J SolutioH.Wen number o f known data = 4 values [By definition] A*y = 0 => ( £ - 1)4* = 0 => (£* - 4C |£ 3 + aC2E2 - *CjE + *C4)y = 0 => (£ * -4 £ 3 + 6£2 - 4 £ + l ) v = 0 Putting .y -yo> (E4 - 4Ei + 6£? - 4£ + 1 ^ = 0 =>

r4-4>»3 + 6>»2-4yi +yo = 0

[Since y0 = t.yi =3,>^ = 9,>^>=?,y4 = 8t]

{•• E nyx = y x+« 1

2 1 4 j E n g w eebw g M m m x n c s - U I

=>

81 - 4 * >>3 + 6 * 9 - 4 > 3 + 1

0

4 y} = 124

=>

=> >*3=31 => fl3) = 31. Ans, Explanation : This differs from 33 i.e., 27 because differencing values arise due to the assumption that/x) = 3 is a polynomials of degree 3 (Since four values are givefi) so that fourth and higher order differences vanish. Ans. Example 3.35 :Flnd the missing values in the following data. x: 2.0 2.1 2.2 2,3 2.4 2.S 2.6 y=f(x): 0.135 0.111 0.100 0.082 0.074 [RGPV June. 2004} Solution. Letyo,y],y 2,yhy*,y$< V6 be given values, in which missing values arey\ andy 4 Here, number of known data = 5 values Asy = 0 => (E ~ I )5y = 0 => ( f 5 - 5C,F» + 5C2& - SCyE2 + 5C4E - 5Cs)y - 0 => (E5 - 5E* + 10£3 - 10/s2 + 5£ - l)y = 0 ...(1) (i) Putting .y = .yo in ( I), we get (£s - 5 £ 4+lO £3-lO£2 + 5 £ -l)y o =0 => ys-$y4+ 'Q w - iQyz + Sy, -yo = 0 [v E ny 0 = y0+J => 0.082 -5>-4 + 10*0.100-10*0.111 + 5 y i-0 .l3 5 = 0 => 5^|-5>»4 =0.165 => y \ - y * - 0.0326 ...(2) (ii) Again putting y - y\ in (1), we get ( £ 8 - 5E* + 10F8 -1 0 E* + 5E - I)*, = 0 =>

=>

-

y6 - 5y6 + lOy* - 10y3 + 5y2 - >i = 0

0 .0 7 4 -5 x 0 .0 8 2 + 1 0 ;^ -1 0 x (0.100) + 5(0. I ll ) - j y , = 0

=> , 10^4 - ^ i =0.781 ...(3) Solving equations (2) and (3), we get 9y4 = 0.8163 =>>4 = 0.0904 i.e.,/2.4) = 0.0904. Ans. and y\ =0.123 i.e.,fil.\)~ 0.123. Ans. Example 3.36 :Find the first term o f the series whose second and subsequent terms are 8, J, 0, -1,0. [RGPVJune2903, Dec. 2007} Solution, Letyo = ? and y\ = 8,>»2 = 3,yj = 0,^4 = - l.ys = 0. Here number of known data = 5 values .'. & y = 0 =* ( £ - l)5,y = 0 =>

(E s ~ 5 C ,^ + 5 C 2£;s -6 C 3£ ? + 5 C 4 « - 6C5)y = 0

=> ( £ 6 - 5 £ 4 + 1 0 F > -1 0 £ 2 + 5 £ -l);y “ 0 (i) Putting y = >5q in ( I). we get {E* - 5£’4 + 10£3 - 10£2 +

- I)y0 =0

...(1) f

I

Nukbocai ANALYSiS4I | 215 =*

,V5 - 5^-4 + 10^3 - \0y2 + 5>i - 3»o =0

=> 0 - 5 ( - l ) + 10(0)-10(3) +5(8)- > 0 =0 = ^ 5 -3 0 + 40 =>>o =>>o ~ 15 i.e.. first term = 15. Example 3.37 :Compute the next three values : x: 0 1 2 3 4 y=f(x): 1 -1 I -I 1 Solution.

[v Eny0 ~ y ^ n] Ans. 5 -

6

[RGPV Dec. 2004J Let >o, y }, >2. >>3, y4, ys, ye, y i be given values in which y 5, >6 and y j are missing. Number of known data = 5 values. Asy - 0 => ( E - \ ? y =0 =>

(EB-5 C ,E 4 + 5C2E 3 - 5C3£ 2 + 5C4£ - 5C5b

=0

=> (E 6 - 5E* + 10E3 - 10E2 + 5E -1 )? = 0 (i) Putting y = >0 in (1), we get ( £ B- 5£ 4 + 10E3 - 10E? +5E - l)y 0 = 0 =>

y6 -

+ 10ya - 10y2 + 5y, - y 0 =0

=> => (ii) Putting>>= V| in (I), we get =>

...(1)

[v E*y 0 =

]

6(1) + 10 (-X) - 10 (1) + 6 (-l) - 1 » 0 > 5 - 5 - 1 0 - 1 0 - 5 - 1 =0 ys = 31.

Vc - b >6 + 10y4 - 10^a

-y , - 0

ye - 5(31) +10(1) - 10(-1) + 5(1) - (-1) - 0 => y 6 - 129. (iii) Puttings- ) h in (I), we get ;y7 - 5y$ + 10;yj - 10>, + 5>s - >3 = 0 => >7-5(129)+ 10(31)-I0(l) + 5(- l ) - ( l ) =0 => ^7-645 + 310- 1 0 -5 - I =0 => >7=351. Thus, next three values are : 31, 129 and 351. Example 3.38 :Find the missing values in the following table: x: 0 5 10 15 20 y: 6 10 . 1 7 Solution.

7

Ans. 25 31 {RGPV June. 2005f

Let yv>y\,yi>>3. >4.>5 be given data if) which>2 and >4 are missing : Number of known data = 4 values, so that A4> - 0 => (E - 1)V =0 ------------ [yA = E -1 ] =>

(E 4 - 4C, E 3 + 4C2E 2 - 4C3E + *C4)y « 0

=>

(E4 - 4 E 3 + 6 E * -4 E + l):y - 0

...(1)

216 |

(i) Puttingy ■ yo in (IX we get ( E * - 4 E * + G E 3 ~ 4 E + l)y 0 =0 =>

>

<

+>0 - 0

=> >4 -4(17) + 6(y 3)-4 (1 0 )+ 6 - 0 => 6y z+ yi = l 02 (ii) Puttingy « y j in (1), we get (E* —4E^ + 6JE® —4E + l)>i = 0 =>

[v £ » y 0 = y ^ J ...(2)

- 4 y 4 + 6^-3 -4 > 2 +>i = 0

=s> 8 1 -4 y 4 + 6(17)-4ya +10 = 0 =» 4>^ + 4^4 = 143 ...(3) Solving equations (2) and (3), we get y 2 * 13.25 and y4 - 22.5. i.e., 'ftlQ)= 13.25 9ni 22.5. > Ans. Example 3.39 {interpolatef(2) from Ike following table: x : 1 2 3 4 5 f(x) 7 13 21 37 and explain why the value obtained is differentfrom that obtained by putting x = 2 in the expression 2 *+ $. Solution. Letya,y\,y 2,y},y 4 be given data in which y ( is missing. Number of known data = 4 values, so that A*y = 0 => (£ -l)V = 0 => (£*-4£3 + 6£2 - 4 £ + l) y “ 0 ...(1) Putting y = yo in (1), we get =>

y 4 -4 y 8 +6y3 - 4 y ,+ y 0 = 0

(v £*yo=ycn-„]

=> 3 7 -4 (2 1 )+ 6(13)- 4 y i +7 = 0 => 4yi *38 => y\ “ 9.5 orX2) = 9.5. A bs . Also sincej{x) = 2* + 5 .*. Exact value at x m 2 Le.,J(2) m 9. But interpolate value ofJ(x) is 9.5 i e , 9.5 differs from the actual value 9 obtained by the expression 2* + 5. The reason for this i s : we have assumed/x) to be a polynomial of degree 3. Ans. Example 3.40 .Find logie 7 and logic U from thefollowing table: x: 6 7 9 9 10 11 12 y-hgnx: 0.77815 0.90309 0.95424 1.000 1.07918 Solution. Letyo, yj.y 2.y 3.y 4.y 5.y 6 be given data in which yi and >>4 are missing. Here number of known data = 5 values, so that A5y = 0 => ( £ - l) 5 y = 0 =*

(E 5 ^ 6 C t£ 4 +6C2£ s - 6 Q S 2 +6C4£ - e C 8)y =0

Nn—atKWM w w u fl j 217 => ( ^ - S ^ + l O E ’ -lO ^ + S g -D y =0 (i) Puttingy * yo in (IX we get

...(1)

(£ 8 - 5E* +10E * - 10E* + 5E - l)y0 = 0 => y& -by* + 10>3 - 10>J + 5yj - y0 = 0 fv & y a = y ^ J => y 5 - 5(1.000) + 10(0.95424) - 10(0.90309) + 5>, - 0.77815 - 0 5y,+y5 e 5.26615 ...(2) (ii) Puttmgy = yi in (IX we get y« - 5 % + i o ^ - i 0 y 8 +5y2 - y i * o 1.07918 - 5ys +10(1) -10(0.95424) + 5(0.90309) - y x = 0 => >i + 5ys • 6.05223 Solving equations (2) and (3), we get y, = 0.84494 Le., tog,0 7 = 0.84494 and y$ = 1.04146 i.e., ^ 1 0 11 = 1.04146. 3.6

...(3)

Ans.

IN T E R P O L A T IO N W IT H E Q U A L IN T E R V A L S

interpolation is the process of finding the intermediate value of a functiony-fix ) from a set of its values at specific points given in a tabulated form i.e., Suppose we are given the following values of y ~j(x) for a set of values of x : x: xo x\ X2 — xH y: yo y\ n ... y» Thus the process of fmding the value of y corresponding to any value of x = x, between xo and x„ is called interpolation. Hence interpolation is the technique of estimating the value of a function for any intermediate value of the independent variable while the process of computing the value of the fistction outside the given range is called extrapolation. 3.7

G R E G O R Y - N E W T O N ’S F O R W A R D I N T E R P O L A T I O N F O R M U L A F O R E Q U A L IN T E R V A L S

Statement: Let yo, y \, y-i. .....y„ are the value o f y - J(x) corresponding to equidistant values of x ~ jcq, ■men,

a'2, ....xm such that x ~x$ + uh^>u~ —j — [A = length of interval] ,

-M )

» ....

n!

•

........(I)

Proof: Let

y = J[x) =/xo + uh) = E?j(xo)

[v x = xo+uh] [Using Shifting operator]

2 1 8 | Cnqb« ^ b^*o MxmEMAncs-IlI

= (1 + A)“>>o

[vo =X*o)]

= U jf o

. . . .

^ ( y - l ) :-

(u -^ i) n!

[Since A" + % , A"+2>b. etc are zero]. Proved. Remarks: (i) Since the formula derived involves the forward differences of y at >>0, it is called Newton's forward interpolation formula. (ii) If three value of y , namely, yo, y i and y% corresponding to x - xo. *1 and xi are given, then Newton's forward interpolation formula is called parabolic interpolation formula.

3.8

GREGORY-NEWTON'S BACKWARD INTERPOLATION FORMULA FOR EQUAL INTERVALS (RGPV Dec. 20021 Statement: y\> y i t ...... yn are the value of y = ftx ) corresponding to equidistant values of x =* X&, x\, X2.........xm such that x, - x, _ j = h for i = 1, 2, 3 , ...... n, and x - x„ = uh x -x n =s> u = —^— . then

u „

y=Ax)= Proof: Let

y ^J{x)

=>

y ~A*n + uh)

, u(u-»-l>

.......+

u(u * l)....(u — ^

1)

...(])

Iv X - Xn =ii/l]

= HyjiXn)

[Using shifting operator]

= ( £ ‘ l )-"/(x?) = (1 - V Y uyn

[ v 7 = l-E -> ]

[ l t “ V > U(U+1) V* i u& + 1'>(u+ ?) V3 ,

I

h

1!

2!

3!

....... Un­

n! [Since V” + lym V"+ 2y„ etc. are zero] t

u _ 1!

y =f(x) » yn + -

u(u + l) u(u + l)....(u + n - l ) + -1 - - V2y n +.... + ----- !— i----------i 2! n!

V»y„.

proved.

N umerical A nalysis- # j 219

Remark : Though it is usually suggested that the forward formula should be used for interpolating or extrapolating y at points near xq and the backward formula should be used for interpolating or extrapolating near xm it is not necessary. Either of the formula can be used for interpolating or extrapolating.? at m y point In feet that the interpolating polynomials o f fix) that occur in the R-H.S.'s of both the forward and backward formulas are identical.

3.9

ERROR IN NEWTON’S INTERPOLATION FORMULAE 1. Error in Newton's forward interpolation formula:

™

- l ) ( u - 2)....■■(»-n )

^ > =

(n •+-1)!

(

y>™ '

X—Xq where xq < c < xBand u - —t — • 2. Error in Newton's backward interpolation formula: « ( « + 1)(« +

r, ,

+ nl m»+i V)

(n +1)! where

xq

<

c

<

x„

*

9}~ *

and« :

h Example 3.41: Estimate the sale fo r 1966 using Newton forward interpolation form ula: Year: 1931 1941 1951 1961 1971 1981 Sale iu thousand: 12 15 20 27 39 52 Solution. Here given interval is equal and h ~ 10. The forward difference table is : Y ear
S a le (y) 12

1941

15

1951

20

1961

27

1971

39

1981

52

Ay 3 5 7 12 13

A*y

2 2 5 1

A*y

0 3 -4

A*y

3 -7

We know that Newton forward interpolation is u .

u ( u - l)

. u (u -l)(u ~ 2 )

A&y

-1 0

220 I e p * N W ^ J4 ro «M tfiC 8 4 iI

1966-1931

A t,* - 1966, taking *o® 1931 and A - 10, so that u * ------^ ---------3 J. Hence (1) becomes: -* * « > -

. (3.5)(2.5K1.5X.5) . „ . (3.S)(2.5X1.5K.5)(-.6). . --------- -- ---------

H

(

'

= 12 + 10.5 + 8.75 + 0.8203 + 0.2734 - 32.34 Thus, the sale for the year 1966 * 3234 thousands. Ans. Example 3.42 .’Estimate the sale fo r the year 1966 using Newton backward interpolation formula : Year : 1931 1941 1951 1961 1971 1981 Sale in thousand: 12 15 20 27 39 52 Solution. Here given interval is equal Le., h ■=*10. The backward difference table is :

Year (x) 1931

S a le (y ) 12

1941

15

1951

20

1961

27

1971

39

1981

52

Vy o U

2

5

*

2

7

5

12

V5y

Vsy

V*y

0 3

3 -7

-1 0

- 4

1

13

We know that Newton backward interpolation is .......

Here,

«=—

-0 )

1966-1981 ------- ^ ----- - 1 . 5 /

Hence (1) becomes:

>1966 =/(1966)= 62 + (-1.5) *13 + (~L^ ~ 5)-x 1 + ( 15)L

x (-4)

( (-1.5)(-.5)(.5)q.5) : ( 7)[ (-1.6)(-.5K.5Xl-5)(2.5)„( ^ 4! 6! = 52 - 19.5 + 0.375 - 0.25 - 0.1641 - 0. 1172 = 32.34 Thus, the sale for the year 1966 - 32.34 thousands. Ans.

NtowErttUtAM mlvs»41t 221 Example 3.43 :The table gives the distances in nautical miles o f the visible horizon fo r the jfcven heights in fe e t above the earth’s surface : x - height t 100 ISO 200 2S0 300 350 400 y -d ista n c e : 10.63 13.03 15.04 16.81 18.42 19.90 21.27 Find the values o fy, when x ~ 218f t and 410f t Solution. Here h - 50 [equal interval] The difference table is : X

100 160 200 260 300 350 400

Ay

y 10.63 13.03 16.04 16.81 18.42 19.90 21.27

2.40 2 .0 1

1.77 1.61 1.48 1.37

A2y

A*y

-0 .3 9 -0 .2 4 -0 .1 6 -0.13 -0 .1 1

0.16 0.08 0.03 0.02

A4y

-0.07 -0 .0 6 - 0 .0 1

(0 At x** 218: Which is near to x&= 200, so using Newton forward interpolation formula: u .

u(w ~l) . .

yz =AX) = yo + Avo+ 2 '\

« (« -!)(« -2 )

y°+ — f i -----

y° +..............(I)

x -x o 2 1 8 -2 0 0 u = —j-— = -----—---- =0.36 and

Here

yo = 15.04, Ayo * 1.77, &?yo * - 0.16, A3>»o= 0.03 etc. Hence (1) becomes:

K ,, -^ 2 1 8 )- 1 5 .0 4 + 0 .3 6 0 .7 7 )+ 2 iS ± «£ (-0.16)

+0.36(- 0.&4X-1.64) (003)+ = 15.04 + 0.637 + 0.018 + 0.001 +.... = 15.696 i.e. 15.7 nautical miles. Ans. QS)Atx~ 410: Which is near the end of the table, so using Newton backward interpolation formula: » -/X ). * V iv *

"here.

+

v = ,„+...................(2,

x -x „

410 - 400 _ ------gj---- 02

Here,x“ 410,x „ -4 0 0 ,yn - 21.27, Vy„ - 1.37, V2^,, = -0.11, V3# , = 0.02 etc. Hence (t) becomes: y* 10 =*410) = 2 1 . 2 7 + 0 . 2 ( 1 . 3 7 ) + ^ ^ ( - 0 . 1 1 ) + .... 21-53 nautical miles.

Ans.

2 2 2 | E m m ta m a M a sh e sm x ic s-III

Example 3.44 :Thefollowing table gives the population o f a town during ike last six censuses. Estimate using Newton’s interpolation formula the increase in (he population during 1946 to 1948. ' fear; 1911 1921 1931 1941 1951 1961 Pop. in thousand: 12 15 28 27 39 52 {RGPV June, 2003] Solution. Here h ~ 10. (equal intervals) The difference table is : Pop. in A> aV A*y A4f(x) Year x t h o u s a n d A 5y y 1911 1921 1931 1941 1951 1961

12 15 20 27 39

3

2 2 5 1

0 7 12 13

52

A II 3 —

•*

3 - 7

- 10

1-------------------.

x - x Q 1946-1911 A t x ~ 1 9 4 6 i v «= - y 1 -------- ^ ------=3.5. Using Newton’s forward interpolation formula : fix) =

. u ( u - 1) +«&*> + ■ 2 j - A2y0 + .......

=^1946)»

+

2!

-(1) +

4!

+ (3-5)(2.5)(1.5)(0.5)(-0.5) (_ 1q) 5!

or fi\946) = 12 + 10.5 + 8.75 + 0.8203 + 0.2734 = 32.34. Again, * -* o 1948-1911 At x ~ 1948: v « = —r 2- “ ------“ ------=3.7. 10 Hence (1) becomes >1948

- /1 9 4 8 )=

10 (3.7)(2.7) 0 . (3.7)(2.7)(1.7)(0.7) . 12+{3.7)(3) + - — ~ — 'x 2 + 0 + ~ — —— —— - — - x 3 2\ 4!

| (3 .7)(2 .7 )q.7)(0 ,7)(~0 ,3) 5! Hence, poiulation increase during 1946 and 1948 19 4 8 ) 1 9 4 6 ) = 34.87 - 32.34 = 2.53 thousands = 2530.

x (-10) = 34.87

Ans.

N umerical A nalysis-11 | 2 2 3

Example 3.45 :Find the cubic polynomial which takes the following values : x : 0 1 2 3 f(x ) 1 2 1 10 Hence or otherwise evaluate ff4). fRGPV Dec. 2001 and Feb. 2010} Solution. T he difference table is : X

0 1 2 3

y = f(x) 1 2 1 10

Ay

A2y

A3y

1t -1 9

-2 10

12

x-xo x -0 Here, h = 1 and u = — r— “ —;— - x n 1 /. By Newton’s forward difference formula, Ax)

u .

u ( u - l) .»

= yo +J7 Ay0 +—gj— A^o+...

x ( x - l) x ( x - lK x - 2 ) Ax) - yo + *Ay0 + A2y„ + ------- J j------- A3y0

[v u = x]

, ! + , ( ! ) . £ < * ^ > (-2) + ^ X- S ^ Z ?1 (12) 2 6 [v y0 = 1, Ay0 = 1, A2y0 = -2, A8y0 = 12] = 1 + x - x ( x - 1) + 2*(x-1X*~2) = 1 + x - x2+ x + 2 x 3-6 x 2 + 4x = 2x3 - 7** + dx + 1 .Ax) =2x3~7x2 + 6 x + 1. .—(I) Ans. Also, put x = 4 in (1), we get A*) = 2 * 43- 7 * 42+ 6 * 4 + 1 = 128-112 + 24 + 1=41. Ans. Example 3.46 :Given sin 45* * 0.7071, sin 50* m 0.7660, sin 55* m 0.8192, sin 60* m 0.8660, fin d sin S2* using Newton*s forward Interpolation form ula. Solution. Lei Ax) = sin x°. The difference table is : X

46 50 55 60

y 0.7071 0.7660 0.8192 0.8660

Ay

&*y

A‘y

0.0589 0.0532 0.0468

- 0.0057 -0.0064

- 0.0007

X —Xt) 5 2 -4 5 A tx “ 5 2 : Here h ** 5, *8 - 45, then a * —^— * — g— = I 4 Using Newton forward interpolation formula:

Ax) - yo.+«Ay0 + or

A2y0 +.....

sin52° - 0.7071 + (1.4)(0.0589) +

(-0.0057) +

(1.4)(0.4)(-0.6)(-0.0007) * 6

= 0.7071 + 0.08246 - 0.001596 + 0.0000392 - 0.788. Ans. Example 3.47 -.Form the table, estimate tke number ofstudents who obtained marks between 40 and 45. Marks 30-40 40-50 50-40 60-70 70-80 So. o f students: 31 42 51 35 31 [RGPV 20011 Solution. To find/40 andX45). The difference table is : M a rk s le ss th e n (x ) 40 50 60 70 80

y=f(x)

Ay

A'y

Aay

A
81 73 124 15^ 190

42 51 35 31

9 -1 6 -4

-2 6 12

37

From the table: A tx * 40=>fl40) = 31.

...(1)

X —Xa 4 5 -4 0 Now, A tx * 45, then u = —^ ~ — = 0.5 /. Using Newton’s forward interpolation formula: M =>

A 4S) . 3 U (0.5)» 42 ,

......

2

, 9»

6

- ’- a? , (-25)

t (.5K -.5)(-l.S)(-2.5)j 24 = 47.87 * 48. ...(2) Hence the number of students getting marks between 40 and 45 =X45)- A * 0) ~ 48 - 31 = 17. Ans,

' ’ NowHBCAt *0*L Y sis-fl1 2 2 5

Example 3.48 :ln an examination the number o f candidates who secured marks between certain limits were asfoUows : 80-99 Mdrks : $-19 20-39 40-59 60-79 17 No. o f Candidates: 41 62 65 50 Estimate the number o f candidates getting marks less than 70. Solution. The difference table is : N um ber o f M arks lean Ay A 'y A4y A 'y candidates}? ^f(x) then (x) 41 19 62 S 103 39 65 -1 8 59 168 - 15 50 -1 8 —33 79 218 17 99 235 We have,

« 19, h * 20, x * 70 f To fmdX70)] u=>

*~«o h

7 0 -1 9 ^ ° 2.55.

By Newton’s forward difference interpolation formula : Ax) -

u ( u - 1) + “ 4y0 +■ 21 * * +

/7 0 ) * 41 + 2.55x62 +

2.55x1.55

*<3) +

2.55x1.55x0.55 x (-18) + 0

2x1 ^ 3x2x1 = 41 +■158.1 + 5.92 -6 .5 2 « 2 0 5 * 2 -6 .5 3 - 19*49 * 198.

Aas.

Example 3.49 :Thefollowing are the marks obtained by492 candidate* in a certain examination : Marks: 0-40 40-45 45-89 $0-55 55-60 60-65 No. o f Candidates: 219 43 54 74 32 79 Find out the number o f candidates: (I) Who secured more than 48 but not more than 50 marks. (Ii) Who secured less than 48 Nit not less thaw 45 mark. Solution.

The difference table is :

M arks less then (x) 40 45 50 55 60 65

No. o f ca nd ida tes (y) 210 253 307 381 413 492

Ay

&*y

*y

* 4y

A 'y

43 54 74 32 79

11 20 -4 2 47

9 —62 89

-7 1 151

222

2 2 6 | E ngw eekh K j M athem atics >111

From the table: Here/40) = 210 and/50) = 307. y y AO J A To fbid f(48) i Here xo = 40, * - 48, h •=5, .*. u = — r— = — r**- = 1.6. n o By Newton’s forward interpolation formula,

Ax) »

=>

+ “ Ay0 + U(| ~ 1- A2y0 + .....

/4 8 ) *210+ 1.6x434

y

>((ll)^

6M

6

^ x(9) *

(1.6)(.6)(-.4)(-1.4) q.6)(.6)(-.4X-1.4X-2.4) 24 120 *210 + 68.8 + 5.28 - 0.576- 1.5964 - 2.3869 * 280. (i) Number of candidates who secured marks more than 48 but not more than 50 marics = /5 0 ) ->(48) =,307 - 280 * 27. Ans. (ii) Number of candidates who secured marks less than 48 but not less than 45 marks = /4 8 ) -X45) - 280 - 253 = 27. Ans. Example 3.50 .’The following table gives the marks obtained by 100 students in Mathematics in a certain examination: M aris : 30-40 40-50 5040 60-70 70-80 No. o f Candidates: 23 35 22 11 7 How many students got more than 55 marks T Solution. The difference table is :

M arks less than (x) 40 50 60 70 • 80

No. o f candidates (y> 25 60 82 93 >00

A» 35 22 11 7

A*y

A*y

A4y

-1 8 -1 1 -4

A n

5

7

To Andf(55) : Here x - 55, xo “ 40, h ~ 10 x -x p 5 5 -4 0 U~ h 10 /. Using Newton forward difference formula :

Ax) = yo+jfoo* - ^y1- A2y0+..... =>

X55) - 25 + ^ x 3 5 + ^ ^ 5^x(-13) + ^ 1 2

)(- )f c f )- x 2 6 ' ‘ (1.5)(.6)(-.5)(-1.5)jt5 24

I

Numerical Analysk-11 I 227 » - 25 + 52.5 - 4.875 - 0.125 + 0.117 = 72.6 » 73'students. Hence number of students getting marks less titan 55 - 73 Therefore, number of students getting marks more than 55 - 100 - 73 - 27. Ans. Example 3.51: From the following data, estimate the number o f persons having incomes between <1)2000-2500, (U) 1000-1700, (U) 3500-400* Incom e: Below 500 SOO-IOOA 1000-2000 2000-3000 3000-4000 No. o f Persons: 6000 4250 3600 1500 650 Solution. First we prepare the cumulative frequency table make equal interval (A - 1000), as follows: Income less than (x): 1000 2000 3000 4000 No. o f persons (y): 10250 13850 15350 16000 Now the difference table is : X

1000 2000 3000 4000

y 10250 13850 15350 16000

Ay

A

3600 1500 650

- 2100 -8 5 0

A 'y

1250

(i) From the table :/2000) « 13850 To find f(2500) : Here x - 2500, xq - 1000, A- 1000.

...(1)

X -X o 2500-1000 ^ en U h 1000 “ K5' .". Using Newton’s forward interpolation formula :

_

Ax) = y o + ~ A y 0

7 1} A>*>+ ........

=> /25O0) - 10250 -t- (1.5) x 3600 +

x (-2100) + 2 6 - 10250 + 5400 787.5 - 78.125 - 14784 students Hence the number of persons having income between 2000 and 2500 -X2500)-X2000) -1 4 7 8 4 - 13850 = 934. (ii) From the table/1000) = 10250 and /4000) = 16000 Now to find/ft) at x - I 7 O 0 Le.f(l700) : Here x = 1700, xo - 1000, h - 1000

^

x -x o 1700-1000 “ = h 1000 ~°*7 Using Newton's forward interpolation formula, Ax) ~ y o +«Avo+

A2y0 +

-g> t f y<>+ .........

x

(1250) ...(2)

Ans. ...(3) ...(4)

228 | ENQiNEcnNo MiWTamncs-IH

i

(0.7)(-0.3X-1.3) (0.7)(-0.3) (1250) (-1 2 0 0 ) + 2 6 = 10250 + 2520 + 220.5 + 56.875 ...(5) = 13047.375 * 13047. r "• Hence the number of persons having income m between 1000 - 1700 Ans. =/1700) -^1000) - 13047 - 10250 = 2797. (tii) Again to flBdJfr) a t x - 3500Le.,ff3500) : Here x = 3500 which is near 4000, so that taking x„ - 4000, h = 1000

*1700) = 10250+(0.7)(3600) +

x-xn 3500 - 4000 = -0.5 h 1000 Using Newton's backward interpolation formula: u=

„ tt(u + l ) _ , Ax) = yn + wVvB+ 2! /3500) = 16000 + (-0.5)(650) +

u(u + 1Xk + 2 ) _ .

------- ***• + •

3!

(-0.5)(0.5)( -850) (-0.5)(0.5)(1.5)(1250)

2 6 ...(6) - 16000 - 325 4 106.25 - 78.125 * 15703.125 « 15703. Hence the number of persons whose income in between 3500 -4000 : Ans. =/4000) -X3500) - 16000 - 15703 = 297. h xomple 3.52 :To obtain the interpolatingpolynomialf[x) satisfying thefollowing data: x: I 2 3 4 y ~ f(x): 26 18 4 1 I f another point x ~ 5, f ( x ) m 26 is added to the above data, will interpolation polynomial be the same as before or different T Explain why ? Solution. The difference table is : X 1 2 3 4

y-f(x) 26 18 4 1

' -8 -1 4 —O

* ’y

A*y

-6 11

17

To find f(x) : Here jco = I, h ~ 1 .t-* 0 i-l it - —r— => » = x - I h 1 Using Newton’s forward interpolation formula .

I hen

Ax> =

‘v * f

J t o . 2 6 - ( * ~ i x - 8)

* ....... +

(

-

6

a7)

N u m erical A n a ly sis -11 | 2 2 9

= 2 6 - 8 ( x - 1) - 3(x* - 3x + 2) + — (x3 - 6 x 2 + l l x -6 ) 6 *7 x3, -2on Ax) - — 0 * 22 + 193 - ^ - x + l, l, .

Ads

Further, if another point x - 5, Ax) “ 26, is added to the data, then ihe difference table a given a s : 1 2 3 4 5

y = /«

Ay

26 18 4 1 26

-8 -14 -3 -25

A*y - 6 11

28

A*y

A*y

17 17

Since, the fourth differences is zero i e.t AV “ 0. hence die interpolating polynomial will i e the same as before. Ans. Example 3.53 : A third degree polynomial passes through the points (0,-1), (1,1), (2, I) and (3, -21, fin d the polynomial. Solution. Theforward difference table is :

2

y « /W -l l l

3

-2

X

0 1

Ay

A*y

9it nu

-2

uQ

-3

A5y —1

X- Xn X ~ 0 u~ — r— = — — - x « 1 By Newton’s difference formula for interpolation, Here

xo = 0, h ~ 1

. x ( x -l ) x ( x - l ) ( x - 2) A*) - >o ■*-*Ay0 — — A2y0 + A3y0 3* 2! 2

(v u = ? j

6

x 3 - 3 x 2 +2x - -1 + 2x - i 2 + x — - l + | l x 2 + f2 +1 - | ) x - l = - - x 9- - x 2 + l 6; I, 6j 6 2 Example 3.54 :Flnd the cubic polynomial y(x) which takes the following values: y(0) - /, y(D - ft y(2) - /, y(3) - 1$, Also fin d y(4). 6

- x - l.

A n s.

230 t ENGM3&ING MATHENAT)CS~III Solution.

The difference table is : X 0

y = /W l

1

0

2

l

3

10

Ay -1

A*y 2

1

6

8

9

x-x^ x-0 h ~ i "* By Newton’s difference formula for interpolation, we have u-

. * . *(*-1) *9 x ( x - lX x - 2 ) . a y(x) - y0 +*Ay0+ ot Aay0 +— — ~ 7 ---- -A3y0 2! 3! 1 + X<-1)+ j f c V (2) + (6) 2 ' ' 6 = 1 - x + x2- x + x3 “ 3x* + 2x“ xj - 2x2 + I. Also,y(4) = 43 - 2 * 4* + I = 6 4 -3 2 + 1 “ 33.

A bs.

A bs ,

M

Exam ple 3.55 :

229249, ^ H x ) * 175212 an4

Given

j 4 f(10) -40365, fin d f(l). Re-write the above data: x; 1-10 4-10 lf(x): 500426 329240 I i (2) (1) By subtracting 0M2)» (2)— (3> and (3)^10), we get x: y = f(x):

1-3 171186

4-6 154028

J

7-10 175212 i (3) 7-9 134847

Now prepare cumulative frequency table: x: y = f(x):

3 171186

6 325214

9 460061

The difference table is : X 3 6 9

[■:«=*]

y 171184 325214 460061

Ay

A?y

154028 134847

-19181

'

Nw u»

cai.

A »* y s »-II j 231

x-xo 1 -3 u = —^— =^~s~ ~ ~

To Rod/IfI): Here x - 1, xo - 3, h = 3

Now using Newton i forward interpolation formula : Ax) = y0 +«Ayo+—^ y 1}A2yo+ ........ = 171186+ (-.667X154028)+ (~’66T},

(-19181)

= 171186 - 102736.7- 10663.5 ■=57785.7. Example 3.56 : The following data : x: 0.51 0.52 0.53 y-f(x): 0.5292 0.5379 0.5465 gpves the values o f the probability Integral f ( x ) = values o fx. Find the value o f ^

0.54 0.5549

0.55 0.5633

£ e~**d x fo r certain equidistant

.

e ~ * d x , using Newton’sforward interpolation

formula. The difference table is :

Ay

y ~f(x) 0.5292 0.5379 0.5465 0.5549 0.5633

0.0087 0.0086 0.0084 0.0084

&*y

A4y

-0.0001 0.0002

0.0003

&'y I s

(x) 0.51 0.52 0.53 0.54 0.55

i i p P ° 8 §

Solution.

' {*.333

1

Ans.

To f in d y at jc - 0.533 Le.,f(0.533) ; Here x = 0.533 , xo = 0.51, h = 0.01

x -V 0.533-0.51 _ “ h 0.0J. 3 Using Newton’s forward interpolation formula : u

a

u (u -l) .,

A x) = y 0 + Y7Ay° — 2! X0.S33) - 0.529 2 ^

!

^

1

° ...... > , g ^ x ( - 0 0 01)

*4

= 0.5292 + 0.02001 -0.0001495 - 0.00001495 - 0.000007848 = 0.549037702 * 0.5490.

\

t

2321 Thus value of lateral at x =*0.533 is 0.5490 e - * d x m 0^490.

'

Ans.

Example 3.57 :Tke pressure p o f wind corresponding to velocity v is given by the following data, Estimate p when v - /S. v." 10 20 30 40 p: 1.1 2.9 4.4 7.9 [RGPV June 2002] Solution. The difference table is : V

10 20 30 40

y ~P =f(x) 1*1 2.0 4.4 7.9

Ay

A*y

* ’y

a0

1.5

2.4 QQ

0.6

-

0.9 ■

To find p * f(v) at v m IS. Le.,f(15) X —Xr, 15 —10 Taking x ^ 15, xo = 10, h - 10 Then u = — 7— = —77.— = 0 5 n Using Newton’s forward difference formula :

Ax) - yo

=,

m

* “^ 2 !~ A*y° * ........

- 1.1 + <0.5X0.9) *

u s, + (° °K -6)(-1.5) x (^> 6) 2

6

i.e., (/>)»- IS - 1.1 + 0.45 - 0.1875 - 0.0375 - 1.325. Pressure p a t v= 15 is 1.325. Example 3.58:Estimatef(42) from d u following data: x: 20 25i 30 35 40 f(x ): 354 332 291 260 231

Ans. 45 204 [RGPV June 2006)

Solution.

The difference table is :

(x)

y= f(x)

20

354

25

332

30

291

35

260

40

231

45

204

Ay -2 2 -4 1 -3 1 -2 9 -2 7

A -y -1 9 10 2 2

A*y

29 -8

T 11

A ‘y

A*y

-3 7 8

45

r NtMHOCAlA*ALYSW-II | 2 8 3

To find/fc> * tx - 42 : Here x * 42 which Is near to 45, so taking % x - x , 42 -4 5 -3 x„ -45 , / .i f * g “ 5 a

0.6

-

Using Newton’s backward interpolation formula: 31 (-.6X.4)q.4)

x(0) 6 ( (-.6X.4XI.4)(2.4) :(S) t (-.6X.4X1.4X2.4X3.4),.(ie) 120 24 Ans. 219. Example 3.59 :From the following data, fin d f(7 .5) 7 8 x: I 2 3 4 5 6 512. 343 y: 1 9 27 64 125 216 Solution. The difference table Is :

f/O ) - 204+(-.6)(-27)+(~ 6*-4 )><(2)+

y l 1 8 2 27 3 64 4 125 5 216 6 343 7 x» —8 512 To find/ft) at jc • 7.5 U.,f(7.5) : X

Here x * 7.5, which is near to x „ - 8,

Ay

A*y

7 19 37 61 91 127 169

12 18 24 30 36 42

*■

x -x .

* 'y

A4y

6 6 6 6

0 0 0 0

6

7 .5 -8

= -0.5.

.-. Using Newton’s backward interpolation formula: fix) « y m+ uVym+

21

v2>» + •

, ______ _ (-0.6X0.6) , 0 (-0.6X0.6X1.5) _ , n X7-5) * 612+(-0.5)(169)+-— — £x42+------ a, — 1x6 + 0 3! = 512 - 84.5 - 5.25 - 0.375 - 421.875.

Ann.

3.10 GAUSS’S CENTRAL DIFFERENCE FORMULA FOR EQUAL INTERVALS We shall develop central difference formulae which are best suitable for interpolation near the middle value of the tabulated set (table). x: X-2 x -i xi x2 y -f(*) y~2 y -i y« yi yi

234 f Engineering Mat^ catk-'s-III

0

Then difference table is : X

ll-D iff.

l-d iffe r e n c e

x 0 ~ 2 h = x_2

y-2 y-i yo

x0 - h = x .l *0 x 0 +h = x 1

V\

x 0 + 2h = x 2

\>2

IIID iff.

rV-Diff.

Ay_z(~ ^ - 5^) Azy_2(=
Ay0( = < ^ )

AVi(=^)

A4y _2

(=
3.11 GAUSS’S FORW ARD IN TER PO LA TIO N FO RM U LA FO R EQUAL INTERVALS U u ( u - l) , 2 Ax) = *> + ^ Ay0 + ~ 2 ! A

(u + l)(u)(u - 1) 3 + ------- 31-------- A y ~l

. (u + l)u (u - l)(u - 2 ) a4„ . , x-Xo -----------4!----------- y~2 + ........where, u= —^— • Remark: This formula is applicable when u lies between 0 and 1 i.e., (0 < « < 1). Example 3.60 :Using Gauss’s forward interpolationformula to evaluate yj$ given thatyjj «*18,4708, y2s - 17.8144, y n ~ 17.1070, y}3 - 16.3432 andy 37 - 15.5154. Solution. The difference table is ? ■x 21 = x_2 2 5 = x _i 2 9 = x0 33 =x, 37= x2

A4y

A 'y 18.4708 17.8144 17.1070 16.3432 15.5154

- 0.6564 -0 .7 0 7 4 r 0.7688 - 0.8278

-0 .0 5 1 0 - 0.0564 -0 .0 6 4

-0 .0 0 7 4 - 0.0076

To find y - f f x ) at x - 30, Le.,f(30) : x -x n 3 0 -2 9 Taking xo = 29, h = 4, x = 30, then u = —^— = — - — = 0.25. Using Gauss’s forward difference formula. u 4.

u ( u - l)

.

. (u + l)u (tt-l)

- 0.0022

NuW-ioCAL Al«LYS*s4l | 235

X30) *>>30= 17.1070 + (0.25X-0.7638) + f f c S f c a & H w n e o a j g w b z a (-o.i 6 24 = 17.1070 - 0.19095 + 0.00529 + 0.0003 - 0.00004 .fto =■ 16.92

or

Ans.

3.12 GAUSS’S BACKWARD INTERPOLATION FORMULA V

'

(u + l)u *o

u .

(u + l)u < u -l) .«

* ~ J ( x ) « % + J j A y - i + i - ^ A » y . 1 + i ------ ^ --------A*y_2

(m+ 2)(u + 1)u(u -1 ) A4y-2 +. 4! where

u-

X -X

q

Remark: This formula is applicable when u lies between - I and 0. Example 3.61 :Find by Gatos's backward interpolation formula the sales o f a concernfor the year 1936, given that 1941 1951 1931 Year: 1961. 1911 1921 27 39 52 Sahfln thousands rupees): 12 IS 20 Solution. Taking the origin at 1941 and h * 10, also x = 1936 x -* o

------- h

1936-1941

[v -1 < u < 0]

~ ----- 10---------- 05

The i ? 'tral difference table Is : X

y

1901=x_4 1911=x_3 1921=x_2 1 9 3 1 =x_! 1941=x0

1951=Xj

12 15 20 27

89 52

Ay

A*y

A*y

A*y

A ‘y

3 5 7 19 A 2 13

2 2 5 1

0 3 — ■“ *ik

3 -7

-1 0

Using Gauss backward formula: (o + l)u Ax) = y o + jy Ay-i + ^ ~ A > y - i 2! A,m

(u + l)u (u —1) A »*»+■ 31

- 3 9 , ( - . S)« (I2) , M x a ) * « > f c & a . M ) A D = 39 - 6 - 0.125 - 0.25 * 32.625 thousands.

Ans.

236 |

3.13 STIRLING’S FORMULA FOR EQUAL INTERVAL

'

* -* o U Then Gauss’s forward interpolation formula is Let

=

—

u“ a u (u -l) (u + l)u (u -l) A*) - y 0 +j j Av0 + 2~ &2y~i + ------- ~ -------A*y-1 3! (u + l)u(u - l)(u - 2) . + ------ — , — A
(u + l)u (u -l)

i

^

3!

i

A3y_2

. (w + 2)(« + l)u(u -1 ) lA...................... + ------------------------- A y. 2 + ..... ...(2)

4!

Taking the mean of (1) and (2), we get . ,

Ax)

U [ Ay0 + Ay-i 1 u2 u (u -l)(u + l) \ A*y.t +A3y- 2 ' - yo+IT[— g— J "iT y“l+ ----- 3 !----- — ■*------- .

u2(ua -1 ) ■A*,y_2 +. 4!

-(3)

which is called Stiriling formula. It is used when « lies between - -7 and ~ . 4 4 Example 3.62 :Employ Stirling fo rm u la to com pute y i 2.2 f r o m the follo w in g table (yx ~ I + logM.stox) x°: 10 II 12 13 14 10syx : 23967 28060 31788 35209 38368 {RGPV Dec. 2005/ Solution. Rewriting: jP : 10 11 12 13 14 y,: 0.23967 0.28060 0.31788 0.35209 0.38368 Taking the origin at xo “ 12°, A= I, x “ 12.2, then

u -

x

-Xq h

12.2-12 1

which is lies between - •7 and — ■ 4 4

0 .2 ,

it

NtMEBtCAl.AMALYSIS-II | 2 3 7

The central difference table is X_2 =10 X_j

0.23967

=11

0.28060

x 0 =12 %

0.31788

=13

0.35209

Xj

Ay

y ~ f(x)

X

0.38368

x2 =14

0.04093

0.03728 0.03421 0.03159

A4y

A3y

A 'y

- .00365

.00058

-.00307

-.00045

- .00262

- .00013

v Stirling formula is

yx =fix)= yo+l i

Ayo+Ay-i

2

u2 +iT

1

u(u2 - 1) A3y_j + A3y_2 3! 2

U2(uz - 1)

4

Hereto = 0.31788, Ayo =0.03421, A>_ i = 0.03728

A V t = - .00307, A3y - ) = - .00045, A3>_2 = .00058, A4>_2 = - .00013. y,2 z - 0.31788 + — ^

.03421+ .03728'

(■2)[(.2): - lj -.00045+ .00058

(,2)2[(.2)2 Y\ 24

.00013)

= 0.31788 + 0.00715 - 0.00006 - 0.000002 + .0000002 = 0.32497. Given 0: CP 5° 10° IS0 tan 9 : 0 0.0875 0.1763 0.2679 Find the value o f tan 16°, using Stirling formula.

Ans.

Example 3.63:

Solution.

Take origin at 15° i.e., xo = 15°, and h - 5 and here/fr) = tan x u=

x -x 0

16-15

0.2.

2tf> 0.3640

25° 0.4663

3

238 I ENGMEEHNe Mathematics-III The central difference table is : X

Ay

y

X_3=0

0

x_2 =5

0.0875 0.0888 0.0916 0.0961 0.1023

0.0875 0.1766 0.2679 0.3640 0.4663 0.6774

X_! =10 x0 =15 xj =20 x2 =25 x3 =30

A*y

0.0013 0.0028 0.0045 0.0062 0.0088

0.1111

A 'y

A4y

A 'y

0.0015 0.0017 0.0017 0.0026

0.0002 0 0.0009

-.0002 0.0009

A‘ y

0.0011

Using Stinting formula, « TAyo+Ay-il

a2

+ ~ ^ T ~ ^ [ A~ ' ' 2~ '3 j,~ ] + — ,

^ A4y_2 + ••••

J^IS) —tan 15° —M C T + f f l g g j J ™ ] + < f - ( .0 0 4 6 ) ( (1.2)(.2)(-.8) ^.0017 + .0017~j | (.04)(-.96) (Q) = 0.2867 (Approx.)

A

3.14 BESSEL’S FORMULA FOR EQUAL INTERVAL The Gauss’s forward formula is m

- ? o + Z * y o + ^ A ’>., , ^

1w » - d A V |

, (u+l)u(u-l)(u- )

2 + -----------J j ----------- A ^ .2 +..„

and Gauss’s backward formula is M

- * +

+ & |2 t 4 V - ,

, (u+2)(u+l)u(u-l) 4I y~2 + ........- ( 2) In equation (2) shift the origin to 1 by replacing a by (« - l) and adding 1 to each ajgument 0, •i- * i......men wc get

Ax) =yt+(«- l)u&y0+

2!

^ +u(u- l)(u- 2)(u- 3)^ 31

N um erical A n a ly sis -11 | 2 3 9

Taking the mean o f equations (1) and (3), we get

u ( u - l) 1 u + (u -1) Ay0 + fix) = r ( y o + y i) + 2!

A2y_! + A2y0 2

(u + l)(u - 1) u (u -l)(u - 2 ) A3y_i 3! + 3!

(tt + l)u(u - l)(u —2 ) l r , -1 ' — - ~ 4j - — - 2 14 y -* + **y-i J + •

or

u (u -l) 2!

/ * ) = y0 + «Ay0

[ A '^ + A ^ p ]

2

u ( « - l ) ( u - | j

+ ---------57---------

(A3>-il

(u + 1)(b)(u - l)(u - 2) A4jy-2 +A*y-i 2 4! which is called Bessel i formula. I It used when u lies between ~r ana — i.e., ~ < u < ~4

4

4

,3

1

3

4

Example 3.64: Employ Bessel fs formula to evaluate y2j , given y2o * 2854, y 24 = 3162, y2g « 3544, y 32 * 3992. [RGPV Dec. 2007} Solution. To find y =»f(x) at x =»25 : Taking the origin at xo = 24, h = 4, x -* o

2 5 -2 4

u - — ,— ------ :— h 4 The central difference table is : X x_1=20 n N> 00

x0 =24 x2 =32

1

4

. . . . .

1

J 3

0.25 which is lies between — and ~ ,

y = /r* ;

dy

A 'y

A*y

2854 3162 3544 3992

308 382 448

74 66

-8

The Bessel’s formula is

f — i)u (u -l) u(u ~ 1) A2?.; +A2y0 ♦ L _ * L -------fix) = >0 + uAyo + 2! 2

240 | Ewoii^ a iBgjil ijiipiin irf i# Putting all values from the table, we get + b 2 5 ) ( . 2 5 K - 7 5 ) x ( _ 8)

^25) = 3162 + (.25) x (382) +

[ 74 + 66 2 L' or >>25 = 3162 + 95.5 - 6.5625 - 0.0625 = 3250.875. Example 3.65 : Apply Bessel’s formula tofind the value off(27.4) from the table : x: 25 26 27 28 29 ffx ): 4.000 3.846 3.704 3.571 3.448 Solution. To findy =fix)vtx = 21A : Taking origin at x = 27, h - I U=

Ans. 30 3J33

x - x * 2 7 .4 -2 7 h ~ = 1 ~ 0-4

1 which is lies between — 4

,3 ~r • 4

The central difference table is :

y

* ‘y

A*y

A*y

A4y

4 *y

4.000 3.846 3.704 3.571 3.448 3.333

—.154 -.1 4 2 -.1 3 3 -.1 2 3 -.1 1 5

.012 .009 .010 .008

-.0 0 3 .001 -0 .0 0 2

.004 - .0 0 3

-.0 0 7

X

x_2 = 25 x_j =26 *0 =27 -2 8 x2 =29 x 3 =30

:. The Bessel’s formula is A x ) = y 0 +uAyQ+

u(u~ 1) I A ^ + A ^ q ] 2!

u(u

3!

[A3Jl J +

(u + l)(u)(u - l)(u * 2) r A
L

/ 2 7.4) = 3.704 + M x ( - .l 3 3 ) + <-4)<°-6) x (.4X-.6X-.1) r 6

i

2

.009 + .01]

2

J

-.003) (1.4)(.4)(-.6)(-l .6) r .004 + (-. 24 I 2

( - .l)(1.4)(.4)(-.6)(-1.6) r +1 — ----- x [-.0 0 7 j = 3.6497. Ans.

Numerical Analysis-11 j 241

3.15

EVERETT’SFORMULAFOREQUALINTERVALS x-xQ Let u = — 7— and u + v = 1 i.e., v = 1 - u h viv2 - ! 2) ^ fix) ~ Wo + ---- —---- A2^

Then,

uiv2 - l 2) ^ - 2 2) ( + ---------- —--------- A y .2

u(u2 - 1 2) a9 u( u2 - 1 2)(u2 ~2 2) 4 + m +- - 3t a V o + —------- ^ -------- - A4^-i + - •

This is known as Everett’s formula. 1 an(1 a 3 It is applicable when u lies between — •j • Example 3.66 :Apply Everett'sformula to fin d the value o f log 337.5, given th a t: x: 310 320 330 340 350 log,ox: 2.49136 2.50515 2.51851 2.53148 2.54407 Solution. Let y - fix) = logj 0 x Tofindffx) a tx = 337.5 : Taking the origin atJCQ= 330 and h - 10,

360 2.55630

337.5-330 ------------ j j ---------0.75

then

1 J -7 3• Which is lies between — 4 4 The central difference table is : X x_2 =310 x_j = 320 x0 = 330 x l =340 x 2 =350 x3 -3 6 0

y = f(x )

Ay

2.49136 2.50515 2.51851 2.53148 2.54407 2.55630

.01379 .01336 .01297 .01259 .01223

-.00043 -.00039 -.00038 -.00036

A*y

A4y

A*y

.00004 .00001 .00002

-.00003 .00001

.00004

Using Everett’s formula is vW - 12) A,

fix) = Wo+ ■ 2,

u(u2 - I2) +m where

v = 1 - u = 1 - .75 = .25

v(v* - l 2)(u2 - 22) . 4

^

^

--------- tfy-z

2 u(us - l 2 )(u2 - 2 *) A. & y * +---------------------- A y- i + — - C )

242 | Engineering Mathematics-HI

Putting the values from the table in (1), we get y(337.5) = .25 x (2.51851) + k 2j M ? g 5 Z 1). x (-.00039) + C2aC^625_-l)(.062_5-4) 120 + (,75)x(2.53148) + —

6

x (-.00038

t (.75)(:5 6 2 5 - l) ( .5 6 2 5 - .4 ) x(00001

6 = .6 2 9 6 3 + .0 0 0 0 2 - .0 0 0 0 0 0 2 + 1 .8 9 8 6 1 + .0 0 0 0 2 + .0 0 0 0 0 0 1 / ( 3 3 7 . 5 ) = 2 .5 2 8 2 8 .

Thus ^

fix) ~ log x

lo g io ( 3 3 7 . 5 ) = 2 .5 2 8 2 8 .

Ans.

3.16 INTERPOLATION WITH UNEQUAL INTERVALS If the values of x are given at unequal intervals, then the definition of differences for equal intervals is not applicable. In such situations, we make use of a new kind of differences, called divided differences.

3.17 DIVIDED DIFFERENCES Let y = fix) be a function. Suppose/xn), fix\), fix-i),.....fix,,) are the (« + I) values affix) corresponding to (n + I) values xo, xi, x2........ x„ of the argument x, where X| -x<),x2- x ] ,......x» -x„_ | are not necessarily equal, in other words there is a case of unequal intervals. The first order divided ’-fi'-v te o f fix) for the arguments xq, x\ is denoted by y[x0, X|] or by f ( x i ) ~ f ( x 0) A f ( x 0) and is defined by xj

Tl>us,

“ 3 Xj - x 0

•

*k

Sim ilarly,

X) - x 0

/ x t, x 2) = J

The second order divided difference ofy(x) is A2

/ x o , x , , x 2) = ^

_. .. Sim ilarly,

-

.

tt ~

f ( Xl > ^ 2)

\

—

/ < * 6 > = ----------- —

AZ

—\

f i x \ ,x 2, x 3) = xf xs / W

^ l)

-----------

/(^2j X 3 ) —/(Xj, Xj)

etc'

x3 - x ,

=

Hence nth divided difference is .... *> = Xii

An

tt~

\

f ( . x l * x 2 .......... X „ )

— / ( X g , X j .......... X n _ i )

Xn n * 0) = ------------------—

-----------------

N um erical A n a ly sis -11 { 2 4 3

The divided difference table is :

y = f(x)

X

Ay v ,

fix o) fix,)

X0 Xl x2

3 .1 8

R E L A T IO N

ftr

A3y

/(X i)-/(x 0) f ix o,Xl ,X2) * 1 "* 0 /(xl ,x2) - / ( x 0,x1)

/(x1,x 2) = ^

f i x 2) fix,)

x3

A*y

)- ^ ) x2 - x ,

x2 - x 0

v v _ /( x 3) - / ( x 2) X3 -X 2

B ETW EEN

/(x0,x1,x2lx3)

/(Xj,X2,X3)

/(x2,x 3) - / ( x 1,x2) X3 -Xj

D IV ID E D

D IF F E R E N C E S

AND

FORW ARD

D IF F E R E N C E S

Statements : If xq, x j,..... x„ are equally spaced arguments, such that xr - xr _ i = h (for r = 1 ,2 ,.... . n), then

Ar/(* 0) =

'
Proof: We have

A /(x 0) =

Also,

A*/(*o> =

f(x 0)

1 ,, v = ^ Af ( xo)- ... (3)

A /(x,)-A /(xo) x^xg -J -A /(x i)-i A/(x0) = A __ __ ».______ 2k

_ A/(xt )-A /(xq) (l.h)(2.h) Similarly

A3/(x 0) =

Finally,

Arf( x 0) =

[v Af(xQ) = f( x x) - f ( x 0)]

A2/(x 0) 1.2 .h2

[ v Using (1) and

=x0 + 2h]

A2f(Xf,) 2 \h 2

and so on. ■

Proved.

3 .1 9 N E W T O N ’ S D I V I D E D D I F F E R E N C E I N T E R P O L A T I O N F O R M U L A F O R U N E Q U A L IN T E R V A L S

Statements : LetX*oXX*i)> f i x „ ) be the values of the function f i x ) corresponding to the aiguments x q , X |,..... . x„ not necessarily equally spaced. Then

(n

+ I)

fix) = /(x 0) + ( x - x 0)A/(x0) + ( * - x 0) ( x - x 1)A2/(*o) + ......+ (x - x Q)... .(x - X„_j )Anf(x0).

2 1 4 | Engineering Mathematics-1JI P roof. By definition : A*, xo) =

/ ( * ) - / ( * o) ^ 7^

A x ) =Ax o) + (X - * o ) /x , Xo)

•••( I

/( X o > x ,) - / ( x ,X o )

Also

Ax, xo, x t) =

or

X x, xo, x i) =

^ f(x ,x 0) - f ( x 0,xj)

=> A x, Xo) = jfa o , X,) + (x - X| )y(x, Xo, X,) Using this value o f ^ x , xo) in (1), w e get

Ax) =Axo) + (x-xo)Xxo,xl) + (x~xo)(x-x,)A x,xo,xi)

...(2)

Proceeding in this m anner, w e can get / X ) = f(X0) + ( X - X 0)Af(x0) + ( X - X 0) ( X - X l )A2f ( x 0) + ............. + ( x - x 0) (x - * i ) ......( x - x „ _ 1 )A“ /(jc 0 ) + 0 ...(3) Since (n + 1 ) values o f fix) are given, => A n*lf { x ) = 0.

Proved.

3 20 LAGRANGE’S INTERPOLATION FORMULA FOR UNEQUAL INTERVALS Statements : Let A xq), A x 0 , ..... A x*) be the values o f the function fix ) corresponding to the {n + 1) aigum ents xo, x j, X2, ..... . x„ not necessarily equally spaced. T hen

y - fix) =

'

yw

( x ~ x i ) ( x - x z )..... ( x - x „ ) 7} c }-----------r x ( x o - x j i x o - x ? ) .....(X o -X * ) ,

( x - X q K x - X z ) ..... ( x - x „ )

K

(Xi - X0)(x l - X2 )......(X i~ X „ ) +

,

1

( x - X o ) ( X - X i ) ..... ( x - x n_x)

(*»-X oX x„-xt )......( x „ - X n. l ) Proof. Since (n + 1) values o f y =AX) are given, so that w e can expressed Ax ) by polynom ial o f the nth degree, w hich is as follow s :

y =Ax) -

- X]XX - x2) .... (x - x„) + A j(x - xoXx - x2) .... (x - x„) + ..... + A ix - x0Xx - X |) .... ( x - x , _ (Xx - x , + j ) ... (x - x „ ) + .....+ / f ^ x - x 0X x - x , ) .......( x - x ^ i ) . ...(1) E quation (1) is satisfied by th e given pairs o f values (xo, yo). ( x j.y i) , {x„,y„). i.e., w hen x = x0, y -yo - U sing this in (1), w e get yo = M x o -x iX x q - x2) ...... (xo - x„)

_____________ 1____________ A° ~ (Xo-X1)(xo-X2)..... (X o -X ,)^0 Similarly putting x = X| andy =y\ in (1), we get ___________ 1___________ A ' = (X j-X oX X i-X z)......(X l~ Xn) yi 311(150 on-

N um erical A n a ly sis -II | 2 4 5

1

An -

Finally

(*» - X*n - *1)......(*n ~ **-» ) Using these values of Ao,A}........ A„ in (I), we get Lagrenge's interpolation formula: _

y

~

W

y *-

(* -* lX * ~ * a )-(* ~ * n )

“ (Xo - X ,X * 0

)

-*2)....(Xo - Xn)

x^o

( x - j q l)) ( x - 3 f e ) ....( x - x ,l ) ;c

(jfl -*o)(^l -* * )—(*j - x n) t

(x -acb )(x -* i)....(x -x ,,.1) } (x, -X o)(*n - x , )....(*,, -x„_,)

Example 3.67 :Given the values

x: f(x ):

Solution,

5 150

7 392

11 1452

13 2366

17 5202

Evaluate f(9), using (I) Lagrange’s formula, (ii) Newton's divided differenceformula. [RGPV, June 201!] (i) Lagrange's Formula : =

(« - X i)(x -x 2) ( x - x 8) ( x - x 4) 3 . (Xo - xx)(x0 - x3)(x0 - x3)(x0 - x4) 0 +... ,

( * - * o X * - * iX * -* 2X *-*3) : (xi ‘-x0)(xi ~ x i )(x4 - x 2)(x4 - x 3) 4

Here xq = 5,xi = 7,X2 = 11, xj * 13, X4 * 17 and yo = 150, y\ =»392, y%- 1452, y 3 = 2366, y 4 = S202. Putting x = 9 and substituting the above values is (1), we get

*01

= ( 9 -7 X 9 - H X 9 - 1 3 X 9 - 1 7 ) A } (5 - 7)(5 -11X 5 - 13X5 -1 7 )

15Q ( 9 - 5 ) ( 9 - l l) ( 9 - 1 3 ) ( 9 - 1 7 ) (7 - 5)(7 - 1 1)(7 - 13)(7 -1 7 )

^ ,

(9-5)(9-7)(9-13X 9-17) -xl452 (11-5X11-7X11-13X11-17) (9-5)(9- 7 ) 0 - 1 1)(9-17) x2366 (13 - 5)(13 - 7)(13 - 11)(13 -17) (9-5 )(9 -7 )(9 -l 1)(9-13) , (17 - 5X17 - 7X17 - 1 1)(17 -13) 50 + ------+--------------3136 3872 2366 +----578 = 810. aiA — 3 15 3 3 5

An*.

246 | Engineering Mathematics-1II (ii) The divided difference table is : X x0 = 5 xj =7 x2 = l l X3 —13 x4 =17

V = f(x)

Ay

* 'y

150 392 1462 2366 5202

121 265 457 709

24 32 42

A4y

1 1

0

By Newton’s divided interpolation formula : Ax) - f(x 0) + ( x - x 0)Af(x0) + ( x - x 0) ( x - x :)A2f( x 0) +...... ...(2) Here, x - 9 and putting the values from the table in (2), we get A9) = 150 + (9 - 5X121) + (9 - 5X9 - 7X24) + (9 ~ 5X9 - 7X9 - 11X1) + 0 = 810. ' Ans. Example 3.68 :Using Newton’s divided differenceformula, flnd/(8), ff9) andf(15) i f x: y =f(x ):

4 48

5 100

The divided difference table is : Af(x) x y = f(*) X0 =4 Vo =48 100 - 48 = 52 5 -4 Xj =5 yj =100 294-100 9 7 -5 x 2 - 7 y2 = 294 900 - 294 x3 =10 ya =900 1 0 -7

7 294

10 900

11 1210

Solution.

x4 - 11 y4 = 1210 xs =13 y5 - 2028

1210 - 900 = 310 1 1 -1 0 * 2028-1210 ---------------= 409 13-11

13 2028

fRGPV Dec. 20061

A‘f(x)

2 0 2 -9 7 01 1 0 -5 310-202 1 1 -7 409 - 310 13-10

A'fix)

21-15 1 0 -4 2 7 -2 1 1 1 -5 3 3 -2 7 13-7"

A4f(x)

l ~ l =0 11-4 ^ =0 1 3 -5

v Newton’s divided difference formula : Ax) = f( x Q) + ( x - x 0)Af(,xQ) +( x - x 0) ( x - x 1)A2f(x 0) +....... ...(1) (i) At x = 8, then (1) becomes : A*) = 48 + (8 - 4X52) + (8 - 4X8 - 5X15) + (8 - 4X8 - 5X8 - 7X1) + 0 = 48 + 208+180+12 = 448. Ans. (ii) Atjc = 9 ,then(1)becomes: / 9 ) = 48 + (9 - 4X52) + (9 - 4X9 - 5X15) + (9 - 4X9 - 5X9 - 7 X 0 + 0 = 48 + 260 + 300 + 40 * 648. Ans. (iii) At x - 15, then (1) becomes ^ 1 5 )= 48 +(15-4X52)+ (15-4X 15-5X 15)+ (1 5 -4 X 1 5 -5 X 1 5 -7 X 0 + 0 = 48 + 572+ 1650 + 880 = 3150. Ans.

N u m e ric a l A n a ly sis* !! | 2 4 7

Example 3.69 :Flnd a polynomial satisfied by (-4, 1245), {-I, J3), (0, 5), (2,9) and (5, 1335), by the use o f Newton's interpolation formula with divided differences. Also findf(l). Solution. Here, let x q = - 4, X[ = - 1, x2 = 0, x ;, = 2, = 5 A n ) m M S , fix ,) = 3 3 ,X x 2) = 5,fiXi) = 9 ,fix*) = 1335.

The divided difference table is :

x x0 - - 4

y = fix) y0 = 1245

Xj=-1

Vi - 33

xz = 0

y2 =5

x3 = 2

y3 = 9

x4 =5

y4 =1335

Af(x)

A’f(x)

33 - 1 2 4 6 -^#04

-1 + 4

-28 + 404 0 +4

+1

0

fix)

10- 9 4

2 +4

2 +2 8 = 1 0 2 +1

9 - 5 =2 5: 9

5 -2

=442

= - i4

13 + 14 5 +4

0

“ ' “ -IS 5 +1

2 -0

^

A4m

5 -0 »

v Using Newton’s divided difference formula : f(x 0) (ac Xo)A/(x0) (jc - x0)(x Xj )A*/(x0) .... Putting the values of divided differences in ( I), we get Ax) = 1245 + (x + 4X- 404) + (x + 4Xx + I)(94) + (x + 4Xx + 1W - 14) + (.v + -*;{x + l).v^r -2X3) + 0 = 1245 - 404x - 1616 + 94x2 + 470x + 376 - 14x3 70x2 - 56r + 3(x* + 3x> bx2 - &x) or fix) = 3X4 - 5x3 + dr2 - 14x + 5. ...(2) Ads. Putx = I in (2), w eget/1) = 3 - 5 + 6 - 14+ 5 = -5 . Ans. Example 3.70 :Use Newton’s divided differenceformula to fin d fix ) from the following data ;

Ax) =

+ -

+

x: 0 1 14 11 f(x ): The divided difference table is .

Solution. X

x0 = 0 Xj = 1

4 5

+

S 6

Asf(x)

A */W

...(1)

6 19 A ‘ f (x )

A4f(x)

11 1*0 1 5 -1 4 2-1

14

x2 = 2

15

x3 = 4

5

x4 = 5

6

x5 = 6

A /(x)

y » /(* )

2 15

-

19

,

5 - 15,- 5 4 -2 6 -5 , 5 -4 <•

6 -5

1 - 3 -- - i -----2-0 ~5 ~ 1 = -2 4 -1 1 +5 *2 5 -2 13-1 6 -4

fi

_2 + 1 = _ l/4 4 -0 2 +2 5 -1 6-2 6-2

l + ( l/4 ) 5 -0 z6 “- 1T = 0

1M

0 - (1 / 4) 6-0 = -(1724)

248 | Engineering M athematics-!!!

v Using Newton’s divided difference formula : = + + x0)(x - x, )A2/(*o)

Ax) f(x0) (x-x0W(x<)) (x-

+(x~ x0)(x - xt)(x-xz)A3/(x 0) + ........ (1)

Putting these values from the table in (1), we get

J(x) = -11 + (x - 0)(3) + x{x - 1){-1) 4-xix- l)(x - 2) ( - i j

+x(x- l)(x - 2)(x - 4K1 / 4) +x(x - l)(x - 2)(x - 4)(x - 5)f— \ 24; = 11 + 3 x - x 2 + x - —(x2 - 3 x + 2) + ~ ( x - l ) ( x 2 - 6 x + 8) 4 4 - — x(x - l)(x - 2)(x2 - 9x + 20) 24

x*

.. . , x8 3x2 x 3X3 2 x3 = l l + 4 x - x 2 ----- + ---------- + ------------+ 2x2 -----4 4 2 4 24 4

5x

3x2 0 x6 x 4 49 , 39 , + -------2 x ------ + ---------- x3 + — xz ----2 24 2 24 12 3 1 3 . 97 ,, 13 , 1 „ = - —- x * + —x 4 ----- x3 + — x2 — x +11 24 4 24 2 6 or

Ax) = “ ^

(*6 - 18*4 + 97*3 ~

+ 4* - 264>-

Ant

Example 3.71: F in d f (10) from the following data:

x: f(x ): Solution.

5 23

3 -13

11 899

27 17315

34 35606

Divided difference table is :

X xo =3 xj =5

y=f(x) -1 8 23

x2 = 1 1

899

x3 = 27

17315

x4 =34

36606

A/(x) 23 + 13 tJ8 5 - 3 =18 899-23 = 146 11-5 17315-899 27-11 35606 - 17315 =2613 3 4 -2 7

146-18 , , 1 1 -3 = 16 1026-146 2 7 -5 2613-1026 , Q 34-11 =69

A3f(x)

A4/M

40-16 ^ 2 7 -3 6 9 -4 0 34-11

0

N u m e ric a l A n a ly s js -II | 2 4 9

By Newton’s divided difference formula.

