EXPANSION OF FUNCTIONS OF
ONE & SEVERAL VARIABLES
OBJECTIVES
At the end of this session, you will be b e able to understand:
Maclaurin’s Theorem
Some Important Expression
Taylor’s Theorems
Function of Two Variables
Total Differential Coefficients
MACLAURIN’S THEOREM: If f(x) can be expanded in ascending powers of x, then
f( x) = f(0) + xf'(0) +
2
x
2!
f''(0) +
3
x
3!
f'''(0) + ...... +
n
x
n!
fn (0 ) + .......
Proof .
Suppose
f ( x) = a0 + a1 x + a2 x2 + a3 x3 + ........................ an xn + ................ (1)
........................an are constant to be evaluated. Where a0 , a1 , a2 , a3 , .. By successive differentiation (1) w.r.t. x, we get
f '( x) = a1 + 2 a2 x + 3 a3 x2 + 4 a4 x3 .......................nan xn −1 +
.......... ...... ( 2 )
f ''( x) = 2 a2 + 3.2 a3 x + 4.3 a4 x2 .......................n( n − 1) an xn −2 + ...........(3) f '''( x) = 3.2 a3 + 4.3.2 a4 x....................... + n( n − 1)(n − 2)an x n −3 + .......( 4) In general, f''( )x = n ( n− 1)( n− 2)............3.2.1 na + te term
containing
x
...........(5)
Now putting x = 0 in (1) to (5), we get a0 = f (0); a1 = f '(0); a 2 =
''(f0) 2!
; a3 =
'''f(0) 3!
n
; .........; an =
f(0)
n!
, ... ....
Putting these values of constants in (1), we get 2
f( x) = f(0) + xf'(0) +
3
x
f''(0) +
2!
x
3!
f'''(0) + ...... +
n
x
n!
fn (0 ) + .......
If we take f(x) = y; f(0) = (y)0; f’(0) = (y 1)0, f”(0) = (y2)0, f’’’(0) = (y 3)0;……..f’’(0) = (yn)0, then the above theorem takes the form as x 2 x 3 x n y y y = (y)0 + x (y1)0 + ( 2 )0 + ( 3 ) 0 + ......... + ( y n ) 0 + ...... n! 2! 3! or
y = (y)0 + xy1(0) +
x 2 2!
y 2 (0) +
x 3 3!
y 3 (0) + ......... +
x n n!
y n (0) + ......
SOME IMPORTANT EXPANSIONS: x
1. Expansion of e . (Exponential series):
Let Let f(x) f(x) = e x Then Then f(0) f(0) = e0 = 1;
f(
n)
( x) =
e x so that f(
n)
( 0) =
e0 = 1, where n = 1, 2, 3, ............
Substituting these values in Maclaurin’s series
f( x) = f(0) + xf'(0) +
x
We get e = 1 + x +
x 2 2!
+
x 2 2!
x 3 3!
f ''(0) +
+ ...... +
x n n!
x3 3!
f'''(0 ) + .......
+ .....
2. Expansion of sin x. (Sine series):
Let f(x) = sinx. Then f(0) = 0, f ’(x) = cosx so that f ’(0) = 1, f ’’(x) = sinx so that f ’’(0) = 0, f ’’’(x) = -cosx so that f ’’’(0) = -1, and so on.
In general, f (x) = sin x+ n
n
1 n nπ so that f (x) = sin nπ 2 2
1
When n = 2m, f (0) = sin mπ m π = 0 and when n = 2m + 1,
1 1 1 n f (0) = sin (2m + 1)π = sin mπ + π = (−1) m sin π = ( −1) m 2 2 2 Substituting these values in Maclaurin’s series, we ge t Sinx = 0 + x.1 + 0 +
or
Sinx = x -
x 3
+
3!
x 3
(−1) + 0 + ...... + 0 + (−1)
3
x 5
− ...... + (−1)
5!
x 2 m+1
m
(2m + 1)!
x 2 m +1
m
(2m + 1)!