Ax) = /(jc0) +(*-JC0)A/(x0) +(x-jc0)(x-x,)A2/(^) + (x - Xo)(* - xt )(x - x2 )A*/(xo J+ ........... ( 1) Putting above values from the table in (1). we get Ax) = - 13 + ( jr - 3)18 + (x - 3 X x - 5)16 + (x -3 X x -5 X x -11)1 = x3-3 x 2-7 x + 8 Differentiate, we get f i x ) = 3x2- 6 x - 7 ...(2) Put jc = 10, in (2) we get/XlO) = 3(10)2 -6(10) - 7 = 233. Ans. Example 3.72 :Findf(x) a polynomial in the power o f ( x - 5 ) from the following data: x: fix ): Solution.

0 4

2 26

3 58

4 112

7 466

9 922

The divided difference table is : X

Xq =0 xi = 2

y */M

A/W

4

26-4 -11 2 -0

26

s * " 26- *

Xj = 3

58

x3 = 4

112

x4 = 7

466

*s -9

922

A‘/W

A*/W

3 -2 112-58 4 -3 466-112 „ 0 7-4 =118 922 “ 466 =228 9 -7

32-11 ---------- =7_ 3-0 54 - 3 2 . , , 4 -2 118-54 , , 7 - 3 ' 16 228-118 ^ 9 -4

n " 7=l 4 -0 16-11 7 -2 22-16 j 9 -3

0 0

Using Newton’s divided formula,

Ax) = f(x0) + (x - x0)hf(xQ) + (x - Xo)(x - x, )A2/(x0) + (x - Xo)(x - Xj )(x - Xj )A3/(X o) + ........ (1)

Putting the values from the table in (I), we get Ax) =4 + (x-0X H ) + (x - OXx- 2X7) + (x - OXx - 2Xx - 3X0 + 0 => Ax) - x* + 2X2 + 3x + 4. -.(2) We know that Taylor’s series in the power of (x - a)

Ax) - ««>+^

« < ■ ) + 6 ^ n o )+ ^ 2 . A o)+....

[Here a = 5]. From (2), Ax) =x3 + 2** + 3x + 4 then f i x ) = 3X2 + 4x + 3.

=> =>

/ 5 ) - 194 f ( 5 ) = 98.

...(3)

2 5 0 | E n g in e e rin g M a th e m a tic s ! II

f i x ) = 6x + 4 => /"(5 ) = 34. f i x ) =6 => f" ( $ ) = 6. /*(*) = 0 Hence (3) becomes :J(x) = 194 + 98(x - 5) + 17(x - S)2 + (x - S'p. Example 3.73 :Find the missing term by using Newton’s divided differenceform ula: x;

Solution.

0 2 1 3 9 — 1 3 y: The divided difference table is : (As given known data) y= /W

o* II O

X

A*/W

4 81 *S/W

l 2

3

1 -0

x2 = 2

9

9 ' 3 =6 2 -1 ’

=4

81

T4 - 2l - 36

Xj =1

X3

Ans.

6 - 2 =2 2 -0 ^ - ^ io 4 -1

-1 0- 2-=2 4 -0

Using Newton’s divided difference formula, 0 xq

Ax) =

)+(x- x W( )+(x~xc)(x-xl)A2f(x0) +(x - X0)(X - Xj )(X -

=>

Xo

)A3f(x0) + ...

Ax) - 1+ - X)+(x- OX*- IX2)+(x- ox*- 1x*- 2X2)

(X 0 2 A x) = l+ 2 x + 2 x (x -l) + 2 x (x -lX * -2 ) = 2x3 - Ax2 + Ax + 1 Put* - 3, we get/3 ) = 2(3)3 - 4(3)2 + 4(3) + 1 = 3 1 . Hence missing term is 31. 1 Example 3.74 :Show that f a ( £ ) - “ o6c
or

The Newton divided difference table is :

a b

i

1 1

a 1

b a _ b-a 1 1 c b _ c-b 1 .1

b c d

A*/W

5

X

ii

Solution.

Ans.

1 c 1 d

As/W

1 ba (-1)2 - J abc 1 be

abed

d c_ * d -c dc

From the table, we get g d ( £ ) - -

•

Ans.

N u m e ric a l A nalysis-11 J 2 5 1

Example 3.75: Use Newton's divided differenceformula tofin d theform off(x), given :

x: f(x ):

0 648

2 704

6 792

8 729

[RGPV June 2006, Dec. 20M(N)j Solution.

The Newton’s divided difference table is : X k *0=°

y = /W 648 704

x2 = 3

729

x3 = 6

792

4 /W

**JW

704-648 2 -0 729-704 o3 -2 ' 792 -729 01 6 -3

2 5 -2 8 _ 3 -0

2 1 -2 5 6 -2

0 -

Using Newton’s divided difference formula, X*) = /(*o) + ( * - *o W O o ) + ( x - x 0)<•* ~ x \ W f i x o) +•

=648+(jc-0X28) +(*-0X*-2X- I)

= 648 + 28x - x 2 + 2 x = - x 2 + 3Ctr + 648. Example 3.76 :Flnd the polynomial o f degree three which takes the values given below :

Ans.

2 1 4 0 5 2 1 1 y Using Lagrange’s formula, (x - l)(x - 2)(x - 4) x l + £ (£ Z 2 K x - 4 ) x l + ^ x - l ) ^ x2 2(2 - 1)(2 - 4) l(l-2 )a -4 ) ^ = ( 0 - l) ( 0 - 2 ) ( 0 - 4 ) X

Solution.

x ( x - l) ( x - 2 ) x5 4(4- l) ( 4 - 2 ) = — (-x 3 + 9x2 - 8 x +12). Ans. 12 Example 3.77 '.Thefollowing table Is given below. Find the Lagrange’s Interpolating polynomialflx) :

x: f ix ) :

0 2

1 3

2 12

5 147

Solution. Here, xo = 0 ,x t = l,x 2 = 2,x3 = 5 a n d ^ = 2 ,^i= 3 ,> ^= 1 2 ,^3 = 147. By Lagrange’s interpolation formula,

(x -x 1)(x -x 2)(x -x 3) t t (x -x 0)(x -x i)(x -x 2) ■^x) “ (x0- x 1)(x0- x 2)(x0- x 3) 0 ...... (*s ~Xo)(*3 -Xj)(x3 - x 2) 3

2 5 2 | E n g in e e rin g M a th e m a tic s-! 1!

_ ( * - l) ( * - 2 ) ( * - 5 ) ( 0 - l) ( 0 - 2 ) ( 0 - 5 )

(x - 0 ) ( x - 2 ) ( * - 5 ) (1-0X 1- 2 ) ( l - 5 ) ( x - 0 ) ( * - l) ( * - 5 ) ( 2 - 0 ) ( 2 - l) ( 2 - 5 )

= (x - lX * - 2 X * - S ) -5

3x(x -2 )(x -5 ) 4

„

(* -0 )(* -l)(* -2 ): (5 -0 X 5 -1 X 5 -2 )

2x(x _l) ( x _ 5 )

. 4 9 (x )(x -l)(x -2 ) 20 = x3 + x2 ~ x + 2. Ans. Example 3.78 :The following values o f the function ffx) fo r values o fx are given : f f l ) - 4,f(2) - S,f(7) * 5,f(8) - 4. Find the value off( 6) and also the value o f x fo r which/fx) Is maximum or minimum. Solution. Here, we have Xo = I j Xj = 2, X2 —7, X3 = 8 andyo “ 4,_yi ” 5 , ^ = 5 , ^ = 4,

Using Lagrange's interpolation formula, we get (* -2 X * -7 )(x -8 ) "x ) ~ (1 - 2 )(1 - 7 ) ( l - 8 )

( x - l ) ( x - 7 ) ( x - 8 ) ::6 (2 - l) ( 2 - 7 ) ( 2 - 8 ) ( x - l)(x - 2)(x~ 8 ) . ( 7 - l) ( 7 -2 X 7 -8 )

( « - l) ( » - 2 ) ( x - 7 ) ( 8 - l) ( 8 - 2 ) ( 8 - 7 )

On simplification, we get -x 2

3

8

* > - - r +r " s 6s

•(1) 3

„

8

34

Put,x = 6 , th e n / 6 )= — + —x® + - = ~

Again, for maximum or minimum of /(*). dy dx =>

x =4,5 and

( d2f ( x ) )

-2x 6

=5.66.

Ans.

- - 0 , gives 3^ _ 2 ^

. 2 =- 2 <0 v /*-4.B Hence, fix) is maximum at the point x = 4.5. Abs. Example 3.79: Given loga 654 « 2,8156, logft 658 - 2.8182, log19 659 “ 2.8189 and logio 661 m 2.8202. Find logit 656. Solution.Given, xo * 654, Xi = 658, X2 = 659, X3 “ 661 and yo = 2.8156, y\ = 2.8182, y2 —2.8189,y j = 2.8202.

N umerical A nalyshJ I I 2 5 3

By Lagrange’s interpolation formula, we have (656 - 658)(656 - 659)(656 - 661) /0 Qt cg, X656) -lo g ,0 (656) - (664 _ 65s)(654 - 659)(654 - 661) t (656-654X 656-6S9X656 - 661) (658 - 654)(658 - 659)(Go8-661)

'

t (656 - 654)(656 - 658)(656 - 661) + (659 - 654)(659 - 668)(659 - 661)

g '

(656 - 654)(656 - 658)(656 - 659) 82Q (661-654X 661-658)(661-6 5 9 ) V ' (-2X-3X -5) ^ . s i s e ) ^ 2^ ) (2.8182) + (2-8189) 5(l)(-2) (-4)(-5)(-7) 4(-l)(-3 ) 2(-2)(-3) (2.8202) 7(3)(2) Ans.

log io (656) = 2.8169. Example 3.80 .-Express the function

3x* + x + J (x -l)(x -2 )(x -3 ) as a sum o f partialfractions by using Lagrange’S interpolation formula. m

3x2 + x +1 = ( x - l) ( x - 2 ) ( x - 3 ) Consider ^ (x) = 3X2 + x + 1 and tabulated its values for* = 1,2, 3, we get

Solution. Given

1 5

x : 3x2 + x +1:

2 15

3 31

Using Lagrange’s interpolation formula, we get #w . ^ M ( 5) + f c M ( 1 5 ) + i ^ M <31) (3 —1) (3 —2) (l-2 )(l-3 ) (2 -1 ) (2 -3 ) t

(jc )

= | ( x - 2)(x - 3) -1 5 ( x - lXx - 3) + ~ (x - l)(x - 2) 2

it

31 15 m &x) ~ ( x - l) ( x - 2 ) ( x - 3 ) *" 2 ( x - l) x - 2 2 (x -3 ) Example 3.81: By means o f Lagrange’s interpolation formula, show that : 0) y<>~ ~{yi

- * / ) - ( > - / -y-s)]

(U) yi ~y» - 0.3 (ys - y .j) + 0.2 (y.j - y ^ ) .

Ans.

2 5 4 | E n g in eer in g M athem atics -111

Solution.

Here yz - f ( x ) i.e. Argument * : - 3, - I, 1,3. By Lagrange’s interpolation formula: ( jc + IX* - IX* ~ 3) ( * + 3X* - IX* - 3) y = f t x \ = ---------—— —£ --------— x v , + -----------—------ —---------- x y ,

y

(-3+ 1X -3-1 X -3 -3 )

3 (-1 + 3X -1-1X -1-3)

( x + 3X * + 1 X * ~ 3 )

+ (I + 3X1 + W - 3 ) y

(x + lX * -lX * -3 j (-4S) ,_3

+

+ (* 1 3 X

”

x

+ 1 X £ -1 )

(3+ 3X3+1X3-1)

(* + 3X* - 1 Xx -3 ) (16)

'

( x + 3X* + 1X*~3). . | (* + 3X* + l X * - l) . 1

(-16)

(48)

^

Put x ~ 0, we get 1 yo-

or

yo =

9

9

1

i 6 -f~3 + i 6 >'-' + 16>'>

l6y*

+ y_i) —^ [O 'j - y i ) - O '- ! -y -i)]

(ii) The argument are x : - 5 ,- 3 ,3 ,5 . Using Lagrange’s furmula: (jc-t-3X*~3X*~S) * (-5 + 3X -5-3X -5-5)

v

(* + 5X* ~ 3X* ~ 5) , 5 (-3 + 5 X -3-3X -3-5)

Proved.

|

3

(x + 5X* + 3X*-5) (x + 5X* + 3X*-3) + ------ —----- —----- - * y-i + ------ —-------------- x + (3 + 5X3+ 3X3-5) (5 + 5X5 + 3X5-3) Putx= I, we get yi = - 0.2 y-s + 0.5 y - i+ y i- 0.3y} or yi - y 3 - 0.3 (y5- > ,3) + 0.2 (yj - jlj)Proved. 3.21 INVERSE INTERPOLATION JRGPV June 2003} The method of determining the value of the argument xt corresponding to the given value of the function when the function lies between two tabulated values, is called inverse interpolation. Thus My =J{x) be the given function then to find the value of x corresponding to the value of y, when y lies between two tabulated values ofy, is called inverse interpolation. The method o f determining the inverse interpolation is as follows : 3.22 LAGRANGE’S INVERSE INTERPOLATION FORMULA [RGPV June 20031 The Lagrange’s inverse interpolation formula is given by ( y - y iX y - y j)-( y - y .)

_ . (y -y o X y -y iM y -y » ) „ ,

N umerical A nalysis-11 | 2 5 5

Example 3.82 :Apply Lagrange’s inverse interpolation to fin d to one decimal place, the value o f x when y = 13.6, given the following. [RGPV Dec. 2002] X y = f ( x ) •• Solution.

40 14.1

35 14.9

30 1S.9

50 12.5

45 13.3

The Lagrange’s inverse interpolation formula is as given below : _

(j'~yi)(y-> 2 >... (y - — y■n) ■■- „J a ■■i ■■ II (yo-y\)(yo - t t ) .... (yo -y « ) j

. (y-jbXy-jfc).... ■■ .... Iiy■y * )

-f-

mm

. ^

1

U -yo)(y1 - vt).....Cxi - y . ) ( y - y 0)Q y-yi).... x

(;y»- ^0)(y, - yi)... (y* - yn-i)

-.(I)

We have, xq = 30, x\ = 35, x 2 = 40, xj = 45, X4 = 50 and yo = 15.9, yj = 14.9, w = 14.1, j-j = 13.3, yA = 12.5. Alsoy= 13.6 Putting the above values in equation (1) we get (13.6 - 14.9)(13.6 -1 4 .1)(13.6 - 13.3)(13.6 -1 2 .5 ), 3Q * " (15.9-14.9)(15.9-14.1)(15.9-13.3X 15.9-12.5)X : (13.6 - 15.9X13.6 - 14.1)(13.6 - 13.3)(13.6 -12.5) „ g5 + (14.9 - 15.9)(14.9 -1 4 .1)(14.9 - 13.3)(14.9 -12.5) X t (13.6 - 15.9)(13.6 - 14.9X13.6 - 13.3)(13.6 -12.5) ^ + (14.1-15.9)(14.1-14.9)(14.1-13.3)(14.1-12.5)X (13.6-15.9)(13.6-14.9)(13.6-14.1X13.6-12.5) ;15 + (13.3 - 15.9X13.3 - 14.9>(13.3 - 14.1)(13.3 -12.5) X (13.6 - 15.9XI3.G - 14.9)(13.6 - 14.1)(13.6 -13.3) x50 (12.5 - 1 5.9)(12.5 - 14.9)(12.5 -1 4 .1)(12.5 -13.3) = 0.45833 - 4.32373 + 21.41276 + 27.7954 - 2.14700 => x = 43.195. Am. Example 3.83 :Applv Lagrange’s formula inversely to fin d a root o f the equation f(x) “ 0, when f (30) = - 30, f(34)= - I3,f(38) - 3 andf(42) «=18, Or given table: X

Solution.

30 -3 0

34 -1 3

38 3

42 IS

f(x ): We have x o ^ O ,* ] = 34, x2 = 38, xj =42,yo = -30 , y, = - 13, >^ = 3, ^ 3 = 18. We are to find-the value of x when y =fix) - 0. Using Lagrange’s inverse interpolation formula (0 + 13)(0 - 3)(0 -18) OA , (0 + 30)(0 - 3)(0 -18) x = (-30 + 13)(-30 - 3)(-30 -18)■x30 + (-13 + 30)(-13-3)(-13-18) x34

(0 + 30)(0 + 13)(0-18) (3 + 30)(3 + 13)(3 -18)

£

(0 + 30X0 + 13X0-3) . g (18 + 30)(18 + 13)(18 - 3)

2 5 $ ^ E n g in eer in g M a ih e m a t ic s -I II

13x3x18x30 30x3x18x34 30x13x18x38 17x33 x 48 + 17x16 x 31 + 33x16x15 x = - 0.7821 + 6.5323 + 33.6818-2,2016 - 37.2304. Hence the required root of the equation/*) = 0 is 37.2304. Example 3.84 :Fbut the value o f x when y - 5, given that:

X y =f ( x ) : Solution.

1.8 2.9

2.9 3.6

2.2 4.4

2.4 S.5

3 0x 1 3 x 3 x 4 2 48x31x15 A n s.

2.6 6.7

We have, xo = 1.8 , Xi = 2.0, x2 = 2.2, xj = 2.4, x* = 2.6, = 2.9,y \ = 3.6, >5 = 4.4, y ^ = 5.5, y 4 = 6.7 and given y = 5. By Lagrange’s inverse interpolation formula

yo

(y - y i X y ~ y 2) ( y - y a ) ( y - y <)

.. ^

(y - y0)(y - y2)(y- ysXy- y*)

x ~ (yo-yi)(yo -y*)(yo-ysXyo- y*)

,

(y-yo )(y-yi)(y-y3Xy- y4). ~ ~ .................................... (yi - jbX** - yiXy2-

,

(y -J b X y -riX y -^ X y -tt)

- y*) 2 Cya- y<>)(ys ~ y \)(^s - yz)(j3 - y*) 1

x3

(y -y o X y -flX y -jfrX y -jfr) (y4 - yo)(y* -yi)(y* -y2)(y* -y*)

4 “ 1}

Putting above values in (1), we get (5-3.6)(5 - 4.4X5 -5.5)(5 - 6.7) (1.8) * (2.9 - 3.6)(2.9 - 4.4)(2.9 - 5.5)(2.9 - 6.7) (5-2.9)(5-4.4X 5 - 5.5)(5 - 6.7) (2.0) (3.6 - 2.9)(3.6 - 4.4X3.6 - 5.5)(3.6 - 6.7) (5 - 2.9)(5 - 3.6)(5 - 5.5)(5 - 6.7) (2 .2) (4.4 - 2.9)(4.4 - 3.6)(4.4 - 5.5)(4.4 - 6.7) (5 - 2.9)(5 - 3.6)(5 - 4.4)(5 - 6.7) (2.4) (5.5 - 2.9)(5.5 - 3.6X5.5 - 4.4)(5.5 - 6.7) (5 - 2.9)(5 - 3.6)(5 - 4.4)(5 - 5.5) ( 2 -6) + (6.7 - 2.9)(6.7 - 3.6X6.7 - 4.4)(6.7 - 5.5) (1.4)(0.6)(-0.5)(-1.7)(1.8) (2 . l)(0.6)(-0.5)(-1.7)(2.0) (1.5)(0.8)(-l.l)(-2.3) + (0.7)(-0.8)(-1.9)(-3.1) t (1.4)(0.6)(-0.5)(-1.7)(1.8) ] (2.1)(0.6)(-0.5X~ 1.7)(2.0) (0.7X-0.8)(-1.9)(-3.1) (1.5)(0.8)(-L l)(-2.3) (2.1)(1.4X0.6)(-0.5)(2.6) (3.8)(3.1)(2.3)(1.2) x=

1.285 2.142 . + —-— 5.498 + -7.1971 3.036 3.2984 3.036 6.5208 2.617. (Approx).

2.293 32.5128 A ns.

N u o m c a i. A m alths-H { 2 5 7

Example 3.85:Apply Lagrange's method to fin d the value o fx when f(x ) “ IS from the given data: fRGPV June 2005 and Dec. 2011J 5 12

* y = f(x ) : Solution.

6 13

9 14

11 16

We have, jro = 5, X| - 6, v> = 9, xj * II, Vo = 12,yi = 13,^2 = 14,>>3 = 16 and giveny - 15. Tofindx: By using Lagrange’s inverse interpolation formula (y-yi){y-y'i){y-yz) x ~

x , ( y - y 0X y -y a X y -y a )

(yo-yiXyo-yzKyo-ys) 0 (* -yo)(yi -ysXyi -ya) 1 ,

(y -y p X y -y iX y -y a ) , (y -y o )(y -y i)(y -y z ) (y2 -yo)(y2 -y i)(y 2 -y 3 ) (y s-y o X y s-y iX y s-y a ) 3 Putting given values from the table in (I), we get (15-13)(15-14)(15-16) t (15-12)(15-14)(15-16) •c “ (12-13)(12-14)(12-16) (13-12X13-14X13-16)

( (15-12)(15-13)(15-16) (14 - 12)(14 - 13)(14 -16) => x =

(15-12)(15-13)(15-14) y U (16 - 12X16 - 13X16-14)

16 9 + — 27 + — 11 =9.125. o

*

Ans.

4

3.23 NUMERICAL DIFFERENTIATION Numerical differentiation is a process where we differentiate a suitable interpolation polynomial in place of actual function when we are given a set of values of that function and find the approxi­ mate numerical value of the derivative (or derivatives) of a function. Working Rule : (i) Fit up an interpolation polynomial to the given set o f values of die function using an appropriate interpolation formula. (ii) Then substitute the values of x to find the derivative (or derivatives) at that point, if desired. dy Remarks: (i) If the values of x having an equal interval and ^ is required near the beginning of the data, we use Newton’s forward formula, and if it is near the end of the data, we use Newton’s dy backward formula, and for the values near the middle of the table, ^ is calculated by means of Stirling’s or Bessel’s formula. (ii) If values of x have not an equal interval, then we use Newton’s divided formula to represent the function.

324 DERIVATIVES OF>>*=fix) BASED ON NEWTON’S FORWARD INTERPOLATION FORMULA (0

1 Ay0 - ~ A 2.y0 + ~ A V o - \ A 4y 0 + £ A V o - ~ A 6;y0 + . " h

2 5 8 | E n g in e e rin g M a th e m a tic s -III

(ii)

1 {*2L\ [dx2jr=xo = h2 A2y0 (d*y) 1 AVo " *3 .

“

A4y 0 - | A5y 0 +

A6y 0 - .

A4;y0 + ^-A3;y0 ~.

325 D ERIVATIVES O F y - J { x ) BASED ON N E W T O N ’S BACKWARD INTERPOLATION FORM ULA: (i)

1 ~ h

[dx,

d 2y ) 1 dx2 j *=*n = h? 1 v 3^ + | v ^ n + l v ^ „

(iii) {dx>l__Xn -

326 DERIVATIVES OE.y ^ /.x ) BASED ON STIRLINGS OR CENTRAL DIFFERENCE FORMULA: (i)

'& ]

(ii)

dy dx L

(iii)

dx,

I

2

1 ( A3y_t +A3y_2 I l f A*y_2 + A6^_3 "

j “ 6^

2

J 301

2

J

■

f A3y .t + A3y_2 ft3

Example 3.86 : Find

dy

x : y: Solution.

( Ay0 +Ay -1

=I

and

I d*y

a t x - 4 , using

1 0

2 1

4 S

8 21

10 27 (RGPV Dec. 2901 A June 2006J

Divided difference table is : X

y

Ay

A 'y

A*y

A4y

*0 =1 Xj = 2 x2 = 4 x3 = 8 x4 =10

0 1 5 21 27

1 2 4 3

1/3 1/3 - 1/6

0 - 1/16

-1 /1 4 4

N umerical A nalysis-II | 2 5 9

By using forward divided difference formula : Ax) = y0 + (* - i)Ayo +(* - 1)(* - 2W y 0 + ( x - l ) ( x - 2)(x - 4)A3>0 +(x - 1)(jc- 2){x - 4)(x - 8)A4y0 + .... =*

Ax) = 0 + < * - l ) . l + < x - l ) < x - 2 ) |+ 0 + ( x - l ) ( x - 2 ) ( x - 4 ) ( x - 8 ) ( ^ } = ( x - l ) + i ( x 2 - 3 x + 2) + ^ - ^ ^ ( x 2 + 1 2 x + 32) 3 144

or

y

= x —1 + —— x + 4 + x4 -15X3 +70x + 120x + 64 3 211 x 4 -1 5 x 3 + ^ - x 2 +120x + 67 V

...(1)

Differentiate : — = 4x3 - 45X3 - 45x2 + —- x +120 ax <3 Ar-4

...(2)

= 252.8.

Again, differentiate (2) :

Ans. d?v

„ 422 - 12x2 - 90x +

^ = 12(4)2 - 90(4) + “

= - 27.33.

Ans.

Example 3.87: Values o fy as plotted against x are :

x: y:

0.5 0.3521

0.75 0.3011

LOO 0.2420

1.25 0.1827

Find ~ a t x = 0.5 dx Solution.

1.50 0.1295 [RGPV June 2002}

Tofindf'(x) ai x - xo = 0.5: Using derivative of forward difference formula.

f& \ , d x ) X~XQ

Ay0 - - A2jy0 + “ A3>0 - -7 A4y 0 + .

••(I)

The difference table is : Here h ~ 0.25 X 0.5 0.75 1.00 1.25 1.50

y 0.3521 0.3011 0.2420 0.1827 0.1295

- 0.051 -0.0591 -0.0593 - 0.0532

A 'y

A’y

A4y

- 0.0081 -0.0002 0.0061

0.0079 0.0063

- 0.0016

2 6 0 | E n g in e e rin g M a th e m a tic s-I II

Hence {I) becomes : 1

f d y]

0.0081 0.0079 0.0016 - 0.051 + --------- + ---------- + ----------

«-o.» ' 0.25 0.25

(- 0.051 + 0.00405 + 0.002633 + 0.0004)

- 0.0439167 = -0.17567 0.25 Hence

= -0.17567

Ans.

Example 3.88: Find — a tx = 1.5from the following table : dx x:

1.5

y • 3.375 Solution.

2.5 3.5 4.0 3.0 7.0 13.625 24.0 38.875 59.0

2.0

IRGPV, June 2008 & June 2011j

To find f \ x ) at x = xo - 1.5 : Using derivative of forward difference formula.

The difference table is : Here h - 0.5 X

15 2.0 25 3.0 35 4.0

...

y 3375 7.0 13.625 24.0 38.875 59.0

From the table, A>>o= 3.625, Atyo = 3.0, Hence (I) becomes: -ii

o-sL

Ay

A2y

AJj

3.62S 6.625 10375 14-875 20.125

3.0 3.75 45 2.25

0.75 0.75 0.75

Af y

0

0

A3y0 = 0.75, A^0 - 0

3.625 - - x 3.0 + - x 0.75 - - x 0 2 3 4

2375 05 = 4.75 _

A ds.

Numerical Analysis-! I | 2$1 Example 3.89 :Findf%x) at x - 0.1 from the following table : 0.1 0.9975

x: y =f(x ): Solution.

0.4 0.9604

0.3 0.9976

0.2 0.9900

Tofindf'fx) at x ** 0.1: So taking xo = 0.1 Theforward difference table is : X x 0 =0.1

0.2 0.3 0.4

y ~ f(x) .9975 .9900 .9976 .9604

Ay

A*y

A*y

- .0075 -.0 1 2 4 - .0372

- .0049 -.0 0 4 8

0.0001

Using derivative of forward difference formula

( *

L

........ ]

Here h = .1, Ay0 = - .0075, A2y0 = - 0049, A*yo = 0.0001, and x0 = 0.1 /'(0.1) = 7TT - 0075- i (-.0049) + 1 (.0001) = -0.0502. O.l A o

Ans.

Example 3.90 (a) : fin d f%x) and f% x) a t x m6, given that

x .* y~ f(x): Solution.

4.5 9.69

5.0 12.90

5.5 16.71

6.0 21.18

6.5 26.37

7.0 32.34

7.5 39.15

Tofin d f%x) andf'fx) at x * 6: So taking xq = 6 X

4.5 5.0 5.5 x„ = 6.0 6.5 7.0 7.5

y 9.69 12.9 16.71 21.18 26.37 32.34 39.15

Ay

A*y

A*y

A4y

3.21 3.81 4.47 5.19 5.97 6.81

0.60 0.66 0.72 0.78 0.84

0.06 0.06 0.06 0.06 0.06

0 0 0

Using derivative of forward difference formula :

2 6 2 | Engem esm N g M a th e m a tic s -! II

or

™

5.19 - i (0.78) + ^ (0.06)1 2 3 J = ae = 2(5.19 - 0.39 + 0.02) = 9.64.

d 2y dx2

and

h2

A2y0 - A3y0 +

A ns.

A*yo +••••

or

A ds. /"(« ) = T T ^ t0-78- 0-06] =4(0.72) = 2.88. \j.Ab Tof i n d f ’ (x) a n d f " (x) a t x - 1.6: So taking x„ = 1.6 and using derivative of backward formula. Here Vy„ =0.281, ^ y „ = -0.018, VV„ = 0.005, V*yn = 0.002, V % = 0.003, V*y„ = 0.002 Then, we have

(£ L .

- i K

S L . ' w

+i v '* 4

[ “ 8,+r

^

v V *+

0,8)tr

+

i v V ' +- ]

(0005)]

+ -x

4

(0.002) +

-

5

x

(0.003)+- X (0.002) 6

= 2.7476. Also,

^

A-*.

= - y f v2>«+ v V (I+ — V4y » + -V 5>'B+ — VV), A L 12 6 ” 180

J

-0.018 + 0.005 + — X (0.002) + ~ x (0.003) + — x (0.002) 12 6 180

(0.1)2 L = -0.638. Hence / " (1.6) = -0.638. Example 3.90 (b): Given that: x : y ♦*

1.0 7.989

1.1 8.403

dv d y Find -j- and — r at x - 1.1. dx dx2

A ds.

1.2 8.781

1.3 9.129

1.4 9.4S1

1.5 9.750

1.6 10.031

[RGPV, Dec. 2011J

N u m e b jc a l A naiy& is-U | 2 6 S

Solution.

The difference table isgiven below : X 1.0

y

Ay

A*y

A4y

A^ _

A*y

A5j >

7.989 0.414 -0.036

8.403

1.1

0.006

0378

-0.002

-0.030

8.781

12

0.004

0348 -0.026

9.129

13

0.002 0.005

0299 -0.018

9.750

15

0.003

0.003 -0.023

9.451

0.002

-0.001

0322 1.4

0.001

0.281 10.031

1.6

Tof i n d f ’(x) a n d f" (x) atx = 1.1: So that taking xo - 1.1 Thenyo = 8.403, AVo = 0.378, 5A*y« = - 0.03. A^0= 0.004, A4>’o = - 0.001. and A5y0= 0.003 Then, we have

(& !.„ ■ ^

4 ^ +j aV«■?

+

■i aV,+"".

($ L \ = 0378 + - x 0.03+ - x 0.004+ - x 0.001+-x0.003 U J„n 0.1 L 2 3 4 5 J = 3.9518. and

1

- r I . ¥ ' *«■*<) W

r*2

*3

+- 4 » - ? a

11 a4

*— 4

= _ L . r - 0 . 0 3 - 0.004 + — x (-0 .0 0 1 ) -

*

L u

12

y:

1.0 2.7183

1.2 3.3201

1.4 4.05S2

1.6 4.9530

137.

1

1 x (0.003)] 6

0.011 0.015 1------- T . . . = -----1— 0.03 —0.004---- —------- -— 0.01 L 12 6 , = - 3.742 dy x:

5 s

Ans.

1.8 6.0496

J

Ans. d*y 2.0 7.3891

2.2 9.0250

2 6 4 | E n g m eer »*g M ath em a tics -111

Solution. Tofindf% x) andf"{x) at x « 2 .2: So taking x„ - 2.2 and using backward difference formula: X

1.0 1.2 1.4 1.6 1.8 2.0 Xn = 2.2

y 2.7183 3.3201 4.0552 4.9530 6.0496 7.3891 9.0250

Ay

A*y

4*3-

A4y

A*y

4 'y

0.6018 0.7351 0.8978 1.0966 1.3395 1.6359

0.1333 0.1627 0.1988 0.2429 0.2964

0.0294 0.0361 0.0441 0.0535

0.0067 0.0080 0.0094

0 0013 0.0014

0.0001

Derivative of backward difference formula : ] 2 ‘"* 3 4 *"* 5 6 [Here h = 0.2, x„ = 2.2, Vy„ = 1.6359, V*yn = .2964. V*y„ = .0535, V*y„ - .0094, V ^ „ = .0014 and V€y„ = .0001]

(* L

■■ A 2-2)= £ 2

1.6359 + - (0.2964) + - (0.0535) + - (0.0094) 2 3 ‘1 -(0 .0 0 1 4 ) + -(.0001) 5 6

= 9.0228. Example 3.92 : Find the value o fffx ) at x ■ .04 from the following table

x: f(x ): Solution.

0.01 0.1023

0.02 0.1047

0.03 0.1071

0.04 0.1096

A n s.

O.OS 0.1122

0.06 0.1148

{RGPV Dec. 2005] Tofin d f'(x) at x —.0 4 : So taking x„ = 0.04 and using backward difference formula

x

y = f(x )

Ay

A*y

0.01 0.1023 0.0024 0.02 0 0.1047 0.0024 0.03 0.0001 0.1071 0.0025 =0.04 0.1096 0.0001 0.0026 0 0.1122 0.05 0.0026 0.1148 0.06 Using derivative of backward difference formula

A 'y

A4y

0.0001 0 -0.0001

-0.0001 -0.0001

V y „ + ^ V 2 y „ + -V ^ „ + . 2 3 [Here h - .01, Vy„= .0025, V*y„ = .0001, V - 0.0001, and

(S L -i[

= .04]

f ( 0 M ) = - ^ .0 0 2 5 + |( .0 0 0 1 ) + ~(0.0001) = 0.255.

Ans.

NdMMCKk'MM^MB-n | 265 Example 3.93 :Flnd the first and second derivatives o f thefunction given below at the point x - 1.2 :

i 0

x: y «* Solution.

2 1

5 8

4 6

3 5

[RGPV June 2003] Tofin d f%x) and f fx) a t x m 1.2 : Which is near to xq - 1. So using Newton’s forward difference formula: u

+•

-.(I)

r

X-Xq

u = —j-— => x * xo + uh. (i.e., xq + uh = 1.2)

where

~ : -•yi $ II

X

1 2 3 4 5 From (1), rewriting

0 1 5 6 8

Ay

A 'y

A*y

A*y

11 4A 11 2

3 -3 1

—o A

10

. , , x u ( u - l) . A*o+uh) =f(x o) +ju Aft*o) + A*rt*o)

u ( u - l) ( u - 2 ) 3, A3/(x0)

^ u ( u - l X « - 2 ) ( u r 3) 4!

(2)

Differentiate (2), w.r.t. u, we get (3u2 -6 u +2) A3/(*o) 6 (4u3 - 12u2 + 22u - 6) 24

...

,

,

+ i -------------—-------------- A4/ ( x 0) ...(3)

Again differentiate w.r.t u, we get h>r(*o + uh) = A2/(x 0) + (u - l)A3/ ’(x0) + (e“2-

x-x0 Putting, A = l ,x 0 = l , u = —^

1.2-1 j

12

AV(*o)

••(4)

= 0.2; and x = xq + uh = 1.2, in equations (3)

and (4), we get (0.4 - l ) x 3

f ( 1.2) = 1+-

[3(0.2)2 - 6 x (0.2) + 2]

6

x (- 6 )

(4 x (0.2)3 -1 2 x (0.2)2 + 22 x (0.2) - 6} x 10 24 Ans. = 1 - (0.9) - (0.92) - (0.853) = 1 - 2.673 = - 1.673.

2 6 6 | E n q m e e b jn g M /u h e n a tk s - I I I

And

/"(1 .2) = 3 + (0 .2 -l)(-6 ) + -—(— ^

2—(— ^ Xa

xlO

= 3 + 4.8 + 0.33 = 8.13. Ans. Example 3.94 :FUut the 1st and Ilnd derivative o f the function tabulated below at point x = 1.1: * fix ) Solution.

1.0 0

1.2 0.128

1.4 0.544

1.6 1.296

1.8 2.432

2.0 4.000

[RGPV June 2009 and Feb. 2010] Since at point jc = 1.1 lies near the 1.0, so we shall use Newton’s forward formula. jw - m

*

^ 2!

* ......

3!

•-(I)

x- a where u - —r— = x - a=>x = a + uh = 1.1 h The Difference Table is :

=>uh

I II

X

1.0 1.2 1.4 1.6 1.8 2.0

Af(x)

A’ fix ) 1 A3f(x)

0 0.128 2.544 1.296 2.432 4.000

0.128 0.416 0.752 0.136 1.568

0.288 0.336 0.384 0.432

0.048 0.048 0.048

A4fix)

0 0

From (I) f a + uh) = f(o) + uAf(a) + —

Z

A2/(o) + —---- --- - 2 —A3f(a) o

Differentiating twice w.r.L V we get h f \ a + uh) = A /( a ) +

Z

A 2f(a ) +

A *f(o)

D

'••(2)

h ? f'(a + uh) - A2/ ( a ) + (u - l)A 3/ ( a ) Here, h = 0.2, a = 1.0, so that u = 0.5 and a + uh = 1 .1 Hence (2) and (3) become :

A M )-a *

o.i28 +

x o .288 + (g ^

l [0.128-0.002] =0.63. 0.2 1 and /'(1.1) = ( 0.2)! [0.288 + < 0.5-1)(0.048)] = 6.60.

^

a s ^ ) k ao48

Ans.

N omemcal A nalysis-!! (2 67

Example 3.9S :A slider in a machine moves along afixed straight rod. Its distance x cm. along the rod is given below fo r various values fo r the time t seconds. Find the velocity and acceleration o f the slider, when t » 0.3 second: t: 0 0.1 0.S 0.6 0.2 0.3 0.4 x : 3013 31.62 32.87 33.64 33.95 3X81 33.24 Solution.

[RGPV Dec. 2004J Here the velocity (dx / dt) and acceleration ((fix i dt*) are to be determined at the point t = 0.3 sec. which is near the middle of the table. Therefore, centra! difference formula will be used. The central difference table is :

Ay

4

30.13 31.62 32.87 33.64 33.95 33.81 33.24

1.49 1.25 0.77 0.31 -0 .1 4 -0 .5 7

o o o o o ! 1 1 1 t

t 3 =0 t.2 =0.1 L i =0.2 to =0.3 tx = 0.4 t2 =0.5 t3 =0.6

*y

y

X

A*y

A4y

A 'y

A(y

-0 .2 4 0.02 0.01 0.02

0.26 -0.01 0.01

-0 .2 7 0.02

0.29

We know that central difference formula for derivatives ; dy)

1 r (Ayp+Ay-i)

2

0

Here

^

2

30

2

2 “ *02,

1 f~(,31 + .77) 1 <.02+ .01) | 1 (.02 +(-.27))] U o t..s "C1)L 2 6 2 + 30 2 J

Also- ( ^ L ,

-

”6

+ A8y_2) [ l (A5y_2 + A5y_3)

h = . 1, tQ= 0.3, Ay<>= .31, Ay_ j = .77, A^ l j = .01, AV-2 = .02, A5>_3 = - .27.

Velocity - /'( .3 ) =

Putting

l (a 3^

+ jjjj (” *125) j = 5.34 cm/sec.

Ans.

h = .1, A2y~i - - . 4 6 , A*y~2 ~ - .01, A^y_3 = .29.

/ ”< • » - c F [ - 46- S (- 01)+^
= - 45.6 cm/sec2.

Ans.

268 I E n g in eer in g MATHKM*ncs-III Example 3.96 :Find x fo r which y is maximum andfin d this value o f y

x: y •' Solution.

1.2 0.9320

1.3 0.9686

1.4 0.9855

1.5 0.9975

0.9996

The difference table is as follows : Xq

Ay

y 0.9320

X

=1.2 1.3 1.4 1.5

0.0316

0.9636 0.9855 0.9975 0.9996

1.6

A*y

44y

A*y

-0.0097

0.0219

-0.0099 - 0.0002 0 -0.0099

0.0120 0.0021

0.0002

Let, y0 = 0.9320 and x0 = t .2, Ayo = .0316, A2yo = - .0097 By Newton’s forward difference formula : u(u -1 ) . „ y =f{x) = y0 + «Ayo t — y° + •*“ y

=0.9320+ (0.0316m)+

2

^ ( - 0.0097) [Neglecting higher differences]

=0.0316+^ —-—j ( - 0.0097)

Differentiate, ^

dy For a maximum put — - 0

0.0316 =j^u-|j(0.0097) =>,,=3.76 u

X —Xn = — 7~^

n

3.76 =

x -1 .2 —

-—

.1

=>

jc

=

1.576.

Ans.

(which is near to x„ - 1.6, so using backward difference formula) Tofind y m^ we use backward differenceformula, x - x K 1.576-1.6 u = — r— -------- :------=> u = - 0.24 n .1 JM *

♦ “ V ,. +

X I - 5 7 6 ) - 0 .9 9 9 6 - ( 0 . 2 4

v y . *2 < ! £ ± » * * k0

.0 0 2 1 )t

= 0.99% - 0.000504 + .0009028 = 0.9996 + 0.0003988 = .9999 approx. Vmax- =0.9999.

2

~ ° 2 4 ) ( - 0 .0 0 9 9 )

Ans.

N u m erica l A n a ly sis -!! | 2 6 9

3.27 NUMERICAL INTEGRATION fb f(x)dx can be evaluated exactly, if/(x) is explicitly defined as a mathematical function and if Jo it is integrable. But sometimes, the function y~f{x) is defined by means of (w + I) paired values (xh y/); i = 0, 1,2, 3,...., n. In such situations, we resort to numerical integration, in which we f^ * replace fix) by any convenient interpolating polynomial P{x) and calculate J P{x)dxt that is f approximately equal to I f(x)dx. The process of numerical integration of a function j{x) is Ja called quadrature.

3.28 NEW TON-COTE’S QUADRATURE FORMULA Let y......yn be the values ofy - J{x) corresponding to x = x q , X | , x 2 .......... . x„ which are equally spaced with step-size h. Then, by Newton’s forward interpolation formula. u (u -l) _ u(u - l)(u - 2) , u„ y = yo+ Y jA yo+ -" 2! - A 2y0 + ------- r ;-------A3y0 +. 3!

X-X0 =>x =xf, + uh= ^dx = hdu n

where, Now

-(I)

u - — r—

f*n

fXQ+nh , r*0+nft f» u ydx = y dx = j o y0 + ^j4yo

u (u - l) —A

u(u - l)(u - 2) A3y0 +3!

y0 +uAyQ+

=h

*" “ + T

hdu [By (l)j

(u2 - u) . 2

(u3 -3 u 2 +2u) , - A2y0 +------------ --------------A 3y 0 +

du

+ I ( t ' “3 + 1 )■&3yo

fn 6 24

3n4

lire3

- 3 a 2 A*y0

2 1 f n e 0 . 35n* 50n3 2n9 + ---------------- + 12n2 A5y0 + ....0 ...(2) '120 [ 6 Result (2) is called Newton-Cote i quadratureformula. We can obtain a variety of special quadrature formulas by putting n = 1,2, j ........ as discussed below :

270 | E m m m v w

M a j h o u t ic s -IH

3.29 TRAPEZOIDAL RULE By putting n = ! in Newton’s-Cote’s quadrature fonnula and omitting terms with A^yo and higher order differences, we get

f * y dx = £ [y, + y21......... so o n . J*i it

Similarly, Finally,

...(2)

P ‘‘ y d x = £[y„_i + y„ ]

Jxn

...(3)

a

f^ydbc = P y d x + P y d x + ......+ (*“ y d x Jx0 *cq Jxi = ^ 0 'o + y i ) + |- ( y i + y a ) + - + - |( y ).- i + y J . [Using (1), (2) and (3)] = j [(y0 + y„>* 2<>i + V i+ ....+ y*-i >]

,..(4)

Which is known as general trapezoidal rule. Remarks: (i) It is to be noted that Trapezoidal rule is applicable for any number of intervals or any number of ordinates. (ii) The total truncation error is Trapezoidal formula is

E = - ^ ( x n - x 0) W ( i \ ) where h - **

and/ " ( ? ) =

1^*^» where xo< tj< x„.

i.e., The errors in the trapezoidal rule is the order o f A2.

3.30 SIMPSON’S ONE-THIRD RULE (OR US) R U L E : By putting n = 2 in Newton-Cote’s quadrature fonnula and omitting terms with AJyo and higher order difference. Assuming that y - fix ) is a polynomial of degree two, we get J% d b c

2y0 +2Ay0 + ^ | - 2

A2y0j .

= M2y0 + 2(y, - y 0) + ^ 0*2 - 2 y t +y0) [Since A*yo = (E - l)2^ =>

f x y d x = ^ (y o + ^ y i + y*).

-2 y , +y0] ...(l)

Numerical AnalVsis-II | 271 Similarly, Finally,

J J y dx = ^ [ y 2 + 4y3 + y4] ........so on 3 ------------------J!

**n-t

= -r[5'»-2 + 4yfl. 1+ y R]

"#> ...(3)

a

f^ y c tc = f* y d x + P y d x + ....+ j*" y d x. «2

Jxg

■«»-*

h h h = J (>0 + 4yi + y2) + - (y2 + 4y3 + y4) + .... + - (yju-2 + 4y2j,_1 + y2J p y d x = ^ [ ( y 0 + y 2*) + 4(yl +y3 +... + y2(l_I) + 2(y2 +y4 + .....+ y2n-2)] -(4) Which is known is general Simpson's 1/3 rule. Remarks: Since ~ r *° = h, where loU /" > (? ) -

3 3 1 SIMPSON’S TH R EE -E IG H T H S (3/8) RULE By putting n = 3 in Newton’s - Cote’s quadrature formula and omitting terms with higher order differences, we get

and

9 9 3 = h 3y0 + - (yi - y0) + ^ (y2 - 2yi + y0) + - (y3 - 3y2 + 3yj - y 0) [Since A2y0 = (E - 1^yo and A3yo = (£ - I )3yoJ . p y d * = -^ (y (,+ 3 y , +3y2 + y3) J*o o Similarly Finally

3* f y dx = — (y3 + 3y4 + 3y5 + y6), and so on o f*“ y d x = ^ ( y „ _ 3 + 3y„_2 + 3y„_, + y j J%-s 8

...(1) -.(2) ••■(3)

2 7 2 | rN W frM M M *W «tlitrK *-H l

JXQ

y d x = [** y d x + f*8 y d x +..... + f *" y d x Jxq Jj* J**.3 = — [(y0 + 3y, + S>2 + y3) + 0-8 + 3y4 + 3>5 + y6) + ..... + CK3«—3 + 3>^_2 + tyin- ) +>’3«)] [Using(I), (2) and (3)]

j ^ y d x = ^oK y 0 +J'3.) + 3 (y i+ y 4 + y 7 + --) + 3(y2+>'5+>'8 + ....) Jxo + 2(^3+ y e +y9+ ....+ yjn)] —(4) which is known as general Simpson i 3/8 rule. Remark : To apply this rule, the interval of integration must be divided into a 3-multiple number of sub­ intervals each of width h.

3 3 2 W EDDLE’S RULE By putting n = 6 in Newton’s-Cote’s quadrature formula and omitting terms with d?yo and higher order differences, we get P y d x = I 6y0 + 184y0 + ^ ( 7 2 - 18)A2y0 + ~ (324 - 216 + 36)A3y0 L 2 6

+J- ( H 2 1 _1944 +792- 108 24{

5

6y0 + 18Ay0 + 27A2y0 + 24A3y0 + ^

10

41

,

If we replace

3 10 °’ ^ lus ^

A*y0 + ^ A8y0 + 10

140

A4y0 + -

A6y0

em>r comra‘tte^ will be negligible and using

Akya = (E~ l)*)^ we get I T y (ir = B ’^ ,0+5;>,1+^ +6y3+y4+5;y6 + y6)

-d )

Which is called weddie i rule for n * 6. Similarly Finally

y d x . |~ ( y 6 + 5 ^ + yt +6y9 +y,0 + 5yu +y2) and so on f* ' 1»-»

=* ~ (y»-« + By,.* + yM^ + 6 y ^ + 1U

f ** y d x = y d x + [*** y d x +...... + (*" y d x Jjfc Jxo J*n-«

+ 5y„_, + yB)

...(2) ...(3)

N u m e ric a l A nalysis-1! | 2 7 3

3h - jq [ ^ 0 + y») + 5^ l + ^7 +>ia + —) + 0 2 + /8

+ >')4

+ ••••)

+ 6(^3 +.V9 + y \2 + ....) +0>4 + >’10 + y\6 + •••■) + 5(vs + yu +>>17--) + 20'6 +y n +yw + ••••)] •■•(*) which is known as Weddle's rule. Remark: To apply this rule, the interval of integration must be divided into a 6 -multipie number {i.e., 6, 12. 18, etc.) of sub-intervals each of width h. Example 3.97 :Fhtd £ j + x 7 *** Av Trapezoidal rule, Simpson’s 1/3, Simpson’s 3/8 and Weddle rule, where the interval is divided into 6 equal parts.

11

*

1O'! | »— •

Sincex„ = x0 + nh for » = I, 2, 3 ,.....„ 6, 2 3 X3=6 Xz=6

4 X« = 6

5 *5 =6

x6 =1

yx =0.6923

Vb =.59016

11

X

1 -0 1 Here, we have n = 6 and h = —g— = —•

0* 11 0

Solution.

[RGPV Dec. 2001/

y=/W

1

Vo =1

y, =.97297

y2 = 9

y3 = .8

l +x*

Using Trapezoidal ru le : J o l + x2

= ^ [( y o +>6) + 2 (y ,+ ^ + .... + >5)] 2 = — [(1 + .5) + 2(.97297 + .9 + .8 + .69230 + .59016)] 12

9.41086 = — —— - .78424.

Ans.

Simpson’s 1/3 rule for n ~ 6 :

f1—1— dx = -^[(;y o+y6)+4(yi +y3+y5)+2(3'2+y4)l 3

J o l + x2

= — [(1 + .5) + 4(.97297 + .8 + .59016) + 2(.9 + .69230)] 6x3 .I

14.13712

--------- — ----- - .78539.

Simpson’s 3/8 rule for n = 6 : • i l , 3h f ----- j d x = — [(>'o+ ^ ) + 3 0 ,1+>'2+3'x+3'5) + 2(y3)] J o I + jc2 8

A ns.

2 7 4 | Engineering Mathematics-!!!

= f i [(1 + .5) + 3(.97297 + .9 + .69230 + .59016) + 2(.8)} 8 6

12.56575 = .7853593. 16 Weddle role for n = 6 :

At

1 , 3h Jo 1+** *** = l o ^ 0 +5:Vl + yi + 6y* +y* +5y* + * ] =

+ 5(.97297)+ .9 + 6(.8) + .69230 + 5(.59016) + .5]

= 0.7854

AO!

x r •1 daxx Example 3.98: Compute the value o f xfrom the formula ~ = j+

JO sub-intervals. When the interval (0, 1) is divided into 10 sub-intervals, n = 10 am

o II y0 =1.0000

x5 =0.5 ys =0.8000

x3 =0.3

x4 =0.4

yt =0.9901

y2 =0.9615

y3 =0.9174

y4 =0.8621

x 6 =0.6 y6 =0.7353

yT =0.6711

oo

1 +x

x2 =0.2

n

1

x, =0.1

o

X

y ■

= 0.1, since x„- x0 + nh for n = 1,2........ . 10.

.*

1-0

1 ii

,

h=

!•?

Solution.

x;

using trapezoidal rule wit,

y8 = 0.6098

x9 =0.9 y» =0.5525

x» =i-o yi0 = 0.5000

By Trapezoidal rule fo r n** 10: j ^ I J'd x = -[(y 0 +>-10)+2(yl +y2 +......+y,)] «

i.e.,

*

[(1-0000 + 0.5000) + 2(0.9901 + 0.9615 + 0.9174 + 0.8621

+ 0.8000 + 0.7353 + 0.6711 + 0.6098 + 0.5525)] - 0.78498. ...(I) r* dx Now, exact value j o j — £ = (tour1x)£ - j Hence n * 3.13992. Since die actual value of it is 3.14159 The error =3.13992-3.14159 = -0.00167.

[Using (I)] Ans. Ans.

N u m erical A n a ly sis -11 I 2 7 5

ft Example3.99 '.Evaluate J p

j by using (i) Trapezoidal rule (ii) Simpson’s 1/3 rule (iii)

Simpson *s3/8 rule (iv) Weddle’s rule and compare the result with its actual value (Taking h “ I). [RGPV Dec. 2002] Solution.

6-0 h —1 i.e., h -------- => n —6

Here,

Since x„ = xq + nh for n = 1,2,3....... 6. x:

Xq = 0

x ,= l

*2 = 2

x3 =3

x« = 4

xs = 5

x6 = 6

1 , = J + x*

Vo=l

Vi ~ 0.5

VZ = 0.2

V3= 01

y, = .0588

y5 = .0385

V6 = 027

(i) By Trapezoidal rule, for n =* 6 : r6 dx h = ~ K^o + >e) + 2(yt + y2 + ^3 + Jo 1 +x2 2

+ ya)]

= - [(1 + 0.027) + 2(0.5 + 0.2 + 0.1 + 0.0588 + 0.0385)] 2 = 1.4108. (ii) By Simpson’s 1/3 rule, for n = 6 : C 7— 7 = ~ f(y0 + Ve) + 4(y, + y3 + y6) + 2(y2 + y4)] Jo l + *z 3 - -1(1 + 0.027) + 4(0.5 + 0.1 + 0.0385) + 2(0.2 + 0.0588)] 3 = 1.3662. (iii) By Simpson’s 3/8 rule, for n=> 6 : f6 = — [Oo + y8) + 3(yi +y2 + y i + y6) + 2y3] Jo l + x2 8 = - ((1 + 0.027) + 3(0.5 + 0.2 + 0.0588) + 0.0385 + 2(0.1)] 8

= 1.3571. (iv) By Weddle’s rule, for n = 6 : f 6 -r ——r = |^Lyo + 5 y i+ y2+6y3+y4+5y5 + y6] Jo l +x 2 10 = 0.3(1 + 5(0.5) + 0.2 + 6(0.1) + 0.0588 + 5(0.0385) + 0.027] = 1.3735. Also

f G dx Jo f +~ g ~ [tan-1 x]3 = tan- 16 = 1.4056 which is exact value.

This shows that the value of the integral found by Weddle's. rule is the nearest to the actual value. Ans.

2 7 6 | E n g in e e rin g Mathematics-111

x l ' *e~*dx, approximately by using a suitable formula.

Example 3.100:Evaluate

{RGPV June 2003 and Dec. 2006, Solution.

Suppose, w e divide the interval (0.5, 0.7) into four equal parts by taking h =

: 4

0.05. N ow th e values o f given function y - x iae~x.

X

y ~ x l1 e~*

x0 = .50

y0 - 0.4288819

x , = x 0 + h - .55

Vl = 0.4278774

x 2 = x 0 + 2h = .60

y2 = 0.4251076

x 3 = x 0 + 3h = .65

y3 = 0.4208867

x 4 = x 0 + 4 h = .70

y4 * 0.4154730

By Simpson's ‘1/3’ rule, for n - 4 : f 07 x 1'2e~xdx = ~r[(y0 + y 4) + 4(yj + y3) + 2y2 ] J 0.5 d =

[(0.4288819 + .415473) + 4(.4278774 + .4208867) + 2(.425!076)|

or

f V v ^ d x = ^ [ 5 .0 8 9 6 2 6 5 ] =0,084271. Am J 0.5 3 Example 3.101: Find an approximate value o f log 5 by calculating to 4 decimalplaces, by Simpson's i/1 e5 dx rule, J o - , dividing the range into 10 equal parts. /RGPV June 2004, Dec. 2008(N)t Solution.

5 -0 Here range of integration (0, 5) is divided into 10 equal parts, i.e.. h = ------= 0.5. Giver 10 function y = 77 — - , then (4* + 5) X

0 11

*

x, = x0 +fc = 0 .5 x 2 = x 0 + 2h = 1 X3 - X q + 3 /l = 1.5 x 4 = x0 + 4h = 2 X5 = Xq + 5 ft = 2.5 Xg = Xq + 6h = 3 := Xq + 7ft = 3 .5 x8 = x0 + 8/1 = 4 x9 =X q + 9ft = 4 .5 X10 = X0 + 10ft = 5

1 4x + 5 y0 = 0.20000 y x = 0 .1 4 2 8 6 y2 = 0 .11111 > 3 = 0 .0 9 0 9 1 y 4 = 0 .0 7 6 9 2 yB - 0 06667 y 6 ~ 0 .0 5 8 8 2 y 7 = 0 .0 5 2 6 3 y 8 = 0 .0 4 7 6 2 y9 = 0 .0 4 3 4 8 y 10 = 0 .0 4 0 0 0 v -

N um erical A n a i y sis -H | 2 7 7

By Simpson’s 1/3 rule, for n = 10: f *10 fl h y d x = —[<3^0 ^xo) + 4(^i +^3 + .... + yg) + 2(y2 + y < + - + 3'8)J *xq A 0.5 3 dx = ~ ~ [2.41515] = 0.40253. j,o 4x+5 ~ 3 Again actual value of the integral is

or

5 = I [log(4x + 5)JS = ~ [log 25 - log 5] = i [log 5* - log 5] J,o 4* + 5 4 4 4 = \ [2 log 5 - log 5] = i [log 5] 4 4 1 0.40253 = - log, 5 4

=>

[Using (1)] Ans.

log, 5 = 1.61010.

Example 3.102:Evaluate

J c o s 0 dd, by dividing the interval into 6 parts. Using Simpson’s

1/3 rule and Weddle ’s rule. Solution.

(RGPV Dec. 2005 and June 2011/

We use Simpson’s 1/3 rule to evaluate the integral, h = ^

7i/2 - 0

6

n 12

y d x = j [ ( ? 0 + f t) + 4 ( ? l+ ? s * f t ) + 2Cfe+y4)] 2C= 0 o II o X

\ x'= V l2

y = 4 cosQ Vo =1 Vl = .98281

X2 =%

y2 = .93060

*3 = %

y3 = .84089 y4 = .70710

x* = V l2

ys = 50874 y& = o

Hence (I) becomes :

f

*/2 VomG Jo

dQ=— .— [(1 +0) +4(.98281 +.84089 +.50874) +2(.93060 +.70710)] 3 12 = ^ x 13.60516 = 1.1873. 36

A n s.

2 7 8 | E n g in e e r in g M ath em atics -111

From Weddle jr rule for n - 6 : ■ \y d x = — [>0 + 5}', +y2 + 6y y +y4 + 5ys + y 6]

= A x JLrj + 5 x 0.98281 + 0.9306 + 6 x 0.84089 + 0.7271 10

12

+ 5 x0.50874 +!

Example 3.103 :Evaluate

= —-x 15.14079 = 1.1892 Ai 40 >5.2 loge x d x by (l)Simpson’s 1/3 rule, (ii Simpson’s 3/S rule. Af

finding the true value o f the integral, compare the errors in the both cases.

. 5.2-4 . . Suppose, we divide the interval (4, 5.2) into six equal parts by taking * = — “— = U.2. Now the values of given function y = log, x are as follows :

L

y - log. x y0 = 1.3862944 yj = 1.4350845

1 1 1

X o* II 45k ©

Solution.

Xj = x0 + h = 4.2 x2 = x0 + 2h = 4.4 x3 = x0 +3h = 4.6 x4 = x0 + 4h = 4.8 x5 = x0 + 5h = 5.0 Xe = Xq +6h —5.2

y2 =1.4816045 y3 =1.5260563 y4 =1.5686159 y5 =1.6094379 y6 =1.6486586

(i) By Simpson's 1/3 rule, for i t a 6 : L y d x = -j[(yo+^6) + 4(>, +y3 +ys) + 2(y2 +y4)] J * 2 log x dx = ^ 1 2 7 .4 1 7 7 0 6 ] = 1.827847.

Ans.

(ii) Simpson’s 3/8 rule, for » = 6 : f ** y d x = ^ • [ ( y 0 + ^6) + 3(y1+y2 +>'4+ 3’5)-,-2(y3)] a

jjoq

f 5 log x dx =

J4

o

[24.371294] = 1.827847.

44

Ans.

N um erical A n a ly sis -11 | 2 7 9

Now exact value: f 6.2 <<1.2 log x dx = I log x. Idx = [x(log x - 1)152 = [5.2 (log, 5.2 - l ) “ 4(loge 4 -1 )] = 1.827847. Hence the errors are : (i) Due to Simpson’s 1/3 rule = 0.000000. (ii) Due to Simpson’s 3/8 rule = 0.000000. Ans. r * dx Example 3.104 :Find the value o f J f — by Simpson fcrule. Hence obtain approximate value o f loge 2. (RGPV June 2006) Solution.

Suppose, we divide the interval (1,2) into 6 equal parts. l

2-1

1

Taking, A = —— = —• Since x„ = x0 + nh, for n = 1,2, 3 , 6 . X

ii~ H x

Xq = 1

Vo = 1 Vi = 6 /7 j?2 = 6 /8

7 /6 x2 = 8 /6 x3 = 9 /6 x4 = 1 0 /6 x5 = 1 1 /6 Xy =

x6

Vz = 6 /9 y4 = 6/10 ys = 6 / 1 1

=2

* 6 = 1 /2

By Simpson’s 1/3 rule for « = 6 : We have P y d * = -7[Cyo+^6)+4(3'i+>'3+y8)+2(>2+y4)]

or

r^ = ± r r 1+n +4f 6 + 6 + M +2r 6 + o i Ji x 18 \ \ 2) U 9 llj U 10J _L 3 18 2

(1434) 693

(108) 80

1 |~3 5736 | 216~ _ 18 L 2 + 693 + 80 _

= — (12.477056) =0.693169. lo Now,

rr22 dx Jj — = [log, x]? * log, 2 - log, 1 = log, 2 x log, 2 = ^593169. [by using (I)]

Ans. [v log, 1=0] Ans.

2 8 0 | E n g in e e rin g M a it c m a t ic S 'III

Example 3.105 :A solid o f revolution isformed by rotating about the X-axis, the area between the Xaxis, the line x - 0 and x 1 and a curve through the points with the following co­ ordinates : \ X

y Solution.

0,00 1.00

0.2S 0.9896

0.50 0.9589

0.75 0.9089

1.00 0.8415

Find the volume o f the solidformed, using Simpson *s rule. Here, h = 0.25 and n = 4 (i.e., 5 ordinates are given) Required volume V of the solid generated is given by h V = f 7iy 2dx = Jt —[ ( ^ + y |) + 4(>f+>!) + 2>’| ] ■0 i) [ v By simpson’s rule for n = 4]

0.25x3.141

[{1 + (0.8415)*} + 4{(0.9896)2 +(0.9039)2} +2(0.9589)2J

= 0 . 2 6 I 8 [ 1 . 7 0 8 1 + 7 . 2 2 1 6 + 1 .8 3 9 0 ] = 0 . 2 6 1 8 * 1 0 . 7 6 8 7 = 2 . 8 1 9 2 . Ans. Example 3.106 :The speed, v meters per second o f a car, t seconds after it starts, is shown in the following table :

t : 0 12 24 36 48 60 72 84 96 108 120 v : 0 8.60 10.08 18.90 21.60 18.54 10.26 5.40 4.50 5.40 9.00 Solution.

Using Simpson's rule, fin d the distance travelled by the car in 2 minutes. If s meters is the distance covered in t seconds, then Since,

ds as

- v => [sjjfg =

cim

vd t

[y / = 2 min. = 120 sec.]

Here 11 ordinates are given i.e., n = 10. Using Simpson’s 1/3 rule : (Here h = 12) i.e.,

s = ^ I(t»o + t»j0) + 4(11! + v3 + v6 + v7 + u9) + 2(v2 + v4 + l>6 + u8)] 12

= — {(0 + 9.00) +4(3.60+ 18.90+ 18.54+ 5.40+ 5.40) , 3 + 2(10.08 + 21.60 + 10.26 + 4.50)] = 1236.96 meters. Ans. Example 3.107: Thefollowing data give the velocity v o f a particle at time t : t(sec): v ( m / sec):

0 4

2 6

4 16

6 34

8 60

10 94

12 136

Find the distance moved by the particle in 12 seconds and also the accerelation at t =2 seconds. fRGPV, Feb. 2016J

N um erical A nalysis -!! | 2 8 1

Solution,

(i) Let S’= distance 12

dS dt " v

i.e. distance

s= jv d t

12 S = jv d t o

... ( 1)

Here 7 ordinates ae given i.e., number of intervals are n = 6. .

b-a

1 2 -0

„

Using Simpson’s 1/3 rule : I Here h = ——- = —- — = 2. From the table, we have: v0 = 4, Vj = 6, v2 = 16, v3 = 34, v4 = 60, vs = 94, v* = 136. f

h

S = J vdt = j [ ( v 0 + v6) + 4(vj + v3 + v5) + 2(v2 + v4)]

o = y [(4 +136) + 4(6 + 34 + 94) + 2(16 + 60)] 1656 _ ____ _

meters.

Ans.

We know that Velocity v Accerelatin a = —------- = — time t

... (2) K

Here at time t = 2 sec. die velocity is 6 m/sec. from the table, hence (1) becomes :

6 Accerelation = — = 3 m/sec2.

Ans.

Example 3.108 :Estimate the length o f the arc o f the curve 2y = x 2from (0, 0) to (1,1/3), using Simpson *s 1/3 rule taking 8 sub intervals. Solution.

The length of arc is given by curve 2y~x?-=$y = x2/2. Differentiate,

dy — =* dx

We know that length of die given are is S -

f ^jl + (d y /d x )2dx Jx^a

=

f ^ l + x 2dx J0

[v

x = 0/ox=

1]

282 | Engineering Mathematics-!II

The values offix) = J \ +x2 since (0, 1) divided into 8 sub-intervals : x: o

ii

X Xj

y = Vl + x*

y0 =i

= .125

y> =1.008 y2 = 1.031

x2 = .25 x3 = .375 x4 = .5 x5 = .625

y3 = 1068

ii

**?

y4 =1.118 y5 = 1.179 y6 = 1-25 y7 = 1.329 y8 = 1.414

Xy =.875 x8 =1

Here, h » - £ - = > h =.125 and« = 8. By using Simpson's rule : we get,

[ 1-Ji+x2dx = ^-[(>0 +>8> + 4Cyi + Js +^5 +yy) + 2(y3 +y4 +y6)] Jo 3 =—

((1 + 1.414) + 4(1.008 + 1.068 + 1.179 + 1.329) + 2(1.03 J + 1.118+1.25)]

= 1.14783. Ans. Example 3.109 :A reserviour discharging water through sluices at a depth h below the water surface has a surface area A for various values o f h as given below :

10 950

h ( ft) : A ( ttq.fi) :

11 1070

12 1200

13 1350

l f t denotes time In minutes, the rate o f fa ll o f the surface is given by

14 1530 dh dt

-4 8 jh A

Estimate the time taken fo r the water level to fa ll from 14 to 10ft. above the sluices. Solution.

dh dt

Given that

-rr *

-448-Jh RJh — — r - ^ d ,' ^

dh C

|

-

■ C

- A

A j! ** h dh

N u m e r ic a l A n a ly s k -M j 2 8 3

i* ■ £ h * w * = / “ / ( A ) * w h e re/* )h : A:

12 1200

yo= - 6 . 2 5 9

yt= -6 .7 2

y 1 = - 7 .2 1 7

13 1350

>>

11 1070

CO t> 1 II

10 960

...(I) 14 1530

y i- - 8 .5 2

Here, n =4 i.e., 5 ordinates are given, Using Simpson’s 1/3 rule, fo r « - 4 :

f U f(h )d h = ~ [(>0 + J j + 4(ji + Js) + 23-2] J 10 3 or

(<)}$ = |[(-6 i2 6 9 + (-8.52)) + 4 (-6 .7 2 -7.8) + 2(-7.217)]

= -[-87.293] 3 = - 29.09 min. [- ve sign indicate the fall of water level] /. The time of water level to fall from 14 to 10ft = 29.00 min. (approx.). Ans. Example 3.110 :A river is 80 meters wide. The depth d (In meters) o f the river at a distance xfrom the bank is given by the following table: x: 0 10 20 30 40 50 60 70 80 d: 0 4 7 9 12 15 14 8 3 Find approximately the area o f cross-section o f the river.\ Solution.

, 8 0 -0 By Simpson’s ‘1/3’ rule, here 9 ordinates are given i.e., n = 8 and » = —g— = 10. /.

Area = P f(x)dx Jio »80 h = I y d x = - [(y0 + ys) + 4(>1 + y3 + y8 + » + 2(y^+y 4 + y6)] Jo 3 = ~ [ ( 0 + 3) + 4(4 + 9 + 15 + 8) + 2(7 + 12 + 14)] u

= — [213] = 710 square meter, t)

[Herej = d, h = 10] Ans.

2 8 4 { Engineering Mathematics-111

Example 3.111 '.Calculate by Simpson’s rule an approximate value o f

x 4d x by taking seven

equidistant ordinates. Compare it with the exact value and the value obtained by using the Trapezoidal rule. Solution.

,

3-(-3)

.

Here the given interval (-3, 3) is subdivided into six equal parts, taking «•------ ~— - L

x0 = -3 x 1 =x0 + h = -2 Xj = x0 + 2h = - 1 x 3 = x0 +3h=0 x4 =x0 +4h =1 x5 = x0 +5h=2 Xg = x0 + 6h=3 1

* X 1! a.

x:

Vo = 81 Vi =16 Vz =1 y3 =0 J/4 =1

Vs =16 V6 -81

(i) By Simpson's 1/3 rule, fo r n - 6 : P y d x * ■^[(3'o+y6) + 4(yi+>3+>’5) + 2(>2+>4)I •>*0 o

•

f ^ d x - i [ 2 9 4 ] . 98. -3 3 (ii) The exact value :

...(1) Ans.

...(2) Ans. (iii) By Trapezoidal rulefo r n - 6 : i * y d x = f[^ > ,o +y6)+2Cy1+>-2 + y3+y4+y5)] fx * d x - -i[230] = U 5 . -3

...(3) Ans.

^

From (I), (2) and (3) this shows that Simpson’s rule gives the value which is much nearer to the exact value. Ans. fx/i Example 3.112 .’Calculate | e* *dx correct to four decimal places, by takingfour ordinates. •0 Solution. Divide the range (0, n t 2) into three equal parts because given 4 ordinates. n/2-0

v Taking

it

h --------------* - -

N u m e r ic a l A n a ly sis - 0 | 2 8 5

v By Simpson *s 3/8 fo r n = 3.

r*'2gsinxfa • , =— 3h ,[(yQ+y3) + 3(yx + y2)] f* Jo

8

= - , - [(1 + 2.71828) + 3(1.64872 + 2.3774)) 8 6 =

[3.71828+ 12.07836]

= ^[1 5 .7 9 6 6 4 ] = 3.1029. Example 3.113 :Find

J

Ans.

e ^ d x , by taking seven ordinates using simpson’s 1/3 rule. [RGPV June 2909]

Given seven ordinates i.e., number of sub-intervals n = 6. x6 - *o 0.6-0 „ Then h = — g - = — t — = 0.1 b

, b

Xj = Xq + h - 0.1 x2 = x0 + 2h - 0.2 Xg = 0.3 x4 = 0.4 x5 « 0.5 x6 = 0.6

y, » e~i = e-° 01 = 0.9900 y2 = e-OM = 0.9608 y3 = «-«•» = 0.9139

ii

1! O

1

y - f(x) = e *2 %

X

ft)1 ok II >— *

Solution.

y4 = e -°16 = 0.8521 y5 = €-° 25 = 0.7788 y6 =
By Simpson *s 1/3 rule for n = 6 : 10° V -^ d x = ^ [(y 0 + ye) +

+ M) + 2(^2 + ^ ) ]

= — [(1 + 0.6977) + 4(0.9900 + 0.9139 + 0.7788) + 2(0.9608 + 0.8521)] 3 = ~

[1.6977 + 10.7308 + 3.6258]

= 0.5351. 2 f „-x/2

Ans. 1

Example 3.114 : Evalute: J e dx, by Simpson’s 1/3 rule, taking four intervals. i Solution. Here n = 4, x0 = 1, *4 = 2.

2 8 6 | Engineering Mathematics-IU

X4 Xn 2“ I I h = - ^ = — -^=0.25 xt =x0 + h= 1.25, x2=jto + 2h= 1.5, x3 =x0 + 3h = 1.75 X

y -f(x )-e -* i

Xq ~ I Xj = 1.25 x2 = 1.5 xj - 1,75 x4 = 2.0

y0 = 0.60653 y, = 0.53526 y2 = 0.47237 yj = 0.41686 y A= 0.36788

By Simpson’s 1/3 rule for n = 4, we have 7

J ydx = -A[ 0 'o+>'4) +4(>'1+>'3) +2>2] *0

f e-*/2 dx = — [(0.60653 + 036788) + 4(053526 + 0.41686) + 2(0.47237)] . 3 = 0.4773. Ans.

r* s in x , Example 3.115:Using Simpsoni 1/3 rule, fin d J -------by dividing (0, x) in 6 equal parts. JC

Solution.

Here, Sub-intervals is n = 6, and h = ainx

x:

x0 =0 Xj =f t 1 6

x2 =2*76 x3 =3*76 x4 = 4*/6 x5 = 5 * /6 .
v - — X

Vo =1 j>! =3 In y2 =2.59009 in y3 = 2 /* y4 =1.29904/* ys =0.6/* Ve =0

Using Simpson’s 1/3 rule, fo r n *=6 : C -^ d x X

= ^ [(y0 + y6) + 4(yj + y3 + o

) + 2(y2 +

)]

N u m e r ic a l A n a ly s is - !! J 2 8 7

7t f , 22.4 7.7426'I = — 1 + ------+ --------18 ^ Jl It ; = 1.852. fK H Example 3.116:Calculate the approximate value o f s in x d x

Solution.

Ans.

(i) by Trapezpidal rule (ii) by Simpson*5 rufa using 11 ordinates. Here given 11 ordinates i.e. number of sub-intervals is n = 10 , w '2 -0 h = ——— = n! 20.

.*. Taking X

y <=atnx

o If X

Vo =0 Vi =0.1564 y2 =0.3090 y3 =0.4540 y4 =0.5878 y5 =0.7071 y6 =0.8090 y7 =0.8910 yt =0.9511 y9 =0.9877 =1.0000

Xj = 7t/2 0 x z =*710 x3 =3?r/20 x 4 = *75 x5 = n ! 4 x6 =3*710 x7 =7*720 x 8 =2*75 x9 =9*720 x 10 = *72

(i) Using the Trapezoidal rule, fo r n *» 10:

«**/ 2

Jo

h

sin xd x

= -[(y o + y io ) +2 (yt+ y2 + .....+*»)] = ~ [ ( 0 + l) + 2(0.1664 + 0.3090 + 0.4640 + 0.6878 + 0.7071

= 0.9981. (ii) Using the Simpson’s rule, fo r n m 10: *>2

h

+ 0.8090 + 0.8910 + 0.9511 + 0.9877)] , Ans.

2 8 8 | E n g in eer in g M athem atics -III

n [(0 + 1) + 4(0.1564 + 0.4540 + 0.7071 + 0.8910 + 0.9877) 60 + 2(0.3090 + 0.5878 + 0.8090 + 0.95 II)} 1.0006. Abs. f> l .1.4 4

Example 3.I I 7 .-Evaluate Solution.

( s in x ~ lo g t x ^ e x )d x approximately using Weddle’s rule correct

to 4 decimals. (Taking h * 0.1). Let J[x) = sin x - log,.* + e*. h =

S ince,

1 .4 -.2 n

1 .4 -.2 n * = 12

.1 =

sin x - l o g ' * + e x y0 =3.0295 Vi =2.8494 y2 =2.7975 y3 =2.8213 y4 =2.8976 y5 =3.0147 y6 =3.1661 y7 =3.3483 ys =3.5598 y9 =3.8001 y10 =4.0698 yn =4.3705 yl2 =4.7042 y =

X

=0.2 Xi =0.3 x2 =0.4 x3 =0.5 x4 =0.6 x5 =0.7 x6 =0.8 x7 =0.9 x8 =1.0 x9 =1.1 Xq

x10 = 1-2 X„ =1.3 xI2 =1.4

Using by Weddle’s fifle, for n =* 12 : f ' f(v )d x = ~ b 0 +5>i +yt +6ys +y4 +5y8 +2>6 + 5y7 + y8 +6y9 J0.2

10

+^io + 5^n +yu] = ^ (0.1)[3.0295 +14.2470 + 2.79J6 + 16.927#;*B.8976 +15.0735 + 6.3322 + 16.7415 + £.55$ + 22.8006 + 4.069*4; ?1.8525 + 4.7042] - (0.03) [ 135.0335] = 4.051, ’ \ " Am.

N um erical A n a ly sis -1 I J 2 8 9 f u it

Example 3. 118 :Using Simpson’s rule, evaluate I * ff Solution.

7 1 / 2 -0

71

-------------------

s i + 3 tin * 9 d 9 , taking n = 6. ,--------------------------

We have, h = — - — = — and y = fl,0 ) = >/l + 3 sin 2 0

Let

x ~ 19

y = J l + 3 sin * e

ii

o

o X

y0 =1

X! = x / 1 2

y, =1.095

x 2 =2^/12

y2 =1.323

x3 = 3/77 12

y3 =1.581

x4 =4>r / 12

y4 =1.803

x 5 =5^/12

> 5= 1-949

x6

-n il

ye

=2

Using Simpson’s 1/3 rule, for « " 6 : f* 2>/l + 3 sin 2 0 dQ = ~ [(y 0 + _)'6) + 4(.y1 +y3 +yi) +2(y2 +>4)] Jo o n = — [(1 + 2) + 4(1.095 +1 .58 1 +1.949) + 2(1.323 +1.803)] 36 = ^ ( 3 + 18.5 + 6.252) =2.422. 36 | Example 3.119 :The follow ing table gives the values o f a function at equal intervals : x: fix ):

0.0 0.3989

0.5 0.3521

2.0 0.0540

1.5 0.129S

1.0 0.2426

A ds.

Evaluate (i)f(1.8) (U)f f l . S ) (IB) J * f ( x ) d x , sta tin g me formula used. fRGPV June 2005}

on.

The difference table is : X

0.0 0.5 1.0 1.5 2.0

y = f(x) 0.3989 0.3521 0.2420 0.1295 0.0540

Ay

A’y

A*y

A*y

-0 .0 4 6 8 -0.1101 - 0.1125 - 0.0755

-0 .0 6 3 3 -.0 0 2 4 0.087

0.0609 0.0394

- 0:0215

290 |

E n g in eer in g M athem atics -U I

(i) Tofin d ffl. 8) : At x = 1.8, so using Newton’s backward interpolation formula _ u(u +1) Ax) = yn +uVy« + - 2!gT 7

u(u + l)(u + 2) „ + — 3!- ------ v

+•

* -* n 1 .8 -2 .0 u = — ;— = — ~ — - - 0.4 0.5

here

A 1.8) = 0.0540 + (-.4)(- 0.0755) + ,

A

(0.037)

(.0394) + ^M X q -g g -6 )g L 6 ) 24

6

0m ^

or / 1 . 8) = 0.078133. Am (ii) Tofindf'(1.5): Taking** = 1.5 and using derivative of Newton’s backward interpolate formula: l - 0.1125+- ( - 0.0024)+- (0.06091) 2 3 0.5 Am

= -0 .1 8 5 4

(iii) Given : y 0 = 0.3989, y x = 0.3521,^ * 0.2420,y3 = 0.1295, .>>4= 0.540 Using Simpson 3r 1/3 rule for n = 4. C

=

+y*^ +4(>l +

+ 20^2)]

f 2f{x) dx - — [(0.3989 + 0.0540) + 4(0.3521 + 0.1295) + 2(0.242)1 Jo 3 0.5

(2.8633) =0.4772.

Ass

Exercise-3(A) 1. Given «o + u&= 1.9243, u\ + «7 = 1.959, U2 + «6 = 1.9823,1/3 + u$ = 1.9956, find 1/4. 2. if «o = 3, u\ = 12, U2 = 8 1, V} = 2000, U4 - 100, calculate A4ho. 3. Form a table of differences the function/*) = x3 + 5x - 7 for x = -1 ,0 ,1 ,2 ,3 ,4 , 5 continue du table to obtain/6). 4. Find the missing values in the following table : x: 45 50 55 60 65 y: 3.0 — 2.0 — -2 .4

N u m e r ic a l A nalysis -11 j 2 9 1

5. Obtain the missing terms in the following table : x: 2.0 2.1 2.2 2.3 Ax): 0.135 — 0.111 0.1

2.4 —

2.5 0.08

2.6 0.07

6. Express 3x* - 4jc3 + 6x2 + 2x + 1 as a factorial polynomial and find differences of all orders.

7. Obtain the function whose first differences is 2x3 + 3X2 - 5x + 4. 8. Locate and correct by means of differences the error in the following table of values. x: 1 2 3 4 5 6 7 8 y: 3010 3424 3802 4105 4472 4771 5051 5315 9. Find the function whose first difference is e*. 10. Given log 100 = 2, log 101 = 2.0043, log 103 - 2.0128. log 104 = 2.0170, find log 102. 11. What is the lowest degree possible for a polynomial ux which takes the following values : u\ = 0, u2 = 3, «3 = 8, u4 = 15, u$ ~ 24, u6 - 35, also find the value of ux. 12. The following table is given. Find ux x: 0 I 2 y: 3 6 11

3 18

4 27

If for another point x = 5, ux - 38 added to above data, will the function be a same as before or different. 13. Prove that: ( 0 /4 ) = /3 ) + A/^2) + A ftl) + A ftl) (ii)X4) =A®)+ 4A/(0) + 6A2A~ 1) + 10A3^ - I) as for as third differences. 14. Evaluate (£2 - 3E + 2) (2x/h + x). 15. Estimate the production of cotton in the year 1935 from the given data: Year*: 1931 1932 1933 1934 1935 Production f(x) 17.1 13 14 9.6 (in millions) 16. Given u ^ = 0.2884, 17. Prove that:

= 0.5856, U49 = 0.6313,1/50 = 0.7620, find the value of U47.

(i) E '" = M+ l s (ii) fxS = ~(A + V) (Hi) l + y = Vl + ^ V 2 (iv) tf(«* + £ -! « )= ft2£)2 _ h zDa + _L h*D* - ...... 18. Prove that: W4 = u0 + 4A«o + 6 A2u. 1 + I0A3u_ |. 19. Evaluate : (i) (E~ 1 A)r3;

1936 12.4

Takingh = \

1937 18.2

2 9 2 | E n g in e e r in g M ath em atics -111

A2*2 (ii) E (x + log x) ’

Taking h

e* e* + e~

(iii)

-

1

Taking h - I

(iv) A cos (a*);

Taking h = 1

Answers-3(A) . u4 = 0.9999557

2. - 7459

3. 239 6.

y

5. J[2.1) = 0.123,X2.4) = 0.09

4. ;-5o = 2.925, yM = 0.225

1 7. - *[4) + 3*t31 +

= 3(.v ]4 + 14 [ x ] 3 + 15M2 + l[x) + I; k * y - 72

8.4151

9. f(x) = - r —.e* -1

+c

10. 2.0086

II. x2- I

12. No change and equal to 2

14. - h

15. 6.6

16.0.4147

(ii) *!

eh _ g-A (iv) 2 sin

(iii)

5

tc ' s

' a x(l + a /,) >

*

'crr ( l - a h)'|

I

to

19. (i)3x2- 3 x + 1,

2

im n fT T F S Exercise-3(B)

I. The area A of a circle of diameter d is given for the following values d: A:

80 5026

90 6362

85 5674

95 7088

100 7854

Calculate the area of a circle of diameter 105. 2. Given: x: fix ):

1 1

2 8

3 27

4 64

5 125

Find/7.5) using Newton’s Backward difference formula.

6 216

7 343

8 512

J

N u m e r ic a l A n a ly s is - II | 2 9 3

3. From ihe following table of half yearly premium for policies maturing at different ages, estimate the premium for police maturing at age 46 and 63. Age: P rem ium :

45 114.04

50 96.16

55 83.32

60 74.48

65 68.48

4. From the following table of values of x anAfix), determine (i)X0.23)(ii)A0.29) x: fix ):

0.20 1.6596

0.22 1.6698

5. The probability integral /(■*) = x: f(x):

1.00 0.682689

1.05 0.7062S2

0.24 1.6804

0.26 1.6912

j

0.28 1.7024

0.30 1.7139

e 'z ^ d x has following values ; 1.10 Q.728668

1.15 0.749856

1.20 0.769861

1.25 I 0.78S700 |

Calculate X I.235). 6. Find the number of men getting wages between Rs. 10 and Rs. 15 from following table : Wages (in R s.): Frequency:

1 0 -2 0 30

0 -1 0 9

3 0 -4 0 42

2 0 -3 0 35

7. In an examination, the number of students who obtained marks between certain limits are as follows : 0 -1 9 41

M a rks: No. of students:

2 0 -3 9 62

4 0 -5 9 • ' 65

6 0 -7 9 50

8* - 9 9 17

Estimate the number of candidates who obtained fewer than 70 marks. 8. Using Newton backward differences formula to the given data, find a polynomial J(x); x: £

ti

1 1

2 -1

4 -1

3 1

5 "l

9. From the following table, find y, when x = 1.84 and 2.4 by Newton’s interpolation formula : 1.7 5.474

II U

*:

1.8 6.050

1.9 6.686

2.0 7.389

2.1 8.166

2.2 9.025

2.3 9.974

10. Find the polynomialX*)» if x: y =fix ) :

3 6

5 24

7 58

9 108

11 174

|

2 9 4 | E n g in e e r in g M a th e m a tic s -] II

11. The following table gives the values of e* for certain equidistant values of x. Find the value of
1.0 0.0

y-

1.4 - 0.016

1.2 - 0.112

1.6 0.336

1.8 0.992

2.0 2.0

0.6 6.896

0.7 8.013

13. Find ><0.543) from the following values of x and y x: y ix):

0.1 2.631

0.2 3.328

0.3 4.097

0.4 4.944

0.5 5.875

14. Evaluate sin (0.197) from the data given below : x: sin ,v:

0.15 0.14944

0.17 0.16918

0.19 0.18886

0.21 0.20846

0.23 0.22798

15. Interpolate by means of Gauss’s backward formula, the polulation of a town for the year 1974, given that: Year: Population (in thousands)

1939

1949

1959

1969

1979

1989

12

15

20

27

39

52

16. Using Gauss’s backward formula estimate the number of persons earning wages between Rs. 60 and Rs. 70 from the following data : Wages (R s.) : No. o f persona: (in thousands)

Below 40

4 0 -6 0

6 0 -8 0

8 0 -1 0 0

100-120

250

120

100

70

50

Answers-3(B) 1. 8666 (ii) 1.7081

2. 421.875 5. 0.783172

3. (i) 110.52567 6. 15

8*fix ) = \ x i - 8x3 + - ^ x 2 - 56x + 31. 1 0 ./ x) = 2t2- 7 x + 9. 12. 0.046 13. 6.303

II. (i) 1.904082

14. 0.19573

(ii) 70.585152 7. 194

4. (i) 1.6751

9. 6.36, and 11.02 (ii) 1.904082 32.345 (thousands)

15.

(iii) 1.904082. 54000.

16.

N u m e r ic a l A n a ly s is -U | 2 9 5

M

M

r mi M j r n x i *

U r m f r I * 5* *

Exercise-3(C) 1. Using the following table, find^x) as a polynomial in x : x: -I 0 3 6 7 Ax): 3 -6 39 822 1611 2. If I) = —3,>{3) = 9,j{4) = 30, and><6) = 132, find the Lagrange’s interpolation polynomial that take the same values as the function^ at the given points. 3. Given the table of values *: 150 152 154 156 12.247 4.

5.

6.

7. .8.

12.329

12.410

12.490

Evaluate Vl55 using Lagrange’s interpolation fonnula. Applying Lagrange’s formula, find a cubic polynomial which approximates the following data : x: -2 - 1 2 3 .yCO: - 12 -8 3 5 Find/(5X by Lagrange’s fonnula for the following data : x: 1 3 4 6 10 y(x): 0 18 48 480 900 The following table gives the normal weights of babies during the first eight months of life : 4ge in months 0 2 5 8 Weight in lbs: 6 10 12 16 Estimate the weight of the baby at the age of seven months : Find by Lagrange's formula the value of tan 33°, if tan 30“ = 0.5774. tan 32° = 0.6249, tan 35° = 0.7002 and tan 38° = 0.7813. Determine by using Lagrange’s formula the percentage number of criminals under 35 years : Age (Year): Under 25 Under 30 Under 40 Under 50 % number o f criminals: 52.0 67.3 84.1 94.4

x2+ 6x -l 9. Express the function (x2 - i)(x - 4)(x - 6) ** a sum

fract>ons-

1#. The following table gives the viscosity of an oil as a function temperature. Use Lagrange’s formula to find viscosity of oil at a temperature of 140°. Temp0 : 110 130 160 190 Viscosity: 10.8 8.1 5.5 4.8 11. Certain corresponding values of x and logio x are given below : x: 300 304 305 307 logio x : 2.4771 2.4829 2.4843 2.4871 Find logio 310 by (i) Newton’s divided difference formula (ii) Lagrange’s formula. GivenXO) = - 18,^1) = 0 , # ' ^ 0,^5) = - 248,/ 6 ) = 0,^9) = 13104, find/x).

2 9 6 | E n g in e e r in g M ath em atics -111

13. The following table gives the values of x and y : jc: 1.2 2.1 2.8 4.1 4.9 6.2 y : A2 6.8 9.8 13.4 15.5 19.6 Find the value ofx corresponding to 12 using Lagrange’s technique of inverse interpolate 14. Obtain the value of t when A = 85 from the following table using Lagrange’s method 14 t: 2 5 8 68.7 A: 94.8 87.9 81.3 15. Find the value of x when y =fix) = 19, given that x: 0 1 2 y=Ax)-0 1 20 16. Find the value of x when y - 18600, given that 56 x: 52 53 54 55 IS 182 y : 19231 18868 18519 1-855 17. Find the value o fx when y -fix) - 6, given that 63 x: 168 120 72 10 y-A x): 3 7 9

Answers-3(C) l.y (x )= x 4 - 3 x 3 + 5x2 - 6

2. x3- 3.*2 + 5x - 6

1 , 3 , 241 _ 4 . ----- x 3 ------ x 2 + ------x - 3.9 15 20 60

5. 160

8. 77.4

1 5 ( x - l)

35(x +1)

13 10(x- 4)

(ii) 2.4786 12. *s - 9x* + I&c3 - x 2 + 9r - I'i 16. 53.7646 17. 147.

3. 12.45 6. 14 lbs

71 70(.r -6 ) 13. 3.55

7. 0.6494

10.7.03 14. 6.5928

11. (i) 2.4786 15. 2.9

Exercise-3(D) >

1. Given 3.8 4.0 x: 3.0 3.2 3.4 3.6 6.672 14.000 y ~AX) : -14.000 -10.032 - 5.296 0.256 Find first and second derivative o f/x ) = at x - 3.0, 2. Given 7.52 7.51 x: 7.47 7.48 7.49 7.50 .206 .198 .201 .203 .195 y ~ A X) '• .193 find dyldx atx = 7.50. 3. Using Newton’s divided difference formula, find/XlO) from the following data : x: 3 5 * H 27 34 A x): -1 3 23 899 17315 35606

7.53 .208

Numerical Analysis-!! { 297 'j• 4. From the table below, for 'vhat value of x, y is mini muni ? Af.;o find ilv.y ■fc *7 4 s 6 x: 3 5 \J._ y: 0.205 0.240 0.262 0.259 '•.250 5. Calculate y'(0) and v' ’(0) from the following table I 4 x: 0 " I 2 J 6 7 y: 4 8 15 6. Find >'(0-5) from the following table : 0 .6 5 0.60 x: 0.35 0.40 0.50 0.55 0.45 : 11 t ! .4 IL 1 .4 8 8 1.506 i .4o~ y r 1.521 7. A rod is rotating in a plane. The following table g iv e s the 0; ra d ia n s ) th ro u g h w hich the rcxi turned for various values of the time i second t: 0 0.2 0.4 0 .6 O.S i.O i.2 6. U. 0.12 0.49 M 2 2 .0 2 3 .2 0 4 .6 7 Calculate the angular velocity and angular acceleration i\- th e :^d w h e n / = 0 .6 se c o n d 8. Find the values of cos 1.74 using values given in the ihie below : x: 1.70 1.74 1.78 1 S2 i.So sin x • 0.9916 0.9857 0.9781 U.v69 [Hint Find first derivative at x = 1.74] 9. The following data gives corresponding values of pressure and specific vo:ume of a super heiueo steam. v: 2 4 6 .8 10 p : 105.0 42.7 25.316 13 .0 Find the rae of change of (i) Pressure with respect to volume when v = 2. (ii) Volume with respect to pressure whenp = 105. 10. Find /'(4 ) from the following data using Lagrange’s formula. 0 r 2 5 i. 0 8 125 Ax)

Answers-3(D) t. 3, 18 6.0.44 (ii)-0.0191.

2. 0.22666 3. 232.87 7. 4,08 radian/sec; 7 radian/sec2 10.48

4. 5.6875,0.2628. 8.-0.19833

5. - 27.1>. 17.-,. 9. (i) - 52.4 S3SPT3G7 JT~..

Extrcise-3(E) log(l T x ~)

1. Use Simpson’s rule with ten equal to prove that J0 ” (i'+x2} 2. Evaluate

e '^ d x by Simpson’s ‘1/3’ rule (taking h = 0.1).

2 9 8 | E n g in e e r in g M ath em atics -111

ft)6 3. Use an approximate integration formula to find the vaJue of Jo yx dx, , using the following 0 values ofy x : x: 0 1 2 3 4 5 6 yx : 0.146 0.161 0.176 0.190 0.204 0.217 0.230 f o.s 4. Use Simpson’s *3/8* rule to obtain an approximate value of J q (1 - 8 x 2)l/2c£r. o *

5. Evaluate

J(ein x +cos x)dx. Correct to two decimal places using seven ordinates.

6. Calculate

yj(l+x~+ x 2)dx. Correct to five places of decimals using eleven ordinates.

7. Find the value of

J3

fe +x 2) *** ^ dividing thp range into eight equal parts.

fS 8. Use Simpson’s ‘ 1/3’ rule to find value of J f(x) dx given x: fix):

I 10

2 50

3 70

4 80

5 100

9. Dividing the interval in 8 equal use Simpson’s 1/3 rule to evaluate 10. Find the value of


x2 log xdx by taking 4 strips.

11. The velocity v of a bike which starts from rest, is fixed intervals of time /, as follows : /min 2 4 6 8 10 12 14 16 18 20 vkm/min 10 18 25 29 32 20 11 5 2 0 Estimate approximately the distance covered in 20 minutes. 12. A curve is given by the table: x: 0 1 2 3 4 5 6 y: 0.0 2.0 2.5 2.3 2.0 1.7 1.5 The x co-ordinate of center of gravity of the area bounded by the curve, the end ordinates and the -Y-axis is given by A x ~

f6

xy dx, where, A is the area. Find x by using Simpson’s rule. A nsw ers-3(E )

2. 0.746832 7. 0.4478 12. 3.032

3. 1.13625 8. 256.667

4. 0.29159 9. 1.61

5. 1.13935 10. 177.483

6. 2.97049 11. 309.33

□

N umerical A n a ly sis - I I I , C orrelation 4.1

and

R egression

NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS Many problems of science and engineering can be formulated in terms of ordinary differential equations with appropriate initial and boundary conditions. When the analytical methods are not of any help to solve differential equations, we shall go in for numerical methods to solve them. It is step-by-step method. The given interval [a, A] over which the independent variable x, is divided into sub-intervals i.e., a = x 0
— ~ fix, _y) with yixo) = yo are obtained one of the following two forms :

4.2

(i) An approximate series for y is ascending powers of x or (x - Xq), from which the values o f y corresponding to specified values of x can be obtained by direct substitution. This category is known as single step method. For example : Picard’s and Taylor series methods. (ii) Approximate value of y. corresponding to only specified value of jc, This category is known as multi-step method. For example : Euler and Runge-Kutta methods. PICARD’S METHOD Consider first-order differential equation : dy t o =fix>y) -0 ) with the initial condition, j^ o ) “yo *•«•>y = yo at x - jto Integrating (1), we get = £ /(*, y)dx =>

y =

+ ^f(x,y)dx

...(2)

Now, we solve (2) by the method of successive approximations, we assure yo as an initial approximation of y and obtain a better approximation _vi by replacing/ in the integrand byyoThen first approximation _vi is :

y] - yo + f f(x ,y0)dx.

3 0 0 | E n g in eer in g M athematics -!H

Similarly, T he second approximation y? is given b> :

y 2 >» S'o + £ f( x , y i ) dx. By repeated this process, the nth approximation y„ is given by :

y„- yo+ £ f(x,yn.i)dx. TJie equation (3) is called Picards iteration formula.

Example 4.0} : I \in g Picard method to obtain y fo r x ^ O .l, i f dy —

Solution.

= / + X)', with y (0) = I.

Here. Ax. y) = 1 + xy, xq = 0 and yo = 1. First approximation : V; ^ -Vo + f

J*o

f ( x , y 0)dx = 1 + f (1 + x. 1) d x Jo

..(I) approximation :

v>- y0 +J.tu f f(x.y})dx

r

JO

1 + x

I + x +

v yx- 1 +x + c-1

* L ' dx 2;

=1 +I'ji+X +X^+yJ<£X X2 -X3 +--. X4 =_I +X+--+ 2 3 8

...(2)

Third approximation : . y 3 = yo + f X f ( x , y 2) d x

JXQ

r

Jo

.

1 + X

X2

It 1 + X + —

2

X4 dx 8J

X3 + — + —

3

[using (2)]

x- X X X X

3 4 5 6 = 1 + X + --- + ---- + --- + --- + --2

W hen, x =

3

8

15

48

0.1, then t , , , (.l)2 (.l)3 J<0!) - 1 + <■•) * 2 * ~ * ><0.1) - 1.105.

(.l)4 (.l)5 (.l)6 g + I T * I T = 1.105. Ans.

N umerical A n a l y sis -111,

Correlation& R^ckession j 3 0 1

Example 4 .0 2 .* Use Picard’s method to fin d approximate value o fy when x = 0.1 given that y = 1 when x —O and

*®3 x + y 2.

{RGPV June 2003j r

Solution.

\ (T.y ! v ^ 7 = f ( x *-v)

Here,/*, y) = 3x +y1, xo = 0, yo = 1 The first approximation :

vi = yo

+ Jrof f(x,yn)dx -- ,vn + *fv«»(3.v + y-W \

= 1 -V f X ( a r + D d r ^ \ * x f —

Jo

2

.

...(0

Second approximation

y-> = >’o + •> f xo

JI

=

r , . . . 3.v- v ! . 3x + I 1 + x + —— ] i dx i i

1 + j" f

r

. 3x2 - 1 + x *•

f 3x' ~ 4.v2 - 5.v * i ; •

o ,, 4 3 "* =1 +X-»•— X + — .V -r — . V : • — .. 2 3 4 2-.

. .12 '

When,x = 0.1, then (2) becomc:.: n = ><0.1) = 1 * (-1) * | u ) ; -

~ l-ly‘ *

~ 1.1264

Hence, M0.1) = 1.1264. Ans. Example 4.03 : Use Picard’s method to approximate the value o f y when x = 0.2 given thaty ~ / when x —0 and Solution.Here,

dy

= x + y , (three iterations). Also fin d error. fix.y) = x+y, jc0 = O.yo = 1

i .. dy _ // j~

\!

The first approximation ; y\ = > '« + [

f ( x , y 0) d x = 1 + f (x + 1)dx

= 1 + .*+*^-.

...(I)

302

| E n g in e e r in g M a th e m a tic s -! 11

Second approximation : y i ~ 1 + J0 f ( x , y \ ) d x

x + ^1 + x + ■ —j

= 1 + X + X 2 + — Xs .

6

Third approximation :

y } = 1 + \ j ( x , y 2)dx = 1 + J* * + j^l + x + x2 + ^ x 3 j dx [Using (2)] = 1 + x + x2 + - x3 + — x4. 3 i 24 When, x ~ 0.2, then (3) becomes : yi =y (0.2) = 1 + (-2) + (.2)2 +

o

( 3) J

+ — - - 1.2427 24

y{0.2) = 1.2427 dy Exact solution : Given —---- y = x dx which is linear differential equation in y. i.e.,

...(4)

I.F. = e 'lldx = e~J. Its solution: =>

y{I.F.)~ Jx (I.F.) dx + C ye~x = jxe-*dx + C =~xe-x - e ~ x + C

=> y = -x -\+ C e x Using initial condition y = 1 whert x = 0, then C = 2 Exact solution, y ~ - x - 1 + 2e* when, x = 0.2 then ><0.2) = - 2 - 1 + 2e° 2 = 1.2428 Error = 1.2428 - 1.2427= 0.0001. Example 4.04 :Obtain Picard’s second approximate solution o f dy x* dx = y T l » M y (0 ) = 0. Solution.

Here.

x2 fix, y) = ' ! " £ - x0 « 0, y0 = 0

First approximation :

y

i = Jo + f* f(x ,y t)d x

...(5) Ans.

N um erical A n a ly sis -111, C o r r ela tio n & R eg r e ssio n | 303 Cx

X2

,

*3

ex

=*>+ JLTr~T I *2dx =—. o ( ^ + l )T4* = Jo 3

...( )

" y1>

Second approximation:

yi = yo +

f' f(x ,y i) dx

* J, = 0+ JI *o- j?f-+—i ** - W *T 6 + + ii I 9

dx [Using (1)]

Putting t =— =>dt =x2dx and t ->0to 3

J

* = Jo**/a? T I A = [ ta n ' l * I , / * = tan_1(*3/3)0* 98 Sincetan-1 0 =0 - -— 3 +— 5-

Xs

l(x 3

Ans.

3 31 3

yi ■

dy Example 4.05: Using Picard's method o f successive approximations to ^ = y* ~ **» x**0,y~ I obtainy in the interval 0 £ x £0.5. (taking h <*=0./). Solution : Here, y) - y 2- * 2* xo~0,yo~ I. First approximation :

tv )’o=U

>>i = >o + J* /(x ,l) dx X® = 1 + fI**(l-x2)dx « l +x-=-.

...d)

dx

[Using (1))

Jo

Secondapproximation

.

3

Q

A

M C

*

7

= 1+ x + x —— — ——x5 + ——x . 6 We obtain value of>
15

63

...(2)

Ans.

304 J Enginleking Mathematics-Ui

Example 4.06 :Use Picard’s method to fin d approximate y when x * 0.1 given that dy y - x y = /,* * e n x = 0 a n d - £ = — Solution.

Here,

•: -

- £ A*, jb x fc - 1 * £ I n r I * r [ 2 - a + x)i

- 1)

= 1+ (X- ± ~ d x -( Xl dx

(l+x)

JoUjc

Jo

- t - (2 !og(l + x) - a*)0‘ - I - x + 2 log (I + x) SvconJ
...(J)

:o;:

■■

rv „ •" - £

, ,

, f * (I - 2x + 2 log(l + x)J j - 1 * Jo 1 + 2 log(l + jc) *

_____ 2x_____ - 1 - r : 1 - 1 + 2Iog(l + x) dx x dx Jo 1 + 2 log(l + x) whici-. is very difficult to integrate. Hence, we have considered only first approximation. Vvnen * ~ ■>!. then (1) becomes: Vi - y (0.1) = 1 -(0.1) + 2!og*(l +0.1) = 0.983. Ans. rU.,rf!'s method, find.approximate values o fy and z corresponding to x a 6.1, 1 + x - 2 f* —

■■.

, iff: - 2. z (0) - } and

dv dz r - a* + s, ~r~ * x - jr . dx ft. ji) 2, zq — * I,


...... *’■^

A*

■— - fix.y, z) = x + z, and -7~. - 4>U,y,z) = x - ) t -

(f-\ tn : U>iiig Picard’s method n - .To *- f f(*,yo>zo)<&'' **0

[Since-jfo>,> ^o) = ^ + 2q = x + 1)

- 2 •*- f (x + l)dx = 2 + x + ~ x2. •»o 2

N u m e r ic a l A n a ly s is - IH , C o r r e l a t i o n f t R e g r e s s io n |

and

z\ = zo + f 4(x,y0,z 0)dx

=1+ jj(x - 4)dx =1-

305

[Since #(x,yo, zo)=x+yo2 = x - 4 ) 4x

+|x 2.

Second approximation : y 2 = yo+

P

f ( x , y l, z l )dx

[v■/•(Xly „ * 1) = x + 2,]

= 2 + | U + 1 -4 X + - X 2 !dx

c(*

•d) z2 = 26 +

J* t ( x ,y l , 2 i)dx

[v ^ (x,y i ,z) = * - y,2]

"v

=1 +hf x- |{2 +x+ — x2 2 J =1- 4x+— x2 -x3 ---- —. 2 4 20

...(2)

When, x = 0.1, then (1) and (2) becomes : n = 2.0852 i.e., ><0.1) = 2.0852 Ans. and Z2 = 0.5840 /.e., 2(0.1) = 0.5840, Example 4.08 :V$lng Picard's method, fin d approximate values o fy and z corresponding t o x - 0.1, given thaty(O) - 1 and z(0) - 1/2 and ^ Solution.

Here, Let,

= z>

- x 3( y + z )

*o =0,>o= l»zo = 1/2 dy =/(x,y,z) = z

dz f a = t ( x ,y t z) = x*(y + z). dx First approximation : >1 = ^ 0 +

f ( x , y 0 ,Z i} ) d x = 1 +

= 1

+Jo

2« = 20 + /* * * ■ * ■ * > * - \ + / '* * (* + j ) *

=1+f • 3x4 = »2 + ----8

306

) E n g in eer in g M athem atics -] 11

Second approximation: 1

2+ 8

1 x 3x6 = 1 + — + ----2 40 and

z2 = *o + Jxo r

dx

-.(I)

»*!><** = 77 2 +

+ Zi)dx

, i + fV

2

Jo

- 2.

^x *

x8

3*B

_ 2 + ~ 8 ~ + 1 0 + ~64~

When,x = 0.1, then (I) and (2) becomes : n = 1.0500008 Le., >(0.1) * 1.05 and z2 - 0.5000385 /.e„ z(0.1) * 0.5. Ans. Example 4.09 :Uslng Picard's m ethod, fin d approxim ate y when x - 0.1 given that “

Solution.

7

+f i ~ + y= (»4«f> = W ,^ = «.l whenx~0.

Let,

dx

d 2y zz> d x2

dz dx

dz Thus the given equation reduces to — + 2xz + y = 0; y(0) = 0.5, z(0) = 0.1 Let,

dy

dz = J(x, y, z) = z\ and ^ = $ (x,

z) = - (2xz + j>) with the conditions

yo = 0.5, zo = 0.1 atxo = 0. First approximation: y\ = yo + JX f*Qf ( x ,y 0,Zo)dx

Jo

or

y x = 0.5 +

and

zi * *o + f* ^ J o - ^ b ) ■>*0

Zq

dx = 0.5 +

J

(0.1)dx = 0.5 + (0.1)x.

...(1)

= 0.1 - Jo (2xzq + y0) dx [v ^x,y0,Z > = -2(xzq + y0)] = 0.1 - JJ (0.2x + 0.5) ctx » 0.1 - 0.5x - (0.1) x2. ...(2) 0

I1

N um erical A n a ly sis -111, C o r r ela tio n & R e g r e ssio n | 3 0 7

Second approximation :

yj or

= Jo + f* f( x ,y i , z ,) dx

y 2 = 0.5 +

z x dx

[v f( x ,y u z{) = zv\

= 0.5 + f * (0.1 - 0.5* - 0. lx 2]d* Jo = 0.5 + (0.1)* and

2

z2 = *o + J*

[Using (2)]

x2 - t^ -x * .

...(3)

3

= 0.1 - £ (2*Zj + y,)dx

= 0.1 - J*[2*(0.1 - 0.5* - 0.1*2) + (0.5 + 0.1*)}dr [by (1) and (2)] = 0.1 - (0.5)* - ^

*2 + ^ Z

O

^

* 4.

...(4)

4

When, X= 0.1, then (3) becomes: ^ = 0.50746667 ».e.,j<0.1)* 0.5075.

4J

Ans.

TAYLOR SERIES METHOD Taylor series method is the simplest method for obtaining the solution of an initial value problem provided all the conditions prescribed are valid at the initial point x - x q . Consider a first order differential equation : dy fa =fa , >0 with y(xo) =y0. -•■(') Let the y(x) is the exact solution of (1) with >
. . . . . . . ^ . (* " *o>2 ... .

y(x) = yo +{x - x0)y0 + ---- —----(* y0"+*o)3 -----—

.

yo + —

...(2)

which is convergent series in* within the interval [*&, *„] the values of y£„ yj, y j,... are known. From (1), we can write / =Ax,y) from which/o can be found i.e., / 0 -flxo, yd)Now, using the fonnula for total derivatives, we can write / ' =f(x>y) =fx +yfy =fx +ffy IV f= A ^ y ) ) when the suffixes denote partial derivatives with respect to the variable concerned. Similarly, / " =» f'( x ,y ) = f„ + f ^ f + f[fys + fyyf] + fy[fx + fyf] ~ fxx + I ffy y + P f y y + fx fy + ffy a n d SO On.

Now, by substituting* = x q andy =yo i n / , / ' , / " , .... we can obtain the constants y^, y'0t y%,... Substituting these values in (2) we can obtain the desired solution of (1).

3 0 6 | E ngineering M athejkatks-III

Example 4.10: Solve

dy

,

„

- 1 - 2 x y , given that y(0) ^ 0 by any method.

/ RGPV Dec. 21

Solution : Given, dy

- 1 - 2xy =>y' - \ -2 x y and xq ~ 0,yo = 0.

By Taylor series :

>o + -——

Jo + —

Since Differentiate : Differentiate :

/ ( x ) - 1 - 2 x y ^ / o ~ 1 -Ix^yo = i [vyo=}< /'(* ) ~ - 2 y ~ 2 x /, then yo" - - 2y0 - 2xayo = 0. /" (* ) = - 2 / - 2xy" - 2? = - 4 / - 2xy”, then yo"’ ~ - 4yd - 2xq/ ' q - - 4. Differentiate : /% x) « - 4y>’ - 2jy”' - 2 /', then y0tv = - 6y0" - 2xQy0"' = 0. Differentiate : yvW, = - 6 / ” - 2xy* - 2 /" , then yov = - Syo'" - 2xdyolv = 32 .... and so on. Hence (1) becomes: Mx) = 0 + x.(l) + 0. ~ < - 4 ) + 0 + ^ - (32) + ... 2x? 4 >
. ”

*

Example 4.11 :Vsing Taylor’s series, fin d y(2.1) correct to 5 decimal places. Given *

dy

=*-

withy (2) - 2. Solution.

x —y

y

X

X

xo - 2,yo = 2, / --------- = 1 ------

Since

/ = 1 - —J

then

yo' =0,

Differentiate:

y" = — + —£ I

then

yd' = ^

then

yd" = “

then

y^ =

y

y" 2y' 2y ---------- + — ---------3- ;

D ifferentiate :

/"

Differentiate :

y" 3y* 6y' y* = - — + — -

X

X

Taylor series expansion about xo = 2 is

X

6y

;

3

< n |

Here

•

N u m e r ic a l A n a ly s k - I I I , C o r r e l a t i o n f t R e g r e s s io n | 3 0 9

When, x = 2.1, then (1) becomes : *2.1) = 2 + (2.1 - 2)(0) + - •I ~ 2)2 ( l / 2 ) + ( — ~ 2)3 ( - 3 /4 ) + ^ ^ 1 ( 3 / 2 ) . ><2.1) =2.00238. Ans. ■4. 12: Using Taylor’s series m ethod to obtain approxim ate value o f y at x - 0.2fo r the differential equation

- 2 y + 3 e * ,y ( 0 ) = 0. Compare the numerical

solution with the exact solution. Given,

[RGPV June 2004j

dy f a = 2y + 3ex;

xo = 0,yo = 0

Since / = 2y + 3e* Differentiate : y" = 2 / + 3ex Differentiate : / " = 2y" + 3e* Differentiate: / ’'= 2 / '' + 3e* and so on. Using Taylor’s series :

=> yo ~ 2yo + 3ex = 3. => yb" = 2yo + 3e° = 2(3) + 3 = 9. => y0'" —2yo" + 3e° = 2(9) + 3 = 21. => yiv = 2yo‘" + 3e° = 2(2!) + 3 = 45

x , ) - * + < » - xM

=>

a

=>

9x 2 7ac3 15x4 ><*) * 3x + ~2~ + ~2~ + ~ 8 ~ +

♦

*-

») - o + *<3)+ f r <9>+ | r <21) + i t <45>+•••

When, x = 0.2, then (1) becomes : >(0.2) = 3(0.2) +

9(0.2)2

7(0.2)3

15(0.2)4 +~ 1 —

or ><0.2) =0.6 + 0.18 + 0.028 + 0.003 = 0.8110. Tofind exact value : Since,

which is linear differential equation in 7 , I .F_ = e -J M * = ^ 2 *

Its solution : =>

X®-21) = J3e*.(e_2r)dx y / r 2* = 3je~xdx =-3e~x + c y =~3<*+ce2x

...(2)

310 | Engineering Mmw nahcs-111 Give initial conditions : y - 0, when x - 0, then 0 = - 3 +c=>c~ 3 Hence y = 3c2*- 3e* v Exact value at x - 0.2 is>(0.2) = 3e%02) - 3e° 2 = 0.8112 £ m r = 0.8112 - 0.8110 = 0.0002.

...(2)J Aa

dy Example 4.13 :Solve by numerical method ^ = * + y * , y (0 ) - J. Find y a tx =»0.1. [RGPV June 2002ft

Solution.

Here,

xo =0,>^= 1 a n d / = x + .y2.

S ince

y

+x

Differentiate : /' Differentiate : /" Differentiate : y» => >0^ Using Taylor’s series :

=>

>>o' = 1.

= 2>/ + 1 => yo" - 2 + 1=3 = 2(y')2 + 2>y' => yo’" = 2 + 2 * 3 = 8. = 4 / y ' + 2 / / ’ + 2>y" = 6 yy' + 2>y” = 6 * 3 + 2 * 8 = 34.

(X - X n) , (X - X n)2 , ( x - Xo)3 „ * x ) = J o + ■ •-y !" - - J o + v- - 2-^ -y g + -v- - y p— J o + -

>
~

JS + § y J o + j j - J g ' + -

[ v xo = 0]

= 1 + x + -—(3) + ~ -(8 ) + -^—(34) + ... 2

3 , 2

6

24

4 , 3

17 , 12

« 1 + X + — X2 + — Xs + --- X4 + ...

When x = 0.1, then (1) becomes X 0 .1 ) =f 1 + 0.1 + | (0.1)* + i (o .i)8 + i | ( 0 . 1 ) 4 = 1.1165.

Example 4.14:Solve by Taylor’s series method

dy

Ans.

dz - x + z , — = x - y* with y(0) * 2, z(0) * /.

Flndy (0.1) and z(0.1). Here,xo = 0 , = 2, zo ~ 1 andy = x+>, t - x - y 1. Then y =x + z=i*>^'=xo + 2o= 1, and / = x - y 2=>zo' = -Jco->'o2 = - 4 y ' = I + z7=> yo" = 1 + z&' = 1 - 4 = -3, and z" « I - 2 > y =*Zo" « 1 -lyay* = “ 3 y " = z" => >o'" = zo" - - 3 and so on. and z"' = - 2(>y' + y 2) ’)T-J» => zo'" » - 2j>oVo" +>'o'23 = 10 and so on. Using Taylor i ieriei y am/ z :

Solution.

(X - Xn) >
, Jo +

(X - Xn) 2

.

(X - Xq)3

2 ? " J o + ------3 j

» Jo +

-(I)

N um erical A n a ly sis -111, C o r r ela tio n f t R e g r e ssio n | 3 1 1 ,

(X - X n )

and

( X - X * )2

2!

zfr) = *o + - !!

When x = 0.1, then (1) and (2) becomes :

*0.1) = 2 + (0.1X1) +

( x - X n )3

,

15

3!

,

+

"( }

,

(-3) + ^ - ( - 3 ) + ....

[v xo = 0]

= 2 + 0.1 - 0.015 - 0.0005 * 2.0845. and

Ans.

2(0.1) = 1 + (0.1X-4) + ^ - ( - 3 ) + ^ - ( 1 0 ) + .... = 1 - 0.4 - 0.015 + 0.001667 = 0.5867.

Example 4.15 :Solve Solution.

d*y

Ans.

dy - y + * — > given y(0) - l t yV) = 0 and calculate y (0.1).

Here, xq = 0,>t> = l,yo' = 0 and / ' = y + x y x>yQ” = >>o + xq>>o' - 1 . Differentiate : / " = / + / + xy" = 2 / + x y ' => yo" = 2ytf + xqvo" = 0 Differentiate : /* - 2 /' +y" +xy"' => yd* = 3yo" + xqVo"' = 3. Differentiate : y v = 4 /" + xy** =>yov = 0. Differentiate : y* = Sy*+ xyv => yo*1= 15, and so on. Using Taylor's series : ( X - Xn)

,

(X - Xn)2

y Q c )- * o + - { i Ly° + When x = 0.1 and putting these values, we get

21

,

+ **'

K0 , , . 1 + 0 + C ^ a ) + 0 * ® j f ( 3 ) t 0 + ^ i ( I 5 ) = 1.005013.

Ans.

14 EULER’S METHOD Euler's method is one of the oldest and simplest method. Let the first order differential equation be dy

~AX»y) with yixo) - yo-

...(1)

Suppose we are to find the value / at n-points besides xo, which are equally spaced with a steplength h. Then, Xj =xo + h, X2 = xo + 2 h ,..... . x„ - xo +nh. First approximation: p d y = P f(x,y)d x Jjo *>*0 Replaclng/x, y) by the approximation/xo, yo), we get

yi -y o = *** f(x*>yo)dx = f(Xo,yQ) p dx

[From (1)]

3 1 2 | E n g bsek n g Mathematics-III

=>

y\ =>^ + (jfi -xo)A*yo)

[vA=*i-*o]

...(2)

Similarly second approximation: yz +/»/(X|,>'i) «* approximation :y„ =y„- j +hj(x„- ,,y „ . ,)■ - ( 3) Remark: To get more accurate result, h should be very small. Example 4.16:Using Euler's method,fin d an approximate value o fy corresponding to jc * 0.1, given dy y - x that - yJr X withy(O) * /. Solution : Here,./fo y) =

y- x

. x0 = 0, y0 = 1.

We divide the interval (xo, x) =(0,0.1) into n - 5 steps i.e., X — Xq 0 .1 ~ 0 h ------- — =>A = — -— =0.02. n 6 Then, xi = xo + h - 0.02, x2 - xo + 2h = 0.04, X3 = 0.06, X4 = 0.08 and X5 = 0 . 1. By Euler's method : yn ~yn -1 + A /(x „- 1, yn- 1), where n » 1,2,3,4,5.

y\

=>0+W xo,yo ) * 1+(0,02^(lTo, = I 02‘ '

= jj "+'*£

^ =>1 + M ^i.Ji) « 1-02 + (0 .02) / 0 .02, 1.02) . ,n n n J 1.02 - 0 .0 2 ) . 1.02 + (0.02)^ ^ — j , 1.0392

« -» *

1 0 3 9 2 + (0 0 2 ) ( i S

^

h

* - « - « fe n > - UW T ♦ »

+ « « . * > - » 0756 ♦ W

) * u m

- 1.0756. (^

S

S

) - 1.0928.

Thus, ys ~y(x = 0.1)= 1.0928. Examplf.ft.V7 : Using Euler’s method solve the differential equation In six steps : * ! Solution.

~dx ^ x +y* y(0) **0, choosing h m 0.2.

Am

[RGPV Dec. 2003}

H ere,/x,y) = x +y, xo ~ 0,yo - 0, h - 0.2 and n - 6 . By Euler’s method: yn =yn- 1+ hAxn- i,y n - 1). where, n = 1,2,3 ,4, 5 , 6 . Then, xj = xq + h ~ 0.2,X2 “ 0.4, xj = 0.6,X4 = 0.8, X5 = 1.0,x 6 = 1.2.

N um erical A n a ly sis -111, C orrela tio n & R e g r e ssio n | 3 1 3

y\ =yo + a/xo,/o) = o +(0.2X0+ 0) =o. y i= y i+ hfixuyO - 0 + (0J2X0.2 + 0) =■0.04. y3 =yi + h fe 2, yi) = 0.04 + (0.2X0.4 + 0.04) = 0.128. >4 =>3 + h ftx 3, / j ) = 0.128 + (0JX0.6 + 0.128) = 0.2736. /5 = /4 + hfix4,y*) - 0.2736 + (0.2X0.8 + 0.2736) = 0.48832. = / 5 + AXx5, / 5) - 0.48832 + (0.2X1.0 + 0.48832) = 0.785984. Hence, y$ =><1.2) = 0.785984. Example 4.18: Findy (2,2) using Euler methodfrom the equation — — xy2 w ttky(2)~ 1. Solution.

Ans.

[RGPV, June 2011[

Here f( x ,y ) = - x y t Xo - 2, /o “ 1, 7bfindy(x) a t x = 2.2 i.e., y (2.2) : Taking rc = 5,

•

x - x0 2.2-2 h - ------- -- — -— = 0.04 ft

J

x0 = 2, Xj = Xo + h = 2.04, x2 = xo + 2A = 2.08 xj = xb + 3h - 2 . 12, X4 =xo + 4h - 2.16,xj = xo + 5h = 2 .2 . By Euler's Method: y„ =y»- 1+ hf(x„ -u y0. i), Where n = 1,2,3,4, 5. Then yi =yo + h.f(xo, yo) = 1 + (0.04) [- x0y5 =/4-»-A./(jf4,>4) = 0.7377 + (0.04) [- (2.16) (0.7377)2] = 0.6907. Ans. 2 x y , y(0) = 0,tokeh- 0.2. [RGPV June 2006} Solution.

Here, xo = 0,

= 0,Xx, y) “ I - 2xy, h “ 0.2 and x - 0.6.

X - Xq 0 .6 -0 =>n- 3 (steps) h --------- - => 0.2 = n n Then, xj = xo + h = 0.2, x2 = 0.4, and xj = 0.6.

314

| E n g in e e r in g M a ih em a tjc s - II!

By Euler’s method: yn = /« - j + hj{x„_ u y » - l), where, « = 1,2, 3. y\ =>o + M*o,yo) = yo + AO - 2xq)>d) = 0 + (0.2X0 = 0.2. Yi =y\ +AX*J,yi)=>i + W - 2 r i/i) = 02 + (0.2X1 ~ 2 * 0.2 x 0.2) = 0.384. >3 + ft fix2t + - 2X2^2) = 0.384 + (0.2X1 - 2 x 0.4 x 0.384) - 0.52256. Hence, -yixy) =><0.6) = 0.52256. Example 4.20 : Solve the equation $x ^

Ans.

+ y* - 2 ~ 0 , y ( 4 ) = 1fo r y(4.1), taking h “ 0.1 and

using Euler's method. Solution.

dy 2 - y3 2 - y2 Given : — = —----- , i.e.,Ax, y) = —----- . xo = 4, /o = 1. A = 0.1, then ax dx ox x j = xo + A — 4 + 0.1 = 4 . 1 .

By Euler’s or Simple Euler’s method : y\ =yo + hfiM,ya) 2-

5x0 Hence,

4.5

y\ =X Jci ) =y (4 -0 =: 1.0050.

Ans.

IMPROVED EULER’S METHOD The Improved Euler’s fonnula is : y„<'> = y«-i + ~ [ f i x K_l ,y n. x) + f(x K, y J \ , wherey„=yn_ i + hf(xn~ h y„. i) oryn<» = y K-1 +

Example 4.21 : I f Solution.

dy

[/(*»-!,y ,-i) + f{x«,y*-\ + /i/(xB_„y,_i)}] for n = 1,2,3,... . = x + y* andy ® 1 a t x m O. Find an approximate value o fy a tx « 0.2 by

improved Euler’s method (taking h m 0,1). Here, /x ,y ) = x + y 2,xo = 0,yo= 1, A= 0.1. Then, X] =xo + h - 0.1, andx2 = xt> + 2h = 0.2. By Improved Euler’s fonnula : /,« ) = y0 + ^ [/(x o .y o ) + / f a . f t ) ] where y x =yo + From(l), we get

= I + (0.1)(0 + 12)= 1;1

/(<'> = yo + ^ ^ [ /( * o » > o ) + /(x 1(l .l ) ]

...(1)

N um erical A na ly sis - I I I , C o rrela tio n & R e g r e ssio n | 3 1 5

= 1 + ^ ~ [ / ( ° . 1) + /(0.1, 1.1)] U = 1 + 0.05[(0 + 12>+ (0.1 +(1.1)2)]= 1.1155. Now,

n (» - y,(,) + ~ [f(xl, y ^ ) + f(xi ,y ^ ] ,

...(2)

where

><2 = y,(,>+ M * ,IJ l(,)) != 1.1155 + (0.1)x0.1. 1.1155) = 1.1155 +(0.1) [0.1 + (!. 1155)2]= J.2499. From (2), we get

^(D = 1.1155 + ^ 1 / ( 0 .1 , 1.1155) + /(0.2, 1.2499)]

Hence,

2 = 1.1155 + (0.05)[{0.1 +(1.1155)2} + (0.2 + (1.2499)2}] = 1.2708. >2(I) = X*2) = X 0.2) = 1.2708.

Example 4.22 ••Solve the equation $ x

dy

Ans.

- 2 - y*; y(4) - I, fo r y (4.1), taking h - 0.1 and using

improved Euler’s formula. Solution.

2 -y * ■. xo = 4,>o= 1, and A= 0.1. 5x Then, x\ =xo + A= 4 + 0.1 =4.1. By Improved Euler’s formula : Here,

J[x,y) =

>i(l) “ y<> + where

...(l)

+ / ( * l Ji)]

y\ =yo +hfixo,>0) = l +(0.1M4, i) 1.005

From (1), we get 3l(i) = 1 + M [ / ( 4 , i ) + /(4.1, 1.005)]

^,(i) = 1 + 0.05 Hence,

4.6

(2- l2) ] [2 - (1.005)2] = 1.0049. 5x4 5x4.1 Ans.

yjW =y(;ri) =>(4.1) = 1.0049.

MODIFIED EULER’S METHOD Consider the differential equation: dy

4i

...(1)

316

| E n g in eer in g M athem atics -

IU

The approximation a re : yn + , = j*

+ ~ » j , + -/< * * , j , ) j ; f0rn = o, 1, 2,

y j = Jo + * /{ * « + ~ » Jo + | / ( ^ J o ) J

i.e.,

yz = Ji + h / jx , + y, + | / ( x , , y , ) j and, 1so on. I z z j Example 4.23: So/ve

= J “ “

» >(ft> “ 1f»ry(0.1), taking h * 0.1 and using modified Euler’s

method. Solution.

2x “ » *o = 0>Jo = I, A = 0.1, then xj = xo + /r = 0.1.

Wen, fix >y) = J

By Modified Euler’s formula : y\ = Jo + =

j*o + “ » Jo +

1 + (0.1)/ jo.05,

J o )|

1+f / (0, i)j ^ s i n c e /(0,1) = 1-

— -

=

1

= 1 + (0.1) f{0.05, 1.05} = 1 + (0.1) |l.0 5 = 1 +(0.1X0.95476)= 1.09548. Hence, y\ =><0.l) = 1.09548. Ans. A'V Example 4.24 : Solve the equation 5x — + y* = 2 ; » 7 fory(4.1), taking h “ f t/ ««
2 - y2 Here,y(x, y) = —----- , xa~ 4,>
1,

/i = 0 .1 , then Xj = xq + A= 4 .1 .

By Modified Euler’s formula : h y, = Jo + h f x o + - , y o + |/(a c o ,Jo )j = 1 + (0 .1 )/|4 + H

, i + M /(4 , i)J

= l + ( 0 .1 )/|4 .0 5 , l + 0 . 0 5 ^ ^ ^ - J j

N umerical A nalysis-111, C o r r e l a t i o n & R e g r e s s io n | 3 1 7

= 1 +(0.1)/{4.05, 1.025} = 1 + (0.1)

2 - (1.025)2 5 x 4.05

1.0047.

Hence, y\ =>(4-1)= 1.0047. Ans. Exam ple 4.25 : Use m odify E u ler’s m ethod to com pute y fo r x — 0.05. Given that dy * , — = * + y , x 0 = O, y 0 = J, result correct upto three decimal places.

[RGPV Dec. 2002] Solution.

Here,./fa y) = x +y, with xq = 0, = I Since compute ><0.05), so we take h - 0.05, then xi=xo + h ~ 0.05. By Modified Euler’s formula :

y\ = y0 + A/|*o + =

1 + (0.05) /

yo + ^f(*o>yo)J

|o +

1+

/(0,

J

1)

[Since X0,

1) = 0 + I = 1]

- 1 + (0.05)f[0.025, 1.025} = 1 + (0.05X0.025 + 1.025) « 1.0525. Hence, y\ =><*)) = .K0.05) = 1.0525. Example 4 .2 6 : Solve by Euler’s modify method ^ Solution.

= io g ,(x + y ),

y (i)

Ans.

=2 at x ^ 1.2 andatx= 1.4

with h “ 0.2. [RGPV Dec. 2004 A June 2009] Here,X*,y) = log* (x + y)f x0 = l,yo = 2 and h = 0.2, then *i = xq + h= 1 +0.2 = 1.2 and X2 = xo + 2h - 1.4. By Modified Euler’s formula: + A /'j* b + ~ , yo + ^ / ( ^ .> o ) J = 2 + (0.2)/j l +

2+

/(l, 2)|

[S ince/l, 2) = log, (1 + 2) = 1.0986] = 2 + (0.2)/{1.1,2.10986} =2 + (0.2) log,(1.1 + 2.10986) = 2.2332. Hence, yi =y(xi) =y( 1.2) = 2.2332. Again,

yi = *

+ |/ ( * j .y ,) |

= 2.2332 + (0.2) f { 1.2 +

, 2.2332 +

/(1.2, 2.2332)J [S ince/1.2, 2.2332) = 1.2335]

3 1 8 j E ngineering Mathematics-III

= 2.2332 + (0.2)/{1.3,2.35655} = 2.2332 + (0.2) log, (1.3 + 2.35655) = 2.4925 Hence, yi =yix2) =>(1.4) = 2.4925. Ans.

4.7

RUNGE-KUTTA METHOD Runge-Kutta method is one of the earliest and the most widely used in numerical methods of solving ordinary differential equations of the first order. The fourth order Runge-Kutta method is most commonly used and is reffered as 'Runge-Kutta method Consider first order differential equation : dy f a ~A*) with >i atxi =Xq + h \ y i H x i ~ ao + 2h and so on. Then we compute, = h Axo, Jo);

*4

k-

6

=HA*o+h, y0+h).

+ 2*2 + 2*3 + k4) wherek is weighted mean of *i, k2, k$ and fcj.

Hence y\ =yo + ki.e.,y(x\)=yo + k. Now to compute>2 i.e., >2 ~ y\ + £: we start from (jci,>i) and replacing the above process, we get (*2» yi) and so onRemarks: (i) MThe Runge-Kutta method of first order is known as Eulers’ method (ii) The Runge-Kutta method of second order is : k\=hfa>,yQ) *2 - v ( * o + f ’» + y ) Then,>i =>0 + where k - k 2. Similarly we can compute yi, >3 ....and so on. “The Runge-Kutta method o f second order is known as modified Euler’s method’. (iii) The Runge-Kutta method of third order: *1 =hAxo,yo) * * = * /( * > + £ .* > + 7 ) *3 =

I + *. >D + 2 * 2 -* t)

* = “ (&l +4*2 +*})• b Then, y\ - yo + k, where k = k2. Similarly we can compute yi, >3, .... and so on. “The Runge-Kutta method of third order is known as Range’s method'.

N um erical A n a ly sis -111, C o rrela tio n & R eg re ssio n | 3 1 9

Example 4.27 .*Apply Runge-Kutta method o f fourth order to solve : dy 10 d x

+y2>'y(°> " 1f° r x * 0 L

fRGPV Dec. 2002}

*

Solution.

Here,

x2 + y2 fix,y) = — r r — > XQ-Ot yQ- 1, and x = 0.1, xu X - Xq

0 .1 -0

[As taking n = 1 i.e., one step]

Then,

h = — — = — ----- =0.1 n 1 x\ = x q + h = 0 . 1 .

We have

*i = M ^ > o ) = ( o . i ) m i ) = 0*1

( 0 + l 2 ''I

n rJ"0-0

k2 = h f \ x o + - , y0 + — J =(0.1)/0.05, 1.005) =

0.1

(0.05)2 + (1.005)2 1 10

J

- 0.01012525

*3 = h f [ y Co + | , Jo + y j =(0.1)/0.05, 1.0050626) =

0.1

(0.05)2 + (1.0050626)* 1 10

0.010126508

*4 = hfixo + h, y0 + *3) = (0.1 )X0.10, 1.010126508) = 0.010303555 Hence,

k = ^(A, + 2*2 + 2*3 + A*) D

= 1(0.060807071) =0.0101345 b y\ ~ yo + k i.e .,y (x \)= y Q + k. => y(0A) ** 1 +0.0101345 - 1.0101345. Example 4.28 .* Using Runge-Kutta method o f fourth order to solve: y '~ x y f o r x - 1.2. Initially x - l , y - 2 ( t a k e h - 0.1) Solution. Here, fix,y) ~xy,xo = !,>>o = 2,A = 0.1. Then =xq + h= M ,andx2 = *o + 2A = 1.2. Hence we compute^) and/2 (i.e., two steps) Step 1 : Starting (x6,>»o): We have ky = hfixo, y0) = (0.1 X^o/c) = (0.1X1 *2) = 0.2 * » + ^ - j - (0.1)^1.05,2.1) = (0.1X1.05^2.1) = 0.2205.

Ans. fRGPV June 2003f

3 2 0 | E n g in eer in g M athem atics-III

= h f ^ X o + ~ , J o + ~ j = ( 0 . 1 ) X 1 . 0 5 , 2.1103) = 0.2215

and

*4 = hftx}i-h,yo + * 3 ) = (0 .1 )/l.l, 2.2215) = 0.2443

* = ^ (* i + 2*2 + 2*3 + *4) • 0.2583. Hence, y t -y o + k i.e.,y\ =X*i) = X l- 0 ~ 2.2583. Step 2 : Starting (X), >j) : We have *i = M x ,,y ,) = (0.1)XU,2.2583)=(0.!XU * 2.2583) = 0.2484 h = h f{ * \ + f > J» + ^ j = (0.1)/(1.15,2.3825) = 0.27399 *3 - * /(* » + f ’ Ji + ^ r ) -(0.1V(1.I5,2.395295) = 0.27546

and

*4

=

A/x,

+

h,y\

+ *3 ) *

(0.1)ffl.2, 2.53376) = 0.3041.

k = !<*i + 2*2 + 2*3 + *4) » A. (1.6514) = 0.27523. Hence,

=>1 + * *.e..><1.2) = 2.2583 + 0.27523 = 2.53353.

Ans.

dy y* - x* Example 4.29; Using Runge-Kutta method offourth order, solve “j~ with y(0) • l a x - 0.2 and 0.4. Solution.

Here,

(RGPV June 2004 A June 200*1

y2 - x2 / x , y) = y + JC2 * *0 = 0, yo = I, taking h = 0.2.

.-. We compute y\ and y 2. Then xj = 0.2 and x2 = 0.4. Step 1 : Starting from (x&, >0) • We have

*, = hfix<>,yo) = (0.2)/0, 1) = 0.2

v /(0. 1) =

«1

k2 = h f ^ + | , Jo + y j = (0 2 ^ 0 .1 , 1.1)**0.19672 *3 = h f [* 0 + ~ , Jo + y j = ( 0 ^ 0 . 1 , 1.09836) = 0.1967

aad

k4 • A/x0 + *,

+ * 3 ) = (0 .2 )/0 .2 ,1.1967) = 0.1891

* = |( * i + 2*2 + 2 ** + *<)

- i[0 .2 + 2 x (0.19672) .+ 2 x (0.1967) + 0.1891] =0.19599 6 Hence,

y\ =yo + k ^ y(0.2) ~ 1+0.19599= 1.19599 * 1.196.