+ ......
+ ......
Similarly, we may obtain the Cosine series: Cosx = 1 -
x 2 2!
+
x 4 4!
−
x 5 6!
+ ....... + (−1)
m
x 2 m (2m)!
+ .....
3. Expansion of log (1+x):
Let f (x) = log (1+x). Then f (0) = log 1 = 0; n
f (x) =
(−1) n −1 (n − 1)! ( x + 1) n n
n-1
So that f (0) = (-1) (n-1)!, where n = 1, 2, 3…..
∴ f ’(0) = (-1)1-1 (1-1)! = 1, 2-1
f ’’(0) = (-1) (2-1)! = -1!, 3-1
f ’’’(0) = (-1) (3-1)! = 3!, iv
4-1
f (0)= (-1) (4-1)! = 3!, and so on. Substituting the values of f (0), f ’(0),f ’’(0), etc. in Mac laurin’s series, we have f (x) = f(0) + x f ’(0) +
x 2 2!
f ' ' (0) + ..... +
x n n!
f n (0) + ....,
x 2 x 3 x 4 x n we get log (1 + x) = 0 + x.1 - .1!+ .2!− .3!+..... + (−1) n−1 (n − 1)!+ ..... 2! 3! 4! n! or log (1 + x) = x -
x 2 2
+
x 3 3
−
x 4 4
+ ...... + (−1) n −1
x n n
+ .....
n
4. Expansion of (1 + x) . (Binomial series): n
Let f(x) = (1 + x ) . Then f (0) = 1; m
f (x) = n(n – 1)(n – 2)….(n – m + 1) (1 + x)
n-m
m
so that f (0) = n(n - 1)….(n - m + 1), where m = 1, 2, 3, ……
∴ f’(0) = n, f ’’(0) = n(n – 1), 1 ), f ’’’(0) = n (n-1) (n-2) and so on. Substituting the values of f (0), f ’(0), f ’’’(0) etc. in Maclaurin’s series for (x), we get n
(1+x) =1 + nx +
n(n − 1) 2!
x 2 + ..... +
n(n − 1).....(n − m + 1) x m + m!
……
Example: Expand the following by Maclaurin’s theorem: (i) tanx (ii) log secx Solution: (i) Let y = tanx. Then (y) 0 = tan0 = 0 2
2
2
y1 = sec x = 1 + tan x = 1 + y so that (y1)0 = 1 + (y)
2 0
= 1 + 0 = 1,
y2 = 2yy1, so that (y2)0 = 2 (y)0 (y1)0 = 2 × 0 × 1 = 0, y3 = 2y1y1 + 2yy2 = 2y1 + 2yy2 so that (y3)0 = 2 × 12 + 0 = 2 , 2
y4 = 4y1y2 + 2y1y2 + 2yy3 = 6 y1y2 + 2yy3 so that (y4)0 = 6 × 1 × 0 + 2 × 0 × 2 = 0, y5 = 6y 2 + 6y1y3 + 2y1y3 + 2yy4 = 6y 2 + 8y1y3 + 2yy4 so that (y5)0 = 0+8 × 1 × 2 + 0 = 16 2
2
and so on. Now by Maclaurin’s we get x2 x3 x4 x5 ( y2 )0 + ( y3 )0 + ( y4 )0 + ( y5 ) 0 +................ y = ( y) 0 + x( y1 )0 + 2! 3! 4! 5!
∴ tan x = 0 + x.1 + = x +
x 3 3
+
x2 2! 2
15
.0 +
x3 3!
.2 +
x5 + ........
x4 4!
.0 +
x5 5!
.16 + ...............