N um erical A n a ly sis -111, C o r r ela tio n & R e g r essio n | 3 2 1

Step 2 : Starting (xt, / | ) : We have *, = h f a u yi) = (0.2)tf0.2, 1.196) = Oil891. *2 = A /( * 1 + ! , y i + ! } = (0.2)/0.3, 1.2906) = 0.1795

k} - h f { x i + £ . yi + ~ j = ( 0 .2 ) M l 1.2858) = 0.1793 and

*4 = hj{x] +h,y\ + kj) = (0.2)/0.4, 1.3753) = 0.1688.

**=!■(*i + 2/% + 2*3 + *4) = 0.1792. D Hence, >-2 = y t + k ^ v(0.4) = 1.196 + 0.1792 = 1.3752. Ans. Example 4J O : Apply Runge-Kutta fourth order method to fm d an approximate value o fy when x = dy • 0.2 in steps o f 0.1 if ~ = x + y 2 gi\>en thaty = / where x**0. Solution.

fRGPV Dec. 2004, Feb. 2010 and June 2011} Here, fix, y) - x +y2, x0 = 0, >-0 = 1, taking h = 0.1. Then xi = xo + * “ 0.1 and x i^ x o + 2h = 0.2. Hence we have to compute y\ and y%in two steps : Step 1: Starting from (xo, >o) : We have *, = h f o o, v0) = (0.1 )tf0, 1) = 0.I [v / 0 , 1) = 0 + 1= = 1] (

h

fc ^

k2 = h f j^xo + - , y0 + — J =(0.1)/0.05, 1.05) = 0.11525

M I = (0.1)/0.05, 1.057625) = 0.11686 k} = hu fA xo + -h , y0 + — aad

*4 = hflx0 + h,y() + kJ) = (0.1 )/0 .1 , 1.11686) = 0.1347 k = i ( * i + 2*2 + 2*3 + *4) =0.1165.

o Hence, y\ =yo + *=>_y(0.1) = 1.1165. Step 2 : Starting(xi,/i), hereX| =0.1 ,>1 = 1.1165 and * = 0.1. We have *1 = M * ,,/i) = (0.1)X0.1,1-1165) = 0.13466

k2=

+|,

+yj =(0.1)/0.15, 1.18383)=0.15514

ki = hu A f l*i + h yt + yMl =(0.1)/0.15, 1.19407) = 0.15758. and

*4 = hAx \ + h,y\ + *3) = (0.1)/0.2, ! .27408) = 0.18233. k = |-(* i + 2*2 + 2*3 + *4) = 0.1571.

Hence,

L

o y^ = i| + k i.e.,y(0.2) = 1.1165 + 0.1571 = 1.2736.

Ans.

3 2 2 | E ngineering M athematics-111

Example 4.31; Use Runge-Kutta method to obtain y when x = /./, given thaty = 1.2 when x = / oji y satisfies the equation : dy A c ~ 3x+ S Solution.

Here,

IRGPV June 200$

A x,y) = 3x + y2, xo = l,yo= 1 2 ,x = 1.1,

so that Then We have

X - x0

1 .1-1

h --------- - =>* = — -— = 0 . 1. [As n = l] n 1 X) = x0 + A= 1 +0.1 = 1.1. i , = hAxo, y0) = (0.1) (3x0 + >o2) = (0.1 ){3 + (1,2)2] = 0.444 k2 = h f ^ x o + | , Jo

=(0.1)^1.05, 1.422)

= (0.1 )[3 x 1.05+ (1.422)23 = 0.517. *3 = * / ( x 0 + | , y 0 +~*-

= (0.1)/(1.05,1.459)

= (0 .1)[3 x 1.05 + (1,459)2] = 0.528. and

*4 = ^ o + A ,Jo + ^ ) = ( 0 .1 ) X l.l, 1-728)

= (0.1)[3 x l.i +(1.728)2] = 0.629. k = i ( * , + 2*2 + 2*3 + *4) = i (3.163) =0.527. b b Hence, yi =>o + ki.e., >(1.1)- 1.2 + 0.527= 1.727. Ans. Example 4.32; Use Runge-Kutta method to approximate the value o f y when x ^ O .l given thaty(O) [RGPV Dec. 2008 (N)J Solution. Proceed as above example. Taking x0 = 0, yo = 1, h ** 0 .1 . Then we get > l =>o + k = 1 + 0.12725 = 1.12725.

dy Example 4.33 ; Solve f a

Aw.

x + y fo r x = 0.5 by using Runge-Kutta Method with x0 - 0,

..>0 = / (take ft *=0.5).

Solution.

Here,

xo = 0, >o = 1, h = 0.5 and/x, >) =

Then

xj = xo + * = C.5.

(* + >)

N umerical A nalysis-11!, C orrelation & R egression | 3 2 3

(0.5^0.25, 1.25)

‘ (0 5>( o.25

’ <0 '6) i 1 * I ‘ 0J3333

*3 = h f \ x o + | , y0 + y ) = (0-5)/^0.25, *4 =

= 0.35294

>o + *J) = 0.5/0.5, 1.35294) (0.5+”T35294) = °-26984.

Hence,

/t = •“ (*i + 2*2 + 2*j + *4) = —(2.1423) = 0.35705. 6 ” 6 yi = yo + k i.e.,y(Q.S) = 1.35705.

Ans.

dy Example 4.34 : The unique solution o f the problem ^ - ~ x y ,y ( 0 ) = 1 is y = e~x i a . Solution.

Find approximate value ofy(0.2) using Runge-Kutta Method. Here, xo = 0, yo ~ 1,fix, y) = -xy, taking h = 0.2, Then X] = xo + h = 0 + 0.2 = 0.2 We have *1 = hfipco, yo)= (0.2)/0, 1) = 0

= (0.2X- 0.1 x I) = -{0.2X0.1) = - 0.02 = k f \ x o + | . y0 + - ~ j = (0.2)/ - (0.2)/0.1, 1 -0.01) = (0.2)/0.1,0.99) = 0.2(- 0.1 x 0.99) = 0.2(- 0.999) = 0.198. *4 - hfix0 + h,y$ + *3) = 0 .2 /0 .2 ,1 - 0.0198) = 0.2A0.2,0.9802) = 0.2(- 0.2 x 0.9802) = 0.2(- 0.19604) = ^ 0.039208 * = - ( * , + 2*0 + 2A3 + *4) = -(-0 .1 1 8 8 0 8 ) = -.019801.

6 6 Hence/1 -y o * ki.e.yy(0.2)= 1 -0.019801 = 0.9802.

Ans.

Since y = er*m solution of given problem, Then atx = 0.2, is exact value = e~< '2',2/1 = 0.98012. Hence, error = 0.9802 - 0.98012 = 0.00008.

Ans.

3 2 4 | E n g in eer in g M ath em atics -111

dy Example 4.35 : Apply Runge-Kutta method to — =x y a, >(1)= 1 to obtain y(1.1). dx Solution. Here, - 1, yo ~ 1 and A= 0.1 then X| = x<>+ A = 1 + 0.1 = 1.1. We have kx = Afco, y 0) = 0.1(1X1),/J = 0-1 l h = A / j x 0 + | , Jo + j ] = 0 . l ( l + ~ ) ( l + ^ j 3 =0.10672 i

1.A h ,y0 + — h )J * n0.1^1 A* + — 0.1V, 0.10672 ky = A /1 x0 + — |^1 + ------ j>3 -0.10684 *4 =W.x0 + h,y0 + ki) = 0.1(1 +0.1X1 + 0.t0684),/3 = 0.11378.

Hence

k = i( 0 .1 + 2 x 0 .1 0 6 7 2 + 2 x 0 .1 0 6 8 4 + 0.11378) = 0.10682 6 ' y } =yo +k i.e.,y(lA) = I +0.10682 = 1.10682.

Ans.

dy Example 4-36 ; Solve the initial value problem ~ ^ = -2 xy* withy(O) * / and A = 0.2 on the intern! Solution.

[0, IJ. Use Runge-Kutta fourth order method. Step 1 : Here XQ= 0,yo = 1, A = 0.2 andjfc, y ) ~ - 2xy2 *i = h fao,y o ) - - 2(0.2X0X 1y = 0 k2 = hf^Xo + | , y0 +

[v

y0) = - 2x<> v02

= - 2 ( 0 . 2 ) (l)2 = _o.04

*3 = A/J^Xo + | , >-o + y j = -2(0.2) ^p j(0.98)2 = _0.38416 *4 = Aftr0 + A, y0 + h ) = - 2(0.2X0.2X0.961584)2 = - 0.0739715

* = - (*i + 2*2 + 2As + *4) = - 0.0384672. b Hence y } =y(0.2)=y0 + * = 1 -0.0384672 = 0.9615328. Step 2 : We have x\ = 0.2, yi = 0.9615328, A = 0.2, we have *1 = A /xj.y,) - - 2(0.2X0.2X0.9615328)2 = - 0.0739636 uA kA k2 = A/^Xj + -h . J 1 + -J-J =-2(0.2X0.3X0.924551)2 = -0.1025754

{

h>

ho ^

*3 = V ^ X , + - . J 1 + y j =-2(0.2X 0.3X 0.9102451)2 = -0.0994255 *4 = A /x i + A, y , + * 3) = -2(0.2X0:4X0.8621073)2 = -0 .118 9 16 6 .

* = ~ (Aj + 2*2 + 2*3 + *<) = _ 0.0994803. b

N um erical A n a ly sis -111, C o rrela tio n & R e g r e ssio n j 3 2 5

Hence y 2 = MO-4) = y, + * = 0.9615328 - 0.0994803 - 0.8620525. Step 3 : y 3 = y(0.6) - 0.7352784 Step 4 : y4 = y(0.8) = 0.6097519 Step 5 : y5 = y( 1.0) = 0.5000073. Ans. Example 4.37 ; Using Runge-Kutta method, solve y"** xy'2- y 2for x = 0.2 corrected to four decimal places. Initial conditions are x =■0, y = /, y '= ft Solution.

dy

Let

y =

So that

y" = ^

dz

= z = A x , y , z ) . say.

[v y '= 2 ]

= xz2 - y 2 = (x, y, z), say.

Here x0 = 0, yo ~ 1. *o ~ 0, h = 0.2 Using Runge-Kutta formulae : we have *l = tyfro, yo, Zo) = 0.2(0) = 0 k2 = =

iJ

h

h U y o + f >Zo + ^ J

[v y -Z =>=o = yo' = 0] /i = h# (xo, yo, 20) = 0.2(- 1) = -0.2 '

h *, n h = **(*> + 2 ,y° + 2 ’Z° + 2, = 0.2 ^(0.1, 1, -0.1) = 0.2

0 . 2 ( - 0 . 1) = - 0.02

[(0.1K -O .1)2 - 1] = - 0.1998

*3 = h*rf ^( x 0 + -*, y o + ^*2, z 0 + ^^ ) = 0 . 2 ( - 0.09 9 9 ) = - 0 . 0 2 *4 = hj[xo + h, y 0 + *3, z0 + /3)

h-h*{*

= 0.2(-0.9791) = -0.1958 l4 = h(xo + A, yo + *3,20 + 6 ) = 0.2(^0.9527) = - 0 .1 9 0 5

= 0 . 2 ( - 0.1958) = - 0 .0 3 9 2

k - ^ (* t + 2*2 + 2*3 + fe4) b

/ = —(/j + 2 fj + 2/3 + /4) b

= - 0 .0 1 9 9 Hence, for x = 0.2,

and

= - 0 .1 9 7 0

we have y =>’0 + *= 1 ~ 0.0199 = 0.9801 /.e.,y(0.2) = 0.9801. y' = z = z0 + / = 0 - 0.1970 = - 0.1970. i.e., y'(0.2) = - 0.1970.

Example 4 3 8 ; .S’o/ve ~

= y z + x;

— = xx + y , g/ven /Aa/ y (0) ■* /; z(0) = - / fo r y(0.l),

z(0.l). Solution.

Here ^

= yz + x =

Ans.

/i(x ,y , z), s a y and ^

= x z + y = f2( x , y , z ) , say

G iven : xq = 0, yo = I, Zq = - 1, and h = 0.1, w e have

k\ = hMx0,y 0, zq) = (0.1)[(1) (- 1) + 0] = - 0.1, l\ = hf2(xo, y0, 20) - (0.1 )[(0X- 1) + 1)] = 0.1.

326

| E n g in e e r in g M a th e m a tic s -IH

uf f ,.2 = hfi i

h A ) + - , Jo + ~ . *o + ~ . J = h/](0.05,0.95, - 0.95)

=(0.1)[(0.95X-0.95)+0.05}=- 0.08525, ( H h h \ /2 « hf2 1x0 + - - ,y 0 + - ± , zo + * , J * h/2(0.05, 0.95, - 0.95), = (0.1)[(0.05)(- 0.95) + 0.95] = - 0.09025. ,, ( h *2 Iq ) kj = */, I x0 + - , y0 + y >*o + J -I - A/i(0.05, 0.957375, - 0.954875) = (0. !)[(0.957375)(- 0.954875) + 0.05] = - 0.0864173, /3

=

*/2 I x0 +

- , y 0 + y

>z o

+ 2“! =*6(0.05,0.957375,-0.954875)

(0.1 )[(0.05X-0.954875) + 0.957375] * 0.0909631. h = */i(xo + h,y0 + kh 20 + 13) = */i(0.1,0.9135827, - 0.9090369) 0.1 01

=( )((0.9T35827X-0.90903&)+ . ]=- 0.073048 U = V i i x o + h , y o + k h z 0 + / j ) = */2 (0.1.0.9135827,- 0.9090369) = (0.1)[(0.1X- 0.9090369) + 0.9135827] = 0.0822679. k = ~ ( k , + 2*2 + 2*g + *4) b

= A [-0.1 + 2(-0.08525) + 2(-0.0864173) - 0.073048] =-0.0860637 6 and

I = —(.h + b

+ 2lg + l4)

= A [0.1 + 2(0.09025) + 2(0.0909631) + 0.0822679] = 0.0907823. 6 Hence, y , = y(0.1) = y0 + k = 1 = 1 and

4.8

0.0860637 0.9 39363 z, =z(0.1)=z0 + /=- 1+0*0907823=-0.9092176.

Ans.

MILNE’S PREDICTOR-CORRECTOR FORMULAE It is a simple and accurate method for solving the differential equation. If we want to find the value of>> using this method, we require four values of x andy namely, X0,xi,X 2,xj andyo,y\,yi, y$ such that h = x<-x<_ j. Consider the differential equation, dy y' = d x or Euler method we calculate y\ - Xxi'), y j - y(x2), y i - .yfo), by using Picard’s method or Taylor’s series method or Runge Kutta method.

N umerical A na lysis -III, C o rrela tio n & R e g r e ssio n | 3 2 7

Next we calculate, /o

= X * o > M

V\ =A*uy\), y'2=A*2,y2\ / } =Ax* yi)-

Then to find >>4 = y(x4) by : 4h >>4 = >o + ~ [2>! - >2 + 2^ 3] o Equation (I) is known as Milne 5’ Predictorformula. Now calculate / 4 =Xjc4, y4) and apply :

Predictorformula :

Correctorformula :

h y+ = y-i + —[y'2 + 4y'a + yi]

...(1)

...(2)

Equation (2) is known as Milne’s Correctorformula. Again apply corrector formula till we get the difference between any two corrected value less than the desired accuracy. Continue this process till we get the required result. Remark: To apply Milne method, we require four starting values of y which are calculated by means of Picard’s method or Taylor’s series method or Eultfr’s method or Runge-Kutta method.

4.9 WORKING RULE FOR MILNE’S PREDICTOR CORRECTOR METHOD Step-1:

First we have to calculated : yo =f (xo,yo) ,yi' = f ( x h y l) , y 2l = /(x 2, y 2)

yi =f(xi,yi)

Step-2:

To find :

Sten -i:

Apply Predictor formula :

y\ = /(*< )•

y* =yo+ — [2y ' \ - y \ +2y'i\ 'Step-4 :

Then obtain the value of v4' = / (x4, yi). Again apply Corrector formula : yi = j’2 + the corrected value o f y* - y (X4). If required, again apply corrected fonnula till to get the difference between any two corrected values less than the desired accuracy. Continue this process til! to get the required result, i.e., to find y s =f(x,) For apply Predictor fonnula : which is

Step-5 : &SP-6:

ys =yi+ y [ 2 / 2- / j + 2 / 4]

Then obtain the value of y i - f ( x j, y 5) Again apply Corrector fonnula : ys = yj + | [ y ,3+4y,4+y'j] which is the corrected value z . y 5=y (x5) and so on.

3 2 8 | E n g in eerin g M ath em atics-1 1 1

dy Example A M : The differential equation ^ r; = 1 + y* satisfies thefollowing sets i f values o fx andy

Solution.

x : 0 0.2 0.4 0.6 y : 0 0.2027 0.4228 0.6841 Compute y (0.8), using Milne’s method. Here, x0 “ 0, xt = 0.2, x2 = 0.4, x3 = 0.6, yo = 0, y } = 0.2027, f t = 0.4228, yy = 0.6841, and h = 0 .2 , then *4 = x$ + h = 0 .8 . Since,/0 =A*o,yo) = I + j*>2 = I y ’l = I +yi 2 = 1.0411 y’z = 1 +y22 = 1.1787 y 'i = I + y s2 = 1.4681. Now using Milne predictor formula : 4h y A = y 0 + Y3 ^2 y i ~ y '2 + 2y^ =0+

4x0. 2

[2 x 1.0411 - 1.1787 + 2 x 1.4681] = 1.0239

Then, >>4' =J{x4,y4) = I + y A2 = I + (1.0239)2 * 2.0480. Using Milne corrector formula : yA = y-i+ |Lyo

+ 4^9 + y\ ]

= 0.4228 + — [1.1787 + 4 x 1.4681 + 2.0480] = 1.0294. 3 Hence, y4 = y(x4) = X0.8) = 1.0294.

Am.

dy Example 4.40 : Apply Milne's method tofin d a solution o f the differential equation ^ = x ~ T in Solution.

the range 0 £ x £ I, fo r the boundary condition y - 0 a tx = 0. First we compute y lf y j, y y Using Picard’s method : Here, / =/{x,y) - x - y 1, x0 = 0,y 0 = 0, taking h = 0.2. So that x 1 = x 0 1 - h = 0.2, x2 = 0.4, x j = 0.6, X4 = 0,8, X5 = 1.0. .•. Using Picard’s method :

[RGPV Dec.

= y0 + £ f ( x ,y 0)dx = 0 + £ x d x = y .

v(2) = y0 + £ f { x ,y ^ ) d x = 0 + £ * “ ( ^ r )

e.

x2

y = T ~ 20 for computing>>,{x) :

dx 20

N um erical

Analvsis-111, Corrioation & Regression |3 2 9

At,

x , = 0.2, then y x =><0.2) =

At,

x2 = 0.4, then y i =>’(0.4) = — r ----------- — — -

A t,

(0.6)* ,i 0 .6 )5 jf3 = 0.6, th e n y j =><0.5) - — 0----------- = 0 .1 7 0 . 4# 20

Hence,

2

20

(0.4)*

i0 .4 )r'

2

20

- 0 .0 2 .

, 2 - ( 0 . 0 2 r - 0.19

yi yi

~ A xi- yi) ~X 2 ~ yz2~ 0-4 - (0.0795 y = 0 .3 9 3 ' ~ A xh

yi) = *3 -

j ’32 = 0*6 - ( 0 .1 ?6 )2 = 0.5690

Now using Milne predictor formula : Ah y 4 = yo + — [2^1 - y i + 2 ^ 3 ] 3

= 0 + 4 * ° ‘2 [2 X 0 .1 9 9 6 - 0 .3 9 3 7 + 2 x 0 .5 6 9 0 ] = 0.3049. 3 H ence,

/ A = / x 4) y A) = 0.8 - (0.3049)2 = 0.7070.

U sing MiJne correcto r form ula :

h,

y 4 = y-> + - b i + 4 ,V3 + y U = 0 .0 7 9 5 + — [0 .3 9 3 7 + 4 x 0 .5 6 9 0 + 0 .7 0 7 0 ] = 0.3046 3 H ence. y A at x = 0.8, is y (0 .8 ) = 0.3046.

Then,

y 4’ = / * 4, y A) = 0.8 - (0.3046)2 = 0.7072.

A gain, using M ilne predictor formula, w e get

y s - y , + Y l 2 A ~ y> + 2 y't ] = 0 .0 2 + 4 X 0-2 [2

3

yi

X

0 .3 9 3 7 - 0 .5 6 9 0 + 2

x

0 .7 0 7 2 ] = 0.4554

= / x 5, ys) = I - (0,4554)2 « 0.7926.

U sing M ilne correcto r form ula : ys = J s + j b s

+4y; + Js 1

= 0 .1 7 6 0 + —

3

Hence,

><*5) =><•)“ 0,4554.

[ 0 .5 6 9 0 + 4 x 0 .7 0 7 2 + 0 .7 9 2 6 ] = 0.4554.

Ans.

3 3 0 | E n g in eer in g M athem atics -!!!

dy Example 4.41: Solve numerically the differential equation — ■*x + y with y (0) = /, Solution.

h] W ine’s predlctor-corrector methodfrom x m 0 to x m 0.4. G >'pn that y1 = x + y andx<> = 0, yo = I. Since / =1 +y => yo1 = *o+yo= l» yit = j ^ ^,0it = 1 + yfli = 2, >

= y

=>

yo"1 = yo" = 2,

>v = => y0IV = 2 and so on. Taking/» “ 0.1, so that X) =xo + A= 0.1, x2 = 0.2, X3-0.3, x4 = 0.4. First we have to find y i,y 2, ya by Taylor’s series :

( x -x 0)2 it ( x - x 0)3 lU y (*) =yo + (x-xo)yol + — — >o + — ^ — >o +•■ / x _ . / \ i . (* | --* 0 ) .. H . (.X, Xq) y(x ,) = y o + (x < -* o )y o ' + — ~ t— y 0 + — j ,—

x >o

where /= 1,2,3. (i) For / =1 : Put Xj = 0.1, and Xo, y0yo1. V , yom and so on, we get y(0.1) = 1 + (0.1) * 1 + ~

“ *2 + ^ - x 2 = 1.1103

Hence y, = 1.1103 (ii) For i - 2, Put x* = 0.2 in (1), we get y (0.2) « 1 +(0.2) * I +

(02)2 „ (02)3 „ ^ - x 2 + ^—^ - x 2 = 1.2427 «

J*

i.e., ^ « 1.2427 (iii) For i - 3, Put x3 - 0.3, in (1), we get

y (0.3) = 1 +(0.3)x 1 + LJCOJ'i2T x2 + ^COJ'I1f ' x 2 “ 13990 i.e., y} = 1.3990 Since y 1 - f (x, y) = x + y. Then, we have y,» = x , +yi = 0.l + 1.1103 = 1.2103 y ix =*2 +yi = 0.2 + 1.2427 = 1.4427 yj' =x3 +>>3 = 0.3 + 1.3990 = 1.6990 Now using Milne predictor formula : Ah >4 2 1 2 3

=yo+-r[ >'i ~y\ + yj]

= 1+ f ( |l ) [2 x 1.2103 -1.4427 + 2 x 1.6990] = 1.5835

III

+ ......

.-0)

N um erical A n a lysis -111, C o rrela tio n & R e g r essio n | 3 3 1

y 4' = x4 + y 4 - 0.4 + 1.5835 = J.9835 U sing Milne corrector formula Hence

h

y4 = y? +'

= 1.2427 +

( 0.1)

1.4427 + 4 x 1.6990 + 1.9835J = 1.5834

H ence y (0.4) - y ( x 4) = 1.5834 Ans. Example 4.42: Using Milne predictor-corrector method, obtain y (0.4) fro m the given set o f tabulated

values and f&L =y2-xK dx x ; 0 y : 1 y' : 1 Solution.

0.1 l.U

1.22

0.2

0.3

1.25 1.52

1.42 1.92

G iven that x0 = 0, x t = 0.1, x 2 - 0.2, x j = 0.3 and h ~ 0 . l T hen Xj = xq + 4 h. Then we have to fin d y4 - y (0.4). A gain given that yo = U = 1-11, y 2 = 1-25, y 0' = h y , ' = 1.22, ^ * = 1.52, and )S = f ( x , y ) = y 2-x> U sing Milne predictor fo rm u la :

y* =jo +

y j = 1.42, * > = 1.92. ...( 1)

4h\

*[2y\ - y \ +2y]]

= I + ^ M [ 2 x 1 2 2 - U 2 + 2 x l .9 2 ] = j .635

y

Since

= / ( x , y)

= y2- x 1

*= ~ y4 u? — y*1 -xr?i

=>

= (1.635)2 - ( 0 . 4 y = 2.513

A gain using M ilne corrector form ula

h 3 = 1 .2 5 + — H ence

[132 + 4 x 1 .9 2 + 2 5 1 3 ]= 1 .64

y ( 0 .4 ) = y4 ~ 1.64

A ns.

Example 4.43 : Using M ilne’s predictor-corrector method, obtain y (4.4),fro m the given set o f tabulated values and Sxy’ + y * - 2 “ 0.

x

:

y : y' : Solution.

G iven x0 = 4.0,

yo = i.o, yo1 = 0.05,

4.0 1.0 0.05

4.1 1.0049 0.0483

4.2 1.0097 0.0467

4.3 1.0143 0.0452

y . 2 ^

0)

y

5x x, = 4 .1 , yx = 1.0049, V = 0 .0 4 8 3 ,

*2 ■4.2, • yi ■1.0097, y l = 0.0467,

x 3 = 4.3 then x4 = 4.4 and h - 0 . I

y i — 1.0143 th en find y 4 = y (0.4)

y 3' = 0.0452.

3 3 2 | E n g in e e r in g M athematics -111

Using the M ilne predictor form ula, we have

y4 =>'o+ ~r[2y* ~y\ = 1.0 +

x A.0483 - 0.0167 + 2 x 0.0452]

*= 1.0187 Since

2 - r — r '—

v'

2 - .v ;

2 - ( 1 .0 l 8 7 ) 2

=> v* - --- ------ - ----- r-----------3Jf4 5 x 4.4

5.t

i.e.. v / - 0.0437 New. vising M ilne-corrector form uia, we have I j - V> I-

hi_j 3

-- .

+ 4 .1*1 + v ’ j

-i ®

[0.0467 + - !0 .0 4 5 2 -r 0.0437] j

■- 1.0187 H cnce

>- (0 .4 ) ~

- 1.0187.

Ans.

Example 4 .4 4 : Using Milne V method to fin d y (0.3) fro m ~”

Solution.

- x * + y 2 .* y ( 0 ) - J. Find the initial

values \m (—0.l),y(0.l) a n d y (0.2) fro m the Taylor’s series method. fRGPV June 2006 and June 2007/ First w e c o m p u te y \, y i , >'3. H ere. >*' -fi x , y ) = x2 + y2, x q = 0 , y 0 = 1 . taking /; = 0 .1. Since / = x 2 + v2 => yo’ = !. D ifferentiate : v" = 2x + 2yy‘ =>>0” = 2 x 0 + 2 * l x l = 2 ; y ‘" = 2 + 2(yy" + / 2) = * ^ ’"(O) - 2 + 2 { 1 * 2 + I2) = 8. .*. By T ay lo r's series : (x - x 0)

)ix) - y0 + or

,

yo +

(* - *o)2 « (x - x 0)3 * 2 1 ^ + 3! -v° + -

itv ) -- 1, + X + X- + -4X 3 3 •+■....

O

4

Put

x = - 0.1, we g et> < - 0.1) = 1 - 0 . 1 + (0.1 )2 + - ( - 0 . 1 ) 3 = 0 9087

Put

X = 0.1, w e getjv(0.1) = 1 + 0 .1 + (0 .1 )- + - ( 0 . 1 )3 = 1.1113.

Put

x = 0.2, w e g e t;< 0 .2 ) = 1 + 0 .2 + (0 .2 )2 + i ( 0 .2 )3 = 1 .2506.

O

4

O

Hence points a r e : x _ i = - 0 . 1 , x 0 = 0 ,x i = 0 . 1 ,X2 = 0-2 ,X 3 = 0.3, a n d /j = 0.1

y - 1 = 0 .9 0 8 7 ,> 0 = l,yi = 1.1113, >5 = 1.2506.

N umerical A n a lysis -III, C o rrelatio n & R e g r e ssio n j 3 3 3

y d =X*o, yo) = *1 + y i - o2 + i 2 = i . yi - / x t . t t ) - * ? + y f = (0 .1 )= + (1.1113)2 = 1 .2 4 4 9 . y 2 =AX2*>'2) = X- + y f « (0.2)2 + (1.2506)2 = 1.6040. U sing M ilne p redictor form ula :

yi = J-i + y l-J'o ~ Jt ■*" “^'2J = 0 .9 0 8 7 + - X —-1. [2 x 1 - 1 .2 4 4 9 + 2 x 1 .6 0 4 0 ] = 1.4371.

y ’i =f(xy v3) = x t * v-

(0 .3 £ • ( 1 . 4 '7 1): = 2.1552.

U sing M ilne correcto r form ula : Vj = y\ + ^ f .v i + 4 v j + v ,] = 1 .1 1 1 3 + y - { l 2 4 4 9 + 4 > 1 .6 0 4 0 ^ 2 .1 5 5 2 ] = 1.4385. H ence.

yj - ) i x j ) =y(0.3)z 1.4385.

Ans.

Remarks: (i) The truncation error in E u le r’s m ethod ib o f O (ir) (ii) The truncation error in m odified E uler’s method is o! (iii) T he truncation error in Runge-K utta m ethod o f fourth order is o f Q(h~) (iv) The truncation error in M ilne's method is o f 0 ( /;s)

4.10

CONCEPT OF CORRELATION The correlation is a statistical tool w hich studies the relationship betw een variables, say x and v, and correlation analysis involves various mctht«dv u n i techniques used for studying and measuring the extent o f the relationship betw een tw o variables. In other w ords, correlation analysis is a stc . '•cu. pv c. -i:.. b. • v- .* can determ ine the degree o f relationship betw een tw o or m ore \ar:?.r*!e--.

4.11

COEFFICIENT OF CORRELATION The degree to w hich th e tw o variables are inter-re hi led i* m easured by a coefficient w hich is called the coefficient o f correlation. The coefficient o f correlation betw een the tw o variables x s ; i d i s generally denoted by r or rn. or p or p (x, y).

4.12

TYPES OF CORRELATION 1. Positive or Direct Correlation : If the increase (or decrease) in one variable, say x results in a corresponding increase (o r decrease) in the oilier variable say then the correlation is said to be positive or direct. For example , the height and w eight o f a grow ing child. 2. N egative o r In v e rse c o rre la tio n : If the increase (o r decrease) in one variable, sav x results in a corresponding decrease for increase) ir, the other variable say v. then the correlation is said to be

3 3 4 | E n g in eerin g M ath em atics-1 1 1

For example, the consumption of electricity and its bill. 3. Linear correlation : If the plotted points (x,,y,) are approximately on or near about a straigh line, then the correlation betweenx andy is said to be linear and see the following figure 4.2 (a) y

A

o ----------------- ** Fig. 4.2 (a) : Linear and Perfect Positve Correlation For example, the correlation between the saving and wealth of a man. 4. Perfectly Linear Correlation : If all the plotted point (x,, yi) lie exactly on a straight line, then the correlation is said to beprefectly linear and see the following figure 4.2 (b):

Fig. 4.2 (b) : Perfect Linear Correlation 5. Perfect Correlation : If the correlation is one variable x is followed by a corresponding and proportional deviation in the other variable y, then the correlation is said to be perfect correlation 6. Perfect Positive Correlation : If the equal proportional changes in the two variates x and are in same direction, then it is a perfect positive correlation and see figure 4.2 (a). In this case r = 1. 7. Perfect Negative Correlation : If the equal proportional changes in the two variates x and y are in opposite direction, then it is a perfect negative correlation and see figure 4.2 (b). In this case r = - I. 8. High Degree Positive Correlation : If the plotted points (x,, y,) fall in a narrow band and the points are rising from lower left hand comer to the upper right hand comer, then there will be a high degree postive correlation between x andy, and see the following figure 4.2 (c). y

Fig. 4.2 (c ): High Degree Positive Correlation

N umerical A nalysis *!!!, C orrelatio n & R e g r e s sio n j 3 3 5

9. High Degree Negative Correlation : If the plotted points (x,, yt) fall in a narrow band from upper left band comer to the lawer right comer, then there will be a high degree negative correlation between x and y and see the following figure 4,2 (d). v *

Fig. 4.2 (d) : High Degree Negative Correlation 10. Low Degree Positve and Negative Correlation : If the plotled points (x#, y,) are not clustered a round a straight line but are widely scattered over die diagram, then there is a very low degree of positive and negative correlation between the variables x and y. 11. No Correlation : If the plotted point (x ,, yi) lie on a straight line parallel to the x-axis or in a haphazard manner, then the variables are not correlated and see the following figure 4.2 (e). In this case r - 0. I

Fig. 4.2 (e) : No Correlation

4.13 PROPERTIES OF COEFFICIENT OF CORRELATION 1. It is measure of the closeness of *fit in a relative sense. 2. Correlation coefficient lies between -1 and +1, i.e., -1 <, r <, I. 3. The correlation is perfect and positive, if r = 1 and it is perfect and negative if r = -1. 4. If r = 0, then there is no correlation between two vaiables and thus the variables are said to be independent. 5. The correlation coefficient is a pure number and is not affected by a change o f origin and scale. 6. It is a relative measure of association between two or more vaiables. 7. Following are the diagrams for various value of r.

336 | E ngin eerin g M athematics -! 11

4.14 COVARIANCE Let (x ,, y,). i = 1, 2 . 3 .......n be a hivariate distribution, when X|, X2 .........x„ are the values of variable x and y i , y i ..... ,y„ those of y. Then Covarience between x and y is denoted by Cov. (x, y) and is defined

as: Cov. (x. y) = i 2 ( x , - x ) ( y , - y )

...(1)

n ivsj

where x -

and y -

Sv

are m eans o f x and y. respectively

Sipcc the above f<'r;r.i.tia lo r the calculations o f covariance is com plicated so we use form ula as below ;

altenative

4.15 METHODS TO S T l DY CORRELATION The various methncK to determ ine whether two variables are correlation or not are as follows : (i)

Scatter ;>r D ot D iaeram method, ^

(ii)

Karl Peam-.i;’? in e ffic ie n t o f C orrelation or C ovariance m ethod,

1

(iii) Spearm an's Rank M ethod.

We shall discuss each o f the methods in tetaii in the following ahed sections.

4.16 SCATTER OR DOT DIAGRAM METHOD This is the s im p le r m ethod to ascertain whether two variables are correlated or not and in case they are, w hat is the nature o f correlation? In this method we use the rectangular coordinate axes to mark a dot corresponding to each pair of x and y values and thus obtain as many points as the num ber o f ordered pairs in the given bivariate distribution. This diagram of dots is called the scatter diagram. B v looking to the trend of the plotted points in the scatter diagram we caR ftmn an idea as to u he: her the variables are correlated or not. The scattcr diagram m ay indicate both degree and the type of correlation. From scatter diagram we can form a fairly good, though rough, idea about the relationship between the two variables. The different types o f correlation are depicted by means of scattered diagrams as shown by figures 4.2 (a), (h). (c), (d). and (e). The advantages and disadvantages

of scatter diagram method are as follows.

A d v an tag es : (i)

It is re a d ily comprehensible and enables us relationship betw een the two variables x and y .

to form a rough idea of the nature of

(iii It is not affected by extreme observations. (iii) ,‘t is not influenced by the size o f exterme items. D isad v an tag es :

i> run a suitable m ethod if the number*of observations is very large. 1ii' Ii enables us to oh: tin an approximate estimating line or line of best fit1by this method. : 1nuijh inc^ur- or corre!atioto where the exact magnitude cannot be know. (i)

It

N umerical A na lysis -III, C o r r ela tio n & De g r e s s io n | 3 3 7

4,17 KARL PEARSON’S COEFFICIENT OF CORRELATION OR COVARIANCE METHOD The correlation coefficient r between two variables x and y is given by First Form : Cov (*,;>>) r (x, y) or r ~ or JV3r( x ) x J V ar(y) Cov (x, y) ax x c y Var(x) = o ,2,

r {x, y) or r Where and Also Second Form ,

•d)

Var (y) = c r /

Var (*) =

n

'L{.xi~x)Z(yi-y)

r or r(x, y) -

.(2)

Where x and y are means of x and y respectively. Third Form : nZxiy i - ( Z x i)(I.yi )

r or r(x,y)

^{n£*i2 - (SXi )2} x {nSy -

(ly i)2J

(3)

This formula (third form) saves a lot of computational labour and it reduces the error due to computation and rouding off. Again if the values of x i's oryi’s are large or involve fractions, then none of (1), (2) and (3) is convenient to use. In such a case use another formula : i.e., short cut method which is given by the followingform ula:

n'Zuivi -(£ u i)('Zvi)_______

r (x, y) or r =

^{»£«<2 - ( E u ;)2} * ^ ^ 2 ~ (£ v ,)2} lu v -or

r =

{S«)(Zv) •(4)

where u, = xf - a and v , = y , - b and o and by being assumed means of x & y series respectively. Note : It is advised to calculate the correlation coefficient by using short-cut method* fonnula only for easier calculations. Remarks: \ . r ( x , y ) = r(u, v) = r 2. r2 is called cofficient of determination. 3, S.D. a ■ where u = x - A ,

A being assumed mean.

3 3 8 | E n g in e e r in g M athem atics -!!!

Example 4.45 : Find the Kart Pearson's coefficient o f correlation between x and y fo r the following data: x : 6 2 4 9 I 3 5 8 y :

8

12

15

9

10

11

16.

Let u, = x, - a = x<-5 and v, =yt - b =yt- 12, where 5 and 12 are assumed means o f x and y respectively. 5

vj

Vi2

—12

Xi

y>

6

13

1

1

l

1

1

2 4

8 12

-3

12 0

9

-1

-4 0

9

4 -4

3

10

-2

-3 -2

12 12

16

1 3

15 9

16 0 9

4

4

9 4

5

11

0

-1

0

0

t

8

16

3

4

12

9

16

Ik, =-2

1? ii *L>

Solution :

13

Iw,v, =53

la,2= 56

Ev,2=56

•

1

16

Here n ~ 8 We have nSujUj - (£u,-) (Su*) yjnZuf -(E U j)2

-(Ei>;)2

8 (5 3 )-(-2 ) (-2)

^8 (56) - (-2)2x ^8 (56) - (-2)5 420 V444 V444 = 420 ~ 444 Ans.

= 0.946 Example 4.46: Calculate the coefficient o f correlation between x and y : 30 29 28 27 28 23 x : y

•

18

20

20

27

21

29

31 27

33 29.

N u m e r ic a l A n aly sis-1 1 1 , C o r r e l a t i o n & R e g r e s s io n j 3 3 9

Nation:

Letw, = x, -2 8 and v ^ y , -2 7 V i-y i-ll

UiVt

u?

V/2

45 7 0 0 -6 4 0 10

25

83

1

0 0 1 2 3 5

-9 -7 -7 0 -6 2 0 2

0 0 1 4 9 25

49 49 0 36 4 0 4

£«, = 5

Ivy = -25

Iu,v, = 60

Sw/2= 65

Ev,2 = 223

= Xi - 28

ui

Xi

y>

23 27 28 28 29 30 31 33

18

-5

20 20 27 21 29 27 29

-1

We have n = 8, nZtfjVj - ( 2 u i)(L v i)

r = r (x, y)

V n l u 'f - a u , ) 2 jn-L vf -(ZV i)2 8 ( -6 0 ) -( 5 ) (-2 5 )

Vs (65) - (5)2 ^8 (2 2 3 )- ( - 2 5 ) 2

=

V495 V1159

= 0.80

Ans.

tample 4.47: Calculate the Kart Pearson’s correlation coeftclent between x andyfor thefollowing data : x : ISO 153 154 155 157 160 163 164 y : 65 66 67 70 68 53 70 63 tation : Let ut = x, - 155 and v,=y,~ 68. Xi

y>

150 153 154 155 157 160 163 164

65 66 67 70 68 53 70 63

Ui = xt - 155 -5 -2 -1 0 2 5 8 . 9 Eh, = 16

v ^y i-6 8

u(v,

ui1

-3 -2 -1 2 0 -1 5 2 -5

15 4 1 0 0 -7 5 16 -4 5

25

Ev, = -2 2

Eu,v, = -84

«2 = 204

4 1 0 4 25 64 81

Vi2

9 4 1 4 0 225 4 i 25 = 272

3 4 0 | E n g in eer in g M athem atics -111

Here n = 8, We have

nZuivi -<£ui )(l.vl) r ~r(x, y) = --------------------------------,J n l u f - ( Z u £)2 x -yj n Z v f - (Zu*)2 =

8 W M * 6 )(-2 2 ) ^8 (204) - (16)2 xyjs (272) - (-22)2 - 6 7 2 + 352 Vl376 Vl692

-3 2 0 = Vl376 V1692 320 1525.84 = -0 .20 9 7.

Example

4.48: From thefollowing data obtain the value o f the correlation coefficient:

n = 10,2> c-140, Ey - 750, £(x - 70>> -- 7*0, S(y - 75^ - 2 /J and £ (x - 10)(y-15)~60. Solution : Let w= x - 10 and v = y - 15. Then Zw = Z (a- - 10) = Z x -n (10) = 140- 10 (10) = 40. Zv = Z (y - 15) = Z y -« (1 5 )= 150- 10(15) = 0. It is given that Z (jr- 10)2= 180,Z (y - 15)2 = 21 a n d Z (x - 10)(y- 15) = 60i.e., Zu1 = 180, Zv2 = 21 and Zwv = 60. , We know that

r(x,y) = r=

An

[ v Z10=I0 [ v £15 = 15

ritu.v, —(Zu;) (Zu,)

■— L*— p - —------ ■—■■■ yjnZuf -(Z u , )2 x ,JriZvf -(Z u ,)2

10 (6 0 )-(4 0 ) (0) / i 0 (108) - (40)2 x yjlO (215) - (0)2 600 V200 >/2150 600 6 V430000 V43 = 0.9151. Example 4.49 : From Me following data obtain the value o f the correlation coefficient: n = 10,£x = S S ,fy = 40, 2k1 = 385, = 792 anrf +y)2 » 947. Solution : We have Z (x + y¥ = 947 Z 2 2 = 947 => Zx2+ 2Zxy + Zy2 = 947 => 385 + 2Zxy= 192 = 947 => Zxy = 185

=>

(x + xy+v2)

An

N um erical A n a ly sis -111, C o r r ela tio n & R e g r e ssio n | 3 4 1

We know that

(by third form)

r=

1 0 x 1 8 5 -5 5 x 4 0 ^ 1 0 (3 8 5 ) - (5 5 )2 x ^ 1 0 (1 9 2 ) - ( 4 0 ) 2 1 8 5 0 -2 2 0 0 V 3850 - 3025 x ^ 1 9 2 0 - 1 6 0 0 - 350

- 350

V 8 2 5 > /3 2 0

V 825x320

Ans. ample 4.50 : From the following data compute the coefficient o f correlation between x and j>. (i) Arithmetic Mean o fx series is 25 ?nd that o fy is 18. (ii) Sum o f the products o f deviations o f x and y series from their respective means = 122. (iii) Sum o f the squeares o f deviation fro m their respective means are 136,138 respectively fo r x series and y series. (iv) Number o f pain o f values ■=IS. station : We are given the following information «= 15, x =25, y = 18. and 2 (x - x ? = 136, E ( y - £ ) 2 = 13 8 , 2 ( x - x ) ( y - y ) = 122.

~ 1 3 6 .9 9 6 ~ 0 8 9 -

Hence the coefficent of correlation is 0.89. Sample 4.51 : From the following data, fin d the number o f items n. r = 0.5, 2xy * 120, o> “ He2 * 90, where x andy are deviations from arithmetic mean.

jifatkra :

A ns.

Let arithmetic mean o fX and Y be X and Y respectively. Then x and y are deviation from arithmetic mean, i.e. x - X - X , y ~ Y - Y Now 2 ( Y - Y Y = 64 n. Also and

U X - X y = Er* = 90, Z( X- X ) ( Y - Y ) = Zxy= 120.

3 4 2 | E ngineering M athematics-1II

We know that correlation coeficient (Second form) : t(X-X)(Y-Y)

r =

V i( x - x ) 2 x V z (y -y )2 120 0.5 = VsiOx V64s

=> Squaring both side, we have

120x120 0.25 = 90 x 64n 120x120 ^ ” 90 x 64 x 0.25 IUHence number of items are n = 10. Ans. Example 4.52 : A computer operator while calculating the coefficient between two variates x and for 25 pairs o f observations obtained the following constants, n - 25, Ex ~ 125, £x* * 650, - 100, 2 p - 460, £xy - 508. It was, however later discovered at the time o f checking that he had copied down f< pairs as (6, 14) and (8, 6). while the correct pairs were (8, 12) and (6, 8). Obtain <1 correct value o f the correlation coefficient. Solution : Given:

Xt Xi y\ y2

Incorrect data

Correct data

6 8 14 6

8 6 12 8

Incorrect date are: Ex = 125. Ey = 100, Ex2= 650, Ey2= 460, Exy =508. Then Corrected Ex = Incorrect Ex - (6 + 8) + (8 + 6) - 125 - 14 +14 = 125 Corrected Ey = Incorrect Ey - (14 + 6) + (12 + 8) = 100 - 20 + 20 = 100 Corrected Ex2 - Incorrect Ex2- (62+ 82) + (82+ 6^) = 650 - 100 + 100 = 650 Corrected Ey2 = Incorrect Ey2- (142+ 62) + (122+ 82) = 460 - 232 + 208 = 436 Corrected Ery = Incorrect Exy - (6 x 14 + 8 x 6) + (8 x 12 + 6 x 8 ) = 508 - (84 + 48) + (96 + 48) = 520. We know that, correlation coefficient •>'

‘

•

_________ flEyy-(Ejf)(Ejr) - (Ex)2} x =

-(E.p)z]

2 5 x 5 2 0 -1 2 5 x 1 0 0 ^{25 x 650-(125)2} x

25 x 436 - (100)2}

.

N um erical A n a lysis -IU , C o rrela tio n & R e o b e ssio n | 3 4 3

Example 4.53 : A computer white calculating the correlation coefficient between the variables x and y obtained thefollowing constants, n -30, 23c« 120,22c2= 600, fy = 90,ty2'=2S0, Hey =356. It was, however, later discovered at the time o f checking that it had copied down two pairs o f observations as X 6 12

z id

while the corrected values were

7

X 8 10

y

12

8

Obtain the correct value o f correlation coefficient between x andy. Solution : Given : Incorrect data Zx = 120, Ey = 90, Zx* = 600, Zy2= 250, Ixy = 356 Also Incorrect values of x and y Correct val ues of x and y X

& 12

X

y

8 10

ii 7

y \2 8

Then : Corrected Zx Corrected Zy Corrected Zx2 Corrected Zy2 Corrected Zxy

= Incorrect Zx - (8 + 12) + (8 + 10) = 120 —20 + 1 8 —118 = Incorrect Zy - (10 +7) + (12 + 8) = ° ’j - 17 + 20= 93 = Incorrect X*2- ( 82+ 122) + (82+ 10') = 600-208 + 164 = 556 = IncorrectZ^-( 102 + 72) + ( 12?4 82) = 2 5 0 - 149 + 208 = 309 = Incorrect Zxy-(8 x 10 + 12 * 7) + (8 * 12 + 10 * 8) = 356 - 164 + 176 - 368. We know that, correlation coefficient r=

nZxy - (Zx) (Zy) ^{nZx2 -(Z * )2} x J{nI.y2 - ( X y f } 30 x 3 68-118 x 93 ^ 3 0 x 556 - (118)2} x ^{30 x 368 - (93)2} 66 a/2756 x >/2391

66 = 0.0257 2567.02

Ans.

Sxample 4.54 : Prove that the Karl Pearson's coefficient o f correlation independent o f change o f origin and scale. •ohition : We know that coefficient of correlation : S iV (T *. - * ) ( y - y ) r = r(x,y) = .. ■

Vz(x-x)2x^Z(y-y)2 Let

u=

x - a

and

y -b k

-(I)

(h, Jfc> 0)

3 4 4 | E n g in eer in g M athemattcs- I II

and a, b being assumed means of x and y series, respectively. =*• x - a + uh and y = b + vk n> x = a + uh and y ~ b + vk Putting these values in equation (1), we get £{(a + u h - a - uh ) (b + v k - b - vk)}

>-= r(x,y)

y/lCa + u h - a - u h ) 2 x

_

i]z(b +v k ~ b - v k ) 2

h k Z { (u -u ) (v-v)} iJh2Z(u - u ) 2 x yjk2Z ( v ~ V ^

Jz(u -u )2 xjz(v-v)2 - r ( u , v) = r

This shows that r (x, y) is independent of the change of origin and scale. Proved Example 4.55: Prove that the Karl Pearson's coefficient o f correlation r lies between - 1 and 1. Le; - 1 & r £ 1. or \ r | £ L Solution : We know that coefficient of correlation

,- r te ,) - - — 22^2<2L aL =_ V£(*«— * )2 x y j U y i - y ) 2 Let x, ~x - a, and>-; - y - b,: i = 1 ,2 ,...... n, Then

f o r i - 1,2.

Zoibi

r =

■Jza2 x J z b 2

r, _

(s°A )2 (Za,! > ab ,2) By Cauchy- Schwarz's inequalitv, we have (IaA )2 *(Za?) (Ebfl (ZQibj)2

£l

=>

( 1 6 ,) r2 <: 1 (Using I)

=»

-1 £ r :£ 1.

Proved

4.18 RANK CORRELATION METHOD

Rank correlation is the coefficient of correlation between different ranks attained by a group oi individuals in two characteristic or attributes. The coefficient o f rank correlations is given by Spearman’s by the following formula. , , =_ r 1=,1 r{x,y)

i

n (n2 ~ 1)

N um erical A n a ly sis -!!!, C o r r ela tio n & R e g r essio n | 3 4 5

where n is the total number of pairs of observations, and d} is the square of the difference of corresponding ranks and d, = (Rank x, - Ranky) or d = Rx - R y TYPE I : Wfaea the R u k are Given : Working R ale: Step 1. Calculate the differences of the ranks, i.e., d — Rx — Ry

Step 2. Square the difference d and write it under column headed by d1. Step 3. Apply the formula, coefficient of rank correlation.

Example 4.56: Two gents X and Yare asked to rank 7 different types o f shirts. The ranks assigned by them are given below : G Shirts E F A D B C 7 3 6 I 4 5 Rank given by X 2 7 6 4 5 2 Rank given by V I 3 Calculate Spearman’s rank correlation coefficient Solution : We know that rank correlation coefficient:

Where d - Rank X - Rank Y Here n = 7. Rank o fX Le; Rx

Rank o f Y Le; R,

d —Rx —Ry

d2

2 1 4 3 5 7 6

I 3 2 4 5 6 7

I -2 2 -1 0 1 -1

1 4 4 1 0 I 1 lcP= 12

Hence (1) becomes 6x12 = , 7 (4 9 -1 ) i.e.;

r = 0.786

3 = 11 4 14 Ans.

3 4 6 j E n g in eerin g M athematics *III

Example 4.57 : Ten competitors In a musical test were ranked by the three judges X, Y and Z in the following order : 7 8 9 2 4 10 3 6 5 Ranks by X : 1 9 1 6 2 4 7 10 8 5 Ranks by Y : 3 7 5 10 3 1 2 8 4 9 Ranks by Z : 6 Using rank correlation method, discuss which pair o f judges has the nearest approach to common likings in music. Solution: ll

!

^

Ranks b yX (Rx) 1 6 5 10

3 5 8 4

3

7

2 4

10 2 1 6 9

9 7

8

Ranks b yZ (RJ 6 4 9 8 1 2 3 10 5 7

d im

d im

ds~

df

d?

4 1 9 36 16 64 4 64 1 1

9 1 1 16 36 64 1 81 1 4

25 4 16

Rjf-Rf Ry~Rt Rx~Ri -2 1 -3 6 -4 -8 2 8 1 -1

-3 1 -1 -4 . 6 8 -I -9 1 2

-5 2 ^4 2 2 0 I -I 2 1

Id ,2=200 I # = 214

Total

Here n = 10 ,

*

d ,>

= 200,

4

4 0 1 1 4 1 S < # = 60

td 22= 2 !4 and W 32= 6 0

(Y — _. r (X Y ) ~ 1 " n ( n 2 - l ) ~ 1 "

r(Y’ Z ) ~ l ~ n ( n 2 ~l )

6 x 200

10(I6 b -T)

40_^£7 = p "" 33 33 0212

165

0297

Since r ( £ 2 ) is maximum, therefore the pair of judges X and Z has the nearest approach to common likings in music. Ans. TYPE I I : When the ranks are not given : In this case we assign the ranks to both the series x and y by giving the rank 1to highest values in both the series (or to the lowest values in both the series) and so on.

N umerical A n aly sis-1 1 1 , C orrela tio n & R eg r essio n | 3 4 7

'Working R ule: Step 1 : Assign ranks to each data of the both series. Step 2 : Calculate d= Rx - R y Step 3 : Square the difference d and write it under the column headed by d1. Step 4 : Apply the formula : , = 1-

n (n2 -1 )

Example 4.58 : The marks obtained by 9 students in chemistry and Mathematics are given below: Marks in Chemistry Marks in Mathematics :

:

35

23

47

17

10

43

9

6

28

30

33

45

23

8

49

12

4

31

Compute their ranks in the above two subjects and the coefficient o f correlation o f ranks. Solution : We first assign ranks to X and Y series giving rank 1 to the highest value in both the series. Marks in Chemistry (X)

Marks In Maths (Y)

Ranks in X (RJ

Ranks in

35 23 47 17 10 43 9 6 28

30 33 45 23 8 49 12 4 31

3 5 1 6 7 2 8 9 4

5 3 2 6 8 1 7 9 4

d = R ^R ,


-2 2

4 4

-1

1

0

0

1

1

1

1

1

I 0 0

Y (Ry)

0 0

lcP= 12 .

Here n = 9,

.

= 12 61d 2 , 6 x 12 _.w„ r - I -------5----- = 1------ - 5— - = u.9 n (n -1 ) 9 (9 - 1)

Ans.

Example 4.59 :10 students got the following percentage o f marks in Mathematics and Physics. Mathematics (X)

8

36

Physics (Y)

84

51

98

Find the coefficient o f rank correlation.

25

75

82

' 92

60

68

62

86

62 1 65

35

58

49

35

3 4 8 | E n g in eer in g M ath em atics-1 1 1

Solution : We first assign ranks to X and Y scries giving rank 1 to the highest value in both the series. Marks in Maths (X)

Marks in Physics (Y)

(RJ Rank in X

(Rj) Rank in Y

d**Rx - R ,


8 36 98 25 75 82 92 62 65 35 Total

84 51 91 60 68 62 86 58 35 49

10 7 1 9 4 3 2 6 5 8

3 8 I 6 4 5 2 7 10 9

7 -1 0 3 0 -2 0 -1 -5 -1

49 1 .1 1 0 9 0 4 0 1 25 1 U 2- 90

Here, n = 10, and

90 \ r => 1----- --------- = 1 - 1 1 0 100 1

Rank of correlation :

n(n - )

( -)

= 1 - ^ = 1 - 0.545 - 0.455.

Ans.

TYPE I I I : When Equal Ranks are given Le.; Tie Case : If two or more items are placed together in any classification with respect to an attribute, i.e; there are more than one item with the same value in either or both series. Then such type problem is solved by assigning average rank to each of these individuals who are put in tie. For example : Suppose an item is repeated at ranks 7, (i.e; the 7* and 8* items are having same values), then the common rank assigned to avarage of 7th and 8* is

7+ 8

g— = 7.5, the ranks which

these items would have been assigned if they were different, and the next rank assigned will be 9. Again, if an item is repeated thrice at rank 3, then the common rank assigned to each value will be = 4.5 i.e; mean of 3, 4 and 5, and the"next rank assigned will be 6. In this way, to find the rank correlation coefficient of repeated rank by the following formula : 6

r' *Where,

correlation factor =

(Id2+correctionfactor) n(n2-l)

m (* 2 - 1)—

Here m is the number of times an item is repeated.

N umerical A n aly sis-1 1 1 , C orrela tio n & R eg r essio n | 3 4 9

i.e.; The modified formula for Tie rank correlation coefficient is given by 6 [ Z
+

( " I - D + ..... ]

• n (» 2 - l ) Where, mlt m2, ...... are the number of times an item is repeated. Example : Front the following table fin d the rank correlation coefficient X : 48 33 40 9 16 16 65 24 16 57 Y : 13 13 24 6 15 4 20 9 Solution : Here in the series X, the value of 16 is repeated thrice say, mi = 3. In series Y the value of 6 and 13 are repeated twice say, m2 = 2 and mi = 2. Tie rank correlation coefficient is given by r ’

...(2)

l ~

4.60

6

...(1)

n (n 2 -1 ) Here n = 10, m\ = 3, m2- 2, m3= 2 To prepare the table : X

Y

Rank in X Rx

Rank in Y Ry

d mRx —Ry


48

13

8

2.5

6.25

33 40

13 24

6 7

5 ± 6 =55 2 3 5.5 10

0.5 -3.0

0.25 9.0

9

6

1

(2+3) 2

-1.5

2.25

16

15

7

-^.0

16.0

16 65 24 16 57

4 20 9 6 19

1 9 4 2.5 8

2.0 1.0 1.0 0.5 1.0 •V

4.0 1.0 1.0 0.25 1.0

(2 + 3 + 4) 3 3 10 5 3 9

i

Total Hence (1) becomes :

1 ^ = 41 j-

6 [41 + i 3 ( 9 - l ) + ^ 2 ( 4 - l ) + i 2 ( 4 - l ) ] r =1

i.e;

10(100-1)

6(41 + 2 + 0.5 + 0.51 _ 990 r - 0.733

264 990 Ans.

19

3 5 0 | E n g in eer in g M athem atics -!!!

Example 4.6! : Find the rank correlation coefficientfo r the following data ; X : 68 64 75 50 64 80 75 40 55 64 ¥ : 62 58 68 45 81 60 68 48 50 70 Solution ; Here in the series X, the value of 75 is repeated twice and vaiue of 64 is repeated thrice, (say m i = 2, m2= 3). In the series Y, the value of 68 is repeated twice, (say m3= 2). Tie rank correlation coefficient is given by

r = 1-

6 [ ^ 2 + U my

- 1 ) + ]]2m 2 (m 2 - l ) + j 7 > m 3 ( " t f - 1 ) ]

•••(I)

n ( n 2 - 1) Here, n - 10, m\ = 2,

= 3,

my = 2

X

Y

Rank in X

Rank In Y

ft. I > 1

Table for Calculation Rank Correlation : tp

(V 62

7

64

58

75

68

(8 ; 9 ) = 8.5

50 64 80 75 40 55 64

45 81 60 68 48 50 70

2 5 10 8.5 I 3 5

(4 + 5 + 6 ) 3

, 5

6

1

i

4

1

i

(7 + 8 ) _ 7 ^ 2 1 10 5 7.5 2 3 9

1

i

1 -5 5 1 -1 0 -A

i 25 25 1 1 0 16

© ii a

68

2^=72

Hence (1) becomes: 6^72 + i r =

J-

2 ( 4 - l ) + i 3 ( 9 - l ) + j |.2 ( 4 - l ) j 10 (100- 1)

. , [72 + 0.5+2 + 0.5] _ 1 ~6 990

N umerical A nalysis-111, C orrelation & R egression | 3 51

4.19 LINES OF REGRESSIONS A line of regression is the line which gives the best estimate of one variable 'x' for any given value of the other variable y . (i) Line of Regression of x on y : It is the equation of line which gives the best estimate for the values of x for a specified value ofy. It is given by

...(I)

x - x = r — (y-y)

wherexandyare means ofx andy series, respectively; axand ay are standard deviation ofx andy series, respectively and r is the correlation coefficient Which are calculated as follows. mean o f x : * = ~ o r n = ~ - where u= x - A (A being assumed mean) Ey Elf y = — or y = ~ »where v =y - 3 (B being assumed mean) -

S.D. of x is

\2 a , = -J— — | — J (where u = x - A ) Euv-

and coeff. of correlation is

r

—

(Eu) ( I v )

n

----------------------------------

The above regression line x on>> i.e; equation (1), can be written as : x = ay + b, where a is the slope and b is intercept of the line. (ii) Line of Regression of y and x : It is the equation of line which gives the best estimate for the values ofy for any specified values ofx. Regression line ofy on x is given by y~ y =r

°x

The above regression line (2), can be written as : y -ax + b where a is the slope and b is intercept of the line.

4.20 REGRESSION COEFFICIENTS Coeffidept of Regression ofy on x : The regression coefficient of y and x is denoted by

and is given by

av where o„ crv are S.D. of x and y series respectively, and r is the correlation coefficient.

...(1)

3 5 2 | E n g in e e r in g M athematics -]II

Coefficient of Regression of x on y : The regression coefficient of x on y is denoted by

and is given by

. _ b*y = r ctv

4.21 PROPERTIES OF REGRESSION COEFFICIENTS (1) The coefficient of correlation is the geometric mean of the coefficients of regression Le.; r

= ±

X b''yx y

Proof: We know that

and

-d ) ...(2)

K> ~ r <*X n , , SU. °y b*>x b» = r c r * r T x = r

Hence

r —± ijb^y x byx

Proved.

(2) If one of the regresion coefficients is greater than unity, then the other is less than unity Le.; If by* > 1 => bv < 1 or if bxy > 1 => byx< ! . Proof: Suppose that byx > 1 . ..•(1) We know that Since

—

r2 - 1 r1 bxy * b^

yjbxy x byx

= bxy* b^ SrS1 <1 <1 (by 2)

..-(2)

[because b^ > 1 => r ^ - < 1] °yx Proved. Hence, if bxy<\. byx <1 (3) Both regressioH'coefneients and correlation coefficient are of the same sign Le.; r > 0, if bxy and > 0, r < 0. rf bxy and b^ < 0. For example : If b„ ~ - 0.4 and b x y -- 0.8, find r. bxy S p < l uyx

Solution : v

r = ± Jbyx *bxy = ± V0.4x 078 = ± 032 r =-0.32

[because by, and b^, < 0] Ans.

N umerical A n aly sis-1 1 1 , C o rrela tio n & R e g r essio n | 3 5 3

(4) Regression coeficrents are independent of change of origin but Rot of scale. (5) Arithemetic mean of the regression coefficient is greater than the correlation coefficient. Remark: Tofind b^ and b^ by the following formulae :

(i) b „ -

------ or 4 , - -----------------------------a ----

(* * -¥ ) a » -g a g a ________ n___

l

- ________n__

n

n

422 PROPERTIES OF LINES OF REGRESSION 1. Two regression lines x o n y a n d y on x always intersect at the means ( x , y ) .

For example : Find mean values of x and y, of two regression lines are : 4 x - 5 y + 33 =0

and Solution: Given

20* - 9y - 107 =0. 4 x - 5 y = - 33 ...(1) 20x- 9y = 107 ...(2) Eliminate x, from (1) and (2), operate: 5 * (1) - (2), we get 20*-25 y = -165 20*- 9y = 107 + - 16y * -2 7 2 y = 17 Putting the value of y in (1). we get * = 13 Hence mean of x and y are 13 and 17 respectively. 2. If r = 0. then the regression coefficients are zero, i.e; regression lines are perpendicular to each other. 3. If r = ± I , then the regression lines become identical, i.e; both pass through the Common point ( * , y ) i.e;jc = * andy = y . 4. The angle between two regression lines is given by a x. Oy \ tan 0 If If

r =0

_2 , _2 a x + °y )

=> tan 0 = «> => 0 = ti/2 => lines are perpendicular 0 = O ^>0 = O =5 lines are identical.

r=±I=^tan

3 5 4 | E n g in e e r in g M athem atics - I 11

4.23 RELATION BETWEEN REGRESSION ANALYSIS AND CORRELATION ANALYSIS Correlation Analysis

Regression Analysis

1. It is relationship between two or more variables. 2. Correlation coefficient r between x and y is a measure of direction and degree of linear relationship between x and y. 3. It is symmetric in x and y i.e; r (x. y) = r(y, x) 4. The correlation coefficient does not reflect upon the nature of variable (independent or dependent variable). 5. It does not imply cause and effect relationship between the variables under study.

6.

It is a relative measure and is independent of the units of measurement. 7. It indicates the degree of association.

8. It has limited application as it is confined to the study of linear relationship between two variables.

1. Regression means stepping back or returning to average value. are mathematical measures 2. bxy and expressing the average relationship between the two variables. 3. The regression coefficient and by* are not symmetric in x andy i e.,biy* b>x 4. Regression coefficients reflect on the nature of variable i.e., which is dependent variable and • which is independent variable. 5. It indicates the cause and effect relationship between the variables. The variabJe corresponding to cause is taken as independent variable. Whereas corresponding to effect is taken as dependent variable. 6. Regression coefficients are an absolute measure of finding out the relatinship between two or more variables. 7. It is used to forecast the nature of dependent variable when the value of independent variable is known. 8. It has wider applications as it also studies non-linear relationship between the variables

Example 4,62 : Find the regression equation o f y and x and the coefficient o f correlation from the following data : Zx~60, Z y -4 0 , X x y - 1 ISO, Ik 2 - 4160, 2 y ~ 1 7 2 0 , n - 10. Solution:

We have.

_ Zx _ 60 _ , x " n "TO " 6: _ Iy 40 . v ' - n ' i o ' 4'

bxy =

Zxly 1160- 6 0 x 4 0 10 n _ (40) 172010

ZytJME 1450 -2 4 0 1720-160

910 1560 = 0.583

N umerical A n aly sis-1 1 1 , C o r r e la tio n & R e g re s s io n | 3 5 5

1150-gP*40 and

b„ = ------------- - — -------------------—

£jc2_(E x)2 n 910 3800

Since

r =+

4 1 6 0 -^ ^ lu

=0.239 x

[v b^Sc. by, > 0, then r >0]

r * + -v/0.583 x 0.239 = 0.373 Ans. Again, the regression equation ofy on x : J ' - j ' -&>*(*- * ) => y - 4 = 0.239 (x -6 ) => _y = 0.239.x + 2.566. Ans. Example 4.63 ; / f regression equation o f x on y : 5x - y - 22 and regression equation o f y on x : 64x - 4Sy —24 are two lines o f regression. Find (i) the mean values o f x andy, (ii) the regression coefficients, (iii) the coefficient o f correlation between x andy, (iv) the standard deviation ofy, i f the variance o f x is 25. Solution : (i) The mean values o fx a n d y , i.e; (jc, y ) lie on the both lines. Sx-y=22 ...(1) 64 x - 45 y =24 ...(2) Multiplying the equation (I) by 45 and equation (2) by 1, we get 2253c -4 5 y =990 64 if -4 5 y =24 on subtract, we get 161 x =966 => jc = 6. Putting the value of jc in (1), we get 3 0 - J? =22=> y =8, Hence means: jc - 6, y = 8. Ans. (ii) The regression equation o fy on x is 64x - 45y ~ 24 => 45y = 64x - 24 _ 64 24 ^ y 45* 45 64 *>i “ 45 [*•’ R.L. y on x is y = by,, x + c] Again R.L. ofx on y is =>

5 x -y = 2 2 5x = y + 22 1

•

22

x “ 5^ T

3 5 6 | E n g in eer in g M athematics -III

=>

b„ = ~

[ v R.L. x in y isx = bzy y + c] Ans

(iii) We have, coefficient of correlation Since

r = ± ,Jbyx x bxy and b„ both are positive, then r will be postive, so that -+

V45 5

= jjr = 0.533.

Ans.

(iv) Since variance of x = 25 o*2 = 25 o a , - 5. We know that

by,

=

r — x

—S

64

-----

.

8

<*y

15

5

— -*— X —~

45

40 = 13.33 .1 « S.D. of y = 13.33 Example 4.64 : I f 4x - 5y + 33 ~ 0 and 20x -9 y —107 are two lines o f regression. Find (i) the mean values o f x andy, (ii) the regression coefficients, (iii) the correlation coefficient, (iv) the standard deviation o f x if the variance o f x is 9, (v) the value o f y fo r x 15J. Solution : (i) The mean values of jc and y, i.e; ( j , y ) lie on both lines. 4* - S y = -3 3 20* - 9 y = 107 Multiplying (1) by 5 and subtracting from (2), we get 16 y = -2 7 2 => y = 17 Putting the value of y in (1) we get x =13. Hence the mean values of x and y are 13 and 17. (ii) Let us assume that R. L. o f y on x be 4 x - S y + 33 = 0 => 5y = 4 x + 33

4 ^ y 5X /. Regression coefficient yon.c : hb>r~ - i5

Ans.

...(I) ...(2)

33 5

*•*

..I

N um erical A nalysis -111, C o rrelatio n & R e g r e ssio n | 3 5 7

Also the R.L. o f jc on y be

2 0 c - 9y = 107 20* =9 'y + 107 *

9

107

20y

20

hxy - - 20 1 O Clearly

4

9

by, x bv = g x — < l

Hence our assumption is correct. 4 by, — g and (iii) Since

9 = 2q *

Ans.

r = ± Jb ^x~ b ^ [ v bn &

3 = + g = + 0.6.

(iv) Since variance of x = 9 => We know that

are positive]

Ans.

0*2 = 9 => ox ~ 3

± - * 5 5 T CTj. = 4 => S.D. ofy = 4. (v) Since the R.L. y on x gives the best estimated value of y for given x, i.e; Put* = 3 in 4 * - 5y + 33 = 0, we get I 2 - 5 y + 13 =0 => y -9. Example 4.65 : l/2 x + 3 y a:7 and 5x + 4 y m 9 are two lines o f regression. Find (i) the mean values o f x andy, (ii) the regression coefficients, (iii) the correlation coefficient between x andy.

Ans.

Ans.

3 5 8 | E n g in e e rin g M a th e m a tic s -H I

Solution : (i) the mean values of x and y, i.e; (it, y ) lie on both lines. 2x + 3 y =7 5 Solving equations (1) and (2), we get

x + 4 y =9

x- = -I?y and^ y- - 5y 17 5 Hence mean of x is -y and mean of y is y . (ii) Let us assume that R.L. of y on x be 2x + 3y = 7 => 3y ~ - 2 x + 1

V

-

2

_

32

x

7

Also, the R.L. of x on y be 5x + 4y = 9 => 5x = - 4y + 9 4 9 x=-gy+5 hb*y = ~ i5 <1 Hence our assumption is correct. Hence of regression coefficients are I.

O yx

-

2

g

A k

8 n t * w jy

(iii) v

r -±

=>

r - - ^6,* x

_

— g

—f i l l A. f15 = -0.73. Hence correlation coefficient between x and y is - 0.73.

[v

& by, both are negative]

N u m e ric a l A n a ly sis -111, C o r r e la tio n & R e g re s s io n | $ 5 9

Example4.fif>: Fora bivariate data the mean value o fx is 20 and the meanvaiue ofy is 45. The regression coefficient ofyo n x is 4 and that o fx on y is 1/9. Find (i) the coefficient o f correlation, (ii) the S.D. ofx, ifS.D . o fy is 12, (iii) the equation o f regression lines. Solution : Given that: (i)V

x = 20, y = 45, by, = 4, r= ±

=x

x byx

r = + ^ |

[v bxy& byx>0, so that r is taken positive]

2 r = + g = 0.667.

(ii) Since

oy - 12

We know that

by i =r —

Ans.

Oy

^

=> => S.D. of x is 2. (iii) R.L.j'on jc :

2 4 - ~x — 4 3 a,

Ox ~2 Ans.

=> => R.L. r ony :

y-y y - 45 y x-x

=>

x -2 0 = |( y - 4 )

=>

12

= V (* " *) = 4 (x - 20) = 4jc -35, = b ^ ty - y )

x = ^ y+ 15.

Ans.

Ans.

Example 4.68 : From the following data estimate the most Ukefy height o f a sister whose brother's height Is 70 cm. Brother's : mean height is 67 cm. with S.D. is 5.5 cm. Sister's : mean height is 65 cm. with S.D. is 2.5 cm. and the correlation coefficient is + 0.8. Solution : Let x = height of brother y = height of sister Then x = 67, y ~ 65, ox = 3.5, oy = 2.5, and r = 0.8 we have to findHL. o fy on x gives the best estimate value of y (height of sister) for given x = height of brother = 70. i.e; y - y =b}, ( x - x ) ...(I)

3 6 0 | E n g in e e rin g M athem atics -111 o ,

Now,

2 .5

= 0 -8 x a s

\ or Hence (1) becomes

y - 6 5 = y (x - 6 7 ) y = 0 .5 7 J ( x - 6 7 ) + 65 y = 0 .57L t + 26.71 P u t.r = 70 in (1), w e g et estim ated height o f sister y = 0 .5 7 ! « 70 + 26.71 = 66.68 cm .

Example

Aw

: Calculate the coefficients o f regression lines and fin d the two lines o f regression from the following data:

4 .6 8

x y Solution :

: 78 89 , 97 69 59 79 68 61 : 125 137 156 112 107 136 123 108. Let u = x - 78 and v = y - J 25, w here 78 and 125 are assum ed m eans o f x a n d y respectively To obtain the regression lines, prepare the follow ing table : x

y

u - x -7 8

v “ y -1 2 5

uv

w1

V2

78

125

0

0

0

0

0

89

137

11

12

132

12 1

144

97

156

19

31

589

361

69 59

112

-9

-1 3

117

81

961 169

107

-1 9

-1 8

342

361

324

79

136

1

11

11

1

12 1

68

-10 -1 7

-2

20

100

61

123 108

-1 7

289

289

4 289

Zx = 600

Zy = 1004

Zu = -2 4

Zv = 4

Zwv = 1500

Z «J = 1314

Zv2 = 20 J2

We have n = 8,

Mean o fx : x -

..

Zx 600 = 75 n -o a

, - zy

1004

_

Mean o fy : y = — * —5— - 125.5 n Coefficent o f regression y on x :

by, f

8

n 13H -

o

N u m e ric a l A nalysis-111, C o r r e la tio n & R e g re s s io n | 3 6 1

1500 + 12 _ 28 1 3 1 4 -7 2 23

l -2 , 7 >and

Coefficient o f regression x on y :

Zut>- ZuZu Z.2 - < ^ n

1512 2010

1500-

( - 24)(4)

8

2 0 1 2 - &L. 8

756 = 0.752. 1005

The line o f regression y o n x :

y - y = by, (x - x ) =>

y - 125.5 = 1.217 ( j c - 75)' => y = 1.217* + 34.225. The line o f regression x a n y : x - x = bv ( y - y )

=> x - 75 = 0.752 i y - 125.5) => x =0.752>'- 19.376. Ans. Example 4.69 : Find the two lines o f regressionsfrom the following data : x : 158 160 163 165 167 170 172 175 177 181 y : 163 158 167 170 160 180 170 175 172 175. Estimate y, when x ~ 164. Solution : Let A = 170, B= 175 be the assumed forx and y series, so that u =x - 170, and v = y - 175

X

u

=x-170

Zw = -12

Iw2= 526

V-2

UV

163 158 167 170 160 180 170 175 172 ; 175

-12 -17 -8 -5 -15 5 -5 0 -3 0

144 289 64 25 225 25 25 0 9 0

144 170 56 25 45 0 -10 0 -21 0

I y= 1690

Zv = -60

•

8

Zx = 1688

144 100 49 25 9 0 4 25 49 • 121

v=y-175

II

-12 -10 -7 -5 -3 0 2 5 7 11

y

W

158 160 163 165 167 170 172 175 177 181

i»*

Zkv = 409

3 6 2 | E n g in e e r in g M a th e m a tic s -I I I

Here n - 10. We have,

-

Z*

Mean o fx : * ~ ~

1688

, , 0<)

~ ~Jq ~ = 168.8

.. , ~ ly 1690 Mean o fy : y = „ = " jq " = 169Regression coefficient y on x : E u a - S a J i1 * - - - ------------ ---------------■----- »fi-

[526- t f i ]

[a .* - fi£ ]

409 - 72 _ 337 =M J9_ 526-14.4

51L6

Regression coefficient x on y :

4og_ (-12) (-60)

- f § '0-756. Regression line o fy on x :

y - y =M*~*) => >>—169 = 0.659 (x - 168.8) => y = 0.659*+ 57.761. Regression line o fx on y : x = ^ 0 - y) => * - 168.8 =0.756 ( y - 169) => x =0.756 >>+ 40.236. Estimated value o fy when x —164 : From equation (1), we have y =0.659 x 164 + 57.761 i.e. _y(164) = 165.837 Example 4.70 : From the following data, obtain the two regression equations : Sales : 91 97 108 121 67 124 51 Purchases : 71 75 69 97 70 91 39

...(1)

...(2)

Ans. 73 61

111 80

57 47.

N um erical A n a ly sis -111, C o r r ela tio n & R eg r e ssio n j 3 6 3

Solution :

Let th e variables x and y, respectively denote the sales and purchases. We prepare the follow ing table : Let u = x ~ 90 and v = y - 70

u ^ x -9 0 v -y -7 0

X

I

1

u1

V?

uv

1

1

1

91

71

97

75

7

5

49

25

35

108

69

18

-1

324

1

-1 8

121

97

31

27

961

729

837

0

0

441

714

67

70

-2 3

0

529

124

91

34

21

1156

51

39

-3 9

-3 1

1521

961

1209

73

61

-1 7

-9

289

81

153

111

80

21

10

441

100

210

1089

529

759

Zu2= 6360

Zv2= 2868

Zuv = 3900

57

47

-3 3

-2 3

Zx = 900

Zy= 700

Z« = 0

Zv = 0

H ere, n = 10. We have,

Mean o f x :

xx = ^_ , n

900 l n = 90

,, Mean ofy:

y- -- —-------700 Jq* -- 70,

10

Regression coefficient y on x : Z uLv n

3900-0 [6360-0]

n

390 = 0.613. 636 Regression coefficient x on y : Zu v -

ZuZ u

O xv---------------------

Regression line o f y o n x : y -y

= M * ~ x)

3900-0 [2868- 0] =

1.36.

3 6 4 | E n g in eer in g M athem atics - ! II

Regression line o f x on y :

x - x ^ b r y iy - y ) => x - 90 = 1.36 (y - 70) =? x = 1.36y - 5.2 Ans. Example 4.71: Find the two lines o f regression from the following data : Age o f husband : 25 22 28 26 35 20 22 40 20 18 Age o f wife : 18 15 20 17 22 14 16 21 15 14. Hence estimate : (i) the age o f husband, when the age o f wife is 19 and (ii) the age of wife, when the age o f husband is 36. (iii) the correlation coefficient between them. Solution: Let x = A ge o f husband y = A ge o f w ife Let u - x - 26 and v = y - 17, w here 26 and 17 are assum ed m eans o f x and y series respectively.

__________________________________ ___

uv

X

u ~ x -2 6

ii *

y

25

1

18

1

i

-1

22

-1 ~A

16

15

-2

4

8

28

2

4

20

3

9

6

26

0

0

17

0

0

0

35

9

81

22

5

25

45

-3

9

18

-1

1

4

20

-6

36

14

22

-4

16

16

v my - 1 7

.

16

-2

4

12

18

-8

64

14

-3

9

24

Ex = 256

E« = - 4

£«* - 450

Ev2

II

Sk


4

15

£

21

36

r-

196

-6

II

14

20

W

40

56

H ere n = 10. We have

_ _ Ex _ 256 1 * T - To"’ 256

Mean o f x :

y

Mean o f y :

y Regression coefficient y on x :

.a

»vv

= S = 1 Z 2 =172 n

10

1 ••

E u v - EuEt> n

» •

v •

n

T O U 8 _ 172.8 450 - L 6 448.4

172-

(-4)(2) .10

450-

< -4 )2

= 0 .3 8 5 .

10

= 78

E uv= 172

N u m e ric a l A n a ly s is -III, C o r r e l a tio n & R e g re s s io n | 3 6 5

Regression coefficient x on y :

Z u u - LuZv brv

172-

(-4 )(2 ) 10

7 8 - ( 2) 1 ' 10 172.8 = 2.23. 77.6 Regression line o f y on x : y - y =b}, ( x - x ) => y - 17.2 =0.385 ( x - 25.6) => y = 0.385* + 7.34 Regression line ofx on y : x -x

-(I)

= bv ( y ~ y )

=> x - 25.6 =2.23 0 - 17.2) => x = 2.23y- 12.76 The age o f husband (x) when the age o f wife (y) is 19: i.e. ; puty = 19 in equation (2), we get x = 2.23 x 19- 12.76 = 29.6 « 30 years. Hence age of husband = 30 years. The age o f wife (y) when the age o f husband (x) is 30 : i.e; Put x ~ 30, in equation (1), we get y = 0.385 x 30 + 7.34 = 18.89 #» 19 years. Hence age of wife = 19 years. Correlation coefficient:

- ( 2)

- ± $ .■yX X = + V0.385x 2.23 = 0.927

by, > 0] Ans.

Example 4.72 : The following data represents rainfall (x) and yield o f paddy per hectare (y) in a particular area. Find the linear regression o f x on y. x ■: . 113 102 95 120 140 130 125 y : 1.8 1.5 1.3 1.9 1.1 2.0 1.7. Solution : Let 120 be the assumed mean for x series;?i)d 1.8 be the assumed mean fory series, i.e.. let u = x - 120 and v - y - 1.8.

366

| E n g o c e rm g M ath em atics-1 1 1

We prepare the following table : u “ * -1 2 0

U1

113 102 95 120 140 130 125

-7 -18 -25 0 20 10 5

49 324 625 0 400 100 25

If

= 825

«M

Ix

X

y 1.8 1.5 1.3 1.9 1.1 2.0 1.7

£k2 = 1523 iy = 11.3

v -j> -1 .8

V2

uv

0 -0.3 -0.5 0.1 -0.7 0.2 -0.1

0 0.09 0.25 0.01 0.49 0.04 0.01

0 5.4 12.5 0 -1.4 2.0 -0.5

Ev = -0.13

Lv2 = 0.89

£i/v = 18

Here n ~ l. We have Mean o f x : Mean o f y ;

—_

x

—

-

Xy n

n

_ 825 st r

= -

1L3

... 117.86. 1.614.

Regression coefficient y on x : Z .u v -

Z u lu

n 18-0.279 0.89 - 0.00241 Regression coefficient x on y : x - x = b iy ( y - y ) => x - 117.86 * I9.965(y- 1-614) => x = 19.965^+ 85.637.

18-

(-15) (-0.13)

0 .8 9 - (^

3)- j

17.721 = 19.965. 0.8876

A ns.

4.24 CURVE FITTING Curve fitting means an expression of the relationship between two variables by algebraic equations. Suppose that we are given n-values*i, x2, ....... x„ of an idependent variable* and the corresponding values yu y i....... . y» of a dependent variable^ depending on x. We try to find a curve that serves as best approximation to the curves ~ f (x). Such a curve is called as the curve of best fit and the process of determining a curve of best fit is called curve-fitting, i.e., The general problem of finding equations of approximating curves which fit given sets of data is called curve-fitting.

4.25 PRINCIPLE OF'LEAST SQUARES Principle of least squares provides a unique.set of values to the constants and suggests a curve of best fit to the given data.

N um erical A n a ly sis - II I, C o r r ela tio n & R e g r essio n | 3 6 7

Suppose we have m-observations (xuyi). (*2.>^)........(x*,.y,,). It is required to fit a polynomial of degree n of the type. y = a + &x + cx2 + ....+ kx” ... (1) of these values. We have to determine the (n + 1) constants a, b, c......... k such that it represents the curve of best fit of (I). In case m = n, we get in genera) a unique set of values satisfying the given system of equations. But if m > n then, we get m equations by putting different values ofx and y in equation (1) and we want to find only the values of n constants. Thus there may be no such solution to satisfy all m equations. Therefore we try to find out those values of a, b, c, ..... k which satisfy all the equations as nearly as possible. We apply the principle of least squares in such cases. Putting x \, x2......... .. for x (1), we get y\ = a + bx i + cxy7 + .....+ toi" y i ~ a + bx2 + cx2J + ....+ fct2" ' ym' = a + bxm+ cxm2 + .....+ kx„" where y i . f t ', .... y* are the expected values ofy for x = xi, x2, ....... x„, respectively. The values y it y2, ........ ymare called observed values of y corresponding to x = xi, x2........ x» respectively. The expected values are different from observed values, the difference y ,- y i for different values of r are called residuals. Introduce a new quantity S such that s = £ 0 - y / ) 2 = Z(y, ~ a - b x , - c x 2- . . . - b c " ) 2 The constant a, b, c ..... . k are chosen in such a way that the sum of the squares of residuals is minimum. 55 dS dS dS Now the condition for S to be maximum or minimum is t - = 0 = — = ~r~ = ....= ~rrda db dc dk On simplifying these relations, we get Ly~m a +blx+.... ,+k'Lxn Ixy = a lx + b lx 2+..... +klx l x 2y = a lx 2 + b lx 3+..... + klxn*2

= number of observations]

I xny = alx" + b lx n*l +.... k lx 2n These are known as (n + 1) Normal equations and can be solved to give the values of the constants a, bf c, k. If we calculate the second order parital derivatives and these values are put, they give a positive values of the function, so S is minimum.

3 6 8 | E n g in e e rin g M a th e m a tic s - HI

4.26 FITTING OF A STRAIGHT LINE Let (*,, y,), i = l , 2....... . n be V sets of observations of related data and y - a + bx be the straight line to be fitted The residual at x = x, is e, = y, ~ f(x,) =y, - (a + bx,), (= 1 ,2 .......... n. S -

n

n

1=1

/=i

... (1)

~ a ~ bx' )2‘

-(2 )

By the principle of least squares, S is minimum 55 =0n and a — 8S = n0. da db

-

Differentiating (2) partially w.r.t. a and b respectively, we get, — = - 2 Y i y ,- a -b x ,) i.e., T ( y ,- a - b x ,) =o da

30 n " — = - 2 £ x,( y i- a - b x ,) i.e., £ 0 ^ ~ ax>~ bx^ 8b ,=1 /=i

and

...(3)

tt

=0

■« <♦)

From (3) and (4), it follows that n n n Y.y> = £ * * * & ;=1 (-1 n

and

ft

i.e.,Iy = m> + £>Lx

...(5)

tl

S r ' v' = a Z x' + x' 2 i-e-, Zxy = olx + * Ir2 /=! >=l /=!

... (6)

These equation given by (5) and'(6) are called Normal&quatiom, can be solved for determining ‘a’ and lb' X] and yt are known. cS dS Differentiating — w.r.t. 'a ' partially and — w.r.t tb' partially, we get ca oo S2S 3 aTi2

„

= 2; and —y = 2 Tx,2 = +ve. *

db2

which are positives Kies. Hence, ‘S' bSMwimum. Now putting the values of ‘a ’1and 'b7 get equation of
N u m e ric a l Analysis-111, C o r r e la tio n & R e g re s s io n | 369

4.27 FITTING OF A PARABOLA *OR SECOND DEGREE CURVE) •• = ( ' f’r •-rx2

L<*»

•••(!)

. : rVe) to be fitted for the V data (x,, yt), i - 1 ,2 ,...... n.

e a , i/u'w. ( -hi Rie r -sidual at jc = x, is

e< y, -/(* » )- y , - (- ^ z 2) r=l z«l

... (2)

By the principle of least squares, S should be minimum for the best vaJues of o, b and c. dS dS j dS = 0, — = 0 and — - 0. da db dc Using (2), these three conditions given in (3), we get Zy = na + bZx + cZx2 Lxy = a lx + bLx2 + d r 3 Zx2y = aZx2 +bZxi +cZx*

here Zx =

... (3)

... (4)

and so on. /=i

These three equations given by (4) are called the normal equations, can be solved for determining a, b and c. Putting the values a, b, c in (1), we get the equation of the parabola of best fit for the given data. Remarks: (i) Normal equations to the straight line y = a + bx are : Zy ~ ma + bZx, where m is number of data points, and Zxy —aZx + bZx2. (ii) Normal equations to the parabola y - a + bx + cx2 are : Zy - ma + b lx + cZx2 £x>> - aZx +bZx2+ c lx i Zx2y = aZx2+ bZxy + cZx* (iii) If the values in the given data are large, we changing the origin and scale by using substitutions. If m is odd i.e., number of data is odd, then we assume. x-(middle term) u ---------------------- > (h = length of interval) If m is even i.e., number of data is an even, then we assume _ x - (mean of two middle terms) M= (A/2) '

3 7 0 | E n g in e e rin g M a th e m a ttc s-H I

4.28 FITTING OF A CURVE OF THE TYPE y?=ab* Let the curve to be fitted be y = ab1. ... (1 Taking logarithm on both sides, we get logy = log a + x.\ogj> or Y = A + B x. ...{2) where Y = logy, A -\o g a and B = log b. Normal equations of (2) are : I.Y = nA + BTx (where n is number of data) ...(3) ZxY= AZx + Bljc2 ...(4) where n is number of given points. Solving (3) and (4), we obtain A = log a and B - log b i..e, a and b are found by taking antilog. Note-: Similarly we can fit the curve y = ax*. 4.29 FITTING OF THE EXPONENTIAL CURVE y = ae** Let the curve to be fitted be y - ae** , ... (1) Taking logarithms of both sides of (1) to the base 10, we get togio y = logio a + bx log,0 e ... (2) This is of the form Y = A + Bx ... (3) wher Y= logio y, A - log,0 a, B = b log)0 e. :. Normal equations of (3) are : Ly = nA + B lx ... (4) XxY =ALx + B lx2 ...(5) 1 Solving (4) and (5), we get A and B from which a and ft can be determine. !

430 FITTING OF A CURVE OF THE TYPE y - a*2 + ~ For obtaining normal equations : The squared sum of errors S dS i.e., ■*

J

2

»V ~ ~ x) ** niinimum.

= 0 given lx , y? ~ alx,* + iEx,

Ixy2 = aZ xt+ blx dS v y( — = Ogives = aZx,+ b l Ou

...(1) 1

2

From the normal equations (1) and (2), a and b are obtained, and as such the desired curve is fitted. v*

N u m e ric a l A n a ly sis -111, C o r r ela tio n & R e g r e ssio n (, 3 7 1

Example 4.73 ; The weights o f a calf taken at weekly Intervals are given below. Fit a straight line using the method o f least squares, and calculate the average rate o f growth per week. A ge: 1 2 3 4 3 6 7 8 9 10 Weight: 52.5 58.7 6.5 70.2, 7S.4 81.1 87.2 95.5 102.2 108.4 Solution. Here number of values is 10, i.e., even, so we changing the origin and scale by put x - (mean of two middle terms) u ----------(h i 2)

-

r

a

1/2

x-55 u = — j - and v = y.

i.e.,

Let the straight line to be fitted be v = a + bu Its normal equations are ; Iv = ma + bZu I w = d a + bZu2

—(I) ... (2) ... (3)

*

v my

* -5 .5 M~ 0.5

uv

1 2 3 4 5 6 7 8 9 10

52.5 58.7 65.0 70.2 75.4 81.1 87.2 95.5 102.2 108.4

-9 -7 -5 -3 - 1 1 3 5 7 9

- 472.5 -410.9 - 325.0 -210.6 -1 5 A 81.1 261.6 477.5 715.4 975.6

81 49 25 9 1 1 9 25 49 81

Sv = 796.2

Z««0

Ii»v= 1016.8

I n 1 = 330

Putting these values in (2) and (3), we get 796.2 - 10 a and 1016.8 = b (330) => a -79.62 and * = 3.081. The required equation of curve of fit is v =(79.62) + (3.081) u

[From(l)]

or y .= 6.162*+4£729. Thus, the average rate of growth per week is 6.162 units.

'if!

Ans.

3 7 2 | E n g in eer in g M athem atics -111

Example 4.74 : Fit a straight line to the following data, taking y as the dependent variable x: I 2 3 4 5 6 .7 8 9 y: 9 8 10 12 11 13 14 16 15 Solution. Since m ean o fx -se rie s is 5 and m ean o f > s e r ie s is 12. [i.e., use changing the origin] Let * = * - 5 and Y = y - U Let the straight line to be fitted be

Y= a+ bX

...(I)

Then its norm al equations are :

ZY = ma + bZX ZXY =aZX+bZX>

... (2) ...(3)

w here m is num ber o f d ata i.e., m - 9.

X

x -S

X

y

\

9

-4

2

8

-3

=

Y - y - 12

'

X2

-3

12

16

-4

12

9

-2

4

4

-2

12

- 1

0

0

1

5

n

0

- 1

0

0

6

13

1

1

1

1

7

14

2

2

4

4

8

16

3

4

12

9

9

15

4

3

12

16

ZXY = 5 7

ZX2 = 60

ZX = 0

<5&

10

w

3 4

tl

XY

Putting these v alues in norm al equations (2) and (3), w e get 0 = 9a + 0 and 57 = 0 + 60 b = > 6 = 0.95 T he required equation o f curve o f fit is

=> a = 0

Y = 0 + 0 .9 5 * y - 12 = 0 .9 5 ( j f - 5 ) [ \‘ X = x - 5 , Y= y~ 12] y = 7.25 + 0.95x Ans. Example 4.75 : The results o f measurement o f electric resistance R o f a copper bar at various temperature fC are listed below t: 19 25 30 36 40 45 50 R: 76 77 79 80 82 83 85 Find a relation R = a + bt, when a and b are constants to be determine. Solution. H ere num ber o f data m = 7. or =>

Let th e given relation to be fitted to the data be :

R = a + bt

N um erical A n a ly sis - !!!, C o r r ela tio n & R e g r e ssio n | 3 7 3

its normal equations are : Z* = ma + bZt ZtR = aZt + bZt2.

... (2) ... (3)

t

R

tR

fi-

19 25 30 36 40 45 50

76 77 79 80 82 83 85

1444 1925 2370 2880 3280 3735 4250

361 625 900 1296 1600 2025 2500

14 ~ 245

Z R ^ 562

Irtf “ 19884

Zt1 = 9307

Putting all values from the table in normal equations, we get 562 = la + 2456 19884 = 245a + 93076 On solving equations (3) and (4), we get a = 70.06, b = 0.292 Hence the requried relation (1) becomes : R = 70.06 + 0.292f Example 4.76 : Fit a second degree parabola to the following data 4 2 x: I 3 1390 1625 y: 1090 1220 Here m - 5 (number of data). Solution. Changing of Origin u = x - 3 and v =

...( 3 ) ... (4 )

Ans. 5 1915

y - 1220

Let the parabola of the curve be v = a + bu + cu2 Its normal equations are Zv = ma + bZu + cZu2 Zav = aZu + bZu2+ cZ«3. Z«2v = aZu2 1 6Za3 + cZu*. y

u = x -3

1 2 3 4 5

1090 1220 1390 1625 1915

-2 - 1 0 1 2 II

X

v=

(y~1220) 5 -2 6 0 34 81 139 Zv - 228

( 1)

u2

u3

u4

uv

u2v

4 1 0 1 .. 4

-8 - 1 0 1 8

16 1 0 1 16

52 0 0 81 278

-104 0 0 81 556

Zu3^ 10

Zu3 •=0 Zu* = 34 Zuv = 411 I « 2v ■=533

3 7 4 { E n g in eer in g M a th em a tk s -JII

Putting all values from the table in the normal equations we get, + A(0) + c(10) = 228 a(0) + /?(lO) + c(0) => A= c(10) + 6(0) Sovling (2) and (4), we get a - 34.6 and c = 5.5 Heace(l) becomes: v = 34.6+ 41.14« +5.5a 2

5a

=411 +c(34)=533.

O '-1220)

or

...(2 ...(J

41.1

-{4!

= 34.6 + 41.1 (x - 3) + (5.5) (x - 3 f

=>

y = 1220 + 173 + 205.5 (x - 3) + 27.5 (x - 3J2 y » 1024 + 40.5x + S7.5*2. Ant Example 4.77: Fit a parabolic curve o f regression o fy on x to the 7 pairs o f values (orfit a secom degree parabola to the following). 4.0 3.5 2.5 x: 1.0 1.5 2.0 3.0 3.4 4.1 2.7 2.0 y: 1.1 1.3 1.6 Solution. Here m = l and changing the origin Suppose

u-

x -2 5 = 2x - 5 and v =y. 0.5

Let the parabola or second degree parabola of the curve be v = a + bu + ew2 Its normal equations are Ev = ma + bZu + cZu2

•d)

Zuv = aZu + bZu2 + cZu3 Zu2v = aZu2 + bZu3 + cZu4.

X

y

u -2 x -5

ym y

1.0 1.5

1.1

-3 -2 -1

1.1 1.3

2.0

2.5. 3.0 3.5 4.0

1.3 1.6 . 2.0 =“ . o 2.7 ■— 1 2 3.4 4.! 3 Z u -0

1.6

;>o ---- -2.7 '3.4 4.1 Iv - 16.2

'

9 4 1 0 1 4 9

uv

u2v

-3 .3 - 2.6 - 1.6 0 2.7 6.8 12.3

9.9 5.2 1.6 0 2.7 13.6 36.9

Z u*-28 Zuv - 14.3

u4 -2 7 -8 : -1 0 i * 8 27

Zu2v~69.9

81 16 1 0 1 16 81 Lu< = }96

N um erical A n a ly sis -U I, C o r r ela tio n & R e g r e ssio n | 3 7 5

Putting all values from the table in die normal equations, we get la + 28c = 16.2 ,..(2) 286= 14.3 =>i = G.5i ...(3) 28a + 196c =69.9 ...(4) Solving equations (2) and (4) we get, a = 2.07 and c = 0.061 Hence (1) becomes: v =2.07 + 0.5U« + 0.061 u* or y =2.07 + 0.511 ( 2 x - 5) + 0.061 (2 x -S ? [v « = 2x- 5 andv =y] or y = 2.07 + 1.022x - 2.555 + 0.061 (4jc2 - 20x + 25) or y =1.04 - 0.193* + 0.243xJ. Ans. Example 4.78: The profit o f a certain company in the xf* year o f Os Ufe are give by x: 1 2 3 4 5 y: 1250 1400 1650 1950 2300 Hiking u - x - 3 and 50v - y - 1650, show that the parabola o f second degree o f y o n x is y - 1140.05 + 7Z lx + 32.1 Sx*. Solution.

( y - 1650) 50 ’ Let the parabola of the second degree of the curve be v - a + bu + cv*. Its normal equation is ' ’ Iv = ma + bZu + cZu2 Zi/v = a lu + bZu2+ cZu3 ZuJv = aZu2+ bZu3+ cZu4. Here m = 5, and given that k = x - 3 and v 1

-1650 50

uv

u2

vu2

16 5 0 6 26

4

2

-8 -5 0 6 13

0 1 4

-3 2 -5 0 6 52

Z u -0

Zv-tf

Zuv ■ 53

Zu3 - 10

Zvu2 —21

X

y

u mx - 3

1

1250 MOO 1650 1950 2300

-2

2 3 4 5

y

... (1)

-1

0 1

1

Putting all values from the table in normal equations, we get, 5a + 6 (0) + c (10) = 6 a (0> + i (10) + c (0) - 53 => b - 5.3 a(10) + 6(0) + c(34' 21 sT

u*

it*

-8

16

-1

1

0 1 8

0 1 16

Z*f .

I 0

Zu4 ” 34

»■• (2)

v-(3) ...(4)

3 7 6 | E n g in eer in g M athem atics -111

On solving equations (2) and (4), we get a = -0.086, c —0.643 Hence (I) becomes: v = - 0.086 + 5.3a + 0.643a2 y -1650 or — —— = - 0.086 + 5.3 (x - 3) + 0.643 (x - t y 50

v =

y -1650 = - 0.086 + 5.3 {x - 3) + 0.643 - = 32.l5x2+ 72.1x+ 1140.05. Example 4.79: Fit a second degree parabola to the following data x: 0 1 2 y: 1 1.8 1.3 Solution. Here m = 5, Changing the origin a = x ~ 2 and v *=y. Let the second degree parabola of the curve be v = a + 6a + ca2. .\ Its normal equation is Zv = ma + 6Za + cZa2 la v = aZa + AZa2 + cZa* Iw2v = aZa2 + 6Za3 + cZa4.

0 1 2 3 4

i 1.8 1.3 2.5 6.3

-2 - 1 0 1 2 Z« = 0

3 2.5

4 6.3

-(I)

uv

u2

U2V

u}

u<

1 1.8 1.3 2.5 6.3

-2 -1 .8 0 2.5 12.6

4 1 0 1 4

4 1.8 0 2.5 25.2

-8 - 1 0 1 8

16 1 0 1 16

Zv = 12.9

Zuv “ 11.3

ii

u =jc- 2

*<*S*b

y

Ans.

11

X

( y - 1650) and v = x -3 50

Zm*v = J3.5 Zu* = 0

Zu* = 34

Putting all values from the table in the normal equations, we get. 5a + 10c = 12.9 106 =11.3 =>6=1.13 10a + 34c = 33.5. On solving equations (2) and (4), we get a = 1.48 and c = 0.55 Hence (1) becomes: v = 1.48+ I.l3u + 0.55w2 or y = 1.48+ 1.13 (x-2> + 0.55 (x -2 )2 [v a = x -2 a n d => y = 1.42 - 1.07x + 0.55x2.

...(2)

...(3) ...(4)

v=y] Ans.

N um erical A n a lysis -111, C o r r ela tio n f t R e g r e s sio n | 3 7 7

Example 4.80 : Fit a curvey * ah* to the following data : x 2 5 4 y : 144 172.8 207.4 Solution. Given y = ab* Taking logm of both sides, we get logio y = logio a + jc logio b :. Its normal equations are : 2 logio > = w logio a + [og,o b Zx I x logio y - log,0 a Zx + log]0 b Zx2

5 248.8

6 298.5 ... ( t)

[here ,7t - 5]

X

y

X1

togi t y

r ogi,y

2

144 172.8 207.4 248.8 298.5

4 9 16 25 36

2.1584 2.2 >>5 ?<168 2.3959 2.4749

+.3168 6.7125 9.2672 11.9795 14.8494

3 4 5 6 Ex = 20

Zx3 <=90

2logi»y “

11 .5 8 3 5

Zxlog,9y “

4 7 .1 2 5 4

Putting all values from the table in the normal equations, we get 5 iogio a + 20 logio 6 = n .5835 ... (2) 20 logio a + 90log]0 b = 47.1254. ... (3) On solving equations (2) and (3), we get logio a ~ 2 and logio b = 0.079 => a - 100 and b = 1.2 [as taking antilog] Hence ( I) becomes :y= 100(1.2)*. Ans. Example 4.81: Fit a curve y = ax* to the following data and estimate y a tx m 12: x : 20 16 10 11 14 y : 22 41 120 89 56 Solution. Given y = ax6 ... (1) Taking logt of both sides, we get Ioge y - log. a + b log, x or Y =A + bX ' ...(2) where Y = \ogey ,A -\o g c aandX= \o^cx Then normal equation of (2) are : ZY = mA + bZX ... (3) ZXY~AZX+bZX> ...(4) m = 5 (No. of data)

3 7 8 | E n g in eer in g M a th em a tk s -III

I

*

y

X - log, x

Y m iog
X?

XY

20 16 10 11 14

22 41 120 89 56

2.996 2.770 2.30 2.398 2.64.

3.091 3.7135 4.7875 4.4886 4.02535

8.976 7.6729 5.29 5.7504 6.9696

9.2606 10.2864 11.011 10.763 10.627

X X - 13.107

Z Y - 20.106

ZX*-34.66

L X Y -51.95

Putting all values from the table in the normal equations, we get SA + 13.1076 = 20.106 and 13.107/«f+ 34.666 =51,95 Solving above, we get A = 10.254 and 6 --2 .3 7 9 4 8 « - 2.38 => log* a — 10.254 a ~ 28491.416 « 28491 Hence (1) becomes y =28491 xx-23* (4) Further, estimates at * = 12 : y bs Example 4.82: By the method o f least squares,fin d the curve y m ax+ bx? that bestfits thefollowing data: x : I 2 3 4 5 y : 1.8 5.1 8.9 14.1 19.8 Solution. Given y = ax + bx2 ( 1) Let (*1»yi) (*2, k ) (*3, yi) ■••• (** y») be the points. Then error or estimate for f* the points (**, y,) is Ei=(yi~ a x ,- bx?). By principle of least squares the values of a and 6 are such that S = ZE? = Ky, -rax, - bx?)1 is minimum. .*. Normal equations are given by

=28491 x(i2)-2«- 76.956«77

A.

dS £ O .a n d ^ -0 da Txty im aXx? + blxj3 and Zx?y, = dZx? * bZx,4 or Zxy * aZx2+6£x3

(2)

1x 2^ = o I x 3 + 6 I ^

...(3)

X

y

xy

X?

x* ’

1 2 3 4 5

1.8 5.1 8.9 14.1 19.8

1.8 10.2 26.7 56.4 99

1 4 9 16 25

1 8 27 64 125

I x y - 194.1

I x * -5 5

Tx> « 255

X*

1 16 81 256 • • 625 Ex* m 979

x 3y 1.8 20.4 80.1 225.6 495 Ix*y * 822.9

N u m e ric a l A n a l y s re -III, C o r r e la tio n & R e g re s s io n | 379

Putting all values from the table in the noimal equations, we get 55a+ 225* - 194.1 225o + 979* =822.9 On solving equations (4) and (5), we get a - 1.52 and * = 0.49. Hence required curve (1) becomes: y - 1.52x + 0.49x2. Example, 4.83 :fita second degree parabola to the following data x: 1929 1930 1931 1932 1933 1934 1935 y: 352 356 357 358 360 361 361 Solution. Here m = 9 (which is odd number) Changing the origin u ~ x - 1933 and v = y -3 5 7 . Let the parabola of the curve be v = a + bu + cu1. Then its normal equations are : Ev = ma + bZu + cZu2 Zvv - aZu + bZu1+ cZu3 Zu2v - aZu2+ bZu* + cZu*.

... (4) ...(5)

An. 1936 360

1937 359

...(1)

X

u -x -1 9 3 3

y

v “y - 357

uv

u*

u2v

u*

u*

1929 1930 1931 1932 1933 1934 1935 1936 1937

-4 -3 -2 -1 0 1 2 3 4

352 356 357 358 360 361 361 360 359

-5 -1 0 3 4 4 3 2

20 3 0 -1 0 4 8 9 8

16 9 4 1 0 1 4 9 16

-8 0 -9 0 1 0 4 16 27 32

-6 4 -2 7 -8 -1 0 1 8 27 64

256 81 16 1 0 1 16 81 256

£ v « //

Z u v -5 1

1m 2 -6 0

Zh2v - - 9

Z u * -0

Zu*“ 70

Zu~0

I

Putting all values from the table in the normal equations, we get 9a + 0 + 60c —11. 0 + 60* + 0 + 0 = 5 1 => * = 0.85. ' 60a+ 0 + 708c = -9. On solving equations (2) and (4), we get ■ ,.. a = 3 and c = - 0.27. Hence fitted parabola (1) becomes: • ;<:• v = 3 + 0.85a-0.27 ••• => y - 357 = 3 + 0.8j$r - 1933) -0.27 (x - $33)* => y = -0.27x^+ J044.67 x - 1010135.08.

...(2) ...(3) ^ ; foy t

« ■

Am.

3 8 0 | E n g in e e r in g M a th em atics -III

Ezercise-4(A) dy I. Using Picard’s method, solution of — = *(1 + x 3y), y(0) = 3. Compute the value hod, obtain the < XO. 1) and y(0.2).

dy

2. Use Picard’s method to approximate the value ofy when * = 0.1, given that — = 3* + y2 with A O )-I. 3. Find the third approximation of the solution of the equation

dy

~

n

y x by

method

when >>= 2 forx= 1. dy 4. Find the fourth approximation of the solution of the equation -fa = 2x - y with y( 1) = 3 using Picard's method.

'

dy

5. Given — = 1 + y2 with><0) = 0, obtain^ as a series in powers of* in the range 0 < x£ 1, using Picard’s method. 6. Solve the following initial value problem by Picard’s method ^

= xey, withXO) = 0. Compute ><0.1), ><0.2) and XI)-

7. Solve the following simultaneous differential equations, using Picard’s method for * = 0.1 and 0.2; t r “ ** + = °» and 2(0 ) = 1. dx dx 8. Using Picard’s method, obtain the second approximation to the solution of d 2y =

„ dy , 1 H + y withX0) = ] ,A 0 ) = 2 ‘

9. Given the differential equation dy — = x z2 + y 2, y(Q) y(0) ~ = 0; det< determine the first three non-zero terms in Taylor’s series expansion dx ax for X*) and hence obtain the value ofXOdy 10. Apply Taylor’s series expansion to the differential equation = 2x + 6y with X0) = 1, obtain X0-1)>XO-2), X0-3) and X°-4) in steps of A= 0.1. dy 11. Solve ^ f=y sin * + cos x, subject to initial condition X0) ~ 0, using Taylor’s series expansion. 12- Given / ~

+ ^ and X0) = I, find the Taylor series X °-1) X 0-2)^ •S'? 13. Using Taylor series method solve y = xy +y*, y(0) =1 at * = 0.1, 0.2,0.3.

N u m e ric a l A n a ly s is -H I, C o r r e la tio n & R e g re s s io n | 3 8 1

14. Solve by Taylor series method of third order the problem/ = (x3 + xy2) e~x,y(0) = 1 to fmdy for x = 0.1,0.2,0.3. ‘ dy dz 15. Solve - z - x , " - y + x withy(0) = 1, z(0) = 1 to get><0.1) and z(0.1), using Taylor’s method.

Answers-4(A) 1.

y(0.1) = 3.005, y(0.2) = 3.020,

2.

><0-1) = 1.12721.

3. y s = 2x - 21og x - ^ (log x)3. 1 5 5 , 7 , 47 , 73 737 4. v = ----- x + — x -----x + — x 2 ------ x + ----- . ' 60 24 6 12 12 120 1 , 2 . 1 , y 3 15 63 6. y(0.1) = 0.0050125. ><0.2) = 0.0202013, y(l) = 0.6487213. 7. ><0.1) = 0.1050, y(0.2) = 0.2200, z(0.1) = 0.9997, z(0.2) = 0.9971. 1 3 v3 v7 9. j r t x ) = y + — + ...;>
11. y = x + — + —— +

12. y(0.l) = 1.0665;y(0.2) = 1.1672. 6 120 13. X0-1)= 1.1167; y(0.2)= 1.2767; y(0.3)= 1.5023. 14. y(0 1)= 1.0047;>(0.2) = 1.01812;><0.3) = 1.03995. 15. y(0.l)= 1.003; z(0.1) = 1.1102. *

Exercisc-4(B) Using Taylor series method do the following problem (1 - 7 ) 1. Solve/ = ysin x + cos x subject tox = 0,y = 0. dx 2. Solve — = x2 + y2 subject tox = l,y = 0 atx = 1,3 dx 3. Find the values of y (l) and y (l.l) for the equation f{x, y ) I 2y - 3e“ where *o = 0, yo - 0. Check the values. 4. Find the values o fy (l.l) and y(1.5) for the equation/ = xym withy(l) = 1 correct to three decimal places. 5. Find the value ofyupto five decimal places atx= 1.02 for the equation — = (xy - 1) with>(I) = 2.

3 8 2 | E n g in eer in g M attcm attcs-III

dx dfi y dy 6. Solve 3 7 = x + y + t and - j- r - x - 1 with y - I . 3 — - -1 .* = 0 at / = 0 . dt d t8 dx

dx dy 7. Solve — = xy + 2t and — = 2ty + x with the initial conditions = l ,y = 1, t = 0. Using

x

Picard’s method do the following problem (8 - 12). dy 8. Find third approximation for the differential equation ~r~ = 2y - 2x2 - 3 with dx m = 2. dy „ y 9. Find a solution upto third approximation of the differential equation = 2 - —, x i ) = 2. dy 10. Find upto third approximation to solve ^ = x + y where y = 0 when x = 0. 11. Find the solution of the diflferential equation

dy

- _ = 1 - 2xy correct to four decimal places for

x = 0.2 given that><0) = 0. dy dz _ 12. Find third approximation of the equation — = 2x + z and ^ = 3xy + x z with >(0)= 2 and z(0) *»0. dy y —x 13. Use Euler’s method to approximate >>when a: = 0.1 given that ^ = y Z x ’

= ^ by taking

A = 0.1.

14. 15. 16. 17.

Solve/ Solve/ Solve / S olve/

= 3X2 +y in 0 <,x <. 1 by Euler’s method h = 0.1 given that><0) = 4. = x +y, XO) = 0 choosing the step length 0.2 fory (l2 ) by Euler’s method. = —j/; ><0) = 1 by Euler’s method for><0.04). =x + y + xy, X0) = 1 forXO.l) by taking h = 0.025, using Euler’s method. dy y -x 18. Use improve Euler’s method to find X01X given ^ - y + x * X0) ~ 1dy 19. We Improved Euler’s method to find X0-5X taking h =0.1 and ^ = y + 8in x >XO)= 2. 20. Use Improved Euler’s method to fineX I-6) if ^

= y 2 ~ ^-.>(1) = 1-

21. Using Euler’s Improved method findX0.2),X0.4) given ^

- x+ I \fy I >

=1

22. Find X 0.1) given y’ = x2+y, X0) = 1 using Improved Euler method. 23. Use modified Euler method to obtain X0-2), X0.4) and XO-6) correct to three decimal places given th a t/ = y - x 2,X p) * 1. 24. Use Euler’s modified method to get X0.25) given that / = 2xy, XO) = 125. Given / = x2 + y2-, X0) = 1 determine X 0.0 and j£0.2) by modified Euler’s method. 26. Solve / = y +x \ y i0) = 1 forX0.02), X0.04) ajid'xO-06) using Euler’s modified method.

N um erical A n a ly sis -111, C o r r ela tio n & R eg r essio n | 3 8 3

27. Using second order Runge-Kutta method find > at x = 0.1, 0.2 and 0.3 given 2 / = (I+jc>2;>(0)=1. dy 2xy + e* 28. If — = x 2 + ygx ' yW = ® solve for> at x.= 1.2, 1.4 using Runge-Kutta method of fourth order. 29. Using Runge-Kutta method of fourth order find y at jc = 1.1, 1.2 given that 2y = 2x3+r>J
+ 2y = x; y(0) = I.

dy 32. Solve — = 1 +

dz

= ~*y for jc = 0.3 given that > = 0, z = 1 at x = 0 by Runge-Kutta

method. 33. Solve

d*y

a dy _ ~ x — ~ 2a?y = 0 given that >(0) = 1, /( 0 ) * 0 for ><0.1) using Runge-Kutta

method. 34. Use Runge-Kutta method to solve / ' - xy + 4y = 0; ><0) = 3; /(0 ) - 0 at x - 0.1. cPy 35. Apply Runge-Kutta method to find y at x = 0.1 given -j^ r = y3; y(0) = 10; y'(0) - 5. dy 36. Solve by Milne’s predictor-corrector method, the differential equation — = y - x 2 with fee following starting values :y(0) = 1, >(0.2) - 1.12186, ><0 .4) - i .4682, ><6)= 1.7379 and find the value of> when jc = 0.8. dy 1 37. Given ^ = - (! + x 2)y* and><0) = 1,><0.1) - 106, ><0.2) = 1.12. ><0.3) - 1.21, evaluate ><0.4) by Milne’s predictor-corrector method. 38. Solve the differential equation / = jc? + y 1 - 2 using Milne’s predictor-corrector method for jc 0.3, given the inital values x - 0, > = 1. The values of > for jc = - 0.1, 0.1 and 0.2 should be computed by a Taylor series expansion.

Answers~4(B) L;F=X + T + i2 0 + ...... 4. >(1.1) = 1.228, >(1.5) = 1.686

2. 0.4158 5. 2.02061

6. x(t) = t + — + .... y(t) = 1 - t + z(t) = -1 + ~ 2 4! 3! 212 ^ ' 4 to A o

7. x(t) =1- t +— +...;>(*) * -1 + ~-t2+- 13- ...

3 .X i)= 1 3 .9 1 ,> (i;i)= i7 .8 7

where * =

d t^

3 8 4 | E n g in e e rin g M a th e m a tic s -!!!

8. y = 2 + x + x2 - — - — x6. X2

X6

X6

X 11

10. y = — + — + 160 + 4400 1l 2i .

9. y = 2x - 2 log x - —(log x )3 O

11. 0.1946 3 40

+— +— x^- +(- — yv —2* + + x^ x ++ x 'x® ' + + — 2 0 x® x’' + i 0 >■z —«3x2 + 4 x4 x + + — 5 x 2 8 Xs x + — x8

13. 1.0928 14. 4.4, 4.843.5.339, 5.9002, 6 . 53825, 7.267078, 8.101786, 9.0589647, 9.10396, 10.25736 17. 1.1448 15. 1.18318 16. 0.9603 20. 1.1766 19. 3.4394 18. 1.0932 23. 1.218, 1.467, 1.737 22. 1.1055 21. 1.2309, 1.5153 26. 1.0202, 1.0408, 1.0619 24. 1.0625 25. 1.1105, 1.25026 29. 2.2213, 2.4914 27. 1.0552, 1.123C, 1.2073 28. 0.1402,0.2705 32. 0.3448, 0.99 31. 0.82342,0.6879. 30. 2.2430, 2.5890 33. 1.005334 34. 2.9399 35. 17.42 38. 0.6148 36. 2.011 37. 1.2797 i •- * r m * itftiilitt ^

^

pi *•*

>>- « *

V

« 1

Exercise-4 (C)

4. 5.

6.

Calculate Karl Pearson coefficient of correlation from the following data, using 20 as working mean for price and 70 as working mean for demand. Price : 14 16 17 18 19 20 21 22 23 Demand : 84 78 70 75 66 67 62 58 60 From the following information relating to the Stock Exchange Quotations for two shares A and B, ascertain by using Pearson's coefficient of correlation how shares A and B are correlated, in their prices ? Price Share (A) iis. : 160 164 172 182 166 170 178 Price Share (B R s.: 292 280 260 234 266 254 230 coefficient between the income and expenditure of a wage earner and comment Find the correli or the resrlt Month Jan. Feb. March April May June July 46 54 56 56 58 60 62 Income 36 40 44 54 42 58 54 Expenditure Calculate correlation coefficient from the following data : N = 10, Ex = 140, Sy f 150,1 (x -1 0 )2 = 180, K y - I5)2 - 215, E (x- 10) (y - 15) = 60 Calculate Ae correlatin coefficient between the following pairs of values : x : 100 ' 110 115 116 120 120 125 130 135 y : 18 18 17 16 16 16 15 13 10 From thc&Howing d$t$» examine whether input of oil and output of electricity can be said to be correlat^f': Input of pit : 6.9 8.2 7.8 4.8 9.6 8.0 7.7 Output of Electricity : 1.9 3.5 6.5 1.3 5.5 3.5 2.2

N um erical A n a ly sis - III, C o rrela tio n & R eg r e ssio n | 3 8 5

7.

Given the following data : Sum of deviation from assumed mean: -1 4 18 Swn ef sqM m of deviation from assumed mean value : 4308 6308 Sum .of predwet ef deviations from their respective assumed mean: 1SO Nuiaier fff pate of observations : 12 CalcVtate the lorrtifaflee coefficient. 8. The coefficient of correlation between two variables x and y is 0.3 Their Covariance is 9. The variance of x is 16. Find the standard deviation of y series. 9. Find the coefficient of correlation between the ranks obtained by 10 students in Mathematics and Physics in a class test are given below : Ranks in Mathematics: 1 2 3 4 5 6 7 8 9 10 Kfeks in PHJrsics: 3 10 5 1 2 9 4 8 7 6 10. 'Em Competitors in a beauty contest are ranked by three judges in the following order: Jwtge x : 9 7 6 2 4 1 5 10 3 8 6 Ju d g ey : S 9 3 5 2 1 4 7 1* 10 5 7 4 ■- 9 Z 3 Judge z : • < 1 11. Calculate fee w flH li Rt ©fctiffWMion rfktot (fee foBbwim dtta by fee method of rank correlation: xi 7| |S 95 7* A M SI SO y: 134 45# 115 l i t 4 148 142 100 12. Compute Rank cfWMwen f»m fee fellofcinj HWe #f HMk Wiiwber ®fSupply (x) and price (y) *: US 134 120 130 124 12t 128 130 127 12$ y: 130 „ 132 Ten competitors in a Nnwty contest are ranked by thwe judges in fee following order: 3 2 8 9 6 10 7 First Judge : 1 5 4 Second Judge : 4 8 7 2 I 10 3 6 5 9 4 Third Judge ; 6 7 8 I 5 10 9 2 3 Use the rank correlation to gaagc which pair ofjudges have the nearest approach to common taste 14.

Following are the scores of ten1sft&nts in a class and their I.Q. Student: 4 10 2 3 7 9 1 5 6 8 Secore: 55 50 35 40 25 65 55 45 85 92 I.Q.: 140 150 130 100 100 110 100 120 140 110. Calculate the rank correlation coefficient between the score and I.Q. 15. Find the mean values of the variables * and y and correlation coefficient between them from the following regression lines : 2y - jc - 50 = 0, 3 y - 2 x - 10 = 0. 16. Regression equations of two variables x and y are as follows : 3ac + 2y - 26 - 0; 6x + y - 3 1 = 0 . Find (i) the means ; (ii) the regression coefficients; (iii) the coefficient of correlation between x andy. 17. The equations of two regression lines obtained in a correlation analysis are ; 3a: + 12y = 9 and 3y + 9* = 46. Find (i) Mean val ues of x and y, (ii) the correlation coefficient between x and y.

3 8 6 ( E n g in eer in g M athem atics -111

18.

19.

20. 21.

22.

23.

24.

25.

The lines of regression of y on x and * on y are given to be y = 5+0.2x and x = 3+.y. Find the means for the two variable x and y. In a regression analysis problem, the following data are given. The regression lines are x + 2y - S = 0 ,2x + 3y = 8 and
*?<>: 1. 4. 7. 10.

r = - 0.954 r = 0.915 r = 0.2949 xandz.

Answers—4(C) 2. 5. 8. 11.

r = -0.96 r = -0.915 ct, = 7.5 * r = 0.9

3. 6. 9. 12.

r - 0.769 r = 0.696 r = 0.224 r = 0.33

N um erical A n a ly sis -III, C o r r ela tio n & R e g r e ssio n j 3 8 7

13. 14.

Second and third judges are in nearest approach to the sense of beauty. r - 0.47.

15.

x = 130,

16.

x =4,

17.

in 11 1*

>=90,

r = 0.866. byx = - | ,

y = 0.333;

r = -0.5.

r = -0.2887.

18. X - 10, y =7. 19. x - \> y=2, ay2= 4, r = -0.866 20. y = 0.6x + 0.4; x — 1.2y+ 1.2. 21. x = 0.64y + 2.63;; y = 0.83x + 51.88. 22. 66 quitals. 23. y = 0.9075x + 41.1375, y - 0.48x + 67.75; and x = 26.93. 24. 1014.7. 25. byx = 1.42; y - 1.42x - 0.532; and y = 141.668 »

* [iM M M H

'f

Ip* n'

r >•**. kto A Vi w

j* '# + * yn

miWiW

Exercise-4(D) 1. 2.

3. 4. 5. 6.

Fit a straight line to the following data: x: I 3 5 7 9 y: 8 15 24 32 40 Taking 1990 as origin for x-series, fit a straight line to the following data showing the production of a commodity in different years in Gwalior. Yearx: 1988 1989 1990 1991 1992 Productiony: 10 12 8 10 14 Fit a second degree parabola to the following data : x; - 3 - 2 - 1 0 1 2 3 y: 4.63 2.11 , 0.67 0.09 0.63 2.15 4.58 Fit a parabola to the following data : x; 2 4 6 8 10 y: 3.07 12.85 31.47 57.48 91.29 Fit a parabola of second degree to the following data : x; 0 200 300 400 y: 1544 1898 2133 2327 Fit a second degree of parabola to the following data : x: -4 -3 -2 -1 0 1 2 3 4 y: 101 104 107 107 108 110 109 110 108 Fit a curve y = ax + — to the following data : x;

1

2

y-

5.43

6.28

3 8.23

4 10.32

5 12.63

6

14.84

7 17.27

8 19.51

3 8 8 | E n g in eer in g M athemattcs-III

8.

b Fit a curve y = ax1+ — to the following data : x x: 1 2 3 y: -1.51 0.99 3.88

4 7.66

9.

Fit a relation of the form R = a + bV +cV2 to the following data, where, V is the velocity in km/hours and R is the train resistance in kg/quintal: V: 20 40 60 80 100 120 R: 5.5 9.1 14.9 22.8 33.3 46.0 Estimate R when V= 90.

10.

Fit a curve of the form y ~ a b x to the following data : *. 1 2 3 4 y: 87 97 113 129

11.

7 193

Fit a curve of the form y = ax* to the following data : x: y:

12.

6 195

5 202

2 43

4 25

7 18

10 13

40 5

20 8

'

60 3

80 2

Fit a curve of the form y = ae* to the following data : x; v.

1 1.65

2 2.70

3 4.50

4 7.35.

Answers-4(D) 1.

7 = 3.55 + 4.05x

2. ,y = I0.8 + 0.6x.

4.

y - 0.34 - 0.78x + 0.99x2.

5. y = 1547.9 + 378.4x- 40x2.

6.

j = 109 + 0.883*-0.2457xJ.

. . 3.03 7 . y = 2.4x + ----- . X

9.

R = -1.5726 + 0.329F + 0.00008K1 an d rta tF = 90 is28.69.

10.

jf = (73.97)(1.68)'.

11. ^ = (78)x‘fl*.

3. y =1.243 - 0.004x +0.22j:2 -

8. y = 0.509X2- — . X

12. y-= (l)e0
a

S ta tistic s 5.1 CONCEPT OF PROBABILITY When an experiment is performed it may result in one of possibly several outcomes. Because of 'chance' factors, we may not be able to predict the outcome precisely. However, we may not be able to offer our opinion about die possible outcomes, by giving a quantitative measure of the ’chance' that a particular outcome would occur. Thus, with every event A consisting of one or more outcomes of an experiment, we may associate a numerical quantity, called the probability of A, denoted by P(A), which will measure the 'change* that the event A will occur. We take 0 < P{A) 5 1, xi'PiA) - O' implies tint there is no ’change’ that A will occur and if 'P{A) - 1' implies 100% chance that A will certainly occur. If P(A) ~ H, then it will mean that there is 50% ’chance1that A would occur. The advantage of defining probabilities of events in quantitative terms is that it now becomes amenable to further mathematical analysis. We can talk of the possible gain to a gambler or ecpected profit from a risky venture. We can calculate the insurance premium to be charges to cover a certain risk. Let A be an event, then probability of an event deponds only on the number of sample points contained in A. We can easily verify that this probability function defined on subsets of sample space S, possesses the following properties: (i) P{A + B) « P(A) + PiB) (ii) if A ^ B = > P(A) <; P(B) (iii) P(S) = 1 (iv) 0 5 P(A) <, 1. Let A be the class of events associated with the outcomes of a certain experiment i.e; A h A2, ........eA, and S bethe sure events. Then the probability function P on A is also called probaility distribution on S. It tells us how the unit probability mass is distributed among different subjects A of S. For example : A die is thrown, then the probability of getting a prime number can be find as follows: Here sample space S = (1, 2, 3,4, 5, 6} n (5) = 6 Let E be the event if selecting a prime number i.e; E = {2, 3, 5} =? n(E) = 3

n(E)

3

i

Probability: P(E) = 7i(S) ” § = 2 '

5.2 RANDOM VARIABLE Let S be the sample space associted with a given random experiment. Then a real valued function X which assigns to each outcome x e S to a unique real number X{x) is called a random variable. In other words a random variable is a real valued function having domain as the sample space associated with a given random experiment.

3 9 0 | E n g in eerin g M athem atics-III

The random variable are two types : (i) Discrete Random Variable: A variable, when real valued function defined on a discrete sample space is called a discrete random variable. For example : The marks obtained in a paper or in a test, number of telephone calles per unit time, number of successes in n-triais, and so on. (ii) Continuous Random Variable: A random variable* is said to be continuous random variable, if it can take all possible values betwen certain limits. For example : Age, weight, height and so on. PROBABILITY MASS FUNCTION: Let X be a discrete random variable. A probability mass Junction (p.m./.) is given by fx ( x ) = P ( X = jc), for all* or, f { x ) = P { x : X fo) = x } It is a function with values between 0 and 1, and whose sum is 1, over all values of x. For example; Toss a balanced coin once. Let X be the number of heads that occurs, then probability mass function (pmf) can be determined as follows : Random Variable :X - { x = 0, 1 } Sample space S = { H, T } => n{S) = 2 Elementary Events : (£) T H

Number of Heads * = { 0 , 1} r =0 JC- I

Probability Mass Function f{x ) =P iX =x) Vi '/j

Example 5.01 : Toss a balanced coin twice. Let X denote the number o f heads. Find the probability mass function o f X Solution : Random Variable: X = {x = 0, 1, 2) Sample space : S = (HH, IT, HT, TH) => n { S ) = 4 Number of Heads * = { 0 , 1,2} jc =

0

x= 1 x =2

Elementary Events : E

Probability Mass Function f i x ) = P i X = x)

TT TH, HT HH

1/4 2/4 1/4

5.3 DISTRIBUTION FUNCTION Let X be a discrete random variable, then distributive function o f * is given by F(x) = P ( X £ jc)= 2 Pi Xi*X CO

such that

p, k 0, and 2 p, = I, wherep, - p

i=l

( jc,)

S tatistics | 391 Again, Let X be a continuous random variable, then distributive function of A" is given by X

F(x)=P(X
I f (x) dx. —OO

Remark: A distributive function F

(jc)

or Fx (x) is also called cumulative distribution function

(cdf).

Example 5.02 zA random variable X has the following probability function : X -x 0 1 2 3 /i p(x)

0

g*

52

52 .

Determine the distributive function o f x. Solution ; We know that distributive function F(x)=P(XZx)=

£

XjfiX

pt

F(Q) = P( X^ Q) =p ( Q) = 0 F ( \ ) = P ( X * ] ) = p ( Q ) + p ( \) = 0 + l = |

F (2 )= /'(X S 2 ) = 0 + | + ! = | F(3) =/>(*< 3) = 0+ | + | + | = r. Example 5.03 xA random variable X has the density function : f(x )=

|

\x > ,

0 £ x £3

1 0 , Otherwise. Find the distributive function. Solution : By definition of distributive function for continuous random variable : X

F( x ) = P ( X £ x ) = J f ( x ) d x

Ans.

3 9 2 | E n g d s e h n g M athemattcs-H I

Properties of Dbtribative fiid ie a : If X is the modom variable, then it holds the following properties: (i) F{x)7> 0, (ii) 0 5 f ( * ) S l , (iii) F ( - 00) = 0 and F (+*>) - I, (iv) F is an increasing and continuous function from the light at every point

5.4 PROBABILITY DENSITY FUNCTION: Let/(x) be a probability function in die interval [a, b], then the probability that the variate value X to lie in the interval [a, b] is given by b P(a <, x <, b) - J f i x ) dx.

a

The functin fix ) is probability densityfunction (p.df.) if and only if, the total probability is unity, i.e., the probability density function (p.df.) possessess the following two properties: (i) f i x ) 2 0 00 (ii) \ f{x)dx - 1, i.e; total probability is unity. -CO For example : whether check the following function p .d f ? f i x ) = fix (I - i ) , 0 S x ^ l 1

For p.d.f., we have

1

J / (x>& = j fix (1 - x) dx 0

0

(x) is the p.d.f. Remarks : [Formulae] L et/ (x) be th e p.d.f. o f a random variable X, w here X is defined in [a, b]. Then

H ence /

( 1)

Arithmetic mean or Expected valve o f x b i.e;

(2)

x or E(x) - j x. f(x )d x

a

The expected value o f x2 :

b E (x2) = J xV (x)dx Q (3 )

First moment about origin : b

m* = £ (x ) = J x .fix )d x .

S ta tis tic s [ 3 9 3

*

Tm» *

($)

>* moment about origi®:r "

b Hz' = £ ( x 2) ' I x\f(x)dx, r

b Hr' = £ ( * 0 = I xr.f(x)dx, a (6) (7)

Mean = \i\ Variance = »**'-

M

<1))^

and standard deviation (8)

t*

moment about mean : 6 Hr * I (X - x y f '(x)dx

W

ft = hj’ - (ftO2, U3 = Hj' - 3ji2V i' + 2 (Hi*)3 M4 = fi4 - 4h3’ (f t1) + 6 ft' (ft1)2 - 3 (ftT and so on. (10) Median : If M is the median, then M 1 \ f(x)dx = g a ~

b

or

f f(x)dx - ~

M £ Mode : mode is the value of x for which/ (jc) is maximum. For finding the mode : / ' (x) = 0 andf ”( jc) < 0 at point x = a. Such point is called mode. (12) Mean Deviation: Mean deviation about mean * is

(11)

b

M.D. = J | x - x |./(x ) dx, a (13) If F (x) is a distributive function, then corresponding p.df. is given by fix ) = £

( F (x)}

le 5.04 : I f f (x) * Kx2,0 < x < 1, has probability density function (p.d.f), determine K and fin d P ( | < x < ! ) and fin d V i f P (X > a) - 0.05. lotion : Since / (x) is a p .d f, so that

b i f(x)dx

=i

3 9 4 | E n g in eer in g M athem atics -III

j Kx*dx = 1

■k"

^ /(x ) is a />.<£/

K= 3.

Since / (x) = 3x2, 0 < x < 1

ro/?nrf/» ( | < x < | )

(i)

we have 1/3

1/2 = J 3x*dx [by (1)] i/s

MT

■ I

-11/3

1 _ J _ = _19^ 8 27 216 '

Ans.

Tofind a : Since P{X> a) = 0.05 (X>

=>

I f (x)dx= 0.05 o 1 J 3x*dx = 0.05

=*

D>y(D)

=> [*3]> = 0.05 => I - o3 = 0.05 => a3 =0.95 => a = (0.95)I/J Example 5.05 : /4 random variable X has the p .d f. : ax, 0£x£l a, l£ x £ 2 /M -a x + 3a, 2£x£3 0, Find (i) value o f a (U) P ( X S 1.5). Solution : (i) Since / (x) is p d -f, so that

otherwise

<30

J f(x)dx =

—«0 1

=>

2

1

3

I ax dx+ j adx + ! (~ax + 3a)dx = 1 0

1

2

Ans.

S ta tistics | 3 9 5

% -j + <**)? + a ( - § “ J + 3o(x)j = 1 ^ + a + ^ (~9 + 4) + 3a = 1 a + 2 a -5 a +6a _ ^ 2a = 1 => a =

Ans

(ii) To find P (X <, 1.5) : We have 1.5

P {X <. 1.5) = J f{ x ) d x

(by defmation)

-0 0

1 L5 = j f i x ) dx + I f{x) dx 0 l 1 lJS = j axdx + I adx 0 1

(by given p.d.f.)

(*)l‘ * | + a (0.5) = « +« 2 2 1’ =a - 2

** Ans. * (v a - - ^)

Example 5.06 : Ffarf the value o f k fo r the p .d .f.: f(x )

j kx\

0 £ x £3,

otherwise, <• 0, and compute P(1 £2). Also fin d the distributive function. Solution : Since /

(jc)

is a p .d f, so that 00

J f ( x ) dx = 1

—00

3

j kx2dx = 1 0

•

,

3 9 6 | ENaNrEMNG MATHEXATtCS-III

|

[27-0]- I

-I Thus p.df.

025x53

/(*> =

i o,

otherwise

To compute P (1 £ x £2) : We have

P ( 1 5 x 5 2) = J f{x)dx i 2

= J J x2«4c

[ty (OI

1 a

•iW = 2 7 18

n

_ .I

27' 7b find distributive Function F(x) : We have, /^(x) = /> (* 5 x ) X

=

/

[ by definition]

/(X > &

= J 4 x2dx o *

K

- b 0,

Hence D.F.

x 27 x <0

=

2_ 05x^3 27 1, x £ 3. Example 5.07: The distributive function of* random variable X is given by

F{x)

0,

Ans.

x£0

F(x) /* x > 3t then (I) find corresponding density function o f random variable X, (ii) compute P(2 < x *3).