(ii)
Let y = log sec x. then (y) 0 = log sec 0 = log 1 = 0,
y=
1
1 sec x
. sec xtanx= tanx so that ( 1y) 0 = 0.
y = sec 2 x= 1 + tan 2 x= x= 1 + 2y1 so that( 2y) 0 = 1 + ( 1y) 02 = 1
2
y3 = 2 y1 y2 so that( y3 ) 0 = 2( y1 )0 ( y2 )0 = 0, y4 = 2 y22 + 2 y1 y3 so that ( y4 ) 0 = 2 × 12 + 0 = 2 y5 = 4 y2 y3 + 2 y2 y3 + 2 y1 y4 = 6 y2 y3 + 2 y1 y4 so that ( y5 ) 0 = 6 × 1× 0 + 2 × 0 × 2 = 0 y6 = 6 y32 + 6 y2 y4 + 2 y2 y4 + 2 y1 y5 = 6 y32 + 8 y2 y4 + 2 y1 y5 so th,a(t
)y = 0 + 8 × 1× 2 + 0 = 16 16,
and so .on
6 0
Now by Maclaurin’s theorem, we get
y = ( y ) 0 + x( y1 ) 0 +
∴
log sec x = 0 + x.0 +
=
x2 2
+
x4 12
+
x 2
( y 2 ) 0 +
2!
x2 2!
.1 +
x6 45
x3 3!
x 3 3!
.0 +
( y 3 ) 0 +
x4 4!
.2 +
x 4 4!
x5 4!
( y 4 ) 0 + ...........
.0 +
x6 4!
.16............
+ ........
Example 2: By Maclaurin’s theorems show that
e cos x = 1 + x − x
2 x3 3!
−
22 x 4 4!
−
22 x5 5!
+
23 x 7 7!
n
+ . .... ... . + 2
2
nπ x n cos . + .. ... ... ... 4 n!
Solution.
Let y= e x cos x. then
( y)0 =
e0 cos 0 = 1
y1 = e cx os x− e sx in x = e (xcos x− sin x), ), so that( y1 )0 = 1, y2 = e (cox s x− sin x) + e (−xsin x− cos x) = −2 e sinx x, so that ( y2 )0 = 0, y3 = −2 e sxin x− 2 e cxos x= −2 e (xsin x+ cos x), ), so that ( y3 )0 = −2, y4 = − 2 e x (sin x+ cos x) − 2e x (c (cos x − sin x) = −4e x cos x = −22 y so that ( y 4 )0 = −2, y5 = −22 y1 , so that ( y5 )0 = −22 , y6 = −22 y2 so that( y6 )0 = 0, y7 = − 22 y3 so tha(t y7 ) 0 = 23 and so on . In general,
ny = (1 + 1)
n
2
cos( x+ ntan −1 1) = ( 2)
n
so that ( yn )0 = ( 2)
2 n
cos( x+ nπ / 4) 2
1 cos( nπ ) 4
Now by Maclaurin’s theorem, we get
y = ( y)0 + x( y1 )0 +
= 1 + x.1 +
x2 2!
x 2
.0 +
2!
( y2 )0 + ............. +
x3 3!
(− 2) +
x4 4!
xn
( −2 ) +
n! x5
2
+
5!
( yn )0 + ................... (− 2 ) +
x6
2
6!
x7
.0 +
7!
2 3 + ... .
1 2n 2 cos π n + ...... n! 4
x n
1 x = 1 + x − − − − + .......2 cos nπ + ........... 3! 4! 4! 7! 4 n! 2 x3
2 2 x4
22 x5
23 x7
n
n
2
TAYLOR’S THEOREM: If f ( x + h ) can be expanded in ascending powers of x, then.
f ( x + h) = f ( x) + hf '( x) +
h2 2!
f "( x) + ..... +
hn n!
f "( x) + .......
Proof.
Let f ( x + h) = A0 + A1 h + A2 h2 + A3 h3 + A4 h4 + .......... + An hn + ..........
......(1)
By succe successi ssive ve differ different entiat iation ion of (1) w.r.t .,., h, we have have f '( x + h) = A1 + 2 A2 h + 3 A3 h 2 + 4 A4 h 4 + ...................