S ta tistics | 3 9 7

Solution : (i) We know that, p.d.f. : /(* > = £

(*-(*)}

xi

ft

1

\ x - l ) 4] A. J dx |, 16 jI I < x i 3 x>3

$<>>•

x5

0,

i.e; p.df.

1
-i y

/(* ) =

1

x>3

0,

-(I)

(ii) We have P(2< x< ;3 ) = ! f(x)dx 2

3

=i

-

T ( x - \ydx

[by 0 )]

_ i F( x - i ) 4 ' "< 1 4 _

= _Lnfi n = l§ 16 * ~ '

Example 5.08 : Ifp .d .f f ( x ) m k *-----J , - ao
Solution : For p .d f., we have

J f{ x ) d x = \

16

Ans.

3 9 8 I E n g in eerin g M a ih em a tics -III

I 1 * ' ( l + x 2) F(x) = P ( * £ x) -

p.d.f. Distribution Function

X

= ! f(x)dx -C O

X

1 = 1 *
[tan-1*]*

-±On>x+± Example

5.09:A frequency distribution is defined by x },

6
( 3(2-xy,

IZXZ2.

Prove that f(x) is a p .d f also fin d mean and the standard deviation. Solution : For p .d f, We have «.

J

1

2

f i x ) d x - \ x 3 d x + J 3(2- x y d x 0

-

M

1r in2 (2-x)4 +3

=>f (x) is a p.d.f. Tofind mean : We have mean

x = ^ l' =£(x)= -
2

=0! x. x*dx +jl x. 3(2— x)3dx 1

2 = J x*dx + 3 I x.(2 -x yd x o i

-M

+3

\x. ( 2 - x ) 4

2

2

l

-4

S ta tistics | 3 9 9

5

-z£ 5

4

*H H l M 1 ,3 6 5 4‘6 11 io-

To find S.D.

We have Hz’ = E (X*) = J x*f(x)dx 1

-CO

2

= J *2. x3<&+ i xJ. 3(2 -x f d x o

i

+3

1

# i;-h

U 3 + 3 [Jv<2 = 6+ 4+ 2 IU

- i + - ——1(0 = 6 4 1 0 |1

-

M

I+ I-JL L i+ Io j-o l 6 4 10L 6V J =

16 + 14 + J10l x16

1 + —+ 3 — 7 = — 6 4 20 10 + 45 + 21 60 !§ -!§ ~ 60 ~ 15=

Hence S.D.:

-4

dx

dx

4 0 0 1ENOweaiH atlwiW W lW H

- W so n

-

-

V15~100 H E

V1500

Ans.

uu Example 5.10 : For the distributive function (dF) : dF ■>y,e~w dx, -co

Prove thatye * ^ , mean ” 0, S.D. “ 4 t , variance * 2, and mean dtviation about mean 1st. Solution : Given

dF -y^er^x dx % -» * " f{x)

[v /(* ) = {F (*)}]...•(!)

Tofind yo : For p .d f,

J f( x ) dx = I [by (l)]

=>

) > 0 ^ fl&f = 1, -00 Since (r| is an even function, so that 00 2 \ y0e^*\dx = 1 0 00 => 2yo 1 e~* dx = I 0 2y0 -1 => =>

[v |x | = jc, jc > 0}

=1

- 2yo [0 - 1] = 1 1 yo = 2*

Hence p.d.f. :

f (x) = ^ e-W

...(2)

Tofind mean: By definition 00

mean = j*i‘ = E (x) = J x . f (x)dx

S ta ti s tic s 14 01

[Because x e-« is an odd function and I odd = 0] -cO Tofind S.D. and Variance : We have \i2' = E (x2) 00 = j x2f(x)dx —g o

00 1 = J *2 “ v xVW is an even fun.] = J x2erxdx 0 = X2 e

[ v \x\ = x, x 2s 0]

V° /o

- J 2*

c

o

CO

0 *+- 2 J xe-*dx o / -* V° 00

--r

t v.x 4 t A - oJ £_1

dx

~ 2 0+ le~*dx o - 2 —

N.

—2 (0 —1} —2

fi2’ = 2. Variance = p2*- (n,1)* = 2 - (0)2 » 2, and S.D. To

° ~ -/Variance

“ >/2< Afecw Deviation about mean : 00 M.D. = J Jx- x ]/(x ) dx

[by def.]

-0 0

=

* 1 J | x - 0 | . ~ «~w dx

=|

J |»|

[v mean x = Oj

402 | EnowBERTng Mathematics-111

1 °° = £ . 2 J |x | e-w abc * 0 00

= J

xer*dx

[ v I*) is an even /?.] [v 1* | = *, x > 0]

-kr-i$* =

0

+

M

— ~ I (0 —1) == 1 Hence M.D. = I Proved. Example 5.11 ; Determine the value o f k, so that the following function represents the p .df. 0, x £ -! fix ) =

* fx + />, - / < x £ 3 4k, 0,

3
Also fin d median* Solution : For p.d.f. : -J

I / (x) dx = 1 3

4

co

J 0 <& + J *(* + 1) a!r + J 4fefc + j 0 -1 3 4 (x + I)?

=1

+ 4k{xh* = 1 •1

|

[16] + 4A(4 - 3) = 1

=>

8* + 4* = 1 k = — * 12'

7b

median : Let M be the median of/(* ) M We have j / (*>a6c = ~

[ by definition}

r

S ta tistics ] 4 0 3

4kM = g + 4*

4.

A M 3

1 1 2 3

M=

|

A / = 2.5 Median - 2.5 Exampte 5.12 : For the distribution Junction (
A ns.

f( x ) ^yoe-^

/. p.d.f. :

[v jto = £

(F(x)]

! f (x)dx = 1

To find yo : For p.d.f.

-<30

00

J yo e~xtt,dx = ! o => =>

-x/a' yo -1/cr

Ia

Hence p.d f. Tofind mean : We have

f (x) - — e-x!°, 0 5 jc < co

•••(I)

Mean = Mi’ = E(x) - J x.f(x)dx

1 o

,-xlc dx -Jl 1/oL o -1/cr

- J e~xladx = - a[e’x/tJ| 0 1 Jo = - o [0 - 1] = a.

4 0 4 | ENGffltemNG M A n e n x n c s -lI 1

Tofind Variance and S.D. : We have c0 Hz' = E (x2) = J x2f{x) dx I x2e~*/adx

£’**/(* ) - ~> e~xh

X 1 Let t - — c =>dt = ~a dx, then

H2' ~ i (t2**2) ^ dt 0 00 = a2 J er*. t2di 0. 00 = tr2 J dt 0 CO

[Since Gama function J e~‘. t^ 1dt 0 fi2’ = o 2 ((3) = or2. 2 ! = 2 ct2.

Herce We know that and 7b

H2' = 2crJ

Variance - ^2' - (Mi1)2 = 2o2 - (a)2 = a 2, S.D. = ^Variance = V ? = ° r (i/ : We have, 1* moment about origin : «0 n ; = J xr. f{x)dx —00

1 00 = — J of. er**dx 0 0 [ Put t = ^

di = Hr = J (toy. e^dt 0

OD

- or. J e~'. r+l~'dt = 0'.

|(r +1)

= r ]
Find median and mode.

(v

£ = J e->r-ldt)

Adj,

S ta tistics j 4 0 5

For p.df. :

1 /(x ) dx = j sin x dx o 2 = - 2 (w **)!

a p.d.f Tofind median : Let m be the median, for which « ! f( x ) dx = 4 (by definition) 2

=> / ( x ) is

TJ 1 sin •x. dx ^

= g1

-[cosx]; =1 - [cos m - IJ = 1 - cos m + 1 = 1 - cos « = 0 cos m = 0

j.

2 Hence median = Tofind mode : We know that x *=a is mode, i f / (*).*» 0 and/* (x) < 0 at x = a. Given

/ (x) = h sin x /

and For mode : Put

(x) - g cos x

/ ' (x) = - ^ sinx /( x ) =0 cos x = 0 cos x = 0 x

Put x =

...d) ...(2)

[by (i)]

2

~

in f (x), we have f

( f ) — 5 si" ( f ) 1 2 <0 is the mode of die distribution.

[by (2)]

~

Hence x

*

te/2

Ans.

4 0 6 | E n g in eerin g M athem atics -111

5.5

THEORETICAL PROBABILITY DISTRIBUTION A theoretical distribution is the frequency distribution of certain events in which frequencies are obtained by mathematical computation. There are two types of theoretical distributions : (i) Discrete theoretical distributions. (ii) Continuous theoretical distributions. some of the discrete and continuous theoretical distribution are : Discrete theoretical distributions: 1. Binomial distribution. 2. Poisson distribution. 3. Power series distribution. 4. Multinational distribution. 5. Negative Binomial distribution. 6. Geometric distribution. 7. Hyper geometric distribution. Continuous theoretical distribution : 1. Normal or Gaussian distribution. 2. Rectangular or Uniform distribution. 3. Exponential distribution. 4. Gama distribution. 5. Beta distribution. 6. Weibul distribution.

5.6

BINOMIAL DISTRIBUTION Binomial distribution is a discrete probability distribution. Let an experiment consisting of n-trials be performed and let the occurrence of an event in any trial be called a success and its non­ occurrence a failure. Lelp be the probability of success and q be the probability of the failure in a single trial, when q - \ - p , so that p + q -\. Probability of r successes in n-trials / ,( r ) = " C / > ', r = 0,1,2...........M or P[X=r\~ "Cr q*~rp r,r^Qt\,2,.................ji where P ( X =r)or P{r) is called binomial probability distribution o f a random variable X. If an experiment consists of n-trials and this experiment is repeated N times, then Binomialfrequency distribution or Expectedfrequency o fr successes is f(r)= N P (r) or f(r)=N.*Crq” 'p rt r= 0,1,2..........n.

5.7

HYPOTHESIS FOR BINOMIAL DISTRIBUTION 1. The variable should be discrete i.e. r=0,l,2,3,4,........etc. and never 1.3, 2.1,3-41etc. 2. The happening of events must be either a success or failure. 3. The number of trials n should be finite and small. 4. The trials or event must be independent. 5. The trials or events must be repeated under identical conditions.

S ta tistics | 4 0 7

5.8

CHARACTERISTIC OF BINOMIAL DISTRIBUTION 1. ft is a discrete which gives the theoretical probabilities. 2. It depends on the parameters p or q and n. 3. The distribution will be symmetric if p - q. ll is skew symmetric or a symmetric if p * q. 4. The statistics of the Binomial distribution are mean = np, variance = npq and standard deviation = Jnpq.

5. The mode of the Binomial distribution is equal to the value of random variable X which has the largest (maximum) frequency. Example 5.14 : For the Binomial distribution, prove that:

Solution.

which is known as recurrence relation fo r the Binomial distribution. By the Binomial distribution, probability of r-successes P(r) = "C, q*~rp r\ r =*0 ,1 ,2 ,..... ji.

=> P(r+1) = " C „ , p r*K.

•••(I)

[ v r - * r + 1]

... (2)

From (1) and (2), we have

P(r+1) "C^q'-'-'p"' n\ q"'-'p'1 r?(«-r)! P(r) ~ "Crq"~rpr “ ( r + I ) ! ( w - r - l) ! « W _ r ! ( w - r ) ( » - r -1)! p _ (w - r ) p ~ (r+ \ ) r \ ( n - r ~ (r + l ) 7 ‘

-\)\q

Hence

5.9

/ ,(r + l)=y

(r+«-'~^> 1) q (#')

Proved.

MEAN, VARIANCE AND STANDARD DEVIATION O F BINOMIAL DISTRIBUTION We know that mean

...(1)

and Variance

-(ji,*)2,

where ji|’ and |i2' are first and second moment about origin, respectively. We have first moment about origin, H,’ = £ r.P(r)

—(2)

4 0 8 j EnGJNEHMNG MATHeKArtCS-III

where / >(r)=:*C,,tf*~^p, is Binomial probability distribution. =>

r»0 V’ **• S I r,~j7—

*-<■ f >

:■ r p(r) = 0 for r = 0]

" v,t = Y --------------- q - 'p '

(*-»)? __ (r-1)!(»-l-r-l)

*>

[v ? + />=!]

H, = «p

...(3)

Thus, mean = jit ~np, i.e., m ~ np. New, second moment about origin: '

=£ r~0

r^ (r) = £

r'* C rq ~ p '

r-0

= £ W r-l)+ r} » C f9- > ' r-0 = J > ( r -1) *C,

[v r J =r ( r~ i) + r ]

r ■*Cr O

'

[By (3)]

r!(« -r)! f

Hr “ !)»(» -1)0* -2 )! _ ^ ,

=«(n-l)p*Y ------ (w-2)! & (r-2)l(«-2-r-2)!

,.f- : p'~z + np

I

• >< ^Sfctasncs j IW

r-2

= « ( « - i ) p 2 ( ? + p r 2+^p = h2p 2 - np2 + np

[•; q +p - \ ]

S p '+ n p Q - p ) ==>

fi2 = n2p 2 + npq

['•'? = !-/?] ...(4)

Since, variance = j^' - n2/>2 + "?*? ~(«p)2

[By (3) & (4)]

= npq. Hence, Variance (ji2) = nP9

(5)

Furhter, standard deviation (S.D.) a = \/Variance or

tr = V«W.

{By (5)]

Example 5.15: In a Binomial distribution the mean and standard deviations are 12 and 2 respectively. Find n and p. Solution.

Given:

mean =np = 12,

... (1)

S. D. - *Jnpq = 2 => *P? = 4. We have,

npo 4 ~ =^

=>

I ? = j-

Since

q + p =l

—(2) [by (I) and (2)]

/> = 1-

2• => /» = ,! - -1 => P = —

2 Now, form (1) np = 12 => n.—~ \2 => « = 18.

Ana. Ans.

Example 5.16: Using the Binomial distribution, fin d the probability o f rotting at most 2 sides in 5 rolls o f a dice. Solution.

Given:

« = 5, P = ~ 5 o

[v Probability o f rolling six is^J

4 1 0 | EN GBeEHNGM AneM xnc8-IlI

By Binomial distribution: P(r) =Cr q"~'p'

..•(!)

P (at roost 2 sixes) = P(r £ 2) = P(r = 0) + P(r = 1) + P(r = 2)

( ? ) +’C’(?) (? )+’C’(f) (?)

54 _ 625 = $r 648 ' Example 5.17 : A coin Is tossed 4 times, what is the probability to getting: 0 ) Two heads, (li) at least two heads. Solution. Given n = 4, let /> =prob. of getting a head in a sinf'; throw of one coin i.e.,

Pw«**d)]

*“ ■

p = xA and q=Vi

ByBinomialdistribution: P(ryCrq*~'pr (i) P (two heads)

...(!)

11 = p (r = 2)= “C, [ ]■ 1 3 . 6.— =4=0.375 16 8

(ii) P (at least two heads) = P(r=2 or3 or 4) = P(r = 2)+P(r ss 3)+P(r = 4)

6 4 ! 11 Ans = — +— + --------=0.6875 16 16 16 !6 AnS' Example 5.18 : The Incidence o f occupational diseases in an industry is such that the workers have a 20% chance o f sitfjfaingfrom U. Find the chance that out o f 6 workers 4 or more will catch the dhmse. Solution: Given; p =20% =*0.2, q = \- p-l-0.2=0.&and n~6, (total workers) By Binomial distribution P(r) = •C .q T p '

- (1)

j S tatistics | 411

P (4 or more catch the disease) = P{r =4 Of 5 or 6) =

P(r=A)+ P(r =5)+P(r = 6)

= 5C4(0.8)5(0.2)4 + ‘C,(0.8)'(0,2)5+ 6C#(0.8)°(0.2)6 = 0.01696 Ans. Example 5.19: A cubical die is thrown in sets o f & The occurrence o f 5 or 6 is called a success. In what proportion o f the sets you expect three successes? Solution,

2 1 p = prob. of throwing 5 or 6 with one die =—= i.e.,

1and ^ q =2p -—

By Bir ..ma] distribution: p(r) = mc r < r Pr

-0 )

/. P (getting 3 successes)= P(r =3)

-• (tm = 56x^=0.2731 3* Hence the required percentage = 27.31 % Ans. Example 5.20 : Ten percent o f screws produced in a certainfactory turn out to be defective. Find the probability that in a sample o f 10 screw chosen at random, exactly two will be defective. Solution.

Given : P ~

1

9 ? = —»n = 10, r = 2

By Binomial distribution: P(r) ="Cr q T 'p ' P (exactly two defective) = P(r = 2) = 10C3

j

j

= 10x9 |r 9_>| _ J — = ] =0.1937. 2x1 VIO; 10x10 2 U 0 J

Ans.

Example 5.21: A bag contains 3 red and 4 blackballs, one ball is drawn and then replaced in the bag and the process in repeated. Getting a red ball in a draw is considered a success. Find the distribution o f X where X denotes the number o f successes in 3 drawn, assuming that in each draw each ball Is exactly likely to be selected.

4 1 2 | E ngwqq *# q M * n « u n c 8 - III

Solution.

3 4 Given: P = ~j> ? = ~ .« = 3, and r = 0 ,1,2,3. Using Binomial distribution:

- o )

Then P(X = 0) = }C0 m

p ( j r = i ) = Jc

n

3TT4f , J ) Vl

[ v n = 3]

i

9 f 4 nJ 717

Thus, the probability distribution of A"is given by : X =0 1 2

"* > -$

« )'

f f i) '

3

(« '

Example 5.22 : Out o f $00 fam ilies with 4 children each, how many fam ilies would be expected to have (i) 2 boys and 2 glris (ii) at least one boy (iii) no girl (iv) at most two girls ? Assume equal probabilities fo r boys and girls. Solution. Given that equal probabilities for boys and girls : 1 P ~~2

1 9 ~2"

[p = probability of a boy]

Also given n = 4, N - 800, using Binomial distribution -

Then expected numbers f(r)~ N .P (r)

0)

... (2)

(i) The expected number of families having 2 boys and 2 girls, / ( r = 2) = 800x4Cz[ j j

= 8 0 0 x 6 x ^ = 300.

A bs.

S ta tistics | 4 1 3

(ii) The expected number of families having at least one boy, =f ( \ o r 2 or 3 or 4) = / ( r = l) + / ( r = 2 ) + /( r = 3 ) + /( r = 4)

Mill}) +‘c,(i)W +‘Ca(i)(i)*+‘c*(iJ = 800x—[4+6+4 +1]~750. 16 J (iii) The number of expected families having no gjris (i.e., r = 4 boys) /(r = 4) = 800x4C4^ ij = 800x^ = 50. = 800

Ans.

Ans.

(iv) The expected number of families having at most two girls, (i.e., at least 2 boys)

= / ( r = 2 or 3 or 4) = /(/• = 2) + /(r = 3)+ /(r = 4)

= 800

- 800 x — = 550. 16 ■

Ans.

Example 5.23: Out o f 800fam ilies with 5 children each, how many would you expect to have (i) 3 boys (ii) 5 girls (iii) either 2 or 3 boys ? Solution. Given that: N = 800, n = 5, and 1 , . 1 1 Probability of having a boy, P=—.then q - \ - p - \ - ~ - ~ . (i) P{3 boys out of 5 children) = P(r = 3) 10

32

= 03125.

The number of expected families =800 x 0.3125 = 250.

Ans.

(ii) /*(5 girls out of 5 children) = P( i£., r = 0 boys)

^ (ir (ij4 — /. The number of expected families = 800 x 0.03125 = 25,

A n s.

4 1 4 | E n q m v b n g M athematjc & -III

(iii) / ’(either 2 or 3 boys) = P(r = 2 or 3) = P{r - 2) + P(r = 3)

= 1 0 x i x i + 1 0 x i x i = — = 0.625. 8 4 4 8 32

The number of expected families = 800 x 0.625 * 500. Ans. Example 5.24 : The probability o f entering student in chartered accountant wlil graduate is OS Determine the probability that out o f 10 students (I) none (U) one and (iii) at least one will graduate. Solution. Given that, the prob. of entering student in chartered accountant will graduate is p = 0.5 /. q = 1- p = 1- 0.5 = 0.5 and n - 10, using B.D. P(r)= *C,
Ans,

(ii) P (exactly one student will graduate) = P(r = 1) = ,flC1(0.5)^,(0.5), = ^ .

Adi

(iii) P (at least one student will graduate) = P(r 2 1) = P(r = 1) + P(r = 2)+ ....+ /*(r = 10) = 1- P(r ~ 0) = 1- Co(0.5)

r«0 (0.5) = 1 - - ^ = 0.99.

Ans.

Example 5.25 : In a bombing action, there is 50% chance that any bomb will strike the target Two direct hits are needed to destroy the target completely. How many bombs are required to be dropped to give a 99% chance or better o f completely destroying the target Solution.

Given: P =

50

1

100

2

Since the probability must be greater than 0.99, if n bombs are dropped, we have

•'

S t a ik t ic s | 4 1 5

- I r c 2 +" Cj +" C4 + .......+" C J 2 :0.99

2 -n-1

2"

zo =>

— r

2:.99

"C0 +*C, +"C2+ ....+'CM=2*]

=£0.01.

100 + 1 0 0 n - 2 " s 0 = > r £100« + 100

By trial, n ~ 11 satisfies the inequality. Hence 11 bombs are required. Am*. Example 5.26: I f 10% o f the bolts produced by a machine are defective, determine the probability that out o f 10 bolts chosen at random (I) one (ii) none (lit) at most 2 bolts will be defective. Solution

10 1 . 1 Given p (defective) = — = i.e., P = —< , 1 9 9 q (non-defective) = 1“ — = — i.e., 9 = — • Also, n = 10, (n is no. of bolts). By Binomial distribution: P W = C r p 'q '( \ Y f 9 YM (i) /. P(one defective) = Pir = 1) ='°C, j^—J

" of e B ) ’ = (9 ,’ = o j8 7 4 ' (ii) /'(none defective) = P(r = 0) \0 / n \10-0 uoj bo,

.. ( I ) A n .

(iii) /*(at most 2 bolts will be defective) = P(r 5 2) = P(r = 0) + P(r = I) + P(r = 2)

= 0.34M+ 0J»74+» C , ^ J ( i J

[by (1) and (2)]

=0.3486 + 0.3874 + 0.1937 = 0.9297.

Ans.

Example 5.27: Assuming that 20% o f the population o f a city are literate, so that the chance o f an Individual being literate is 1/S and assuming that 100 investigators each take IQ Individuals to see whether they are literate, how many investigators would you expect to report 3 or less were literate ? Solution.

1 4 Given: p = — and? = —>* = 10,usingB.D., P(r)=nC,
- " & li

•

e

r - T

, v ,i ? ' +' “Ci

B

M

s ' - i s j ) '

4 't = ( j j K0.8)5+ 2(0.8)* +1.8(0.8) + 0.96] = 0.2097[0.512+1.28+1.44 + 0.96] = 0.2097 x 4.192 = 0.8791 Required number of investigators = 0.8791 * 100 = 88 (nearly). Aas. Example S.2S : Find the binomial distribution whose mean is 4 and variance is 3. Also fin d its mode. Solution.

Given: mean = np = 4

... (1)

and variance = npq = 3 Dividing (2) by (1), we get

... (2)

S ta ti s tic s { 4 1 7

Thus the required binomial distribution is Ans. Now mode of this binomial distribution = integral part of (np + p) - integral part of f 4 + —J , integral part of 4.25 = 4.

Ans.

Example 5.29 : The probability that a bomb droppedfrom a plane will strike the target is 1/S. I f six bombs are dropped, fin d the probability that ft) exactly two will strike the target, (ii) at least two will strike the target. Solution.

1 Given that P - ~ and n = 6. 4. q = \, - p = ,\ —1 - — H y 5 5 Using B.D.

P(r) ='C,

p ' , r = 0,1,2, ..,ji

(i) P (exactly two will strike the target) = P(r ~ 2) Ans. (ii) P (at least two will strike the taqjpt) ;= P(r 1 2) = P(r = 2 or 3 or 4 or 5 or 6) = P(2) + P( 3) + />(4) + P(5) P{6) v £ P(r) = 1 r»0

= 1- [P(r = 0) + P(r = 1)]

S=l= 0.345.

= 1-[0.26?1 +0.3932] Ans.

418 ( Engmeermg Mathematics-!11 Example 5.30 : The sum and the product o f the mean and variance o f a Binomial distribution are I and 128. Find the distribution. Solution.

Given:

np + npq = 24 and (np)(npq) = 128

==»

« p (l

[ v mean = np and variance = np

+ q ) = 24 and n1p zg ^ 128 24

and

(np)2q = 128 ~ ~ 1 9 = 128 1+9 /

[by (1

=>

576 (FT^F 9

=>

9q = 2(1 +2? +•./)

=>

2q2- S q +2=0 => (2 q -l)(q -2 )= 0

=>

q=

I hen

I 1 l> i - <7^ i - —-=>/> = —

^<*q = 2 ( U q f

[But here = 2 is not possibi

From (1), we get „ = _24x4 __ ^

„ - 32.

f i iY Hence Binomial distribution is (? + P f - 1 ^ + J J *

Am

Example 5.31: Suppose A and B are two equally strong table tennis players. Which o f thefollowk two events is more probable (a) A beats B exactly in 3 games out o f 4 or (b) A beats B exactly in 5 games out o f 8. Solution.

1 Given that P =
P(r) = *Cr q*~rp f.

S ta tistics I 4 1 9

(a) Prob. of A beating B in exactly 3 games out of 4 , \3

—(ij(0

[V r = 3. » = 4]

= 4 x i x i = i = 25% 2 8 4

...(I) w

(b) Prob. of A beating B in exactly 5 games ou; of 8 [ v r = 5, n = 8]

= 5 6 x l x — = — = 21.875% . 8 32 32

...(2) K'

Clearly from (I) and (2), that first event Is more probable. Ans. ample 5.32: A set o f 8 biased coins was tossed 256 times and the frequencies o f heads obtained are given by the following table: Number o f heads (x) 0 1 2 3 4 5 6 7 8 Frequencies (f) 2 6 24 63 64 50 36 JO J Fit a binomial distribution to this data, ution. Here n = 8 and x = r ~ 0, 1 , 2 , ......8. 9* Since Mean =m = — W _ 2xQ+6xl+ 24x2 + 63x3 + 64x4 + 50x5 + 36x6 • -Uv /-t-lx8 2 + 6 + 24 + 63 + 64 + 50 + 36 + 10 +1 0 +6 + 48 + 189 + 256 + 250 + 216 + 70 + 8 _ 1043 256 256' Also

I / = W = 256

1043 Since mean of B. D. m =np~ — — 6JO

M -S S p-

- 0.5093 2048 9 = l - p = l - 0 . 5 0 9 3 = 0.4907

t" = * l

420 I ENGINEERINGMaTHEMAT1CS-IH Binomial distribution is (q + p )' = (0.4907 +0.5093)* - = *Cr(0.4907)*'(0.5093)'. Now expected frequencies for jr=r=0,l,2,...............,8 are : f( x

=

r)=N. "C, q -'p'-,

for x - 0.

/ ( 0)=256x 'C0 (0.4907)* (0.5093)° =0.861.

for j :

=

1,

/(l)=256x*C, (0.4907)’ (0.5093)' =7.145.

for x

=

2,

/(2)=256x aC2 (0.4907)6(0.5093)2=25.956.

for x

=

3,

/(3)=256x %

for x

=

4,

/(4)=256x SC„ (0.4907)4(0.5093)4=69.903.

for x

=

5,

/(5)=i56x*C, (0.4907)1(0.5093)5=58.042.

for x

=

6,

/(6)=256x ’C4 (0.4907)2(0.5093)6=30.121.

for x

=

7,

/(7)=256x8C7 (0.4907)' (0.5093)7=8.932.

for x

~

8,

/(

(0.4907) 5 (0.5093/ =53.88.

8)=256x *C„ (0.4907)° (0.5093)* =1.1589.

A ds

Example 5.33 : The following data are the number o f seeds germinating out o f 10 on dampfitterf « 80 sets o f seeds. Fit a binomial distribution to these data X : 0 1 2 3 4 5 6 7 8 9 10 Total F : 6 20 28 12 8 6 0 0 0 0 0 80. Solution.

Here n =10

N - I f - 80. Ifx mean ~

m

~

y

~

0x6 + 1x20 + 2x28 + 3x12 + 4x8 + 5x6 _ 174 _ 87 go ” 80 ~~40 87 = 40

Since

mean of B.D, m =

=3*

/> = - * ! = 0.2175. 400

Then

q = 1-/?= 1-0.2175 =0.7825.

np^>—

np

[• « = 10]

Hence the binomial distribution to be fitted to the data is = 80(0.7825 + 0.2175)10 [V f ( r ) = N ”CTqn rp'\

Statistics | 421

or

/(* =

r)= 80x ,oCr(0.7825),<>'' (0.2175)',

r = 0,l,2.........,10.

Thus, expected frequencies of x = r = 0,1,2,.............10 are respectively 6.9, 19.1, 24.0, 17.8, 8.6, 2.9, 0.7, 0.1, 0, 0, 0. Ans. Example 5.34; In litters o f 4 mice the number o f litters, which contained 0,1, 2, 3, 4 females were noted. The data are given below No. o f female mice 0 1 2 3 4 Total No. o f litters : 8 32 34 24 5 103 I f the chance o f obtaining a female in a single trial Is assumed constant, estimate this constant o f unknown probability. Find also the expectedfrequencies. Solution.

Here n ~4 .

N - I f = 103. L/x 8x0 + 32x1 + 34x2 + 24x3 + 5x4 mean ~ m ~ ^ ~ 8+32+34 + 24+5 32 + 68 + 72 + 20 192 , ----------------------- — = I .864. 103 103 n = 4 and np = mean = 1.864.

=>

/> = ^ 1 = 0 .4 6 6 . 4

Then

q = !-/> = 1-0.466 = 0.534.

[ v mean of B.D., m = np\ [v « = 10]

Hence the expected frequencies are the respective terms of the binomial expansion of = N .(q +p)n ~N .RCt q*~r p '. or

/ ( r ) = 103x-'C,(0.534)*" (0.466)' for r = 0,l, 2,3,4.

Ans.

Example 5.35 : Fit a Binomial distribution thefollowing data and compare the theoreticalfrequencies with the actual ones. x: 0 1 2 3 4 5 f: 2 14 20 34 22 8. Solution. Given : n = 5, N = E/= 100. Zfx 0x2 + 1x14 + 2x20+3x34 + 4x22 + 5x8 mean = ">= — =---------------------- —

.’. Since,

mean of B.D, m - n p ^ n p - 2.84

4 2 2 | E n g in e e r in g M a th e m a tic s - 1 II

p ~ 0.568. i'hen

q ~ \ —p ==> ^ = 1-0,568 => q = 0.432,

! lento theoretical or expected frequencies are A x = r) = N.(q + p y = N. *C, q*~'p' = 100 xs Cr(0.432)w (0.568)', for r = 0,l,2

,5.

Thus, theoretical frequencies of x - r - 0,1,2,3,4,5 are respectively 1.5, 9.89, 26, 34.2, 22.48, 5.9.

5.10

POISSONDISTRIBUTION

Am j

The Poisson distribution was discovered by a French Mathematics Simesi Denis Poisson ii 1837. It is a discrete distribution and is very widely used. Poisson distribution is a limiting foq of the Binomial distribution in which n, the number of trials, becomes vsiy large and p, th probability of the success of the event, is very-veiy small such thar me.tn m - n p is a finit quantity. .-. Probability of r-successes.

3

e~mmr P(X = r)= — “ , r

.v ....................

Also theoretical or exported frequencies is g~m (ri

f ( X - r ) = N..... .

r!

where m = mean and N is number of trials.

The following are the statistical measures of the Poisson distribution. (i) Mean - np to m. (ii) Variance ~ m or np. (iii) Standard Deviation 0

^

.

(iv) Moment measure of skewness (? i)= ' ^ p

(v)Moment measure of kurtosis (Yj ) = ~ . rrt Some examples o f Poisson distribution : 1. The number o f deaths in a city in one year by a rare disease. 2. The number of printing mistakes in each page of the first proof of a book. 3. The number of defective screws per box of 100 screws.

S ta tistic s |

5.11

423

CONDITIONS UNDER WHICH POISSON DISTRIBUTION IS USED 1. The random variable X - r should be discrete i.e., r = 0. 1,2,... n. here n i» i.i-jic. 2. The happening of the events must be of two alternatives such as succeo. *>■-' ; jlure. 3. It is applicable in those cases when the number of trials n is very iarj:c u p .! probability of success p is very small but the mean m is finite. 4. p should be very small (close to zero). Ifp -* 0, then the Poisson distribution is J-shaped and imimodal.

Example 5.36 : Prove that Poisson distribution as a limiting case o f Binomial distribution, when n -* oa Solution.

Binomial distribution

424 j Engineering Mamematics-UI Since n is very large so that L im

since

..... f l - ^ l j = ( l - 0 ) ( 1 - 0 ) ..... ( 1 - 0 ) « 1 a „ d

v lim fl-—)"*=<;

= e'm. and

XJ

Then (1) becomes : . e mmr p ( x ~ r ) = — ~ , r = 0, 1, 2 ,...,«

which is probability function of Poisson distribution.

5.12

-

Proved.

M EAN VARIANCE AND STANDARD D EVIATIO N O F TH E POISSON DISTRIBUTION For the Poisson distribution e mrn' P ix= r)= ~ 7 r >

r =0,1,2..... «

We know that mean and variance = ^ 2' - ( n , ) 2,

where |ij' and u2’ are first and second moment about origin. F o rr=0,1,2,..,,oo first moment about origin.

... ( 1)

ScMtsnc^ 142S

From (1) mean = jti' = m.

Now, second m om ent abovr origin, r 2 P (r)

[Since r 2 = r ( r - l ) + r ]

r=0

^

,

,.e 'mrnr

mO

r■

V'

e~Mmr

^>0

' *

AC

m

1!

__3

_4

1!

2!

j W m m ‘ + — + — ■•+.

- ‘ - I ( r - 2)1 r=t , m = m 2e~m 1 + —

[w = mean]

,

2!

+«

+m

[by (3) and (4)3

« « " /+ «

• (5)

jij =m +m

From (2), variance

t

t - ( h, )j [by (4) & (5)]

- m 2 + m -(m )2 = m.

(R G P V Feb. 2006 A June 2008 (N )]

Thus, variance = m Further, standard deviation,
E xam ple 5 .3 7 : Show th a t in a Poisson distribution w ith u n it m ean, th e m ean deviation about the m ean, is 2/e tim es the standard deviation. Solution.

By Poisson distribution P (X ~ r ) = ? - ^ f- ,

r\

>"=0,1,2......oo.

Given that mean m =1. X f\x -m \ We know that mean deviation about mean,

I

/ k - 1!

r*0________ Xf

N

- I f and x=r]

... (1)

4 2 6 | Engdeehng MxnEMAnc&-lIi

Here

/ —P‘r')-e~m

and

Z /= I

J

r\

/•!

[ v « = 1]

r\

[■.* total frequencies - 1]

r-'”> I

Hence (I) becomes : * > -l|

u jx - i

) 0 - I |) l l - l [ | | 2 - l | | |3 - l j t 0! 1! 2! 3! 1+ ( y ) + < y ) + (iz i)+.

2!

3!

4!

=e

, 2 1 3 1 4 1 1 + ----- —■-i----------- ^ ---------- K 2! 2! 3! 3? 4! 4!

-e

, 1 1 1 1 1 1 1 + ——— + — —— + — -— + .

1! 2! 2! 3! 3! 4! = 2e "

Since standard deviation of Poisson deviation S.D . =Jnt=-J\ =1. Hence (2) becomes: M .£).=—.! = —(S.Z>.). e e Exam ple 5.38 : F or the P oisson distribution, prove that

P(r+1)*

m

P(r).

(r + 1) W hich is know n as recurrence relation fo r Poisson distribution. Solution.

By Poisson distribution e _m

P (r)= _

r=0)1>2t>

replacing r-»r+ l. P{r +1)=

e mr*1 (r+1)!

... ( 2 )

Iv « = 1]

Proved.

S ta tistics | 4 2 7

Now,

P (r +1) e"m mr*> . r \ P{r) (r + 1)! e - m r

r! (r + l)r!

m

r+ 1

Hence*

Prvoed.

Exam ple 5.39: F ind the probui !!'• y th a t a t m ust 5 d efectivefu ses w ill b efo u n d in a box o f 200fuses, i f experience sho- > th a t 2 percent o f such fu se s are defective. Solution.

G iven :

p -2 Vo

- i t r 02- "■ *” mean m = n p =>

m= 200 x .02 => m = 4 .

Since n i» lt«ge so that using Poisson distribution g-m

p (r >----- —

• r = 0, l , 2, ........... 2 00.

... ( 1 )

/. P (at m ost 5 defective) = P (r £ 5) - / » ( 0 ) + /*(!)+/* (2 )+ P(5)+P(4)+P(5)

-e

4 42 4J 44 4s +— II 2! 3! 4! 5!

1 ^— 4--*'—+ —

[ by ( 1 )]

=0.0183[1 + 4 + 8+10.6667 +10.6667 + 8.5333] =0.0183x42.8667=0.7845. Exam ple

Ans.

5.40: In a certain fa cto ry turning razor blades, there Is a sm all chance o f 0.002fo r any blade to be defective. The M odes are in packets o f 10. Use P otsson's distribution to calculate th e approxim ate num ber o f packets containing no defective, one defective and tw o defective blades respectively in a consignm ent o f 50,000packets.

Solution.

Given, p =0.002, n=10, N =50,000 m -np

= 10 x 0.002=0.02

By Poisson distribution _ .r e„-K» m P { X = r ) = — — , ,-= 0 ,1 .3 ,.......... 10

(i) / >(n o d e f ic tiv e ) = /’ ( r = 0) =

® 1 = 0 .9 8 0 2

4 2 8 | E nginhew ng M axhocattcs-III

Hence number o f packets containing no defective blades =N.P(r=0)=50, 000x0.9802=49010.

Ans.

(ii) P (one defective) = P(r=l)=e~001.— ^ -= 0 .0 1 9 6 0 4 Number o f packets containing one defective blade =M />(r=l)=50,000x0.019604 =980.

Ans.

—^—=0.00019604

(iii) P (two defective) ~ P (r ~ 2)- e -0®

/. Number of packets containing two defective blades =Ar./>(r=2)=50,OOOx0.00019604=9.8«10.

Ans.

Exam ple 5 .4 1 : I f th e probability th a t an Individual su ffers a bad reaction fro m a certain injection is 0. 001, determ ine the probability th a t o u t o f 2000 individuals (i) exactly 3

(li) m ore than 2 individuals

(iii) none

(iv) m ore th a t 1 individuals

w ill su ffer a bad reaction. Solution.

Given

p =0.001 (which is very small)

n =2000 (which is large), then m - n p => m* 2000x0.001=2 Using Poisson distribution: P (X = r)= e m~ ^ , r = 0 ,l, 2.................2000.

(i) P (exactly 3 suffer a bad reaction) ~ P ( r = 3)

3!

8e_I 4 r = 0 . 180. 6 3e2

(ii) P (more than 2 suffer a bad reaction) = />(r> 2)= l - [ i >(r= 0 )+ i>(r= l)+ />(r=2)] = 1 - e~m + m e

1!

m V *‘ i -i 2~. J. .2 "1 =1 - r — _e e 2!

2V .

= 1 -4 = 0 .3 2 3 . e (iii) P (none suffers a bad reaction) = P(r= 0 )= e '" = l/e 2=0.135.

Ans. Ans.

S tatistics | 4 2 9

(iv) P (more than 1 suffers a bad reaction)

= />(r>l)=l-[/’(r=:0)+i,(r=l)] =

_m mle 'm

1-

■1-

1!

' I 4___ 21 y

Ans.

= 1 - 4 = 0 .5 9 4

e

Exam ple 5.42: A car-hire fir m tw o cars, which it hires out day by day. The num ber o f dem ands fo r a car on each day is distributed as a Poisson distribution w ith m ean 1.5. C alculate the proportion o f days on w hich neither car is used and th e proportion o f days on which som e dem and Is refused. Solution. Given mean m = 1.5, i _r

Poisson distribution ^ ( r ) —

m r\

r =0,1,2,3,.

(i) The proportion of days when neither car is used i.e.,

P (r = 0)

m 0!

=

Ans.

e‘l 5=0.2231. (ii) Since total cars = 2 demand is.refused when r £ 3. Hence, the proportion of days on which some demand is refused = />( r s 3 ) = l- [ F ( r = 0 )+ /,(r= l)+ P (r= 2 )] e

„0 -m rrt — +e

0!

2 ffi1 . mr7i — +e —

1!

2!

1+0-5)+ ( I S ? =1-0.8087

=0.1913.

Ans.

Exam ple 5.43: A telephone sw itch handles 600 calls on th e average during a rush hour. The board can m ake a m axim um 20 connections p er m inute. Use P oisson distribution to estim ate th e probability th a t the board w ill be o verta xed during any given m inute. Solution. Given that mean number of calls per minute

4 3 0 | Enchnehong M /m o w ic s -IIl

fit' Poisson distribution P(r )= e

r = 0 . ' 2,............600.

P (using 0 to 20 callas per minute) =P(r-Z2Q) =P(r=0)+P(r=l)+ ...... + /V = 2 0 ) »

„<■

“5 *

7*T*

“

20 -io V * r-0 §

_r

m

7T

-

0 )

Thus P (the board will be over-taxed during any given minute) " P (when the calls are more than 20) = P ( r> 2 0 )= l-P (r5 2 0 )

*»0

Ans.

fr-

~\

Exam ple 5.44 : I f 3% o f th e electric bulbs m anufactured by a com pany are defective, fin d the probability th a t in a sam ple o f 100 bulbs exactly fiv e bulbs are defective. Solution.

Given that

3

=0.03, n=100

/. mean

m=np=100x0.03

=>

m = 3.

By Poisson distribution ^ ( r) ~ e

irf

r= 0,1,2,......... 100.

/. P (exactly five bulbs are defective) = P (r= 5 ) = e „ ( 3 ) l = 0.04979 > 2 4 3 = 0 1 0 0 8

5!

120

An8,

Exam ple 5.45 : A m anufacture know s th a t th e condensers h e m akes contain on an average 1% o f defective. H e p a cks them in boxes o f 100. W hat is the probability th a t a box picked at random w ill contain 4 or m ore fa u lty condensers T Solution.

G iven th a t/> = ^ ~ = 01,

By Poisson distribution:

»=100. Mean =/n=«p=100x0.01=I.

S tatistics | 4 3 1

P (4 or more faulty condensers)

=/>(r£4)=P(4)+/>(5)+....+ iT.,;0) =1 -[/>(0)+ P(\) + P( 2) f /’(.I)]

=

1-


e l

1!

2!

---- + —— +

0!

e-'

4- —

3!

- • - j ' ;+H

=1 ——=1 -0.981=0.019. 3e

Ans.

Exam ple 5.46: A S killed typist on routine work kept a record o f m istakes m ade p er day daring 300 w orking days M istake p er day :

0

1

2

3

4

5

6

N um ber o f d a y s :

143

90

42

12

9

3

1

F it a Poisson distribution to the above date and calculate expected (or theoretical) frequencies Solution.

1 4 3 x 0 + 9 0x1 + 4 2 x 2 + 12x3 + 9 x 4 + 3 x 5 + 1x6

Mean = « = -

300

0 + 90 + 84 t- 36J- i 5 + 6 267 300

m --

or

By Poisson distribution: e~mmr H

’

r = 0 ’1-2’............6>

The expected (or theoretical) frequency for r success f( r ) = N P ( r ) ^ N .^ - ~ ~ - . r = 0,l,2....... 6

for

r =0

/(0)=NP(0)=-^ .C— ”i2.^ -=300x0.411=123.3«123 - ^ — ^ =300x0.365 =109.5 «1 10

for r = 1,

/(l)= A y (l)= ^

for r = 2,

/(2)=A7><2)=

300e“°*9(0.89)2 -=300x0.163=48.9*49 2!

for r = 3,

/(3 )= ^ ( 3 ) =

300e~°*9(0.89)* =300 x 0.048 =14.4 «14 3!

[V N = 300]

432 | B iQiBBBjiin Mwwamaics-ni fo rr = 4,

/(4)=jW >(4)=-— - ” p ^ -=300x0.011=3.3 *3

fo rr = 5,

3Q^ ^ 5,^ .- - ) -=300 x 0.002 = 0.6 »1 .f(5)=NP(5)= x-

for r = 6,

/ ( 6 ) = M>( 6)=—

«300 x 0.0003 « 0.09 * 0

Thus expected (or theoretical) frequencies are : r: 0 1 2 3 4* 5 6 /,: 123 110 49 14 3 1 0. Exam ple 5.47: The frequency o f accidents p er sh ift in a fa cto ry Is given in the fable below A ccident p e r s h # (x) :

0

1

2

3

4

Ans.

Tpftl

Frequency ( f ) : 192 100 24 3 1 320. F ind the corresponding Poisson distribution and com pare w ith actual observation.

„ , ^

w

l£ c

Mean =mf

0x192+1x100+ 2 x 24 + 3x3 + 4x1

—

1«+fdo724 ; t ; i —

=

= — =0.503 320 . e '"m r By Poisson distribution, ^ \ x ~ r) -----—* r =0,1,2,3,4. The corresponding probabilities for r success o /x

-« « (0.503)'

P i r ) * * * 3” - ~

r=0,l,2,3,4.

i.e., Probabilities a re : 0.605, 0304, 0.076, 0.0128, 0.0016.

Am .

Now the corresponding expected frequency for r success / ( r ) = JVJ»(r)=320x«-05® i 2 ^ L , r = 0,l,2, 3, 4. r\ i.e., Corresponding frequencies are : 193.6,

97.3,

24.5, 4.1, 0.5.

A»s.

Exam ple 5 .4 8 : F it a P obson’s distribution to th e fo llo w in g calculate theoreticalfre q u e n c ie s ; r:

0

1

f:

122

60

2 15

3

4

2

1

S tatistics | 4 3 3 „ ,

,

Solution.

w Zfx 0 x 122 + 1x60 + 2x15 + 3x2 + 4x1 M ean = « = — ------------------------ — ---------------------2*/ 2w

:.

60 + 30 + 6 + 4 200

N = Z f= 2 0 0 ]

100 _1 _ Q s 200

2~

‘

The theoretical frequencies of r success f { r ) = N P ( r ) = N . e - ^ = 200x«f<’ » x < ° ^ l ; r = 0,l,2,3,4.

For r - 0,

/ ( 0)=200xe-<,J1 2 ^ ! = i21.3*121

For r * 1,

/ ( I ) - 2 0 0 x ^ s^ ^ - 6 0 , 6 5 * 6 1 .

For r = 2,

/( 2 ) - 2 0 0 x « ^ £ ~ 9 l- lS .2 5 * lS

For r - 3,

/( 3 ) = 2 0 0 x e ^ M l= 2 .5 3 * 3

For r = 4,

/(4 )* 2 0 0 x e ^ & 2 l* 0 3 l6 « 0 .

•!

Thus, theoretical frequencies are : r: 0 1 2 3 4 /,: 121 61 15 3 0. Aos. Example 5.49: The m ortality rate fo r a certain disease is 7 in 1900. W hat is th e probability fo r ju s t 2 deaths on account o f th is disease in a group o f400 ? Solution.

Given:

7

p ------- = 0.007 and n * 400. F 1000

mean, m = np * 400x0.007 - 2.8.

By Feisson distribution, probability of r-success ^ ( r ) = ~ r = 0 , U ........... 400. r\

P{2deaths) = P(r = 2) = ~ —

= 0. 2384 = 2 3 .8 4 %

A ns.

434 | Engrcehing Mathematics-III Exam ple 5.50 : I f x is a Poisson variety su ck th a t P (x = 2 ) - 9 P ( x ' m4) + 90P (x'*6), f i n d the m ean o f x. Solution.

By Poisson distribution with mean m is x = 0,1,2...... .co.

P(x) = Since,

P(x = 2) = 9 P(x = 4 ) + 9 0 P(x = 6) e~mm 2 —

e~"m4 = 9 x —

e"m 6 + 9 0 x —

=>

, 9m 2 1=— +.

=>

, 3 2 1 4 U - w + 2 m = > » 4-3 m l - 4 = 0

=>

(m 2 + 4)(m2 -1 ) = 0

=>

m = ±1

[*.* m 1 - -4 is rejected]

=>

m - 1.

Ans.

12

90m 4

6x5x4x3

E xam ple 5 .5 1 : S ix coins are tossed 6,400 tim es. U sing th e Poisson distribution, w hat is approxim ate probability o f getting six heads x tim es. Solution.

Given: n - 6400. v Probability o f getting a head in a throw o f a coin =

Then, Probability of getting six heads in a throw of six coins, P = ^ ~ ~ -

mean = m - n p = 6400x— = 100. 64 By Poisson distribution, probability o f r success, i>( r ) = l ^ l ( r = 0, 1,2 ........

r!

e mm‘ e-m (\Q 0y P (six heads x tim es) = P{r = o r ) -------------------------------

Ans.

S tatistics | 4 3 5

5.13

NORMAL DISTRIBUTION It is a theoretical and continuous probability distribution in which the relative frequencies of a continuous variable are distributed according to the normal probability law. In other words, it is a symmetrical distribution in which the frequencies are distributed evenly about the mean o f the distribution. Normal distribution is a limiting form of Binomial distribution under the following conditions: (i) h, the number o f trials is infinitely large, i.e., «-><». (ii) neither p for q) is very sm all I.e., p and q are fairly near equally. A random variable x is said to have a normal distribution with mean ‘n’ and standard deviation ‘or’ if its probability density function is given by /(* )= — «W2 «

•00
The probability density function with mean zero i.e., h = 0 and standard deviation
-00< x < 00.

o V 2 rt

Normal distribution was first discovered by British Mathematician De-Moivre in 1733. Normal distribution is also known as G aussi an distribution The total area under the normal curve is equal to unity and the percentage distribution o f area under the norm al curve is given below as shown in the figure. (i) About 68% o f the area fells between h - cr and (i + a . (ii) About 95.5% o f the area fells between - 2o and (i + 2o. (iii) About 99.7% o f the area falls between p - 3cr and n + 3a.

68.27% • 95.S% •

4 3 6 | E n g jn eew ng M athem atics-1]I

5.14 STANDARD NORMAL DISTRIBUTION A random variable z w hich has a norm al distribution w ith m ean - 0 and a standard deviation cr = 1 is said to have a standard normal distribution. Its probability density function is given by / ( z ) = - ?L = -e ',J/2,

— oo < 2 < oo,

V2 rt It is denoted by N (0, 1). In short, standard norma] variety is w ritten a s S.N. V. The area under any norm al curve is found from the table o f a standard norm al probability distribution show ing th e area betw een th e m ean an d any v alue o f the norm ally distributed random variable. For a given valu e o f and an d
fi

CT The purpose o f standardization o f norm al distribution is to enable us to m ake use o f the tables o f t

the area o f th e standard curve / ( * ) =

^

for various points alo n g the x-axis.

The standard norm al distribution is also know n as Unit Normal Distribution or Z-Distribution. The standard norm al cu rv e helps u s to find th e areas w ithin tw o assigned lim its under the curve. The areas b etw een th e standard norm al curve draw n at tw o assigned lim its a and b w ill give the proportion o f cases fo r w hich th e values o f z lie betw een a and b. Thus the area betw een two assigned lim its a an d b und er th e standard norm al curve w ill represent the probability that Z will be betw een a an d b. It is denoted by P ( a £ Z £ b ) .

z = -3

z =- 2

z- 0

4

t

0.6827

0.9545 0.9973

z = +l

z = +2

z = +3

S tatistics | 4 3 7

5.15

PROPERTIES OF NORMAL CURVE The normal probability curve with mean p and standard deviation o has the following properties: 1. The equation of the curve is 1

,

-OOSX^OO.

and it is bell-shaped. The top of die bell is directly above the mean p. 2. The curve is symmetrical about die line x = p and x ranges from - » to +«. 3. Mean, mode and median coincide at x *

p as the distribution is symmetrical.

4. .A'-axis is asymptote to the curve. 5. In Normal distribution: Arithmetic Mean = p and Variance * a 2 6. The points of inflexion of the curve are at x = p + a, x = p - a and the curve changes from concave to convex at x = p + o to x= p~ cr. 7. The mean deviation from the mean in normal distribution is equal to 4/5 of its standard deviation, S. All the odd moments about the mean are zero i.e.,

=0.

9. The maximum ordinate lies at the mean i.e., at x = p. 10.

The curve o f normal distribution has a single peak, i.e., it is a unimodal.

11 .The two tails of the curve extend indefinitely and never touch the horizontal line.

5.16

M ETHOD TO FIND THE PROBABILITY W HEN TH E VARIATE IS NORMALLY DISTRIBUTED Let X be a normal variate with mean p and standard deviation or. Suppose we want to find the probability that a randomly selected value f o r X will lie between a and b i.e., P(a < X < b ).

Step /. Convert X into a standard normal variate by die formula ...{0

O

Step 2. Find the limits of Z corresponding to the limits o f X . ry

tf-P

When

X = a then Z --------

t Put X ^ a in (1)]

When

~ b-p X - b then 2 = -----a

[ P u tX = b in ( l) }

4 3 8 | E ngineering M athematics-III

Step3. Thus

P{a< X< b) = p [ ^ ^ < Z < t t ^

Step 4. From the normal table find the probability that Z is between (a - n )/o and (6 - \i)/a Exam ple 5.52 : A certain type o f wooden beam has a m ean breaking strength o f 1500 kgs and a standard deviation o f 100 kgs. F ind th e relative freq u en cy o f a ll such beam s whose breaking strengths are betw een 1450 and 1600 kgs. Solution.

Let X be the breaking strength. Then we are to find P(1450< X < 1600) Let A"be a normal variety with mean p = 1500 and standard deviation a = 100. Then standard normal variety N (0,1) : ^ _ X -y. _ X

cr

500 100

When

X = 1450, then Z = !i f f

= -0,5

When

X = 1600, then

Thus

P ( \4 5 0 < X <1600) = i ,(-0 .5 < Z < l)

lUU

= P (-0 .S < Z < 0) + i>(0 < Z < 1) = 0.1915 + 0 3 4 13 = 0,5328

[From normal table]

=> 53% of the beam has the breaking strength between 1450 kgs and 1600 kgs.

Ans.

Exam ple 5.53 :Prove th a t th e po in ts o f inflexion o f the norm al curve are x = ±
Let the equation of normal curve be

...d)

y = y 0e*’/2*’ We know that points o f inflexion are given by d 2v

d*y

Differentiating (1) w.r.t. V, we get

S ta tistics j 4 3 9

dx3

a*{

a 2)

Now, for inflexion points :

d 'y ■- 0 => '

xJ ^

d x2

v /

V

A lso a t x = ± o , d 3y l d x >* 0. H ence th is show that points o f inflexion are at x = ± a .

P roved.

Example 5.54 : Prove that in a normal distribution, all the moments o f odd order about the mean are vanish Le., zero. Solution.

By definition o f m om ents about th e m ean u o f odd order :

H ^ , = ) (x-M )*V
W h e r e ,/ (x) is p.d.f. o f norm al distribution i.e.,

/ « = — c tv!2rji- e W

'” ‘

'
I

= — pL=

- «>

[by ( 1 ) ]

'

f ( x - M y + '- e ^ r f'^ d x

Putting,

= - J _ ]

( o O ^ '.e ^ . o d t ,

v m i

= 0; since t u *xe~'>12is an odd function, so that

T h u s,

J oeW = 0 A ds.

4 4 0 1 E n g m b h m o M * n tE M x n c s4 U

Exam ple 5.55 .* Prove th a t in a norm al distribution, a ll the m om ents o f even order about the meat is

= (2n - l)(2 n - 3)------- 5.3.1 — a u . Solution.

Let f ( x ) be p.d.f. of normal distribution i.e., 1 /(* ) =

( 1)

We know that moments o f even order about the mean jt is = } ( x - n ) u f(x )d x

[by (1)] =— J—

rt

Putting,

T (x - p . f '. e ^ '^ d x . J

1-- - ——=>dx = a d t

ii - —7- — f

^

{at) 2”.e~*>n.o d t

o V (2 « )i

J t 7Me -1,2d t

=

2o> ' ^ 0

since t ^ e ^ ' 1is an even function, j even - 2 J even -
0

Now putting /2 = 2 z or t - y f o j z , '■ dt - d z i ^ ( 2 z ) , we get ^

= 1 L - f z<",/2>e ' d z . V7t ;

2*o*"rf J l

f z(*+V '<£r vk

5

v J e"'zN' xdz ~ V(N)

S ta t o u c s { 4 4 1 Replacing n by (n - 1), we get 2- ‘a

^

1^1

- (2)

2J

i^ r T

From (1) by (2), we have

[v r*=(x— i)r(x-i>]

Now proceeding as above processes, we get m ,= (2 » -l)(2 » -3 )....5 .3 .1 .o 2j\

Proved.

Exam ple 5.56 : To drive the norm al distribution as a lim iting case o f B inom ial distribution, when p -q . Solution.

We know that in a Binomial distribution frequency of r-success is m

Since

... ( 1 )

= N.*Crq” p r 1

1

P = 9 => P = ~ and ? = -

[because p + q = \]

Suppose that n is an even integer, i.e., n = 2k, k being an integer. Hence (1) becomes: m

Replacing r - y r + 1, we get

f ( r + 1)^ fir) u Cr

(2 t)l(r)K 2 t-r)l _ 2k - r (2k - r ~ l)!(r + 1)!(2£)! r + 1 '

... (2)

442

| E ngineering M athematics -!!!

The frequency o f r successes will be greater than the frequency o f (r + 1) successes if / ( r + l)//(r)< l

2 k - r < r + \ => r > k ~ 2

- (3)

In a sim ilar w ay the frequencies o f r successes will be greater than the frequencies o f ( r - 1) successes i f .

1

r < k + —. 2

... (4)

From (3) and (4), w e observe that i f * “

< ' , < * + ^ , t h e frequency corresponding to r

successes w ill b e m axim um . C learly r - k is the value o f the success corresponding to w hich th e frequency is m axim um . Suppose it is Then from (1),

—

'( a '- 'i s a i r

Let yx b e th e frequency o f ( i +

jc)

successes then w e have (2 *)!

x )K k -x )\ k lk l y0

(it + x ) ! ( £ - x ) !

k ( k - \ ) ( k - 2 ) ...( ,k - x + \) (k + x )(k + x - ! ) ....( £ + 1)

... (6)

[by (5) & (6)]

or

Taking log o f both sides, w e get

logf e ) ' K 1' l ) +log( , ' l ) +-"+log( ' - T i l

- Iog(,+i)+log(,.l)+...+Iog(,+i) N ow w riting expansion for each term and neglecting higher pow ers o f x/k as k is very large), w e get l o g f ^ U - r t f + 2 + 3 + . . . ( * - ! ) } - - i - 0 + 2 + 3 + ...( x - I ) + x }

U J

*

*

S ta tistics | 443

= - j { l + 2 + 3 + ...+ (x -l)} ~ ~ [By using A.P.] X.

or

X

X3

I O g |^

y. ... (7)

y .-y + ~ * n

Since in Binomial distribution,

o = -Jnpq

*

9

1

1

=>

t r = npq => a 2 = 2k.—.—

=>

2^ = * .

[v n = 2k]

2 2

Hence (7) becomes: y t =y*e

i.e.

■

f { x ) = y 0e~’3'2*3.

Which is normal distribution.

A ns.

Exam ple 5.57 : F or the norm al curve y - M

Solution.

>—o ^ l 2 x

F ind m ean and standard deviation. Since / (x) is p . d f , then oo

mean = J x . f ( x ) d x

J__

-I

dx = o -\/2 dt. i

[Putting / = — J*

.xdx.

cw2

— j i — r f e ''1lta^/2 + n )a -j2 d t o>/(2 n ) i V ’ 1

2a 1 J e~ * td t+ \iB 'j2 J e~*dt\ -«© *« J <8 j s i n c e J e^tdt-- O and

J e~,1dt =Jn

4 4 4 | Engineering MATHFJumrs-III

X0 +

Mean = 1

x n o V 2 x-v/ir = ji

cV 2n

Thus mean of normal curve is p Further, by definition of variance

A ds.

oo

Variance =

J

( x - mean)5.f ( x ) d x

= f(x -/< )2- 4 — a>/27t

dx

[ v Mean = jx]

[Putting t = ^~ r-= > d x = a ^ 2 d l .] av2 Variance

= — 1—

=

ctV2 tc i

,

f

W ( 2n ) i

{to42 \ a^2dt '

'

| e t2dl =—n - \ i

(te * ltdt

'

'

On integrating by parts, we get Variance =

2a 3 >/jr 2a

—x>/n 2

V J e'^dt^yfH

= 0 ‘.

Hence Variance of normal curve is o2. Ans. Exam ple 5 .5 8 : The m ean deviation fro m the m ean o f the norm al distribution is tim es its standard deviation. Or F ind th e m ean deviation fro m m ean fo r norm al distribution.

S tatistic# j 4 4 5

Solution:

By Normal distribution

m

- tWZJl

<»

We know that, mean deviation from mean V M.D. = £

\x -\i\f(x )d x

IP“ ' * M.D. = £

J

I 'T a ,\.- ^ - e - f .^ 2 d ,

U le -'d t

=

du Put u= t2=> du=2tdt =>tdt- —

2 a >/2 r® du ct f e - T M D - ~ ~ 7 T ‘° e 7 = 7 S r h l -2 o

yj2%

* (0.8)o approx 4a ~ — approx

„ Proved.

Exam ple 5.59 : In a norm al distribution 31% o f the item s are under 45 and 8 % are over 64. F ind the m ean and standard deviation o f th e distribution. Solution.

3 \% of items are under 45 => Area to the left is 0.31. But area right from this point is (0.50-0.31) =0.J9. (See figure)

4 4 6 | E ngwehbng Mathematcs-III L et X be th e ran d o m variety, w hich is norm ally d istributed w ith m ean ji and standard deviatio n o .

Z » -Z , T hen, 2 -

X ~ \n

2T-0

Z>

is a standard norm al variable (S.N.V.)

T he S.N.V. co rresponding to X - 45 and X - 64 are as below W hen X = 4 5 , then Z = i l l E = _ z i. (Say) c

•••(!)

W hen X = 6 4 , th en z = — ^ a

...( 2)

= Z , , (Say)

From the figure it is obvious that P (0 < Z < Z i )= 0 ,42= > Z j = 1.405 and

P ( - Z , < Z < 0 ) = 0 .1 9 = > />( 0 < Z < Z , ) - 0 . 1 9 .

=>

Z, = 0 .4 9 6

[From the norm al table]. [by symmetry] [From th e norm al table],

Substituting th e values o f Z\ and Z j in ( I ) an d (2), w e get 4 5 -n

6 4 -n

■= -0 ,4 9 6 => 4 5 - - 0 .4 9 6 c

= 1.405

6 4 - n = 1 .4 0 5 a

... (3)

... (4)

S olving (3 ) and (4 ), w e get a = 1 0 , n = 4 9 .9 6 = 5 0 (ap p ro x .) i.e., S.D . = 10 a n d m ean « 50.

Ans.

S ta tistics | 4 4 7

Example 5.60 : The m ean height o f 500 students is 151 cm , and the standard deviation is 15 cm A ssum ing th a t the heights are norm ally distributed,fin d how m any students have heights betw een 120 and 155 cm . ? Solution.

N o. o f students = 500

.'. N = 500

M ean, p = 151 cm , and a = 15 By standard norm al variable z =

Standard norm al variable

x .-n 1 2 0 -1 5 1 -3 1 = -*------= ------—-----= -7 7 - 0

15

'•

15

W hen x2 = 155 cm,

xi - \ i _ 1 5 3 -1 5 1 Standard n o n n al variable *

o

.

15

4 _ Q

27.

15

^(120 < jc < 155) = / ’(-2.07 < z < 0.27) = P ( - 2.07 i z £ 0 ) + /»(05zS 0.27) = P (0 Z z£ 2 .Q 7 ) + P ( Q l z Z 0 2 7 )

= 0.4808 + 0.1085 = 0.5892.

[by norm al table]

T he required no. o f students =0.5892x500 = 294.

A bs.

Exam ple 5 .6 1 : The distribution o f w eekly wages o f 500 w orkers in a fa cto ry is approxim ately norm al with the m eans and standard deviation o fR s. 75 and R s.15. F ind th e num ber o f workers who receive w eekly wages: (i) M ore than Rs. 90 (ii) L ess than R s. 45. = 500 i.e., N = 500 Solution: G iven: N o. o f w orkers M ean, ft = 75 an d S.D. a = 15 x -\i

B y standard norm al variable z= ~ a

.. .

(i) W hen, x - 9 0 : T hen standard norm al variable jc- * i

z=-

9 0 - 75

15

o

15

P (m ore than Rs. 90) = P(x> 90) = P(z> 1) = 0 . 5 - / >( 0 < z < l) = 0 .5 -0 .3 4 1 3 = 0.1587

( 1)

4 4 9 1 E ngineers* ; M ajhem atks -III

Hence number of workers who receive weekly wages more than Rs. 90 ■ 0.1587x500 = 79.35*79 workers 2= £ liU 4 5 -7 5 = _ 2

(ii) Whem, x = 45 : Then

o

15

P (less than (Rs. 45) - P (x <45)=P(z < - 2) = 0 .S -P (-2 < z< 0 )

= 0.5 - 0.4772 = 0.0228 Hence No. of workers who receive wages less than Rs. 45 = 0.0228 x 500 = 11,4'» 11 workers Ans, E xam ple 5.62 : A sam ple o f 100 dry battery cells tested to fin d th e length o f life produced the fo llo w in g results : M ean x - 1 2 hours, standard deviation a - 3 hours.

Solution.

A ssum ing the data to be normality distributed, w hatpercentage o f battery cells are expected to have life (I) M ore than 15 hours (ii) L ess than 6 hours (iii) Betw een 10 en d 14 hours ? Here x denotes the length o f life of dry battery cells.

x -jf x -12 The S.N.V.; * « a 3 (i) When x - 15, then z = 1. P (more than 15) - P (x > 15) = P{z > 1) - P { 0 < z < c o ) ~ P ( ( ) < z < 1) = 0.5 - 0.3413 = 0.1587 - 15.87%

Ans.

„

6 -1 2 (ii) When x = 6, then 2 = —~— = - 2

P (less then 6) = P (x < 6) = P (z < - 2) X«0

Z --1

Z-2

= P ( z > 2) - P ( 0 < z < oo) ~ P (0 < z <2)

= 0.5 - 0.4772 = 0.0228 * 2.28%.

A bs.

10-12

(iii) When x = JO, then 2 = — -— = -0.67. 3 14-12

when x = 14 then z ~~

=0.67.

Z = -.6 7 Z =0 Z =.67

S tatistics | 449 P (B etw een 10 & 14) = P (10 < x < 14) = P ( - 0 .6 7 < z < 0.67) = IP (0 < i < 0.67) = 2 x 0.2487 = 0.4974 = 49.74%

A n s.

Exam ple 5.63 iA ssum ing th a t the diam eters o f1000 brass plugs taken consecutively fro m a m achine fro m a norm al distribution with m ean 0.7515 a n . and standard deviation 0.0020 cm ., how m any o f th e plugs are likely to be rejected, i f th e approved diam eter is. Solution.

G iven range of diam eter are : x, - 0.752 - 0.004 = 0.748 cm and

x2 = 0.752 + 0.004 = 0.756 cm M ean jx = 0 0.7515 and S.D. «r = 0.0020 cm T he S.N.V. is r =

x -n

A t Xi *= 0.748, then

0.748 - 0.7515 = -------i t t : ------ = -1 .7 5 .

Z — 1.75

Z~«

Z -2 2 S

0.002

0 .7 5 6 -0 .7 5 1 5 = 2.25. A lso at x2 = 0,756, then z2 = ■

0.002

P (x 1 <

X < X2)

=

P

(Z| <

2<

Z2)

= /> ( - 1.75 < z < 2 .2 5 )

= P (0 < z < ~ 1.75) ■+■.P (0 < z < 2.25) [From nonnal table)]

= 0.4599 + 0.4878 = 0.9477. .’. N um ber o f plugs likely to be rejected = 1000 (1 - 0.9477) = 1000 x 0.0523

[ v Shaded are* = (1-0.9477)]

= 52.3 s> 52.

A ns.

Example 5.64: A m anufacturer knows fro m experience th a t the resistance o f resistors be produces is norm al with m ean ft = 100 O hm s and standard deviation a m 2 O hm s. W hat percentage o f resistors w ill have resistance between 98 O hm s and 102 O hm s ? Solution.