...........(2)
f "( x + h) = 2 A2 + 3.2 A3 h + 4.3 A4 h2 + ...................
...........(3)
f "'( x + h) = 3.2.1. A3 + 4.3.2. A4 h + ........................
............(4)
........................................ ........................................ Putting h = 0 in (1), (2), (3) and (4), we get
f ( x) = A0 : f '( x) = A1 : f ''( x ) = 2 A2 : f '"( x) = 3.2.1. A3
⇒
0
= f ( x); A1 = f '( x); A2 =
1 2!
f "( x); A3 =
1 3!
f "'( x) and so on.
Substituting these values in (1) we get the Taylor’s theorem as
f ( x + h) = f ( x ) + hf ' ( x ) +
h2 2!
f " ( x ) + ........... +
hn n!
f " ( x ) + .......
...(5)
Cor.1. Putting a = 0 in (5) we get
h2
f (a + h) = f (a) + hf ' (a ) +
2!
f " (a ) + ........ +
hn
f " (0) + .........................
n!
Corl.2 Putting a = 0and h = x in Cor.1we get the Maclaurin’s theorem
f( x) = f(0) + xf'(0) +
x 2
f"(0) + ........ +
2!
hn
f"(0 ) + ..................
n!
Cor.3 Putting x = h and h = a in (5), we get
f ( a + h) = f ( h) + af '( h) +
a
2
f "( h) + ............ +
2!
xn n!
f n ( h) + ......
Cor.4 Putting h = x - a in Cor.1, we get
f ( x) = f [a + ( x − a )] = f (a ) + ( x − a ) f ' (a ) +
( x − a ) 2 2!
f " (a ) + .......... +
( x − a )" n!
f n (a ) + .........
Example: Expand log sin(x + h) in powers of h by Taylor’s Theorem Solution:
∴
f ( x + h) = log sin( x + h) f( x) = log sin x. Futher f'( x) =
1 sin x
cos x= cot x,
f " = − cos ec2 x, f '"( x) = 2 cos ec2 xcot x, ............................. ......................... By Taylor’s Theorem, we get f ( x + h) = f ( x) + hf h f '( x) +
h2 2!
f "( x) +
⇒ log sin( x+ h) = log sin x+ hcot x−
Example: Show that log( x + h)
= log h +
x h
h3 3! h2 2!
−
f "'( x) + ............. cos ec x+ 2
x 2 2h 2
+
2h 3
x 3 3h 3
3!
cos ec2 xcot x+ ..................
− ...............
Solution: Since we are to expand log ( x + h ) in power of x, therefore we are to use the from given in Cor3. putting x for a in Cor3, we get.
f ( x + h) = f ( x) + xf '( h) +
x 2 2!
f "( h) + ................
(1)
Now f ( x + h) = log( x + h)
∴
1
f ( h) = log h; f '( h) =
; f "( h) = −
h Put Putti ting ng the thesevalu sevaluein ein(1) we get get log( x+ h) = log h+
x h
−
x2 2h 2
+
x3 3h3
1 h2
; f ''( h) =
2 h3
, ... ..... .... .
+ . .... ..... ....