G iven th a t m ean p = 100 Ohms, standard deviation ct = 2 Ohms x -p By standard norm al variable is 2 = ~ 9 8 -1 0 0 , W hem x = 09, then z, = ----- :----- = - l . 2

W h e n x = 102, then

102-100

e l.

4 5 0 | Engineering Mathemahcs-III

P (98 < x< 102) = P ( z \ - - \ < z < t 2 ~ \) = P ( ~ ) < z < 1) = / > ( - 1 < z < 0) + / >(0 < z < 1 ) - 0.3413 + 0.3413 = 0.6826. Thus, th e percentage o f resistors having resistance betw een 98 ohms and 102 ohms. = 0.6826 x 100 = 68.26%

Ant

Exam ple 5 .6 5 : In a sam ple o f1000 cases, th e m ean o f a certain test is 14 and standard deviation is 2.5. A ssum ing the distribution to be norm alfin d (i) H ow m any students score betw een 12 and 15 T (ii) H ow m any score above 18 ? Solution.

1000, mean \i = 14, o ,= 2.5.

G iven that

The S.N.V. is z = ^ CT 1 2 -1 4

(i) W hen x = 12, then.

= -

o

W hen x - 15, th en , zi =

2.5

1 5 -1 4

0.8

= 0.4

2.5

P (12 < x < 15) = P ( - 0 . 8 < z < 0.4) « P ( - 0 . 8 < z < 0) + P (0 < x < 0.4) = 0.2881 + 0 .1 5 5 4

[From norm al table]

= 0.4435. .*. T he required nu m b er o f students = 1000 * 0.4435 = 443.5 * 444. IO

(ii) W hen x = 18, then, z t *

| A

— = 1.6 2.5

P (score above 18) = P (x > 18) = P (z > 1.6) - L eft a rea - A rea betw een 0 and 1.6

Z - 0

Z - 1.6

F 0.05 - i* (0 < z < 1.6) *= 0.5 - 0.4452

[From norm al table] j

= 0.0548. .v T he required n u m b er o f students = 1000 * 0.0548

“ 54.8 * 55.

Ans.

S ta ti s tic s | 4 5 1

Exam ple 5.66 :The lifetim e o f a certain kin d o f battery has a m ean o f 300 hours and a standard deviation o f 35 hours. A ssum ing th a t th e distribution o f life tim es, which are m easured to th e nearest hour, is n o rm a lfin d the percentage o f batteries, which have lifetim e o f m ore than 370 hours. Solution.

Let x be a random norm al variate m easuring the life tim e o f batteries H ere m ean ji= 3 0 0 ,cr= 3 5

x —ii x - 3 0 0 T he S.N.V. is z = -----~ = - ~ o 35 170 —^00 W hen x = 3 7 0 , t h e n z = ± ^ ^ = 2 .

P (M ore than 370) = P ( x > 3 7 0 ) = P ( z >2)

z -o

= left area - area betw een z = 0 and z = 2

z -2 * -3 7 0

z = 0 and z = 2 * 0 . 5 - P (0 < z < 2) = 0.5 - 0.4772 = 0.0228. H ence percentage o f batteries having life tim e m ore than 3 7 0 hours = 0.0228 x 100 = 2.28%

Exam ple 5.67 :F it a norm al curve to the fo llo w in g data length o f line (in cm ): 8.60 8.59 8.58 8.57 8.56 frequency: 2 3 4 9 10

8.55

8

8.54 4

8.53 1

Ans. 8.52 1

<*-n)3

Solution.

Let th e n o n n a l curve be : y -

... ( 0

H ere iV = 42. X

f

u * * x - A Taking A = 8.56

u^

/«

f » 2

8.60

2

0.04

0.0016

0.08

0.0032

8.59

3

0.03

0.0009

0.09

0.0027

8.58

4

0.02

0.0004

0.08

0.0016

8.57

9

0.01

0.0001

0.09 ,

0.0009

8.56

10

0

0

0

0

8.55

8

- 0.01

0.0001

-0 .0 8

0.0008

8.54

4

- 0.02

0.0004

- 0 .0 8

0.0016

8.53

1

^0.03

0.0009

-0 .0 3

0.0009

8.52

1

- 0 .0 4

0.0016

- 0 .0 4

0.0016

0.0060

tfu = O l l l

T fit2 « 0.0133

2 /= 42

4 5 2 | E n g in e e rln g M a th e m a tic s-III 0 .1 1

,

mean (n) = ^ 4" ^ r "*I,56+' 42_

=8.56+0.0026=8.5626=8.563(approximate) and

S.D . (ct) =

J IfuY i 0.0133 r0.11V l ^ J j Vl 42 A42 J

2/

= yjUO•000317 - 0 - 0000069}] = 0-0175cm . H ence from (1) the equation o f n o n n al curve to be fitted is

y - / (x) = 9,8 « « « < * * » » .

A ns.

5.17 RECTANGULAR DISTRIBUTION OR UNIFORM DISTRIBUTION; A random variable x is said to be a continuous rectangular or

uniform distribution o v e r an interval (a, b), w here b > a, i f its p .d f. is given by :

1 b-a

b-a'

fix)

0, D istrib u tio n

Function

of

a
Rectangular Distribution :

The p .d f. o f rectangular distribution is given by

fix)

1 6- a ’

a
0,

otherw ise

Then distribution function o r {C.D.F.) F (x) is given by

Fix) = P i X Z x ) X

=

.

j f(x)dx

-
-J

1

b- a

dx

b - a ixYa - x ~a b-a T hus, d.f. 0,

F(x)

/{ * )

x< ,a

x-a b-a *

a
1,

x St b.

S tatistics | 4 5 3

5.18 MEAN, S.D. AND VARIANCE OF RECTANGULAR DISTRIBUTION The p .d .f o f rectangular distribution i f given by ,, .

/(

jc)

1

~

a
T----- >

b-a

. 0,

otherw ise

To fin d m ean:

...( 1 )

We have, 00

I x.fix)dx

M ean = |A[’ = E(x) =

-<0

b = I (T ^ ) *

f by w i

—l— f“ ] (6 - o )

[ 2 ^

1 (b2 - a z )

2(6-o) 2

=

•(2 )

To fin d S.D. and Variance : T he second m om ent about origin : at>

I x2.f(x)dx

Mi’ -

6 1 = J x \ (& T ^ ) dx

[ by ( 1 )]

__ L_ [63- a3] (6 - o )

3

( b - a ) (6 2 +, a_2 +ab) “ ( b - a ) ■ ---------------T = •! ( a 2 + 62 + ab).

...(3 )

We know that Variance = jij' - (u i1)2 = |

(a2 + 6^ + ab) -

= g ( a 2 + 6* + ab) - ^ (a2 + 62 + 2ab)

[by (2 )]

4 5 4 | E n o n b e h n g Mathematics-111

- a 2 + b2 - 2 ab 12 _ (b-a)2

[v * >

12 and S.D.

a]

o' = V V a r ia n c e

yfl2.

(b -a ).

5.19 MEAN DEVIATION OF A RECTANGULAR DISTRIBUTION: T he p.d.f. o f rectangular distribution is given by 1

a
b-a

fix)

0,

othew ise

•d )

We know th at m ean deviation ab o u t m ean is 00 M .D . =

}x - m ean \ f ( x ) dx

J

x -

(a + b)

_____ L _ f (fc~ a> I

P u t/ = x -

( a + b)

and lim its are : t => -

x-

[by def.]

1

( b - a ) dx (a + b)

( o + 6) , [ v m ean = — ^— ]

dx

dt - d x

=>

(b-a) ^ (b-a) to

■>K>M .D .

-

(b-a) 1

(b-a)

j

W dt

(b-a)

2

(±±) 2 (b-a)

? J

^

[ v jf| is an e v e n /* .]

S

2

(fr-o) ?

.

I

tdt

= (6 - 0)

tatistics

| 455

[ v |/| = /, if (b - a) > 0 ]

(6 -o )

= __2_

1 r a ]—

( b - a ) ' 2 I Jo 1

(b -a )*

4

(b -a ) • ,

0 -°>

4 ’ Exam ple 5.68 : A variate X has the rectangular or uniform distribution with p .d f. L given by :

1 r

0 < x <100

100 0, otherw ise Com pute (i) P [ X > 60} and (ii) P [20 * X * 4 0 ] m

-

Solution : (i) We have too

P [X> 60] = J f ( x ) dx 60 &

^

100

= I

10 0 *

[v by given/(x)]

100

~ 11 000 W l* J60

= 0.4 (ii) We have, 40

P [20 5 x 5 40] = J f( x ) d x 20

40 = 1 20

=

A

100 *

1 f ]40 10 0 ' XJ20

20 100 =

0.2

A ns.

4 5 6 ( E n g in ee r in g M a th e m a tic s -H !

5.20 GAMA DISTRIBUTION: (i) A c o n tir io u s ra n d o m v a ria b le X is s a id to be a gam a distribution w ith param eter

X > 0, if its p.d.f. is given by : e -V -'

0 <

jc

< oo,

X >0

~ W

/(* > «

0,

othew ise

J e~zxx~'dx. 0 (ii) The p.d.f. a gam a distributin w ith tw o param eters a > 0 and X > 0, is given by Where

[(Ij =

° f i x ) o r f { x , a, \ ) =

0
g^ayX-i •

a > 0, X > 0

Iw

0,

‘

otherw ise

5.21 DISTRIBUTION FUNCTION OF GAMA DISTRIBUTION The p .d .f o f the gam a distribution is 0 < x < c o , A_ > 0

fix)

m 0

; •••(J)othe

We have, distributive function o r C.D.F. o f / (x) is given by

F (x) = P iX < x) = 1 f i x ) dx -00

x( e - '.x x-' F i x ) - J - r ^ r - dx

[by (l)]

1(A)

o

1

[by definition]

j e~*xM dx

W) x

1 H ence d .f

f(X )

F{x) =

/

o 0,

e~* xK~'dx,

0 < x < oo otherw ise.

5.22 MEAN, S.D. AND VARIANCE OF GAMA DISTRIBUTION: T he p.d.f. o f gam a distribution is given by

e~*.xk‘‘ /(* )=

.

m 0,

0 0 otherw ise

..( l)

S tatistics | 4 5 7

To find mean : We have M ean = fit’ = E (x) <30

= J xf(x)dx —00

7 e 'V '' = \ x 0 RT) I

[by (l)]

dx

00

-T—» J e * x* '- l dx m

o

[v

CO « = J e*. J f'1
[ v |( » + 1) = « [(n)3 ■ W = X. 7o fin d S.D. and variance : T he second m om ent about origin : <50

ji2’ = J x * .f(x ) dx 7

i
-

,

[by (1)3

* [by def. o f G am a function]

[v |(n+l) =«((w)] = X2 + X. We know that V ariance ( n 2) = to ' - (Mi')2 V ariance ( |i 2) = X

K + X - (V = X

4 5 8 | E n g m eem n g M a tv e m a tk s -III

and S.D.

a — -^Variance

This show s that mean and variance are equal fo r the gama distribution.

5.23 BETA DISTRIBUTION OF FIRST KIN D: T h e ra n d o m v a r i a b le X is s a id to b e Beta d istribution o f fir s t k in d , i f its p.d.f. is given by 1 -5 ( a , p ) ■xa~l ^

_

f ( x ) or f i x , a , P) =

0,

0 < x < 1 and a , 0 > 0 otherw ise.

W here B ( a , 0 ) is a B eta function. I R e m a rk :

(1 - x p -' dx

B ( a , P) = f o

fwl¥) |( a + P)

5.24 COMMULATIVE DISTRIBUTIVE FUNCTION OF BETA DISTRIBUTION OF FIRST KIND: The p .d f. o f B eta distribution o f first kind is given by : 1

5 < a rp)

/(* ) =

0 - ^ l.

0,

0 < x < 1, a, p > 0 otherw ise

By definition o f distrubtive function

F i x ) =P{ X<. X) “

/ fix)dx -ao

= I T t k $ xa~ ' v ~ x r '

dx

= 5 ( ^ p ) I r -1 0

*

[by (i)]

Hence d.f. is X

F{ x ) =

j /*-1 (1 - f ) M dt, 1 ;

0 £x < I x £ 1.

N ote : The above cum ulative distributive function f i x ) is called Incomplete Beta Junction, which is giving by equation (2 ).

S ta tistics | 459

5.25 MEAN S.D. AND VARIANCE OF BETA DISTRIBUTION OF FIRST KIND: The p.d.f. o f B eta distribution o f first kind is given by 1 3 < a ,p )

/< * ) =

0 < x < 1; a, P > 0

0,

...( 1 )

otherw ise

To fin d mean : We have M ean - fi\ = E (x) 00

= J x/(x) dx i? ( a ,p )

[by definition]

j x**'-» (1 -JL )M dx 0

^ p j * B

[by ( 1 )]

( a + i, P)

( v J x^~l (1 - x ^ d x - - B ( l , m)]

V B (/, m) =

( a + p ) . a [(a )

[v

[(a ) ( a + p ) |( a + P) a a +P

|(/ + m)

(n + 1 ) - n («)]

- ( 2)

To fin d S.D. and Variance : T he second m om ent ab o u t origin H2

=j

x>f { x ) dx

*(a,p) jJ0 x2. jf“-'

(1 - x ) ^ - ‘ dx

= 5 (^ ~ p ) * B ( a + 2, P)

l( a + P + 2)

[by (1)]

[by def. o f B eta function]

4 6 0 | E n g in eerin g M athem atics -U I

IjEH*---- (g+l)«|(5 R o)

(a + (S + IX a + P ) ^ a + P )

CT^-X, l(

1

)'

a (a + 1)

••(3)

(a + p)(a + p + l) We have Variance (n 2) = M2 - (Mi')2

a (a + l) (a + p)(a + p + l)

a Va+p

[by 2 & 3]

<*• a +1 a a + p j .a + p + 1 a + p « f(a + p) ( a + 1)- a (a + p + l) " a+p [ (a + p) (a + p + l) ap

(a+ p) (a + p + l) S.D. (
and

ap

1

(a + p) V(a + p +1) 5.26 BETA DISTRIBUTION OF SECOND K IN D : The random variable X is said to be Beta distribution o f second kind, if its p .d f. is given by

.

0 -1

1

S M

/(* ) =

'm

S

0

;

0 < i < » a n d a , p >0

;

otherw ise

...(I)

R e m a rk : I f w e put y ~ r -— , then distribution ( I ) becom es, Beta distribution o f first kind.

Exampla 5.69 : Prove that total area (total probability) o f Beta distribution o f second kind is 1. Solution :

The p.d.f. o f B eta distribution o f second kind is given by : 1

fix) =

00

B < a , P ) '( i + x )a+P

otherwise

0,

We have

Total area =

=

[by definition}

J f(x)dx

0}1 B (a, P) (1 + x)a+p

...(1)

dx

[by (I)]

S ta tistics | 4 6 1

a-1 * -

-

j

B ( a ,p ) J ( l + x)a+P

dx

1 1 1-y ~dy [Putting y = -:— - => 1 + x = ~ = > x = —~r~ =>dx = 2 and lim it y~-> 1 (w h e n x = 0) 1+x y y y to y —* 0 (w hen x = 00)] Total area - _ _ L _ B ( a ,p )

J (1

p .:

1 I -d y V j Y ^ 'U 2 J

f fiz z ) 1 I y )

y 5-1
B ( a , P) 0 1 b7

1 (I - > ) a"‘ dy

I

B ( a ,p ) 0,

1

—

B ( a , p)

p ( a , P)

[ v B (/. * ) = J x '- ‘ (1 - or> -1 dx] 0

Hence total area = 1.

5.27 MENA, S.D. AND VARIANCE OF BETA DISTRIBUTION OF SECOND KIND: The p.d.f. o f B eta distribution o f second kind is given by : .'*-1

1

0 < x < co and a , p > 0.

S ( a , P ) '( W . - c f -3

/(* ) =

otherw ise

0 ,

...( 1 )

To fin d mean : We have m ean = ji|' = E (A) 00

=

1 * ( a ,p )

*

1+x

X =

1- y

x a_*

[ * ' ( l + x )a ^

1

[Putting y =

[by definition]

i x f (x) dx

*

dx

dx = - —7 dy and limits o f v -> 1 to 0] v"

4 6 2 | E n g in e e r s M athem atics-I 11

l

? (i - y ) a.(y )a+>

i

B ( “a , pe) ) = ^

a

f

&

0 -y Y * x~' dy

' p j x B (P - 1, a + 1)

W )W )

,.2

[by def. o f B eta function]

](a + p >

l(P —1) - a ( a ) l(a X (P -l) (P -D

■

a

w hen P > I

P -l' Hence

a >*» “ F T

meai

To fin d S.D. and Variance : The second m om ent about origin is given by : 'GO Hz' =

/ x » / ( r ) dx -
_ |

1

^ a -1 [b y (1 )]

5 ( a , p ) x2- ( i + x ) a+p *

£ ( a ,p )

o ( l + * ) a+p

1 l —V _■» [Putting y = — — =>x= —— => a&c = —A a y and lim its o f y A * y y2

1 to 0]

S ta ti s tic s | 4 6 3

’ B lk j)

‘ B (h >

= B ( a,p) xB0 - 2 . a + 2) j f o + P) .. R F 5 j 1( ^ 2 ) [ ( * jR M

l(a + P )

( a + l ) a l a |(p - 2 ) “

|( a ) 0 - 1) 0 - 2 ) |0 - 2 )

, _ <*(<* + !) Mz ( p - l ) 0 - 2 )

m " (3)

We have Variance (jij) = n 2' - ( u i1)2

+ " < p -l)(p - 2 ) g f (g + l) "(p -l)L (p -2 )

f a U -lJ a | (P tfJ

g (a + p -l) ~ ( p - l ) z (p-2) _

and S.D.

[by 2 and 3]

-(4)

a — ^ V a ria n c e

_

1

/a (a + p - 1)~

" 0 -D V

( P “ 2)

5.28 EXPONENTIAL DISTRIBUTION: A random variable X is said to be an exponential distribution w ith param eter X > 0, if its p.df. is given b y :

/«- f Xr": 0;

xS0

otherw ise

...( 1 )

464 |

E nginkfrtng

Mathematks-W

The graph o f exponentai) distribution is

R a m srk :

The total area o f exponential distribution is unity.

S olu tio n :

We have,

1 f{x)dx

[by definition]

=oj Xe-^dx

[by distribution]

total Area =

.-Xx

~x

5.29 DISTRIBUTIVE FUNCTION OF EXPONENTIAL DISTRIBUTION: The p.d.f. o f exponential distribution is given by

Xe-**;

/(* > =

x2Q

0;

otherw ise

...( 1 )

By definition o f distribution function p r C .D .F .: X

F(x)=P(X<,x)= 0

! f(x ) d x *

1 f ( x ) d x + J f( x ) d x 0

—eo

= 0 + J Xe^xxxix

[■j i 1] :J_

[by 0)1

S tatistics I 4 6 5 Hence distributive fu n c tio n :

0, 1-

F (*) =

jc

er*',

<0

jc > 0,

1,

X> 0

JC = ao

Graph o f distribution Function is

5.30 MEAN, S.D. AND VARIANCE OF EXPONENTIAL DISTRIBUTION: The p.df. o f exponential distribution is given by :

m

_

j

le -* * ;

i.

0 ;

xzo,

.

otherw ise

X > 0. ...( 1 )

To find mean : We have,

M ean = ji|* = E

=

( jc)

xf(x)dx

[by defh.]

! x. Xe-^dx

[by ( 1 )]

/

o \00 = X

JC.

-X

oo

-Xx

-S ^ -d x /o

0 + I e -^ d x o

-X

i~Xx -X

- X ( ° - !>= X Hence

1 jii = m ean = ^

••(2 )

4 6 6 j E n g in eerin g M athematics -III

To find. S.D. and Variance : T he second m om ent about origin

J x\f(x)dx

-

= J x2.Xe~** dx

[by (i)I

- X J x2.e~k*dx

= X

x 2. e

0”

-x L

; 0 + 2 J xe~u dx 0

= 2

00

I 0

„

' Hence

\

X* ( 0 ~ ° ~

I *

■ _2 _

/0 _

X2

2 “ X2

...(3)

We have V ariance fi2' - ^ 2^ - (Mi')2

2

-* -« r and

S.D. : a - 1/ V a r ia n c e ~ T

[by 2 & 3]

S tatistics | 4 6 7

5.31 MEAN DEVIATION ABOUT MEAN OF EXPOBEBTIAL DISTRIBUTION: The p.df. o f an exponential distribution is given by /< * ) =

Xe-**; 0

x'i.O and X > 0

;

otherw ise

- .( I )

We know that m ean deviation about m ean is given by M .D . =

J

l x - m e a n |/ ( * ) d r

[by ( 1 ) and mean ~ jj-]

= J |Xjc —1[ er^dx 0

Puting Xx = t => Xdx - dt and limit t -> 0 to «>, w e have

go M .D . =

Since

I '- H

f i \ t - \ \ e-

=

f1 i

X

°o

] (1 - i) e~ld t + J (* - 1) e ' d t o

l

= X [ - ( 0 - D + ( O o - ( 0 - 0) - ( * r O r ] X [1 + (
*r , Xr ‘

[by

2)

468

| E n g in e e r in g

Mathematics-411

Example 5.70: A random variable X has an exponential distribution w ith p.
.

I[ 0,

i f x SO . Com pute th e probability th a tX is not less than 3. A lso fin d , m ean and standard deviation and prove th a t coefficient o f variation is unity.

Solution : (i) Com pute P (X K 3) : We have

P (X * 3) = P (X > 3) 00 = / f(x)dx

3

'

00

= J 2 e-2* dx

3

= - [0 - e~*] = (ii) To fin d m ean : We have,

m ean = fi|' = E (x)

S tatistics | 4 6 9

—

5 (0 -1 )

_ I 2

(iii) To fin d S.D . :

We have,

1

H:’ =

-C O



t

= J x2. 2e-2* cfr

= 2 J x'er^dx 0

3- e

-2x

-2 x V

- ! 2x S ~ d x

"2Jo o

=2

2 0 - 2 J x e 2xd x ( - 2) o

CO

= 2 / x
~2

4 7 0 | E n g in e e rin g M a th e m a tic s-!!!

- j er^dx o

e - 2* )*

- 2 Jo = - 1 (0 - 0 We know that, S . D . : a - ^jj-2 ' - ( m ') 2

■

f W

=

/H I V2 4

- UH =I ~ V

and

4 ‘ 2

Variance = ^

A ns.

To fin d Coefficient o f Variation :

We have that, coefficient o f variation =

„_,

1/2 = 172=1.

^ved.

Example 5.71 : The Incom e ta x X , o f a m an has an exponential distribution w ith p .d f. is given by

f(x )

0

;

x <0

I f incom e ta x is levied a t th e rate o f S% , w hat is the probability th a t h is Incom e exceed Rs. 10,000 ?

Solution : If income exceed Rs.

10,000, then income tax will be : Income tax = 10,000 * = 500.

S tatistics | 4 7 1 H ence tax o f a m an will exceed Rs. 500. required p ro b a b ility :

P (X > 500) =

\ f(x)dx

500 00

=

7 e~*IAdx

J

[by given p.d.f]

600 4 aO

i [ l -*n y I 4 1 -1 /4 500 = _ (0 - e-[” ) = e~]2S

A ns.

E x a m p le i5.72 : Suppose Ike life o f mobile batteries is exponentially distributed with parameter

X —.001 days. What is the probability that a battery will last more then 1200 days ? S olution :

The p.d.f. o f exponential d is trib u tio n :

fix)-

Xe~^ \

x >0

0

otherw ise

;

...( 1 )

w here X, = 0.001 days.

P (battery will last m ore than 1200 days) = P ( X > 1200)

=

I f ( x ) dx 1200

J

hr^dx

1200

( * rV

[using ( 1 )]

J\2< '1200

=_ (0 _ {j-XxIMO)

: ^-0.001*1200 e~12 « 0.301

[X = 0.001] A ns.

4 7 2 | Engineering Mathematjcs-JH

5.32 TEST OF SIGNIFICANCE The statistical co nstants for the given population such as m ean (p), standard deviation ( a ) etc. are called the parameter's. Sim ilarly, the statistical constants for the sam ple draw n from the given population such as m ean ( ic ), standard deviation (j) etc. are called th e 'statistics'. In general, the population param eters are not know and th eir estim ates given by the corresponding sam ple statistic are used. It is necessary to test th e statistic to see w hether their difference is significant or not. Such tests are called th e 'tests o f significance'. It can be used to com pare the characteristics o f tw o samples o f the sam e type. For applying th e test o f significance w e first setup a hypothesis.

5.33 NULL AND ALTERNATIVE HYPOTHESIS A hypothesis w hich is tested for possible rejection under the assum ption that is true is called null

hypothesis, and is denoted by HoIn other w ords, a null hypothesis Ho that there is no variation exist betw een variables or set of given observations.

For example, T he null hypothesis H0 m ight be that there is no relationship betw een tw o measured phenom ena. The hypothesis w hich is accepted, w hen the null hypothesis has been rejected is called the altenative hypothesis, and denoted by H) o r HA For example-, in a clinical trial o f a new drug. Then, w e have Null hypothesis Ho : there is no difference betw een the tw o drugs on average i.e; the new drug is no better on average, than th e current drug. A lternative hypothesis H\ : the tw o drugs have different effects on average, i.e; the new drug is better on average, than the current drug. The purpose o f hypothesis testing is not to question the com puted value o f the sam ple statistic but to m ake a judgm ent about th e difference betw een that sam ple statistic and a hypothesized population param eter.

534 LEVEL OF SIGNIFICANCE The probability level below w hich are reject the hypothesis is know n as the level o f significance. T he level o f significance usually em ployed in testing o f hypothesis are, 5% and 1%. W h a t, if we test a hypothesis at the 5% level o f significance? T his m eans that w e will reject the null hypothesis, if the difference betw een th e sam ple statistic and population param eter is so large o r a larger difference w ould occur, on th e average, only fine o r few er tim es in every 100 sam ples when the population param eter is correct. A ssum ing the hypothesis is correct, then the significance level indicates the percentage o f sam ple m ean that is outside certain limits. In this section we shall study the follow ing four tests o f significance only, viz : (i) x 2 - test, (ii) t - test, (iii) F - test and (iv) Z - test.

5.35 CHI-SQUARE TEST OR x2- STATISTIC The C hi-square test is a very pow erful test fo r testing the significance o f the discrepancy between the actual (or observed) frequencies and the theoretical (o r expected) frequencies in a sample, ft w as given by prof. Karl P earson in 1900 and is know n a s " Chi-square test o f goodness offit". The C hi-square is a letter o f th e G reek alphabet w hich is w ritten as %2.

S ta ti s tic s | 473

Let f 0 and f be a se t o f observed an d expected frequencies o f a class interval o r cells. Then Chisquare distribution is defined as ifo-fe)2

- ( 1)

X2 W here 'Lf„-'Lft = N = total frequencies

Another Form o f %z distribution :

k /w .r l Since

X2 = 2

I

I

[/„ *

I - I

'•

{£-% ♦/.[ f 2 /o fe 4

f 2 *0 -2 N +N

or

<
[v i/0 = i / , = ao

...(2 )

-AT

5.36 PROPERTIES O F x2 DISTRIBUTION : (i)

The values o f %2 are alw ays positive, i.e; x 3-c u rv e is alw ays positively skewed.

(ii)

T he low est valu e o f %2 is zero and the highest value is infinity.

(iii)

I f degree o f freedom v = 1 , then x ^ u r v e will be y = y 0e~*

, w hich is the exponentail

distribution as show n the follow ing figure

(iv)

I f v > 1, then x *-curve is tangential to X -ax is at the origin and is positively skewed.

474

( E n g in e e r in g M a th e m a tic s - ! ! ]

(v )

T he probability P th at the value o f y} from a random sam ple will exceed xi >s gives by

P = J ydx. *0

537 CONDITIONS FOR THE VALIDITY OF x*-TEST: (i)

T he m em bers o f sam ple should be independent.

(ii)

C onstrains o n th e cell frequencies should be linear.

(iii)

Total frequencies N should be reasonably large say, greater than 50.

(iv)

As x ^ -te st is based w holly on sam ple data, no assum ption is m ade about the population values (or param eters).

(v )

T he expected frequencies should not be less than 5, If it is less than 5, than adjacent it.

(vi)

x 2-test

w holly dependent o n degress o f freedom (do/.) v.

538 USES OR APPLICATION OF x*-TEST: (i)

(ii)

Test o f goodness o f f i t : C h i-s q u a re s ta tis tic m ay b e u se d to d e te rm in e w h eth er tw o curves are fitted good o r not. Test o f independent o f attributes in a contingency table test for a specified standard deviation.

(iii)

Test fo r a specified standard deviation i.e ; it m ay b e u se d to te s t o f p o p u latio n variance.

(iv)

C om pare a nu m b er o f frequency distributions.

Remarks on the x*-Test : 1. C ells : I f given data are arranged in the com partm ents, w hich is called cells or class interval and frequency o f a cell is called cell frequency. 2. Contingency Table : Let a group having N persons is divided into tw o attributes A and B. A having m classes say, A lt A 2........A , ......... and B having n classes say, B u B2, Bi t ......... B„ and

Am

/oij are observed frequencies o f celles belonging to classes A, and Bj.

A \B

B,

B 2 .... ......

A|

f\ l

f \ 2 •— ...... A j .........

/.

a2

fix

h i ...... ......f u ......... ....... f u

fi

A

fix

fa

A*

fm\

fm2 ......

Total

Ft

Fi

Bj ....... ......

...... ...... f j

..... .....

B„

.......... ....... U

Total i

f. fm

Fj

.......

....

Fn

N

S ta ti s tic s | 4 7 5

3. Calculation of Expected Frequencies : C onsider the 2 * 2 contingency table :

A \B

B,

b2

A, a2

/.

h

h

U

N, n2

Total

Hi

Na

iV (G rant Total)

Total ^

r

The expected (o r theoretical) frequency o f cell 11 is given by

t\\ f e\

N

Expected frequency o f cell 12 is : ft\2

-------- J T

Expected frequency o f cell 21 is n

Mi

3*n 2 N

Expected frequency o f cell 22 is :

f

U22

- ^ 4 * ^ 2

N

4. D egree o f fre e d o m (d o f) : T he num ber o f independent variates is called the num ber o f degree o f freedom (dof) and denoted by v. (i)

In Binomial d istrib u tio n : dof v = n - 1, (w here n is total sam pling, for exam ple x : 0, 1, 2, 3 , 4 then n = 5 - 1 = 4 )

(ii)

In poisson distribution : dof v - n - 2

(iii)

In N orm al distribution : dof v = n -

(iv)

3.

For m * n contingency table. d o f v = (m - 1 ) ( n - IX w here m = num ber o f row s, an d

n - num ber o f colum ns.

W o rk in g R u le o f x 1- T est : Step 1. C onsider the Null Hyp*,

s is H 0 : N o association exists betw een the attributes.

Alternative Hypothsis H \ : A n association exists betw een the attributes. S tep 2. C alculate the ex pected (o r theoretic) frequency / „ corresponding to each cell.

476 | E n g in e e rin g M a th e m a tic s-IH Step 3. Calculate x 2-statistic by the formula.

and calculate the degree o f freedom v. S tep 4. See the value o f x 2 from the table, i.e; value o f d o f v, as calculated in step 3 .

at u.% level o f significance and for the

N ote : If no value o f a % is m entioned, then take 5% i.e; a ~ .05 S tep 5. D ecision : (i)

If calculated value o f x~ < tabulated value o f x i >• Then accept the null hypothesis H0.

(ii) If calculated value o f x 2 ^ tabulated value o f xi,* T hen rejected null hypothesis llQi.e; accept the alternative hypothesis H\.

Example 5.73 : From the table given below, whether the colour o f so n ’s eyes is associated with that of father's eyes ? Given that the value o f x 2fo r / d o f at 5% level o f significance is 3.841. Eye Colour is so n ’s Not Light

Light

Eye colour

Mot Light

230

148

in father

Light

151

471

Solution : S tep 1 : N ull H yothesis H0 : T he colour o f the son’s eyes is not associated with the colour o f father's eyes. S tep 2 : C alculation o f Expected Frequencies f < :

Light

Total

Not Light

fo\\ = 230

f , n = 148

378

Light

foi\ - 151

622

Total

381

619

N « lOUO

iv

N ot Light

i!

G iven the O bserved freq u en cies/,// are :

The Expected frequencies will be : N ot Light

N ot Light Light

Light

A"

378 x 381 1(X)0

...

/<21 '

381x622 _ 10 0 " 237

.

}' n

378x619 10 0 0 619x622 10 0 0

_

S ta tis tic s | 4 7 7

S tep 3 : C alculation ol ' x : -stutisiu- • We have or

^ (fo ■ fc ?

x- ■=

2 = < 2 3 0 -1 4 4 )*

( 1 4 8 - 2 3 4 ) 2 (151 - 2 3 7 ) 2 O O A 144 234 2O 3l3 7’7 - 51.36 + 31.61 + 31.21 + 19.21 = 133.29. 1 A A

(471 - 3 8 5 ) 2 g g g

and degree o f freedom V

- (w

- (2

I)

I)

1)(2

I)

[ 2 ^ 2 c o n tin g e n c y ta b le ]

-- I. Step 4 : The calculate value o f X2 at 5% levels o f significance and for 1 d o f is 3.841 i.e.. Xno5.i “ 3 84 I S te» S : Decision : C learly calculate value o f x ’ - 133.39 ^ tabulated value o f Xnovi - 3 84 ro the Null hypothsis is rejected i.e. there is an association betw een thfe colours o f eyes o f son's and colours o f eyes o f father's. A ns.

Example 5.74 : In a sample survey o f public opinion, uswers to the questions fi) Do you drink? (ii) Are you in favour o f local option on sale o f liquor ? are tabulated below : Question Yes

No

Total

Yes

56

31

87

No

18

6

24

Total

74

37

UI

Can you infer whether or not the local option on the sale o f liquor is dependent on individual drink ? (Given that the value o f z 2fo r d o f at 5% level o f significance is 3.841.) S olution :

Step I : Null Hypothesis : The option on the sale o f liquor is not dependent {or not associated) with the individual drinking. S tep 2 : Calculation of expected or theoretical frequencies f e The expected frequenie.s. corresponding to the observed frequencies are calculate as follows Expected frequencx o f cell 11 is 87x74

4 7 8 | E ngin eerin g M athematjcs-111

E xpected frequency o f cell 12 is

f< " = - t i t

5=29

E xpected frequency o f cell 21 is _ 74x24 ~ 111 ~ E xpected frequency o f cel! 2 2 is

x

U i~ S tep 3 : Calculation o f

We have

24x37 m

- 8.

statistic :

x: “ ^

! ^ ~ >e

-(1

ifo-fef

fa

f.

56

58

4/58

31

29

4/29

18

16

4/6

6

8

4/8

U

z !. m

Total H ence

, o _937

Value o f j } = 0.957

A lso degree o f freedom

(dof) v - ( « - 1) (n - 1) = (2 - 1 ) ( 2 - 1 ) = 1.

S tep 4 : T h e tabulated value o f %2 at 5% level o f significace and fo r 1 d o f is 3.841 is 3.841 X0.05,1 = 3.841, Step 5 : Decision: C learly calculated value o f x2 = 0.957 < tabulated value o f x2o.os.i =3.841. => die N ull hypothesis is accepted. => S ale o f liq u er is n o t dependent o r not associated w ith the individual drinking.

A ds.

S ta ti s tic s 14 7 9

Example 5 .7 5 : SOstudents selected at randomfrom 500students enrolled in a computer crash programme were classified according to age and grade points giving the following d a ta : Age (in years)

Grade Points

21-30

above 30

Total

up to 5.0 5.1 to 7.5 7.6 to 10.0

3 8 4

5 7 8

2 5 8

10 20 20

Total

15

20

15

* li *

20 and under

Test at 5% level o f significance the hypothesis that age and grade point are Independent (Given that Xaos,i = 5.99J). S tep 1 : N ull H ypothesis H0 : T he age and grade point are independent. S tep 2 : Calculation o f expected frequencies f t : T he expected frequencies are as follow s : G r a d e P o in ts

20 a n d u n d e r

2 0 x 10 50

5.1 to 7.5

15x20 , 50 “ 6

2 0 x 20 50

7.6 to 10.0

15x20 50

2 0 x 20 50

upto 5

a b o v e 30

2 1 -3 0

o * w 11 UJ , 6

. 4

15x10 50

8

15x20 , 50 =6

'

15x20 50

. 3

.

From the ab ove table, w e see that som e values o f expected frequencies are less than 5, so that to am algam ate these cells to its neighbours (i.e; m erge 1 & U rows). T hen new obseved an d expected frequenies table a re : E x p ected fre q u e n c ie s ta b le (ft) : 20 a n d u n d e r

2 1 -3 0

above 30

I b ta l

upto 7.5 7.6 to 10.

9 6

12 8

9 6

30 20

T otal

15

20

15

Z II g

Solution :

20 & u n d e r

2 1 -3 0

ab o v e 30

Total

upto 7.5

11

12

7

30

7.6 to 10.0

4

S

8

20

T otal

15

20

15

N = 50

G r a d e P o in ts

O b se rv e d fre q u e n c ie s ta b le {f0} : G r a d e P o in ts

4 8 0 | E n g in e e rin g M athematics-111

§ t s n l : Calculation o f x2-statistic : We have,

x2 ~

nr

v» X

9

i ( 1 2 - 1 2 )2 12

( 7 - 9 )2 9

( 4 - 6 )2 6

( 8 - 8)2 , ( 8 - 6)2 8 8

= 2 .222 . A lso degree o f freedom . (dof) v = (m - 1 ) (n - !) = (2 - 1) (3 - I) i.e;

[ v A s new table]

d o.f. • v = 2

S tep 4 : T h e tabulated value o f x 2 at 5% level o f significance and for d o f v = 2 is 5.991 i.e.. X&.m 2 = 5.991. S tg P -5 : Decision : C le a rly c a lc u la te d v alu e o f x 2 = 2 .2 2 2 < ta b u la te d v alue of XSoj.2 = 5.991 => Null hypothesis ts accepted. => the grade points and age are independent o f each other.

Ans.

Exam ple 5.76 : In experiments on pea-breading, mendal obtained the following frequencies o f seeds:

R ound and

W rinkled and

R ound and

W rinkled and

yellow

Yollow

Green

Green

315

101

108

32

Total

556

Theory prtodlcts th a t the freq u en cies should be in proportion 9:3:3:1. E xam ine the correspondance betw een theory and experim ent G iven th a t th e value o f x 1fo r 3 d o f at 5% level o f significance is 7.81S. S olution : S tep t : Null Hypothesis H0 : there is a correspondence betw een theory and experiment. : Calculation o f Expected frequencies ( fj : G iven frequencies in proportion 9 : 3 : 3 : 1 Total sum o f proportion = 9 + 3 + 3 + 1 = 16, therefore (i)

9 E xpected frequency o f Round and Yellow seed is = j g * 556 » 313

(ii)

g E xpected frequency o f W rinkled and Yellow seed is = j g * 556 * 104

3 (iii) E xpected frequency o f R ound and G reen seed is - j g * 556 ~ 104 (iv) E xpected frequency o f W rinkled and G reen seed is =

* 556 * 35.

S ta tis tic s | 481

Step 3 '.Calculation o f %2-statistic :

H ere

f 0 = observed or actual frequencies f e = expected o r theoretical frequencies

r , _ (3 1 5 -3 1 3 )2 (1 0 1 -1 0 4 )2 (1 0 8 -1 0 4 )2 (3 2 -3 5 )2 X 313 104 104 35 ^ 4 9 , 16 , 9 3 1 3 104 104 3 5 c A lso th e degree o f freedom v = n - l = 4 ~ 1 = 3 . S tep 4 : T he tabulated value o f x 2 W 5% level o f significance and for d o f v = 3 is 7 .8 15 X005.3 ~ 7.815 S tep 5 : C alculated value o f %2 = 0.5 < tabulated value o f x 5.os,3 = 7.815 => N ull hypothesis is a accepted. there is very high degree o f correspondence betw een theory and experim ent. A ns. Example 5.77 : A servey o f 320fam ilies with S children each revealed the follow ing distribution :

No. o f boys :

5

4

3

2

1

No. o f g ir ls :

0

I

14

56

3 88

4

No. o f fam ilies :

2 no

0 5

40 12 Is this result consistent with the hypothesis that the male and fem ale births are equally probable ? (Given zks, *-s * 10.07). Solution :

Step 1 : Null Hypothesis Hq : M ale and

fem ale births are equally propable.

S t e a l : Calculation o f Expected frequency f t : From th e Binom ial distribution, th e expectedfrequency:

f (r) = N. "C, q ^'p ”',

r = 0 , 1, 2, 3, 4, 5.

H ere N = 320, p = q = ^ (equal probability) and n = 5. For r = 0, then (1) becom es : / , ( 0 ) = 3 2 0 * 5 C 0 ( ! ) 5 ( ! ) ° “ 10 F or r = 1, then / , ( ! ) = 320 * ^

( I ) '( I /

=50

F or r = 2, then

f t (2 ) = 320 x JC2 ( i )

(i)

= 100

f (3 ) - 320 x SC 3

(!)

=100

For r = 3, then

4 8 2 1ENQMemiG M/utcm/otcs-III For r = 4, then / , (4) = 320 x !C4 ( i ) 4 ( i ) ' = 50 For r ~ 5, then / . (5 ) = 320 x SC5 ( | ) ° ( | ) 5 = 10.

Hence expected frequencies for r = 0, 1,2, 3 ,4 , 5 or 5 ,4 , 3 ,2 , 1,0 a re : / , - 10, 50. 100, 100, 50, 10 : Step 3 : C alculation o f z 2-sta tistic : we have

= Z fe

/.

/,

14

10

56

50 100

110 88 40 12

< f* - f.W

< /;-/.)* '

1.60 1.72

100 50

36 100 144 100

1.44 2.00

10

4

0.40

1.00

Total

%2 = Z

f(/o -/« )* ! ’ = 7.16 fe J

Also degree o f freedom for BD . v " 6 - 1 = 5. S eto4 : The tabulated value o f %2 at 5% level of significance and for dof v - 5 is 11.07 i.e, $ 05,5 ■ it .07 StenS : D ecision: Clearly calculated value of %2 * 7.16 < tabulated value of xS.05,5 * 11.07 => Null hypothesis is accepted => the male and female births are equally probable.

Ans.

Example 5.78: In 60 throw s o f a dice,fa c e one turned up 6 tim es, fa c e tw o or three, 18 tim es, face fo u r or fiv e 24 tim es and fa c e six, 12 tim es, Test a t 10% level o f significance, i f the dice is honest, it being g ives th a t P ( z * > 6*25) ~ 0,1 fo r 3 degrees o f freedom . Solution : Step 1 : N ull H ypothesis Hq\ The dice is honest Step 2 : C alculation o f expectedfrequenies f , : Given that the observed frequency are as follows : Face o f dice x or (r) : 1 %or 3 observed frequency f a

:

6

18

4 or 5

6

24

* 12

(N = 60)

S ta ti s tic s J 4 8 3

Under the assumption of Ho that the dice is honest, so that expected frequency for each face is /.W

“ N x g - 6 0 x i = 10 .

[Because prob. of turning up any one of the numbers J, 2,3,4, 5,6 is g]. i.e. x or r : 1 2 or 3 4 or 5 6 expected frequency f t 10 20 20 10 Step 3 : C alculation o f £ s ta tis tic : Wehave

, - E

— s-

Tkble for x2-statistic : Face of dice x or r

/.

fe

10

2 or 3 4 or 5 6

6 * 18 24 12

Total

tf-6 0

I

u 16 0.2 0.8 0.4

20 20 12 JV*60

x2= £ '

\ i f o ~ f e )2 fe r

3-0

Thus,

Calculated value of %2 * 3.0. Also degrees of freedom v = 4 - 1 =3. Step 4 : Since given that tabulated value of v3at 10% level of significance and for 3 dof is 6.25 i.e. xS.i,v»3 “ 6.25. : D ecisio n : Clearly calculated value of x2 “ 3.0 < tabulated value of XS.i.3 - 6.25. => Null Hypothesis is accepted => The die is honest Ans. am ple 5.79: C alculate th e expected frequencies fo r the fo llo w in g data presum ing the two attributes C ondition o f hom e

C ondition o f ch ild

G ear

D irty

dea n

70

50

F airly clean

80

20

D irty

35

45

UsecM*quaretestat5%lmldf9ignyteancelosm6 whethcrthetwoattributesareindependent (table value o f£ a t 5%for 2 dcfit 5,991, endfor 3 dofis 7,915 and fo r 4 d o f Is 9.488).

4 8 4 | ENQBoramo M m h em m k s -111

Solution i

Step 1 : N ull H ypothesis H q : No association between the attributes. Step 2 : C alculation o f expected frequency ( fj o f each c e ll :

Clean

Dirty

f o : 70

f o ' 50

120x185 300 /0 :8 0

Fairly Clean *e

Dirty

Clean

185x100 300 fo : 35

,

1

f t =

, Total

185

„ = 4 6

115x100 300 fo * 45

1

/ • - 1SIZ om = 49*33

115x120 ~ m " / . : 20

Total

115x80 ____ 300 = 30-67 115

120

100

80 N = 300

Step 3 : C alculation o f x 2 -sta tistic :

_ ( 7 0 - 7 4 ) 2 . (50 - 46)2 . ( 8 0 -6 L 6 7 )2 . (2 0 -3 & 3 3)2 X2 “ - 7 4 - ’+ — 4 6 - + ~ L67— + “ W ‘ t ( 3 5 - 4 9 3 3 )2 (20 - 38.67)2 4933 + 30.67 •» 0.2162 + 0.3478 + 5.4482 + 8.7657 + 4.1627 + 6.6954 = 25.636. Also degree o f freedom (d.o.f.): v = (m „ I) (n - 1) = (3 - 1) (2 - 1) = 2. Step 4 : Since the tabulated value o f %2 ** 5% level o f significance and for dof v = 2 is 5.991, i.e; xfl.ov * 5.991. Step 5 : D ecision : Clearly, calculated value o f x2 * 25.636 tabulated value of X&.05, 2 = 5.991 => the Nufl Hypothesis is rejected. => there exists and association between the attributes. Ans. Example 5.80 : The fo llo w in g fig u re s show the distribution o f digits in num ber chosen a t random fro m a telephone d irecto ry: D ig its: 0 1 2 3 4 5 6 7 8 9 F reuency: 1026 1107 997 966 1075 933 1107 972 964 853. Test a t 5% level w hether th e digit* r m y b f Ufken to occur equally freq u en tly in the directory. (T he ta b le w in e o f fo r 9 degree o f freedom m 16.919)

S tatistics I 4 8 5

Step 1 : Null Hypothesis H o : The digits occur equally frequently in the directory. Step 2 : C alculation o f expected frequency ( fj : V Given that total observed frequencies N * l f Q“ 10,000. Expected frequency for each digits 0, V, 2 , ..... 9; f.{ r)~ N * p { r)

10,000 x ~

I* p (r) = to ]

for r = 0, 1, 2,..., 10.

* 1000;

Step 3 : C alculation o f -sta tistic : f2 *-

We have,

I

xorr

f.

/. .

0 1 2 3 4 5 6 7 8 9

1026 1107 997 966 1075 933 1107 972 964 853

1000 1000 1000 1000 1000 1000 10Q0 1000 1000 J m .

N = 10,000

N = 10,000

(fo -fe )* fe

0.676 11.449 0.009 1.156 5.625 4.489 11.449 0.784 1.296 21.609

.

k / w , ) 2! ■jk—58.542 le

Also degree o f free& ^ v ^ 10 - f 1* 9. Steiy 4 : Tfife tabulated vafue o f x* a?5% level of significance and for dof v = 9 is 16.919, i-e;$o5,9 16.919: Step 5 : D ecision : Clearly, calculated value of x2 * 58.542 * tabulated value of xios.9 = 16.919. => the Null Hypothesis is rejected. Ans.

=> the digits do not occur equally frequently. Exam ple 5.81: A d ice is tossed 120 tim e* w ith th e fo llo w in g re su lts: N um ber turned u p : F reuency:

1

2

$

4

5

30

25

18

10

22

Test th e hypothesis th a t th e dice is unbiased (fits , s m 11*07)

6 15

Total

120

4 8 6 { E fin n rm a

-V

ig

M athem atics- i n

Solution :■ Step I ; Nutt Hypothesis Hq : The dice is unbiased oat. Step 2 : Calculation o f expectedfrequency : Here N 31120. On the basis of Hq, the dice is unbiased* so that the expected frequency / , ( r ) - N p ( / 0 - 120 » | - 20; for r - 1 ,2 ,....6 i.e; expected frequencies for each turned up r * 1,2 ,3 ,4 ,5 ,6 is 20. Step 3 : Calculation o f ^sta tistic : We have

... ( 1)

Table for x2-«tatistic Cf . - f #

(f o - f e )■ fe

20 20 20 . 20 20 20

100 25 4 100 4 25

5.00 1.25 0.20 5.00 0.20 1.25

tf»120

—

X2 - 12.90

x o rr

/.

f.

1 2 3 4 5 6

30 25 18 10 22 15

Total

AT-120

Also degree of freedom v = 6 - 1 =*5. Step 4 : The tabulated value ofx2 at 5%level of significance and for dof v=5is 11.07, i.e; Xtas = 11.07. Step 5 : Decision: Clearly, calculatedvalue of x2* 12.90 K tabulated value of xS.os.j * 11 07. =>the Null Hypothesis is rejected. => the dice is a biased one. Ans. Example 5.82 : Front the adult male population c f seven large cities random sample giving 2 x 7 condnguency table cf married andunmarried mem ra given below were taken. Can it be said that there is a significant variation among the cities In the tendency of men to marry T City A B E C D F G Total Married 133 164 155 106 123 153 146 980 Unmarried 36 57 40 37 55 33 294 36 Total 169 195 221 143 208 156 182 1274 (At (2 —1) (7 —1) d.f. Hike 12.6) Solution : Step 1 : Null Hypothesis H0 : There is no significant variation among the cities in the tendency o f men to many.

S tatistics 1 4 8 7

Step 2 : Calculation of Expected frequency: On the basis of Null hypothesis the expected frequencies are: 980 Expected number of married people in City A * J2 7 4 x *69 = 130 980 Expected number of married people in City B * J2 7 4 x 221 =

170

980 Expected number of married people in City C * J2 7 4 * ^ = 15® Expected number of married people in City D ~

980

x

“

=

980 Expected number of married people in City E ~ J2 7 4 x 208 - 160 Expected number of married people in City/*=

980

1274

x 156 ~ 120

980 Expected number of married people in City G m J2 7 4 x 182 = *40 Similarity the Expected frequency are 39, 51,45,33,48,36 and 42. T a b k for expected freqieacy f 4 Married 130 170 150 110 160 Unmmarried 39 51 45 33 48 Step 3 : Calculation of -statistic :

120

36

We have, /. 133 164 155 106 153 123 146 36 57 40 37 55 33 36 N= 1274.

-

-

A

V .-f#

130 -THT 150

9 36 25 16 49 9 36 9 36 25 16 49 9 36

110

160 120

140 39 51 45 33 48 36 42 N** 1274

fe 0.069 0.212

0.167 0.145 0.306 0.075 0.257 0.231 0.706 0.556 0.485 1.021

0.250 0.857 X2 = 5.337

140 42

4 8 8 | E n g m ee m n g M a ih em * iic s -H1

m m i • Th* tabulat&d value of x2 a* 5% level of significance and for v = (2 - 1) (7 -1) = 6 dof is 12.6 i.e., xfl.os, 6 * 12.6 . Step 5 i Decision : Clearly calculate value of x 2 * 5.337 < tabulated value of X2 * 12.6 => the null hypothesis is accepted. => these is no significant variation among the cities in the tendency of men to marry. Example 5.83: For 2 * 2 contingency table

(o + b + c + d ) ( a d - be ) 8 z - ( a + b )fr+ ,rf(b+d) ia + cy

prove that,

Solution : Given the observed frequency (fa) table :

Total

a c

b d

*+-c

b+a

Total a+b c +d

i

|

N=a + b + c +d

Table for expected frequencyf e :

(a+6)(6+d)

(o + 6 )(a + c) a+b+c+d (c+
o + 6 + c+ d (c+d)(]b+d) o + 6 +c+d a w . ) 2!

Wehave,

E

r l

(o+ 6 )(« + c)i 2 a+b+c+d J (q+ 6)(q+ c) q+6+c+d

[As four terms]

- V ' [q2 + ofr+qe-fad-q 2 - a c - a b ^

mV

(q + 6) ( q + c ) (q + 6 + c + d)

(a d -b e)2______ ( q + 6 ) (q + c) (in- 6 + c +d)

[As four terms]

Now open the summation for four terms, we have r2 -

(qd-fcc) L_r____ i b)(c * c) (q + 6 +c+ d) L(q+W

J*

(q + b)(b+ d)

(c+d)(a + c)

(c+d)(b+d)

S tatistics 1 4 8 9

{a d -b e)2 ff b+d+a+c \ .f b+d + a + c 11 [ l( a + 6 Ko+ c)(b+d) r U' l(c+ c + ddXXoo+ + ccXb+d)!} (a + b + c + d ) Ll(a+&Xa+c)(t>+d)J )(6 -(a d -b e y

[(a + b )(o + c)(b + d ) + fc + d X « + c )(6 + d )] _______ c + d

+ a + b _______ 1

(a+b)(c+d)(b* d)(a +c)J

( a + b + c + d ) (a d -b c)2 (a + b)(c+d)(b+d)(a+ c)

Proved

Example 5*84: Prove thatfor a 2 * n contingency table:

x ^ fi+ ft wherefi, f 2 are twofrequencies in a subgroup and Nb rows. Solution : Table for observed freqaeacy / , : ----- 1-----------

are the marginal sums of two

Marginal sum (Total)

-

h

Marginal sum (Total)

N,

»•••

f\

N* N, + N2

fi+ fi

The expected frequency corresponding to f , : ( / . + / 2) * i n, +n2 ( / l + / 2) * 2 n,+ n2 ' We have

%2 = £

^ y -^ -

Taking value of chi-square for the first column are :

{ f i+ fi) N \ Nt +N2

C/i + f i ) ^ 2 n x+ n 2

[ v by (I)]

4 9 0 1 En g m k r m g M ath em atics-!!!

iVi(/i+/2) W + N 2) W2(/, +f 2XN1+N t ) =

1v (a " 6)2 = <*'

(A + /*)W + iV 2)

,XM M AW ,

n ,n

2 \& __-^-T 1 2 LM N t l (fl+ fl)

Hence the value of chi-square for 2 * n table is

/T >

Proved.

/ l + /2

Example 5.85 : /» jfte accounting department o f a bank 100 accounts are selected at random and examinedfor errors. Thefollowing results have been obtained. No. o f Errors 0 1 2 3 4 5 6 No. rtfAccounts 36 40 19 2 0 2 1 DoestUs information verify that the errorsare distributedaccerdlng to Poisssonprobability tow 't (tbs, s m 11.07). Sobttkm ; Stop 1 : Null Hypothesis Ho : The errors are distributed according to Poisson probability law. S teal : Calculation of expectedfrequency (£): We know that the expected frequency of poisson distribution : e~m mr fA r)= N - — where

’

0, 1, 2,3,4, 5, 6 .

N = y ~ 100 Zfr m = mean = - Qx 36-flx40-h2xl9 + 3x2 + 4xQ-hSx2-f6xl m~ 36 + 40 + 19 + 2 + 0 + 2 + 1 _ 100

100 " *•

Expected frequency: , 1°

100x0367

S ta tistics | 491

37

For /• = 1 :

^ (1) = 100 x r i. ffi!. *

For r = 2:

/, (2) = 100 *

Forr=3:

f. (3) - 100 x e->. S i-

Forr~ 4 :

/ , (4) » 100 x e~K

For r ~ 5 :

o )5 f t (5) * 100 * e~l. "jj" = 0.306 » 0

For r -

f, (6 ) * 100 x

:

6

^ « 18 [2

(i)4 14

0)‘

«6

«2

* 0.051 « 0

Step 3 ; Calculation of £ statistic : We have

X2 =

-< l) /* Since same expected frequencies axe less than 5, so that the frequncies for r =3,4,5 , 6 have been pooled together (see in table). No. of errors ( r )

Obseved frequency ( f , )

frequency ( / , )

f'

0

36 40 19

37 37 18

0.027 0.243 0.055

1 2

3 4

0

5

2

6

1 J

Total

2

6

1

iV* 1 0 0

1

1.125

2

[ 5

1 8

0

0

//=

100

J

X2 = 1.450

Step 4 : The tabulated value ofy* at 5%level of significance and fordofv-6 -1 =5 is 11.07 i.e., .*$.05, j —11.07.

4921Enumiuumg MxiwMxncs-fll

S b a i : Decision: Clearly calculated value of x2 = 1-45 < tabulated value of xl.os, 5=11.07. => the null hypothesis is accepted. => the errors are distributed according to Poisson probability law.

539 /-STATISTIC AND/-DISTRIBUTION: The statistic t was introduced by W.S. Gosset in 1908. Let x\7x2, jc* be the values of random sarnble of size n, drawn from a nonnal population with mean ji and variance a 2 (unknown). Then we define the /-statistic as : * where

-

S* = O T l)

(X'~ * y

1 £ Jc = sample mean - — Ex,. n i=l Here S is the standard deviation of sample with degrees of freedom v - n - 1, when n < 30. If we calculate t for each sample, we obtain the sample distribution for t. This distribution known as student's t-distribvtion is given by

and

....................... H ) Where y0 is a constant, so chosen that the total area under the curve is 1.

Remark: When random sample of size n is large, or [when population variance are known] that /-statistic define as:

where

a2 = ^

2

(xt - x Y -


S m sn c t | 493 5 .4 0 C O M P U T A T I O N O R W O R K I N G R U L E O F l J H S T R f f i U T l O N :

(a) To test the significance of a mean (snail sample) Step 1 : Consider the Null Hypothesis ffo: There is no Significant difference is between the sample means and population means. Alternative Hypothesis H% : There is significant difference in between the sample means and population means. Step 2 : Calcualte /-statistic : [when variance is not known] where

- (^ i)

When variance is known: then where

o 2 ** “fw I ( x - x ) 2. Step 3 : Level ofSignificance or Tabulated value of t : See the value of / from the table i.e.» value of t at a% level of significance and for the dof v. St£&4 : Decision: (i) If calculated value of |/j< tabulated value off. Then accept the null hypothesis H0. (ii) If calculated value of J/|Af tabulated value of t. Then reject the null hypothesis Ho. (b) To test the s&rificaace of dtffcreacc hetweea two sample weans : Let xu x2, xHi and y u yi* «•« y *2 be two independent random samples with means x and y and standard deviation S\ and $ from a normal population with the same variance. Suppose we have to test the hypothesis that the population means |ij and (i2 are the same, for it we calculate /-statistic as follows : 1

SSzH p* j22_ S

i ( n 1+n2) ’

S2 - ------- ---------- s----- ■* Hi + n2 - 2 with degrees of freedom v - n\ + t*i - 2 . When variances cr? and aj are known : Then 2 2 „ »i<*i + *2<*2 »1 + fl2 - 2 * Remarks : When variance b unknown : 1. For dof v and 95% confidence limits for p are given by : c J1 = 7 ± to.03 v>* “7*» where

"•'l>

4 9 4 1 E n o o c b b m q M athem atics -III

2. For dof v and 99% confidence limits for \i are given by: g M “ 5 ± /o.ei *» x "7“ Vit When variance is known : 1. For dof v and 95% confidence limits for p are given by :

<

|i * jp ± 4mh« x

4n- 1 2. For dof v and 99% confidence limits for ji are given by: i i ssx ± y i » x

a

Example 5.86 : Find the student's t-statisticfor thefollowing variable values in a sample : -4 ,-2 ,-2 ,6 ,2 ,2 ,3 ,3 , taking the mean of the universe to be zero. Solution : We have /-statistic : / Where

...d)

=

Z ( * -x )2

...(2)

Here n = 8 , and mean of universe p - 0. Serial No.

X

1

-4

2

-2

3 4 5

-2 2

6

2

7

3 3

8

x- X

( * - 3c)1

- 4.25 -2.25 -2.25 -0.25 1.75 1.75 2.75 2.75

18.0625 5.0625 5.0625 0.0625 3.0625 3.0625 7.5625 7.5625

Zx« 2

49.5000

vT o x - Mean - ~ = f =0.25 71 o and

S~

1 » -l

= J i p = 2.659 i 7

[by (2 )]

From (1), we have (0 2 5 -0 ) V8 . 2.659

Ans.

Svftnsnc* ( 4 9 9

Example 5.87 : A sampoo manufacturing company was distributed a particular brand of sampoc through a large number of retail shops. Before a heavy advertisement Campaign, the mean sales per sampoo was 140 dozens. After the campaign a sample of 26 sampoo was taken and the mean sales figure was found to be 147 dozens with standard deviation 16. Can you consider the advertisement effective 7 (Given /***2s m 1.708) Solution : Given: n - 26 (small samples) mean * y * 147; S.D. Le.» ct= 16 (variance are known), and degrees of feedom v *=n -1 * 26 - 1 * 25. StgflLl : Null hypothesis Hq : There is no difference in between the sample means and population means, i.e., the advertisement is not effective Le., H q : * 140. \A : Calculation of/-statistic : I We have

t = ——

—-

[when S.D. a is known]

_ (147-140)^26 ----------- 16----= 2.19. Step 3 : The tabulated value of / at 5% level of significance and for v * 25 dof is 1.708. i.e., fo.os, 25 = 1.708. Step 4 : Decision : Clearly calculate value of |f| - 2.19 K tabulated value of *0.03,25 = 1*708. => Null hypothesis Hqis rejected. => The advertisement is effective. Ans. Example 5.88 : Ten objects are chosen at randomfrom a population and their heigktsarefound to be in inches 63, 63, 64, 65, 66, 69, 69, 70, 70, 71. Discuss the suggestion that the mean height in the universe is 65 inches, given thatfor 9 dof the value o ft and 5% level of significance is Z262. Solution : Given n » 10, (small smaples) Universe mean : \i = 65 and variance is not known and degrees of freedom v * = n - l ” 9. Step 1 : Null hypothesis Hq: There is no difference between the mean height of sample and universe i.e., p » 65. Step 2 : Calculate of /-statistic : We have Where

t'

...(1)

71-1

...(2)

4 M | E n g b c h m g M athematics-IR

Zx mean x ~ n

and

670

67.

10

(x- x r

X

16 16 9 4

63 63 64 65

-4 -4 -3 - 2

66

- 1 ,

69 70 70 70 71

1

4 4 9 9 16

2 2

3 3 4

Z(x- jc)^ = 88

670 From die table, then equation (2) becomes *

9 = 3.13 inches

Hence (1) becomes t = ( 6 7 - g y i Q =2Q24 S s p j : The tabulated value of t at 5% level of significance and dof v = 9 is 2.262 i.e., ^0.05,9 = 2.262 Step 4 : Decision: Clearly calculated value of j/j312 .0 2 4 < tabulated value of to.05, 9 - 2 .262. => the null hypothesis Ho is accepted. => die mean height of the universe is 65 inches. Ans. Example 5.89: A salesman is expected to effect an average sales ofRs. 3500. A sample test revealed that asparticular salesman had made thefollowing sales. Rs. 3700,3400,2500,5200,3000and 2000. Conclude whether his work is belowstandard or not Using 5% level of significance of t and 5 dof is 2.015.

~

Solution : Given

S ta ti s tic s

j 497

n - 6 , (small sample)

Mean of sample:

3700+ 3400 + 2500 + 5200 + 3000 + 2000 J - ------------------------g------------------------

or

3c =

= 3300 and n - 3500

Here variance is not known and degrees of freedom v = w- 1=5. Step 1 : Null Hypothesis Ho : the salesman is up to standard. Step 2 : Calculation of /-statistic : We have

t = -—

Where

S2 *

...(1) ...(2 )

71—1

3700 3400 2500 5200 3000 2000

1* 1

&

x —jg

Sale in Rs. x

400

160000

100

10000

- 800 1900 -300 - 1300

640000 3610000 90000 1690000 S ( x - x >2 = 6200000

Ix = 19800 From the table, we have S1 => Hence (1) becomes:

= 1240000 o 5 = 1113.55 Rs.

[by (2)]

(3300 - 3800) s ’ na55 or / * - 0.44 | / 1 = 0.44 Step 3 : The tabulated value of t at 5% level of significance and dof v = 5 is 2.015 i.e., *0.05. 5 = 2.015. Step 4 ; Decision : Clarly calculate value of \ t | = 0.44 < tabulated value of *0.05, 5 = 2.015. => the hull hypothesis is accepted. => the salesman is upto standard. Ans.

4 9 8 | E n g in eerin g M athem atics -11 I

Example 5.90 : A random sample of size 16 has as mean is 53. The sum of the squares of the deviations taken from the mean is 1501 Can this sample be regarded as taken from the population having 56 as mean ? obtain 95% and 99% confidence Umits of the mean population. (For v * 15, taoi * 2.95 and tAB5 * 2.131). Solution: Given that n = 16 (small sample) Sample mean 3c = 53 and n = 56. Also

I (x - x ¥ - 150 and variance is not known. S2 =

n~l

= io. lb -l

=> S = y/lO => 5 = 3.162. The degrees of freedom v = n - J = 16 - 1 = 15. Step 1 : Null Hypothesis Hq : Population mean is 56 i.e., H0 : \i = 56. Step 2 : Calculation of /-statistic : We have,

/ =

...(1)

(53-56) Vl6 _ 3.162

=>

| / 1 - 3.79.

Step 3 : Given that tabulated values for dof v = 15 are :

(i) ^o.os, is = 2.131 (*0 *o.oi, t5 = 2.95. Sten 4 : Decision:

(i) Since calculated value of |/j = 3.79 K tabulated value of ^ 05,15 = 2.131 => the null hypothesis is rejected. => the population mean is not 56 at 5% level of significance. (ii) Since, calculated value of |/1 = 3.79 K tasulated value of fooi. 15 = 2.95 => the null hypothesis is rejected. => the population mean is not 56 at 1% level of significance. Again, we known that 95% confidence limits for ft (dof v = 15) are given by

S ta ti s tic s | 4 9 9

H = 5 3 ±2.131 x

H = 53 ± 1.684 50.316 < ^ < 54.684. 99% confidence limits for p (dof v = 15) are given by M“ x ± fo.oi x ~T= V/i = 53 ± 2.95 x 3.162 = 53 ±2.33, Ans.

50.67 < n < 55.33.

Example 5.91; A random sample of size 7from a normal population gave a mean o f977.51 and a standard deviation o f 4.42. Find 95% confidence interval for the population mean, (given that * - 2.447) n * 7, (small sample) Solution : Given that x = 977.51 and a = 4.42 (known) and dof v = « - 1=6. We known that 95% confidence limits for population mean ji for 6 dof. are given by

= 977.51 ± 2.447 x

442 V6

[*♦* fo.05,6 ~ 2.447]

= 977.51 ±4.416 973.09
Ans.

Example 5.92 : The nine items of a sample had thefollowing values: 45,47, 50, 52,48,47, 49, 53, 51. Does the mean of the nine items differ significantlyfrom the assumed population mean of 47.5 ? Given that

Where P is the area to the left of the ordinate at t

5 0 0 | E n g in eerin g M athem atics-111

Solution : and degrees of freedom JC 45 47 50 52 48 47 49 53 51

n - 9 (small sample) H - 47.5 and variance is not known v - n - 1 =8. x~ x

(x - x ) 2

-4.11 -2.11 0.89 2.89 -1.11 -2.11 -0.11 3.89 1.89

16.8921 4.4521 0.7921 8.3521 1.2321 4.4521 0.0121 15.1321 3.5721

Zx ~ 442

E(jc- x P = 54.8889 2* = 442 x = n = —9 =49.11

and

n -1

o

S = V6.8611 = 2.62 By /-statistic

= (x~\i)yfn S

_ (49.11-47.5) >/9 2.62 = 1.84. From the given data let us interpolate the value of P for / = 1.84 is as follows : fo r /=1.9 P = 0.953 / = 1.84 for / = 1.8 P = 0.945 /= 1.8 difference = 0.1

difference = 0.008

difference * 0.04

(Diff. in P) (Diff. in calculated value of t ) Difference -----------------Diff. of t as given--------------

S ta tistics | 5 0 1

Thus required value of P for P « 0.945 + 0.0032 * 0.9482 = 0.95 Thus *0.95 t - Area to the right of / = 1.84 = !-/> ,.i.m= 1 -0.95 = 0.05. Hence /’ (I/I* 1.84) - Left + Right = 0.05 + 0.05 = 0.1 which is greater than 0.05 | / 1 * 1.84 > 0.05 => Chance of getting value of t is greater than 0.05. The value of / is not significant => Sample may be a random sample which is drawn from the normal population with mean 47.5. Example 5.93 : The average number of articles produced by two machines per day are 200 and 250 with standard deviations 20 and 25 respectively on the basis ofrecords of 25 daysproduction. Can you regard both the machines equally efficient at 1% level of significance ? (Given that t6.0j,*s * 2.58) Volution : Given that fli » 25, xl = 200, S.D. C| = 20 n2 ~ 25, x2 ~ 250, S.D. a2 = 25 We have

p = "l°l + »2°rl _ 25 x (20)2 + 25 x (25)2 25 + 25 - 2

S = V533.85 = 23.1 Degrees of freedom v = n\ + n2 - 2 = 48. Step I : Null Hypothesis Ho : both the machines are equally efficient i.e., Hi = H2Step 2 : Calculate of /-statistic :

5 0 2 | E n g in e e rin g M a th e m a tic s-! 11

-50 23.1

x 3,54

- - 7.65 => | / 1= 7.65 Step 3 : The tabulated value of / at 1% level of significance and dof v = 48 is 2.58 i.e., *o.oi, 4* = 2.58. Step 4 : Decision : Since calculated value of |/| = 7,65 K tabulated value of fo.oi, 48 = 2.58. => the null hypothesis is rejected. => the two machines are not equally efficient at!% level of significance. Aas. Example 5.94: Two horses A and B were tested according to the time (in seconds) to run a particular track with the following results: HorseA 28 30 32 33 33 29 34 Horse B 29 30 30 24 27 29 Test whether you can discriminate between two horses♦ (Given that t&os, n “ 2.20). Solution : Given that «i = 7, n2 - 6 _

mean of A :

2 * _ 219

x ~ nx ~

31.29

7

169 28.17. 6 Degress of freedom v = «t + »2 - 2 = 11. Step 1 : Null Hypothesis Hq : No discriminate between two horses i.e., Ho: Hi = *i2. Step 2 : Calculation of /-statistic:

mean of B :

y = *2

t=

We have

x -y

€

n|»2

Table for calculation of S X

(x - x )

(x - X?

y

y- y

(y-y)2

28 30 32 33 33 29 34

-3.29 - 1.29 0.71 1.71 1.71 -2.29 2.71

10.8241 1.6641 0.5041 2.9241 2.9241 5.2441 7.3441

29 30 30 24 27 29

0.83 1.83 1.83 -4.17 - 1.17 0.83

0.6889 3.3489 3.3489 17.3889 1.3689 0.6889

31.4287

t y = 169

Lx - 219

;

26.8334

S tatistics |

Since

503

Z ( x - * ) 2 +Z ( y - y ) 2

„

n l + n 2 ~

2

3L4287 + 26.8334 7+6 -2

S = VS!296 = 2.30 Hence (1) becomes /=

(31.29-28.17) '7x6 7+6 2.30 .

5.608 • 2.30 = 2.44. Step 3 : The calculated value of t at 5%level of significance and for dof v = 11 is 2.20 i.e., *0.05, n = 2.20. Step 4 : Decision : Since the calculated value of |/| 2.44 < tabulated value of t = 2.20. => the null hypothesis is rejected => there are dicriminate between two horses at 5% level of significance. Ans. xample 5.95 : Below are given the gain in weights in kgs of cowsfed on two diets x andy. Diet x : 13 14 10 11 12 16 10 8 11 12 9 12 Diety: 7 11 10 8 10 13 9. Test at 5% level of significance whether the two diets differ as regards their effect on mean increase in weight (given that n «*2.11) olution : Given that: n\ ~ 12, n2^ l and degrees of freedom v = nx+ n2 - 7 = 17. =

bT

= 1 2 ' = 11S

Step 1 : Null Hypothesis H0 : two diets do not differ significantly i.e., Ho : Hi: Step 2 : Calculation of /-statistic :

504

| E ngin eerin g M a ih em a h c s -III

Table for calculation o f S : X

x -

13 14 10 11 12 16 10 8 11 12 9 12

x

(x -x ?

2.25 6.25 2.25 0.25 0.25 20.25 2.25 12.25 0.25 0.25 6.25 0.25

1.5 2.5 - 1.5 -0.5 0.5 4.5 -1.5 -3.5 -0.5 0.5 -2 .5 0.5

( y -y ) 2

7 11 10 8 10 13 9

-2.71 1.29 0.29 - 1.71 0.29 3.29 -0.71

7.3441 1.6641 0.0841 2.9241 0.0841 10.8241 0.5041

Sy = 68

23.4256

S ( * - i ) 2 + S (y -J )' -------n ^ n t - 2

_

Smce

y-y

t

53.00

138

y

*

5a00+2a4256 12 + 7 - 2 « 4.4956

S = V4.4956 =

2.12

From (1), we have 212

>.71) 112x7 V12 + 7

- 1.77. Step 3 : The tabulated value of *at 5%level of significance and for dof v = 15 is 2.11 i.e., *0.05.15 = 2.11 Step 4 : Decision : Since calculated value |*| = 1.77 < tabulated value o f *0.05, 15 = 2.11. => null hypothesis is accepted. => the two diets do not differ significantly. Ans.

S tatistics

| 505

5.41 FISHER’S F-TEST The F-test was first originated by a great statistician even agriculturist prof. R.A. fisher. The test is also known as Fisher’s F-test or simply F-test It is based on F-districution. The F-test refers to a test of hypothesis concerning two variances derived from two samples. The distribution of variance ratio F with vt and v2 degres of freedom is given by :

where y0 is so chosen that the total area under the curve is unity.

5.42 F-STATISTIC: Let xu *2, ....* x„u and y x%y i , ....y*i be the values of two independent random samples drawn from the two normal populations with mean iii and n2 and variances ai2, a22 respectively, then we define variance ratio F as :

when

and jc, y are the sample means, with vi = n\ - 1, and v2 = Note : The value of F-statistic is always greater then L

- 1 are degrees of freedom.

5.43 ASSUMPTIONS IN F-TEST The F-test is based on the following assumptions : (i) The values in each group should be normally distributed. (ii) The error should be independent of each value. (iii) The variances within each group should be equal for all groups. 5.44 COMPUTATION OF F-TEST: We have the following steps : Step 1 : Null hypothesis Ho : ctj2 = a22 Alternative Hypothesis H\ :
5 0 6 | E n g in e e rin g M a th e m a tic s-! 11 S 2

or

F = — , when S22 £ Si2. Sf

Step 3 : Choose tabulated value of / ’-statistic at a% level of significant with degrees of freedoms vj and V2. Step 4 : Decision : If calculated value of F < tabulated value of F^*\, v2) Then accept null hypothesis Hq. If calculated value F \ tabulated value of i(vi, V2> Then reject null hypothesis Hq. Example 5.96 : Random samples are drawn from two populations and the following results were obtained: Sample X : 20 16 26 27 23 22 18 24 25 19 Sample Y: 27 33 42 35 32 34 38 28 41 43 39 37 \ Find variance of two populations and test whether the two samples have same variance. (Given that F&0Sfor 11 and 9 dof is 3.112). Solution : Given that wj = 10 and n2 - 12 with degrees of freedom v = n\ - 1 = 9 and V=

- 1 — 11.

Step 1 : Null Hybothesis Hq : Let CTj2 = 022 i.e., the two samples have the same variance. Step 2 : Calculation of F-statistic : We have to find Si2 and S22 : Sample X y

-2 -6 4 5 1 0 -4 2 3 -3

4 36 16 25 1 0 16 4 9 9

27 33 42 35 32 34 38 28 41 43 30 37

0

120

y -y

(y-y)2

-8

6 8 -5 2

64 4 49 0 9 1 9 49 36 64 25 4

0

314

-2 7 0 -3

-

1 3

-7

0

Ex = 220

(x-x)2

tl -f* to.

20 16 26 27 23 22 -----18 24 25 19

x- x

V:

X

Sample Y

I S ta tistics

Here

n\ = 10, -

=

wj - 12

420 - . a = 35 12 ^ »2 314 £ (y -y )* = 28.5 11 *2 - l

« 220 =

x

nx 10 SQc - x)2 Si2 = fli~ 1 Hence F

I 507

= £2_= gas Si2 " 103

22* 120

= 133

2.14

[V

S22 > S ,2 ]

Hence calculated value of Fis 2.14. Step 3 : The tabulated value of F at 5% level of significance for the dof v = 11 and 9 is 3.112 i.e.t Fo.os - 3.112. Step 4 : Decision : Calculated value of F = 2.14 < tabulated value of F0.05 = 3.112 => the null hypothesis H0 is accepted. => the two samples have the same variance. Example 5.97: in a test given two groups o f students drawtffrom two normalpopulations, the marks obtained were asfollows: Group A : 18 20 36 50 49 36 34 49 41 Group B : 29 28 26 35 30 44 46 Examine at 5% level, whether the two populations have the same variance, (Given that Fe.83(*,6) “ 4,15). Solution : Given that: n\ = 9, and n2 = 7

and

jjc * “ »i

333 9

-

238

SL

37 34.

Step 1 : Null Hypothesis H%: ctj2 = a^, Le., two populations have the same variance. Step 2 : Calculation of F-statistic: Group B

GroupA x 18 20 36 50 49 36 34 49 41 Ex = 333

x -x -1 9 -1 7 -1 13 12 - 1 -3 12 4

(x-x)* 361 289 1 169 144 1 9 144 16 1134

y 29 28 26 35 30 44 46

= 238

y-y -5

-6 -8 1 -4 10 12

(y-y? 25 36 64 1 16 100 144

386

5 0 8 | E ngin eerin g M athem atics-111

We have

Z ( x - x ) 2 1134 . _ Si2 ---------nj -1;— = —8Q— = 141.75

Clearly

z(y-y)2 n2 - 1 Si2 > S22 Si2 =

238 6 = 64.33

S{ 141.75 F = c2 64.33 *^2 i.e., F “ 2.203 Step 3 : The tabulated value of Fat 5% level of significance and for degress of freedom 8 and 6 is 4.15, i.e., Fqm = 4.15 Step 4 : Decision : Since calculated value of F ~ 2.203 < tabulated value of Fo.os = 415. => the null hypothesis Ho is accepted. => the two population have the same variance. ( Ans. Example 5.98 : In a laboratory experiment, two samples gave thefollowing results: 111611

Sample

Size

Sample mean

Sum o f squares of deviation from the mean

I

10

15

90

i

12

14

108

Test the equality ofsample variance at 5% level ofsignificance. (Given that />, n (U.os) - 2.90) Solution : Given that: = 10 n2 = 12 jc

y - 14

~ 15

Z ( y - y)2 = 108 Z { x - xY =90 Step 1 : Null hypothesis Ho . C\2 = o22 i.e., two samples have the same variance. Step 2 : Calculated of F-statistic : '

We have and Clearly

£ ( s - s-?\2 ) 5,2 = Hi -1

90 9

=

10

108 z(y-y)2 = 9.82 n 11 n2 - 1 S22, then c2 10 = 1.018 f ~ -i-*22 =

sS

982

Step 3 : The tabulated value of F at 5% level of significance for dof 9 and 11 is 2.90 i.e., Foos = 2.90.

S ta tistics |

509

Step 4 : Decision : Since calcualted value of F ~ 1.018 < tabulated value of Fo.os ~ 2 .9 0 .

the null hypothesis //<> is accepted. the two populations have the same variance.

Ans.

5 . 4 5 F I S H E R ’S z - D I S T R I B U T l O N

l^t x \ , x 2, ...... .v„i and y i% y2.......yn2 be the values of two independent random samples'with estimated variance S{2 and S22. Then Fisher’s /-distribution is the statistical distribution of halfof the logarithm of an F-distribution variate: z ^ |l o g « F

i.e .,

S1 = 2o log. ^ S^V ,I; where S\2 > S22. With degress of freedom vt ~ n\ - 1 and i>2 = n2 - I. It was first described by Ronald Fisher in 1924. The probability density function of /-distribution is defined as follows : u\!2 y

-

------------- --------- -- X --------------------------- -.----------- r j ~ ; - o o < z < OC

a t e .* )

Remark: With the help of equation (1), all the applications of F-distribtution may be regarded as the application of z-distribution also. Example 5.99 : Two gauge operators are testedfor precision in making measurement One operator completes a set of26 readings with a standard deviation of 134 and the other does 34 reading with a standard deviation of 0.9%. What is the level ofsignificance of this difference ? Given that for dof v, * 25 and v2 m33, the value of z*os * 0.306 and z+oi *=0.432. Solution : Given . = 26 and n2 ~ 34 S.D. at = 1.34 and o2 = 0.98 Since => => =>

, Z(x~x)2 , U y~ y)2 Oj2 - ---------- and o22 - ----“----rt, n2 (1-34)2 -

and (0.98F =

I ( x - x f - (1.34)2 x 26 and 2. ( y - y Y = (0.98)2 * 34 I. (x- x)2 ~ 46.68 and I iy - y? = 32.6536

5 1 0 | E n g m e h b n g M athem atics-111

Step I : Null Hypothesis H0 : The difference between variances is not significant i.e., O!2 =
and Clearly

Si2 >

we have

= | log, (1.8770) = 0.3148. Step 3 : The tabulated value z at 5%and i % level of singificance and for dof 25 and 33 are zoos —0.306 and zo.ot36 0.432. Step 4 : Desicion : (1) Since Calculated value of z = 0.3148 K tabulated value of £o.os= 3.06 => the null hypothesis Ho is rejected. => the difference between variances is significant at 5% level of significance. (ii) Since calculated value of z = 0.3148 < tabulated value of zo.oi ~ 0.432. => the null hypothesis H0 is accepted. => the difference between variances is not significant at 1%level of significance. Ans. Example 5.100 : Test whether the two sets of observations: I : 17 27 18 25 27 29 27 23 17 n : 16 16 20 16 20 17 15 21 Indicate samples drawnfrom the same universe. (The value o f z at 5% levelfor 8 and 7 degrees o f freedom Is 0.6575J. Solution : Given that /i, =9 J»2 = 8 with dof V| = «2 - 1 = 8 and v2 = *2 - 1 = 7.

S ta tistics |

5 ll

Step 1 : Null hypothesis Ho : Two population variances are same i.e., H0 : ai2 = O22. Calculation of z-statistic: 11 observation

I observation X

x- X

17 27 18 25 27 29 27 23 17

-6.33 3.67 -5.33 1.67 3.67 5.67 3.67 -0.33 -6.33

y

y-y

40.0689 13.4689 28.4089 2.7889 13.4689 32.1489 13.4689 0.1089 40.0689

16 16 20 16 20 17 15 21

- 1.625 - 1.625 2.375 - 1.625 2.375 - 0.625 - 2.625 3.375

184.6001

141

(x- xY

210

0>- y Y 2.6406 . 2.6406 5.6406 2.6406 5.6406 0.3906 6.8906 11.3906 37.8748

We have _

S'

_ 184.0001 = , ,

£ (*-3i f

n i- 1

8

23

w . a t B i . 2 z * 6 a - M107 n2 - 1 Since

S\2 > S22 ,2 r l ^i 5

z = i lo&

2

= 1 10& (sfiW ) = \ log, (4.2508) = 0.7236. Step 3 : The tabulated value of z at 5% level of significance with dof 8 and 7 is 0.6575. i.e., 20.05 = 0.6575. Step 4 : Decision: Since calculated value of z = 0.7236 K tabulated value of zoos= 0.6575. => the null hypothesis is rejected. => the two variances are not same. Ans.

5 J & I E n q w e e o n g M athemattcs-III

Exercise-5 (A) 1.

Find whether the following function is a p d.f. r f x _ J *» 0 < ; x s l J W “ |2x, 1 £ x £ 2 ’

2.

A continuous random variable A"has />.<£/ /(x ) = 3x2»0 ^x ^ 1 Find the value of ‘a \ if P(X<>a) = P ( X > a). 0, x <2 •^■(3 + 2x), 2 < x < 4 8 0, *>4

3.

If

4.

Prove that/ (x) is a p .d f Also compute P (2 < X < 3). A random variable X has the p.d.f.

f ix ) -

x,

0 £ x <1

f i x ) _ •{2 - x , 1 £ jc < 2 0, x 2: 2

Find distributive function or cumulative distributive function (c.d.f.) of X. A random variable X has p.df. :

/far3*' *>o jo , 6.

is o

Find (i) the value of k (ii) compute P (1 < X < 2) and P (X> 3). A random variable A'has a p.d.f. x 0 1.5).

S ta tistics ) 5 1 3

7.

Find the mean o f the follow ing p .d .f.

fix)

8.

0,

otherwise.

For the continuous distribution : dF = y0 (x - jc2) dx, 0 £ x <, 1. Find

9.

mean, median and mode.

If a random variable X has p.df. fix) = ^

-aZxZa.

Find first, second, third and fourth moments about origin. 10. A p.df. function defined by — (3 + x)2, - 3 < * £ --! fix ) =

X

( 6 - 2 * 2 ),

~ ( 3 - a:)2,

- 1 <

x

< 1

1 < x £ 3.

Find mean, variance and standard deviation. 11.

1

Vox ^ p.d f f {x) -

Find mean, median, and mode. 12. A distributive function F (x) is defined as 0

, x <>2

1

x >£

F (x)-

Find p.df. of x. Also find mean.

Answers-5 (A) 1. 2. 3.

* 7 No., because J f(x) dx = « * 1. a * (0.5)l/3. 4/9

514 \ E n g in e e rin g M a th e m a ttc s-III

fo

, x<0

4.

F(x) = 2* 1

5.

k -3 ,

6.

I 3 15 8 ’ 8’ 32 ‘

2

l * x<2 , xZ 2

P ( I < A*< 2) = e*3 - e** and /> (A' £ 3) = e~9.

13 5 8.

_y0 = 6 , mean =

median =

mode = ^

9.

m' = o,

10. 11.

mean = 0, variance = 1, S.D. = 1. mean = mode = median - 0 .

^ ,M 3 ’ = o, h4'=

,_ x+3 12. a = , mean = 5.375

Exercise-5 (B) For a p.d.f f(x) /(* )=

s - G <* <6 [ 0, otherwise Find the value of k and show that/ (x) is a rectangular distribution. Also fmd mean. For the rectangular distribution ; f i x ) ~ fjj~ , - a < x < a

Find \i2, ^3 and m (i.e., 2nd, 3rd & 4thmoments about mean). For the rectangular distrubiton : f(x\ = \ 2k [ 0,

10-fe
otherwise

Find mean and variance. A lso find distribution function

F (x).

S ta tistics I 5 1 5

4.

Given the Incomplete Beta function: x

3 (I fn)

B, (/, m) * J jc^1(1 - *)"-1dx and lx (/, m) = 5.

then prove that Ix f/, tti) * 1 - /lHr (m, /) The income tax of a man is exponentially distributed with the p.d.f is given by 1 --x/3 /<*) = 3* What is the probability that his income will exceed Rs. 17000 assuning that the income tax is levied at the rate of 15% on the income whose Rs. 15,000 ? The life time X in hours of a T.V. tube of a certain type obeys an exponential distribution with X = 0.001 hours. Find (i) P (X> 1000), (ii) P (700 ZX<> 100) Find the value of *c\ so that / ( jc ) be a p.df. of/ ( jc ) = ce-2*, x > 0, then prove that (i) fiQ O M(Jffc), (ii) Cwfficient of variation is 1.

Answers-5 (B) k - b ~ a, mean = ^ (a + b). 2

4

|«2 ^ "jp

® IT*

mean = 10, variance =

and

jc- ( 1 0 - £)

d.F.

F(x) =

10-/t
xZ 10+£. -100 6.

(i) 0.368, (ii) 0.129. j* **A f* |»

m

>

m

Exercise-5 (C) 1. 2. 3.

Find the Binomial distribution whose mean is 5 and variance is 10/3. Find the Binomial distribution whose sum of mean and variance is 4.8 for 5 trials. In a Binomial distribution, the sum of the mean and variance for 5 trials is 1.8. Determine the distribution.

516 | Engineering Mathematics-III

4.

5. 6.

If 10% of the rivets produced by a machine are defective, find the probability that out of 5 rivets chosen at random : (i) None will be defective, (ii) One will be defective, (iii) At least two will be defective. If the probability that a newborn child is a male is 0.6, find the probability that in a family of 5 children there are exactly 3 boys. Three percent of a given lot of manufactured parts are defective. What is the probability that in a sample of four items none will be defective. [Hint.

7.

/>(0) =4 C0(O.O3)°(O.97)4 = 0.885 ]

The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 pass in the examination. [Hint.

p= 0A,

<7=0.6

/»(*;>4)=P(x = 4)+/>(jc=5)+P(jc=6). =C4(.6)4(.4)2 +C5(0.6)5(0.4)=C6 (0.6)6=0.5443]. 8.

Five coins are tossed 3,200 times. Find the frequencies of the distribution of heads and tails and tabulate the result. Calculate the mean number of successes and standard deviation. [Hint.

9.

In a particular market 40% of the consumers prefer readymade clothing. A sample of n = 5 consumers is to be drawn. Give the probabilities of having 0,1,2,3,4 and 5 preferring readymade clothes. [Hint. Probability of r consumers preferring readymade clothes out of a sample of 5=5CX(0.4)r(0.6)5_r. Put r =0,1,2,3,4 and 5.]

10. The incidence of occupational disease in an industry is such that the workers have a 20% chance of suffering from it. What is the probability that out of six workers chosen at random four or more will suffer from the disease. [Hint.

Here p=0.2, ^=0.8, n=6. p(r ;>4)=C4(0.2)4(0.8)2 +C5(0.2)5(0.8)+C6(0.2)6].

S ta ti s tic s | 517

11.

A sample of 10 pieces was examined out of a large consignment which has 5% defective pieces. Give the probability of 1 defective in the sample of 10. [Hint Here

12.

10 0.05, ?=0.95, n=10 />(r= 1)=C,(0.05)(0.95)9].

In 100 sets of ten tosses of an unbiased coin, in how many cases should we expect seven heads and three tails. [Hint Here Probability of head

»$=—,»=10 #=100.

ioo— Vj rI —I n 3' Required number of cases »100xC7l 13.

Fit a Binomial distribution for the following data and compare the theoretical frequencies with the actual ones: x : 0 I 2 3 4 5 / : 2 14 i0 34 22 8 14. In a Binomial distribution, for n = 5, if P(r - 1) = 0.4096 and P(r - 2 ) - 0.2048, then find the value of p. 15. Ten coins are tossed 1024 times and the following frequencies observed. Compare these frequencies with the expected frequencies. x : 0 I 2 3 4 5 6 7 8 9 10 / : 2 10 38 106 188 257 226 128 59 7 3 16. Seven coins are tossed and the number of heads are noted. This experiment is repeated 128 times and the following distribution is obtained: x : 0 1 2 3 4 5 6 7 Total /: 7 6 19 35 30 23 7 1 128 Fit the binomial distribution considering the coin is unbiased. 17. Fit a Binomial distribution for the following data and find the expected frequencies of the chance of machine being defective is 1/2. x : 0 1 2 3 4 5 6 7 / : 7 6 19 35 30 23 7 1

Answers-5 (C)

518 | E n g in e e rin g M a th e m a ttc s-III

4. 5.

(i) 0.5905, 0.3456.

9-

C,(0.6)r(0.4)5*', r = 0,1,2,3,4,5.

1°*

(ii) 0.3281, 6. 0.885.

(iii) 0.0815. 7.0.5443.

8. Mean = 2.5, S.D. =1.118.

C a { 0.2)4 + C5(0.2)5(0.8)+(0.2)6.

/ ] V° 12. i°0xC7|j J 10

11.

,0C, (0.05)(0.95)9.

13.

/(r) = 100x5 Cr(0.432)5~f(0.668)r, r - 0, I, 2, 3, 4, 5.

14.

I P=?

15.

1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1.

16.

128(0.517+ 0.423)7; 1, 8, 23, 36, 33, 19, 6, I

17.

1 ,7 ,21,2,5,21,7,1.

Exercise-5 (D) 1.

2.

If a random variable has a Poisson distribution such that P (1) -P (2), find (i) Mean of the distribution (ii) P{4). 2 Suppose that X has a Poisson distribution. If P{X - 2) ~~P{X = I), find, (i) P(X = 0)

3. 4.

5.

(ii) A ^ = 3).

A certain screw making machine produces on average 2 defective screws out of 100, and packs them in boxes of 500. Find the probability that a box contains 15 defective screws. The probability that a man aged 35 years will die before reaching the age of 40 years may be taken as 0.018. Out of a group of 400 men, now aged 35 years, what is the probability that 2 men will die within the next 5 years ? Suppose a book of 585 pages contains 43 typographical errors. If these errors are randomly distributed throughout the book, what is the probability that 10 pages, selected at random, will be free from errors ?

S ta tistics |

519

Using Poisson distribution, find the probability that the ace of spades will be drawn from a pack of well-shuffled cards at least once in 104 consecutive trials. If m and prdenote by the mean and central rth moment of a Poisson distribution, then prove that am In a normal summer, a truck driver gets on an average one puncture in 1000 km. Applying Poisson distribution, find the probability that he will havie *. (i) No puncture (ii) Two punctiires in a journey of 3000 kms. j Six coins are tossed 6400 times. Using Poisson's distribution find approximate probability of getting 6 reads x times. A manufacturer knows that the razor blades he makes contain on an average 0.5% of defectives. He packs them in packets of 5. What is the probability that a packet picked at'random will contain 3 or more faulty blades ? A source of water is known to contain bacteria with mean number of bacteria per cc equal to 2. Five 1 cc test tubes were filled with water. Assuming that Poisson distribution is applicable, calculate the probability that exactly 2 test tube£ contain at least* 1 bacterium each. Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in a sample of 10 tools chosen at random (i) Exactly two will be defective and (ii) More than one will be defective, by using Poisson distribution. Between the hours 2 P.M. and 4 P.M., the average number of phone calls per minute coming into the switchboard of a company is 2.35. Find the probability that during one particular minute there will be at most 2 phone calls. [Hint

n/ , e mm' e~235 x(2.35)r p(r) = — T ’= ----------------- » r = °»X' 2»3*

P(X 2) = P(X = 0)+P(X = 1) + P(X = 2), =
, 2.35 (2.35)'v2 1+

-------- * +

1!

2!

= 0.583].

If 5%of the electric bulbs manufactured by a company are defective, use Poisson distribution to find the probability that in a sample of 100 bulbs, 5 bulbs will be defective. [Hint

5 2 0 | E n g in eerin g M athem atics-!!!

15.

16. 17.

18.

19.

20.

21.

The number of accidents in a year attributed to taxi drivers in a city follows Poisson distribution with mean 3. Out of !,000 taxi drivers, find approximately the number of drivers with (i) no accidents in a year, and (ii) more than 3 accidents m a year. The probability that a man aged 50 years will die within a year is 0.01125. What is the probability that out of 12 such men at least 11 wilt reach their fifty-first birthday. A machine produces large quantities of hems and past experiment shows that on an average it produces 1%defective items. A sample of 20 items in drawn from a large number of Hems. Use Poisson distribution to determine the probability of 2 defectives. Fit a Poisson distribution to the following data and calculate theoretical frequencies : Deaths: 0 1 2 3 4 5 6 7 Frequencies: 305 365 210 80 28 9 2 1 Fit a Poisson's distribution for the following data: x: 0 1 2 3 4 5 6 7 8 9 10 y: 103 143 98 42 8 4 2 0 0 0 0 From records of Prussian Army corps kept over 20 years, the following data was obtained showing the number of deaths caused by the kicks of a horse. Calculate the theoretical Poisson frequencies : Number ofdeaths 0 1 2 3 4 Tbtal Frequencies 109 65 22 3 1 200 Fit a Poisson's distribution for the following data: x: 0 1 2 3 4 5 Tbtal f: 229 211 93 35 7 1 576

Answers-5(D) 1.

(i)2

(ii) 2/3
2. 3.

(i) e-4 0.035.

(ii) 4e~*. 4. 0.01936.

8.

(i)

(ii) (415)
11. 12.

0.3459. (i) 0.18395

(ii) 0.2642,

5. 0.4795.

6. 0.865.

13. 0.583.

14. 0.1823.

(ii) 353 nearly 15. (i) lOOxe-3 = 49.8, 17. 0.0164. 18. 301.2, 361.4, 216.8. 86.7. 26.0, 6.2, 1.2, 0.2. 19. 107, 141, 93, 41, 14, 4, 1, 0, 0, 0, 0. 20. n = 109, 66, 20, 4, 1. 21. 227.26, 211.35, 98.28, 30.46, 7.08, 1.31.

16. 0.9916.

S tattstkb i 5 2 1

5.

Ex Find the probability that the standard nonnal variate lies between 0 to I 5. Find the area under the nonnal curve for (a) Z ~ 1.64, (b) Z *• 1 32. (c) Z ~ 0.56 (d) Z - J 54. Find the area to the (a) right of Z *■0.25; (b) left of Z * 1.96; (c) left of Z = - 1.96; (d) between Z = 0.4 and Z * 0.6. A large number of measurements in normally distributed with mean of 65.5 cm and a standard deviation of 6.2 cot. Find the percentage of measurements that fad between 54.8 cm. and 68.8 cm. In an intelligence test administered to 1,000 students the average score was 42 and standard deviation 24. Find
=*0.333.

P(Z > .333) = 0.5-0 1304- 0.3696. Expected number of students exceeding a score of 50 = .3696 x 1000 = 369.6*370. [Hint (b) WhenXhes between 30 and 54, then Z lies between - 0.5 and 0.5. Probability of having students with score between 30 and 54 = 0.1915 + 0.1915 - 0.3830. Expected number of students to score between 30 and 54 = 1.000x0.3830 = 383. 100 1,000

[Hint (c) Probability of getting top 100 students ------- = 0.1. Standard normal variate having 0.1 area to the right = 1.28. X -42 24

Also 6.

>X = 72.72 « 73

In a distribution exactly 7% of the items are 35 and 89% are under 63. What are the mean and standard deviations of the distribution ? [Hint

! 4 8 ^ 3 5 -n = I.48a

a

... (0

/ 'j

Z =

= j .23 ^

a

63 - n = -1.23o

Solve (1) and (2) we get ji = 50.3, o = 10.33 ].

t

.. (2)

5 2 2 | E n g in eerin g M athem /u tc s -III

7.

A sample of 100 diy battery cells tested to find the length of life produced the following results: |A=12hours, a = 3hours. Assuming the data to be normally distributed, what percentage of battery cells are expected to have life (a) more than 15 hours, (b) less than 6 hours (c) between 10 and 14 hours. [Hint (a)WhenX= IS, Z = ^ j^ * = l Area to the right of Z = 1 is 0.5-0.3413 = 0.1587. Percent of battery cells having life more than 15 hours = 0.1587 x 100 = 15.87%. [Hint (b) When X = 6, Z =

= -2

Area to the left of Z is -2 = 0.5 - 0.4772 = 0.0228. %of Battery cells having life less than 6 hours = 0.0228 x 100 = 2.28%. [Hint (c) When X = 10, Z = —~

=-0.67 When JIT=14, Z = -

2 =0.67

Area betweenX - 10 to X ~ 14 is twice the area to the left of Z = .67 = 2x0.2487 = 0.4974. 8.

% of Battery cells having life span between 10 hours and 14 hours = 49.74%]. The mean lifetime of 100 Watt light bulbs produced by Larsen and Turbo is 200 hours. It is known that the standard deviation is 20 hours. Assuming that the life time of light bulbs are normally distributed, what are the probabilities that a single 100 Watt light bulb extracted fromthe production lot will (a) bum out between 180 hours and 210 hours ? (b) bum out for a time greater than 250 hours. [Hint (a)WhenX= 100,then Z = - —

20

°^ = -land vhen*= 210 then Z = 2-

°

20

= 0.5

Area between Z= - 1 and z * 0.5 is 0.3413 + 0.1915 = 0.5328. Required probability = 0.5328. [Hint (b) When X = 250, Z = — - - = 2.5.

9.

Required Probability= Area to the right of Z equals 2.5 = (0.5 - 0.4938) = .0062] In a sample of 1,000 the mean weight is 45 kgs with standard deviation 15 kgs. Assuming the normality of the distribution, find the number of items weighing between 40 and 60 kgs.

= /X-0.333 <; Z £ 1) = P{r0.333 < Z < 0) + P(0 < Z < 1) = 0.1293 + 0.3413 = 0.4706. Required number of items = 1,000 x 0.4706 = 470.6 » 471].

S ta tis tic s | 5 2 0

10.

In a sample of 120 workers in a factory the mean and standard deviation of wages were Rs. 11.35 and Rs. 3.03 respectively. Find the percentage of workers getting wages between Rs. 9 and Rs. 17 in die whole factory assuming that the wages are normally distributed. 11. Fit the equation of the best fitting normal curve to the following distribution: x : 0 \ 2 3 4 5 / : 13 23 34 15 11 4 12. Fit a nonnal distributions to the following data: Mid point of interval 100 95 90 85 80 75 70 65 60 55 50 45 Frequency 0 1 3 2 7 12 10 9 5 3 2 0 13. Fit a normal distribution to the following data: Variable 60-62 63-65 66-68 69-71 72-74 Frequency 5 18 42 27 8

Answers-5(E) 1. 2. 3. 4. 6. 7. 8.

0.4332 (a) 0.4484;
II.

„ , ioo -(it)1 y= f(x )= -rrre^ V3.4*

12. 13.

(d) 0.4382. (d) 03811. (c) 73.

10. 75.1

Mean (*i) = 71.2, S.D. (a) = 9.95; Theoretical frequencies : 0.2, 0.6, 1.8, 4.1, 7.3, 10.1, 10.7, 8.9, 5.7, 2.9, 1.1, 0.3. Mean (p) - 67.5, S.D. (cr) = 2.9; Expected frequencies : 2.55, 13.43, 13.10, 18.64, 4.81. VMftx mi

f. * *r

>- X. '

Ml

«

h*

*,

'

Exercise-5 (F) 1.

A sample of 300 students of Under-Graduate and 300 students of Post-Graduate classes of a University were asked to give their opinion toward the autonomous colleges. 190 of the UnderGraduate and 210 of the Post-Graduate students favoured die autonomous status. Present the above data in the form of a frequency table and test at 5% levels, the opinions of Under-Graduate and Post-Graduate students on autonomous status of colleges are independent (Table value of chi-square at 5% level for 1 d.f. is 3.84).

5 2 4 | E ngin eerin g M athem atics -III

2.

3.

4.

5.

6.

7.

A certain drug is claimed to be effective in curing colds. In an experiment on 164 poeple with cold, half of them were given the drug and half of them given sugar pills. The patient’s reactions to the treatment are recorded in the following table. Test the hypothesis that the drug is not better than sugar pills for curing colds. Helped Harmed No effect Drug: 52 10 20 Sugar Pills: 44 12 26 (Table value of chi-square at 5% level for d f 2 is 5.99) In a survey of 200 boys, of which 75 were intelligent, 40 had skilled fathers, while 85 of the unintelligent boys had unskilled fathers. Do these support the hypothesis that skilled fathers have intelligent boys. Use Chi-square test. (Values of x2 for 1-degree of freedom at 5% level is 3.84). Four dice were thrown 112 times and the number of times 1, 3 or 5 were as under : Number of dice showing 1,3, or 5: 0 1 2 3 4 Frequency: 10 25 40 30 7 (Table value of chi-square at 5% level for 4 d.f. is 9.488). To test the efficiency of a new drug a controlled experiment was conducted where in 300 patients were administered the new drug and 200 other patients were not given the drug. The patients were monitored and results were obtained as follows : Total Condition No Cured effect worsens 60 300 Given the drug: 200 40 200 50 Not given the drug: 120 30 500 no Total 320 70 Use x2 test for finding the effect of the drug. (x0o$, 2 = 5.99) The following table gives the classification of 100 workers according to sex and the nature of work. Test whether the nature of work is independent of the sex of the work. Skilled Unskilled Male 40 20 Female 10 30 (Use xi.05.4 * 3.84) There is a general belief that high income families send their children to public schools and low income families send their children to Government schools. For this 1000 families were selected in a city and the following results were obtained. Income Public Schools Govt Schools Total Low 100 200 300 High 500 200 700 Total 600 400 1000 Use Chi-square test to determine whether income level and the type of schooling were associated. (Xoos. 1 = 3.84).

S ta tistics J 5 2 5

8.

Calculate chi-square for the follow ing data :

Class :

9.

10.

11.

12.

13.

14.

15.

A B C D Total 8 29 44 15 4 10.0 Jn : 7 24 38 24 7 100 f' Calculate the value of t in the case of two characters A and B whose corresponding values are given below: A 16 10 8 9 9 8 B 8 4 5 9 12 4 The table below are for protein tests of the same variety of wheat grown in two districts. The average in District I is 12.75 and in District II is 13.03. Calcualte / for testing the significance between the means of the two districts : Protein results District 1 12.6 13.4 11.9 12.8 13 District II 13.1 13.4 12.8 13.5 13.3 12.7 12.4 [Given that /oos. to ~ 2.228] Ten cartons are taken at random from an automatic filling machine. The mean net weight of the 10 cartons is 11.8 kg and standard deviation is 0.15 kg. Does the sample mean differ significantly from the intented weight of 12 kg ? (Given that for /0.05, - 2.26). A fertiliser mixing machine is set to give i 2 kg ofnitratefor every quintal bag of fertiliser. Ten, 100 kg bags are examined. The percentages of nitrate are as follows : II, 14, 13, 12. 13. 12, 13, 14, II, 12. Is there reason to believe that the machine is defective ? Value of t for 9 degrees of freedom is 2.2621. Ten students are selected at random from a college and their heights are found to be 100, 104, 108,110,118,120,122,12-4,126 and 128 cms. In the height of these data, discuss the suggestion that the mean height of the students of the college is 110 cms. (Given /005.9 = 2.26) Talcum powder is packed into tins by a machine. A random sample of 11 tins is drawn and their contents are found to weight in kgs as follows : 0.44. 0.51, 0.49, 0.52, 0.45, 0.48, 0.46, 0.45, 0.47, 0.45 and 0.47 Test if the avarage packing can be taken to be 0.5 kgs. (/0 05. 10 = 2.28) Samples of the types of electric bulbs were tested for the length of and the following data were obtained Type I Type II Sample No. 8 7 Sample Mean 1234 hours 1036 hours Sample S.D. 36 hours 40 hours In the difference in the means sufficient to warrant an inference that type I is superior to type II regarding the length of life ? (Given /0.<)5,15 = 2.131)

1 2 6 | E n @m beh in g M athematics-111

16. The standard deviations calculated to n two random samples of size 9 and 13 are 2 and 1.9 respectively. May the sample be regarded as drawn from the nonnal populations with the same standard deviation. (Given Aw(s. «> * 3.51) 17. TWo sample are drawn from two normal populations. From die followong data test whether the two samples have the same variance at 5% level: Sample I 60 «S 71 74 76 82 85 87 Sample II 61 66 67 85 78 63 85 86 88 91 (Given f 9.9s (9. 7) * 3.68) 18. Following results were obtained from two samples, each drawn from two different populations A and B . Population A B Sample I II Sample Size n\ «* 25 rt2 = 17 Sampled s.
Opinion on autonomous status and level of graduation are independent Computed %2 = 1.622, and Drug is not better than sugar pills x2- 0-^8, Skilled fathers have intelligent boys. All the four dice are fair. (%2 * 1.845)

5.

6. 7.

x2 ~ 2.434, Drug is not effective. x2 ~ 16.666, The nature of work does not seem to be independent of the sex of the worker. The level and type of schooling are associated.

7. 9.

x2 = 6.77 tA= 1.66;/a = 0.85

S ta ti s tic s | 5 2 7

10.

/ = 0.85, No difference.

11.

t - 4; the sample mean differs significantly.

12.

t = 1.464, there is no reason to belive that the machine is defective.

13.

t - 1.937; the mean height of the students can be taken as 110 cm.

14.

/ = - 3.43 i.e., | f | = 3.43; No

15. / = 9.35, Type I bulb is superior to type II. 16. F ~ 1.15; the samples can be regarded as drawn from the population with the same standard deviation. 17.

F= 1.468, the samples have the same variance.

18. F - 2.205; variance of brand A is not more than the variance of brand B. 19. F = 1.415, the two populations have the same variance. 20.

z

= 0.086. □

5 2 6 | E ng in eerin g M a ih e m a iic s -III

TABLE AREA OF A STAMJAMBNORMAL DISTRIBUTION As entry in At total*Is Dmpmpmrtkm m&r «bt /I ottav am i wfelckii brtwn t ■ I ml t piMhc / I valmoft Aw wfcwrftiw nMnined / I bytym etry. z

.00

.01

... — .03 .02

0.0 0.1 02 03 0.4

.0000 .0398 .0793 .1179 .1554

,0040 .0438 .8832 .1217 1591

.0000 .0478 .0871 .1255 .1628

0.5 06 0.7 0.8 0.9

.1915 .2257 .2580 .2881 .3159

.1950 .2291 .2611 .2910 .3186

1.0 1.1 1.2 1.3 1.4

.3413 .3643 .3849 .4032 .4192

1.5 1.6 1.7 1.8 1.9

—^ -----£ .04

.05

.06

.07

.08

.09

.0120 .0517 .0910 .1293 .1664

,0160 .0557 .0948 .1331 .1700

J0199 .0596 .0987 .1368 .1736

.0239 .0636 .1026 .1406 .1772

.0279 .0675 .1064 .1443 .1808

.0319 .0714 .1103 .1480 .1844

.0359 .0753 .1141 .1517 .1879

.1985 .2324 .2642 .2939 .3212

.2019 .2357 .2673 .2967 .3238

.2054 .2389 .2703 .2995 .3264

.2088 .2422 .2734 .3023 .3289

.2123 .2454 .2764 .305! .3315

.2157 .2486 .2794 .3078 .3340

.2190 .2517 .2823 .3106 .3365

.2224 .2549 .2852 .3133 .3389

.3438 .3665 .3869 .4049 .4207

.3461 .3686 .3888 .4066 .4222

.3485 .3708 .3907 .4082 .4236

.3508 .3729 .3925 .4099 .4251

.3531 .3749 .3944 .4115 .4265

.3554 .3770 .3962 .4131 .4279

.3577 .3790 .3980 .4147 .4292

.3599 .3810 .3997 .4162 .4306

.3621 .3830 .4015 .4177 .4319

.4332 .4452 .4554 ..4641 .4713

.4345 .4463 .4564 .4649 .4719

.4357 .4474 .4573 .4656 .4726

.4370 .4484 .4582 .4664 .4732

.4382 .4495 .459! .4671 .4738

.4394 .4505 .4599 .4678 .4744

.4406 .4415 .4608 .4686 .4750

.4418 .4525 .4616 .4693 .4756

.4429 .4535 .4625 .4699 .4761

.4441 .4545 .4633 .4706 .4767

2.0 2.1 2.2 23 2.4

.4772 482! .4861 .4893 .4918

.4778 .4826 .4864 .4896 .4920

.4783 .4830 .4868 .4898 .4922

.4788 .4834 .4871 .4901 4925

.4793 .4838 .4875 .4904 .4927

.4798 .4842 .4878 .4906 .4929

.4803 .4846 .4841 .4909 .4931

.4808 .4850 .4884 .4911 .4932

.4812 .4854 .4887 .4913 .4934

.4817 .4857 .4890 .4916 .4936

Z5 2.6 2.7 Z8 19

.4938 .4953 .4965 .4974 .4981

.4940 .4955 .4966 .4975 .4982

.4941 .4956 .4967 .4976 .4982

.4943 .4957 .4968 .4977 .4983

.4945 .4959 .4969 .4977 .4984

.4946 .4960 .4970 .4978 .4984

.4948 .4961 .497! .4979 .4985

.4949 .4962 4972 .4979 .4985

4951 4963 4973 .4980 .4986

4952 .4964 .4974 .4981 498fc

3.0

.4987

\

4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 >4990

E xam . P a per s | i

a e . (Third Stouter) EnMinttM, Ji m , 2008 eacft UniftfrooMlpidsory. AH questions ifnifrl is aiMmatytic function and find its derivative. Ans. (Refer Page 22) f(°)

=

Aw. (Refer Page 59) C

Or (a) Evaluate the following integral using Cauchy’s integral formula; 4-3z dz z(z l)(z 2) c where C is the circle jzj = 3/2.

I

(b) Prove that:

l*2x

sio 2 0

2 ll

i «

. #j

= p r(a-V j -b ) where 0 < 6 < a.

Ans. (Refer Page 71) Ans. (Refer 93)

A2 , . A2sin(x+ h) ~£ + )+ Esin(x+ h) Ans. (Refer Page 202for a Similar Pmtetm) (b) Use Bessel’s formula to find the value ofy if x - 3.75, given: (a) Evaluate:

x y 25 24.145 3.0 22.043 33 20.225’ 4.0 11644 45 17.262 5.0 * 16J047; Ans. (Refer Page 239for a Similar Problem) Or (a) Find the first, second and third derivatives of the function tabulated below, at the pintx= 1.5: X is 2D 25 3.0 35 4.0

m 3375 7-0 13.625 24.0 38*75 59.0

Am. (Refer Page 260) (b) A body is in the form of a solid revolution. The diameter D in cm of the sections at distances x cm from one end are given below. Estimate the volume of the solid:

%|

E N G J^& O N G M A neM /m cs-U l

X..-. . D i t * . f i * i V/fi: *»*»*$,' v ~ * H* .v . • .-i< bni s VJ im SjO 6.0 s 'l ■’ ■ « .%« V. v •W *“ ■■ 8.75*' sir-1..- ‘ C * 10.0 625 55 12.5 15.0 4.0 nm> r *w jvt maiiwnw rrwmiy (a) By using Newton-Ramphson method, find the root o f x ^ x - 9 * 0 , which is near tox~ 2 correct to three places of decimal. Aas. (Rrfer P*$t 143) 2Jx+ 6 y - z m 85 6x + 15y + 2r - 72 x'+ < 0 )= i AM.(g^er (b) Using Runge-Kutta method o f fourth order, solve

dy dx

v.

50/ jbr a Similar Problem)

y^ —X* ---- r with v(0)" 1 atx “ 0.2, 0.4. y*+x Am. (Xtfer Page 329)

Unit-IV

4.

(a) Find the rank o f:

' 6 4 10 16

1 2 3 4

3 8 % -1 9 7 12 15

(b) Show that the set S ® {(1,2, IX (2,1, OX (1, - 1,2)} forms basis of V^JX). Or (a) Show that the mapping/: V3 (K) -> defined as below /(a , A, c) ■ (c,
Exam. P apeib | 3

5.

[_• Unit-V (a) Prove that every Hermitian matrix # ean be represented in the form P+IQ, where P and Q are real symmetric and skew-symmetric metric? and phow th at: H * H is real if PQ “ - QP, t* 1 -1 2 -1 (b) Using Cayley-Hamilton theorem find the inverse 6f the matrix: A 1 -1 2 Or 3 -I I - 1 5 -1 to diagonal form. Hence (a) Find the matrix P which transforms the matrix A 1 -1 3 calculate if1. (b) Reduce foeanonical form and find the rank imd the signatwe o f the quadratic form : q ■ 1x2 -2 > ,2-3 * 2 - 4 yz+ 6zx

B.E. (Third Semester) EiM riM tjuit D c^ 20N (N) Mathematics-Ill . Note : The question paper ,is divided into five tmits. Each unit carries an internal choice. At­ tempt one question fiom each unit All questions carry equal marks. Uaift'I

I.

(a) Find the imaginary part oftheanalytic function/OO whose real part is x?-Ixy* +3x2 -3 y 2. Ans. (R tfer Page 2 1 ) *2+/ (b) Evaluate J ^ (2 jc+ + 1)<* along the two paths: . CO * ■ /+ U y - 2 f i - 1.

2.

(ii) The straight line joining ( 1 - 0 and (2 + /). A ns. (R efer Pmge 56 ) Or

1 1 (a) Find the image of the infinite strip ^
3.

Ans. (Rtfer Page 42)

f2*— ^ — * ■r ?mmmta > b > 0 Ans. (Refer Page 81) fo r a sim ilarproblem . Jo a + b cos0 Va2 -&2 Uatt-II (a) Use Newton’s divided difference formula to find a polynomial hi jc, given: jc 0 2 3 6 f(x) 648 704 729 792 Am*. (R efer Page 251) (b) Find the approximate value o f log. 5 by caicu&ing to four decimal places using Simpson's l/3rd rule for

f* dx

dividing the range into 10 equal parts. Or

Aas. (R efer Page 276)

4 | ENG1NEEHB4G MATHEMATfCS-lII

4.

(a) (i) Prove that:

( #

H

?

A ns, (R efer Page 204)

:

00 Estimate the missing termin foe following table: x: 0 1 2 3 4 /(x ). 1 3 9 ■ 81, Asa (Refer Page 213) (b) Find the first and second derivatives of the ahead function tabulated at the point x * 1 : x: ■ 1 3 5 7 9 f ( x ) : 85.3 74.5 67.0 60.5 54.3 Ans. (Refer Page 260) for a simalar problem. Unit-Ill 5.

(a) Use Newton-Raphson method to find a root Of die equation co sx -x 2 = x correct to four decimal places. Aas. (Refer Page 141)for a similar problem. (b) Use Runge-Kutta method to approximate the value ofy when x"m0.1 given that y (0) = and gfg 3*+ y

Or ^

Ans. (Refer Page 322)

(a) Solve the following system of equations by Jocobi’s iterative method correct to three decimal places: 10x-5>>-2z = 3 Ax - IO y + 3z = -3 x+6.y + IG* = - 3 Ans. (Refer Page 173) (b) Using Euler's method solve the differential equation for y at x = 1 in five steps: dy “£=jc2 + y2, j<0) = 1 Ans. (Refer Page 312)for a similar problem. 7.

Untt-IV (a) Apply Simplex method for solving the folk) ving L.P.P.: Maximize: z = 7x, + 5x2 Subject to :

x, + 2x2 56 4jc, + 3x 2 5 1 2

x ,,x2,£ 0 (b) Solve the following transportation problem: 01 02 Oi . Demand 8.

A 5 10 7 8

ih 3 7 5 12

JH 6 12 8 13

D< 5 4 -4 6

Supply 15 11 13

(a) Hie following table shows the efficiency of each operator handling different machines in terms of profit:

J E xam . P apeh » I 5

-* Machines Operators 4 I M hi

IV

9.

Mi

m2

M i-

m4

47 38 32 46

32 48 42 40

46 40 36 42

44 38 30 1 46

(b) Find the dua] of the following L.P.P: Minimize: z * x, + x2 + x3 Subject to; xt - 3x2 + 4x3 = i Xj - 2x2 £ 3 2x2-x3i 4 Xj.Xj ^O x3 is unrestricted in sign. »»it-V (a) Obtain the steady stale equations for the queueing model (M/M/1): (oo/FCFS). (b) Find the transition matrix corresponding to the following transition diagram:

Or (a) One type of aircraft is fqund to develop engine trouble in 5 flights out of a total of 100 and another type in 7 flights out of a total 200 flights. Is there a significant difference in the two types of aircrafts so far as engine defects are concerned ? (b) Write short notes on the following: (i) Sampling (li) Randomsmapling (iii) Factorial design (iv) Tfegpchi loss function B.E. (Third Semester) Examination, Jine, 2009 Matkematics-ni Note: Attempt all questions. All questions carry equal marks. One complete question solve at one place. 1. (a) Show that the Polar form of Cauchy-Reimann equations are : du 1 dv dv 1 du 10.

61 EwawcMBwam nK M tm xm ?r ' toVtfrr

+A~ -du ~+ $0? frl 58$

«■'' ‘.'-.-ir- • 1i A‘J

421 +z + 5 ^ ----------- - (
co f(35>

■;

m

Abs. (ReferPage 63) Or

(a) Expand:

m

1 (* - !)(*^2) ^ * e ,^ on W> 2.

A it (Refer Page 101)

cos26d6 2jw 2 f 2*_________________ j-(oJ < 1) Am. (Refer Page 89) *> l-2acos0+ a2 I (a) Show that the nth difference ofa polynomial of degrees will be constant and all (»+ l)thand higher ofder difference arc low. A*%. (Refer Page 200) lx= 1.1:

(b) Show that:

2.

X .

fix)

1J5 "'

6 0.128 0544 L296 2432 4jOOO

' 12

U 1j6

U 2J0

Ans. (Refer Page 266)

Or (a) Find;

3.

r

e*dx

by taking seven ordinates using Simpsoffe 1/3 mle. Ana. (Refer Page 285) (b) Define forward, backward, central difference operator and shift average operator also. (a) Solve the following by Euler modified method: ^ = log(x+y), y(0) * 2 at x » 1.4 with h « 0.2. A n. (Refer Page 317) (b) Find the roots of equation xe* = eosx using Regula-Falsi method, correct to four decimal places. Or Abs. (Refer Page 131) (a) Solve by Gauss-Seidel method: lOx, -2x2 -* j -x 4 * 3 -2xj + I0x2 - x 3~x4 «15 -x, - x 2+ 10*3 -24x4 » 27 -x, - x 2 -2xj + 10x4 m-9

Ana. (R efer Page 182)

V- v

B cmi. P apgbs { 7

(b) Giventhe values of u (x, y ) on the boundary'of the square in bloow. Evaluate foe function ‘ ‘‘i (x,$ safl&fymg foe Laplace equation ^ *0 aMhe Pivotal poWofthis figure. i© i

1000

1000. 1000

aooo

500

aooo

behB the lower

combined). Tbe belt A requires* fimcy buckle and only 400 per day are available. There are 700 buddes a day available for belt B. Deferable foe optimal product mix by graphical method, (b) Solve the following transportation problem: Destination Di Supply Da D* Oi 19 7 50 10 30 Sy Source 30 40 70 9 60 S2 70 20 40 8 18 S, Demand

5

8

14

7 Or

(a) Fourjobs are to be done on four different machines The cost (in rupees) o f producing fth job on the Jth machine is given below:

Machine Mx M2 M3 M4 A Job Ji A Ja

15 17 14 16

U 12 15 13

13 12 10 II

15 13 14 17

Assign the jobs to different machine so as to minimize the total cost

| E n g in eerin g M m h e m a iic s -ID

(b) Solve by Simplex method: Minimize: x f -3 u c j + 2 x 3

Subject to:

*rti .

3jcj ~ x 2 + 3x3 5 7 -2 x j + 4x2 512 ~4X| •+3x2 + 510 X|, Xj, x3 £0

(a) In a railway marshalling yard goods train arrive ata rate of 30 tram per day. If the distribution of arrivals is the Poisson and that of service time is exponential with an average 36 minutes,1 then find: (i) Mean queue size. (ii) The probability the queue size exceeds. If the input of trains increases to an average 33 per day what will be the change in (i) and (ii) ? (b) The state transition matrix’for retentions gains and -losses of firms A, B and C is given below. Using this matrix determine the steady state equilibrium condition: . Form To A B C A 0.700 0.100 0.200 B 0.100 0.800 0.100 C 0.200 o.ioo 0.700 ! Or (a) Write the advantage and disadvantage of Robust design method. (b) The number of units of an item that are withdrawn from inventory on a day to day basis is a Markov Chain Process in which requirement for tomorrow depend on today’s requirement A one day transition matrix is jpven below. Number of units withdrawn from inventory : Tomorrow 5 10 12 5 0.6 0.4 0.0 Today 10 03 03 0.4 12 0.1 03 0.6 (i) Construct a tree diagram showing inventory requirements on two consecutive days. (ii) Develop a two day transition matrix.

E xam. P a fb k } 9

B.E* (Third Semester) Fiawlasrton, Feb, 2010 Matfceraatks-IU Note : Attempt five questions in all selecting one question from each unit All questions cany equal marks.

U nft-I J.

(a) Show that the function e* (x cos y - y siny) is harmonic and find its conjugate. Abs. (Refer Rage 24) 2:

-

•

where C is the circle Iz 1*3. (b) Evaluate f ----*c ( z - lX * -2 ) 2.

Abs. (Refkr Page 76)

Or (a) Find the bilinear transformation which maps the joints * = l , i , - 1 onto points w * 4 0, - i. 1 Abs. (Refer Page 25) r2/% COS20dB (b) Evaluate ! ~—r----- r— r* (tf2< t.) Jo l-2tfcos6+
Abs.

Page 89)

U nfc-II 3.

(a) (i) Prove that:

Abs. (Refer Page 204) (ii) Obtain the function whose first difference is 2X3 + 3x* - 5x + 4.Abs. (Refer Page 211) dy (b) Find ~ at x * 1.1 from the following table:

X

’ TF 12 1.4 1.6 1* 2.0

4.

y 0

0.128 0544 1296 2.432 ' 4.00

Or (a) Find the cubic polynomial which takes the following values x : 0 1 . . 2 3 f(x) : 1 2 1 10 Hence or otherwise evaluate/(4).

Abs. (Refer Page 266)

A bs. (R efer Page 223)

1 0 1EmMHOHQKwMAncs-ill (b) The following tabfe gives the vetority vofapartjcleattinK;;;

-

r

2 4 6 8 10 12

! v (m W .i v " ■' . *■' ' *6 16 34 60 ' 94 136

Find the distance moved by the particle in 12 seconds and also the acceleration at/-»2 second. Ass. (Refitr Page 28$)

Unit-III 5

6.

(a) Find a real root of the equation jc log* x ~ 1.2, correct to five decimal places by NewtonRaphson method. Ans. (Refer Page 146) (b)Sotvetheequdiom: Sx+2 y + z * 12 x + 4 y + 2 z * 15 x + 2 ^ + 5 z * 20 by Qmm Saidel method. Ans. (Refer Page 185) Or (a) Solve the equations: 3x + y + z » 4 x+2y+2z * 3 * • 2x+y+3z * 4 byCroutVmethod (b) Apply Runge-Kutta method to find an approximate value of >»for x * 0.2 in steps of 0.1, if *dy — “ X+^.^CO)* 1.

Ans.(ReferPage321) U r it-IV

7.

(a) Solve the following L.P.P., using Simplex method: Maximize z*5xj + 3xj Subject to : x\ +x2 £ 2 5xi + 2x2 5 10 3xi + 8x2 5 12 xi,x2 it 0

11

*’*L

(b) Solve the foi O m s u i ^ ^ ..A” ‘ I 11 ni

6 4 I

a 9 2

4 8 6

14 12 5

Required

6

10

15

31

,

8

Available

* ir

i

*' v

(a) Using duality solve die following problem: Maximize: •z **0-7 x, + 0.5 x 2 Subject to : x i ^ 4 , x2 ^ 6 . xj + 2xi * 20 2&1+X2 £ 1 8

.

*>' >■ ■

.

X |,X 2 * 0

(b) A firm plans to begin production of three new products on its. three ptftst*. H eiptt cost of producing / at plantj is as given below. Find the assignment that Minimises the total unit cost: Plant ’ Product

1 2 3

1 10 18 6

2 8 6 4

* 3 12 14 2

U n it -V

9.

10.

(a) Obtain the steady state equations for the (M j M11) : (oo /FCFS) qticuing model. (b) Write sfaoit notes on the following: (i) Taguchi toss function (K) Markovian process Or (a) There is congestion on the platform of railway station. The trains arrives at the rate of 30 trains per day. The waiting time for any train to hump is exponentially distributed with an average of 36 minutes. Calculate the mean queue size and the probability that queue size exceeds 9. (b) Write short notes on the following: 0) Factorial design (ii) Robust design method.

1 2 | E n q b « e8 b « 3 M a th e m a tic s-!!!

B.E. (Third Semester) E uniaitkm , J u e , 201! Mathemattes-III Note: Attempt all questions by selecting any two part*\fi*« e*ch Questions. All questions carry equal marks. (a) Find the imaginaiy part of the analytic function who^e/eal part is : 1. Ans. (Refer Page 21) x 3 -3 ^ -3 * a -3 ^ (b) Evaluate: f _ £2z... dz k < * + l)4 Ans. (X ^ r

where C is the circle 12 j * 3. (c) Show that: r*«__£ *0 o 2 + cos0 (d) Show that the transformation co “

.

2

2it ^

'

6/)

Ans. (Refer Page 81)

2*+3 ^ changes the circle x1 +>*-4z * 0 into straight line Ans. (ffgfer

4* + 3 ~ 0, (a) Prove that: (i) e * > » l + A

5??

Ans. (Refer Page 198) Ans. (Rtfer Page 204)

09 e* (b) Find/(9 ) from the following table: x 5 7 11 13 17

/<*) 150 392 1452 2366 5202

Ans. (Rtfer Page

>

(c) Find — atx * 1.5 from the following table: x 1.5 2.0 2.5 3.0 15 4.0

y 3.375 7.0 13.625 24.0 38.875 59.0

A bs. (R efer Page 260)

• A EXAK. PAFtMfi | IS (d) Evaluate:

"V m/2 .

^ *'* *’*' -*" viV/

fU

: . . t,

... \ . .

,.*■

(i) Rising Simpsons - ~ rule.

i: . i

u,

, • • . .

(ii) Using Weddle’s rule. Aim*.(Refer Page 277) (a) Using Newton-Raphson method find the real root of the equation 3x * cos x + 1 correct to five decimal places. Abs. (Refer Page 141) (b) Solve the following system of Aquations using Gauss-Seidel method : 27x + 6 y -* * 85 6x + \Sy + 2z “ 72 x + y + 54z « 110 Ans.(ReferPage 179) dy ‘ . (c) Findy (2.2) using Euler’s method from the equation xy*9y (2) * 1. Ass. (Refer Page 313) (d) Apply Runge-Kutta method to find an approximate value ofy for x » 0.2 in steps of 0.1 if dy

-r£ pcx+^a, given thaty* 1 whenx~0.

Ans. (Refer Page 321)

(a) Solve the following L.P.P. by Simplex method: Max: z - 3r|+2r2 Subject to: X| + Xj £ 4

x j-x j £i 2 X |,X 2 fc'O

(b) Solve the following L.P.P. by Graphical method: Max.: z * 5xi + 7xj Subject to constraints: xj + x2 S 4 3xt + 8x2 £ 24 10bC| + 7x2 £ 35 xn x2 * 0 (c) Find the initial basic feasible solution of the following transportation problem by Vogel’S approximation method: Wvehouse

Factory

W, F, 19 F2 70 F, * 40. 5

w, 30 30 8 8

w, 50 40 70 7

w4 10 60 20 14

7 9 18 34

minutes. If be mpaks scts in oif4fryta which ti&y come maad if ike arrival of sets Is approxi­ mately Potoon within averagfrfjats of 10 per 8 hours perday, what is repairman's expected idfe time each day? How manyjobs are ahead of the average set just brought in j — (c) Write short note on the foUowbgg: N^ toetioii R o^dw gnoK ttxxL ?f, U t (Third Stme&er) fcuwhurtlee, Dee, 2011 ii* fiom each Unit All question carry equal marks! (Jlitt-I (a) Shew that the frootiong^x3-la y 1 is harmonic andlind the corresponding analytic function afthisas the*Wpait ( Ass. (Refer Page 20) fh) jjl^ lTed flflrivn ^airTij theorem. Aaa. (3Iq^tr A ge 5*> Or (a) Find the b l t & i e a ^ j s u q a s U iv*-1 to /, 0, - / respectively. > 1 Abs. (Ifgfcr Page 35) *

A

V

f oosxr « integral formula, evaluate the integral1 L 7' ’n ;r .’n where Cis the circle .*; • . • • • ' . ; ’■.* . -iy ;vv’■' U"■ ‘VV Ans. (K g tr/to ? * 77) UttU-H (a) Prove with usual notations that: AD " log (1 + A) * - tog (1 - V) (b) Expressy-2X3 - 3X2 + 3Lc- 10 in fectoriri notation form. Or (a)G|ventiiat: x y . 1.0 7.989 1.1 8.403 8.?8l U ■/* 1.3 9.129 9.451 1.4 F.5 9.750 1.6 10.031

Find ffe dx

1j6»

Aaa-fref&Fage Iff) Abs. (Refer-Page 210)

A bs. (R tfe r Page 242)

*?«■

(b f

m *J»d*> fSttd 4 # k fa e N lfJr

“r; .v.'.'f.'.>'J £* o*. >:$v,,.w:. y. ' ijuT •■ ■•'?* •"A/ -c. HJt'". if '■** ■’ f2 ** •*•*'• V* ,r>' ' r .1;ruc •; *••: v -^ V 13 ‘ 9 14 11 16 U n lt-m •^v (a) Ffcid a nal coot of fee equation ** - 2x - £ * 0, by ffegnM yai d+cinidptocet. \y ■ (b) Perform ofPkwd’s netted to find aaj^xadflnle value profefem-: * '. - • ' t ■; ' V >. A ■• *|V V: i ., : J -* ♦ * > < » -* • « , ' *

'

S N ljfce fhw»

• ■*> *‘'f,^‘ . **,v,~ . •■»*

;

efft* M ia

Am * Or (a) Solve thesystem ofequations by Qqm*G*toi ftyeffooafeod: . i . ., 27* + 6 > - z » | 5 n-2z - 72 .?' > + * + 54* « 110, ,, . ^ (b) By using NeftftteRaphson mated, find* rpatf** of*ie S ^ p t t a i i ^ w f e f e f c li. n^ar to * » 2#Mitect to three deotal places, " IMIttMdW W M V (a) Find th v M v f ftr<owing L.P.P.: Mf*%jr+zx - 3 y + 4z *>.5 '• .. x + 2y 5 . 3 2x-ii4 x,^,z £ 0 (b) Four jobs are to be demon four (Sfiefeat xaacktnes. The cast 0n rupees) oCfKofecmg Alt job toyth machine is gh*a below: V m kim — ►

M i

*!

i

ii

u

IS

in

41 13

IV

*2

30 33 25 30 —------

•

if # 2i * 24

f 'H f

-

^ ------

31 21

If..

1

r 'W f T - ' ■Stauncei st S2 Si Demand

" A

*>2

*3

o4

19 70 40 5

30 30 8 8 ,

50 40 70 7

10 60 20 14

7 9 18 34

(b) Solve the following L.P.P. by Graphical method : Maximize : z = 3x + 2y Subject to: 2x+y £ 5 x+y £ 3 x,y * 0

U iiit-V (a) Write advantage and disadvantage of Robust design method (b) in a railway marshalling yard goods train arrive at a rate of 30 trains per day. If the distribute of arrivals is Poisson and that of service time is exponential with an average 36 minutes, the find: (i) Mean queue size (ii) The probability that queue size exceeds. Or (a) Describe the differentitiMifference equations for the queuing model (M/M/1) : (oo/qo/FCFS (b) Write short notes on the following: (i) Taguchi loss function (ii) Factorial design.

The subject Engineering Mathematics is the backbone of a degree in any field of engineering and occupies a prom inent place in the schemes and syllabi of technical institutes and universities. While fram ing the content and m atter of the book special attention has been paid to the em erging demand of engineering students in the field of mathematics. The book in its present form covers topics and examples,, both solved and unsolved exercises, useful for students’ practice and explanatory purposes. KEY FEATURES OF THE BO O K

An atte m p t has been made to assure th a t the book caters the need of engineering students. Worked examples and graded exercises are used throughout to develop ideas and concepts. The form at of this book makes it an easy companion fo r the basis fo r a complete m odular course in Engineering Mathematics. The book covers following topics for the second year engineering student

• • • • • •

Functions of Complex Variables Numerical Analysis Interpolation Correlation and Regression Theoretical Distribution Testing of Hypothesis

Dr. D.K. Jain did his M.Sc. in Mathematics in the year 1991 from Jiwaji University, Gwalior. He has obtained M.Phil. Degree in the year 1992. He has also cleared the UGC JRF-NET Exam for Lecturership and was awarded Junior Research Fellowship for his Ph.D. degree from Jiwaji University in the year 1997 on Special Function and its Applications. He has published several research papers in national and international Journals and anthologies besides having authored a full series of books for the Engineering students on engineering mathematics, useful for the Undergraduate as well Post-graduate students. He has a Post-graduate teaching experience of over Sixteen years and is presently teaching in the Department of Applied Mathematics at Madhav Institute of Technology and Science, Gwalior, a renowned technical institute of Madhya Pradesh.