FUNCTION OF TWO VARIABLES:
expand f ( x + h , y+ k ) in powers of h and k, Taylor’s Theorem for function of two variables . “To expand f in case f case f (x , y) and all its parietal derivatives are continuous in a certain domain of the point (x, y)” Taking f (x+h ,y+k) as a function of one variable, say x i.e.x varies while y remains constant then expanding by Taylor’s theorem we have
∂ f ( x , y + k ) h 2 ∂f ( x , y + k f ( x + h, y + k ) = f ( x, y + k ) + h + + ......(1) 2! ∂v ∂x 2 Further, expanding each term on the right hand side of (1) by Taylor’s theorem taking y as variable and x as constant, we have
∂ f ( x, y) k2 f ( x, y) ∂ ∂ f ( x, y) f ( x + h, y + k) = f ( x, y) + k h f x y k . . . . . . ( , ) . . . . . . . + + + + + ∂ y ∂x ∂ y 2! ∂ y 2 ∂f ( x, y) h2 ∂ 2 f x y k ( , ) . . . . . . . + + + + ............ ∂ y 2! ∂ y 2 ∂ f ∂ f 1 2 ∂ 2 f ∂ 2 f 2 ∂ 2 f ⇒ f ( x + h, y + k) = f ( x, y) + h + k + h + 2 hk + k 2 + ...... 2 2! ∂ ∂ ∂ ∂ x y x y ∂ ∂ y x Or in symbolic from, we get 2 ∂ f ∂f ∂ 1 ∂ 1 ∂ f ( x + h, y + k ) = f ( x, y) + h + k f + h + k f + ... .. + h + n! ∂ x ∂ y ∂ y 2! ∂ x ∂x
"
∂ k f + ...... where ∂ y
f = f(x ,y) n
n n ∂ ∂ ∂n f f n( n− 1) n − 2 2 ∂ n f n ∂ n −1 n ∂ f . +k f = h + nh k n−1 + + ..... + k n h h k n n −1 2 ∂ ∂ 2! x y ∂ ∂ ∂ ∂ ∂ ∂ y x x y x y
[by Binomial Theorem]
We now give an accurate statement of Taylor’s Theorem for function of two variables. If f (x, y) th possesses continuous partial derivates upto n order for all points (x,y) in the region ( a ≤ x ≤ a + h, b ≤ y ≤ b + k ) , then we have 2
∂ 1 ∂ ∂ ∂ f ( a + h, b + k ) = f ( a, b) + h + k f ( a, b) + h + k f ( a, b) + .......... 2! ∂ x ∂ y ∂ x ∂ y ∂ ∂ .. .. + h k + ( n − 1) ! ∂x ∂y 1
n −1
n
1
∂ ∂ f ( a, b) + h + k f ( a + θ h, b + θ k ) n ! ∂x ∂y where 0 < θ < 1
If u = f( y ); then to show that du =
Give u = (y); then u + δ u
∂u ∂u dx + dy : ∂ x ∂ y
= f ( x + δ x, y + δ y ).
∂ f ∂ f ∴ δ u = f ( x + δ x, y + δ y ) − f ( x, y ) = f ( x, y ) + δ x + δ y + ...... − f ( x, y ) ∂ y ∂ x ∂u ∂u = δ x + δ y to first order of approximation (replacing f by u) ∂ x ∂ y ∂u ∂u dx + dy. Thanking limits, we have du = ∂ x ∂ y TOTAL DIFFERENTIAL COEFFICIENTS:
If u = f ( x, y) wh w here x = φ ( t) an a nd y = φ ( t) , then we know that du = But
∴
du =
∂u ∂u dx + dy. ∂ x ∂y du
.dt
dx dx =
dx
dt dt du ∂u dx ∂u dy = . + dt ∂x dt ∂y dt
.dt ,
dy dy =
Again if u = f = f =(x, y) y) where x where x = φ (t φ (t 1 ,t 2 )then )then
and
∂u ∂u ∂x ∂u ∂y = + ∂t1 ∂x ∂t1 ∂y ∂t1 ∂u ∂u ∂x ∂u ∂y = + ∂t2 ∂x ∂t2 ∂y ∂t 2
dy dt
.dt .........(1)
Example: Expand f ( x, y)
= x2 y + 3 y − 2 in power of (x - 1) and ( y + 2 ) by Taylor’s theorem.
Solution: By Taylor’s theorem
f ( x, y ) = f (a, b) + {( {( x − a ) f (a , bx ) + ( y − b) f (a ,by )} +
1 2!
{(x − a )2 f (a ,b xx )
+ 2( x − a)( y − b) f (a , b )xy+ ( y − b )2 f (a , b )x}y +
1 3!
+ 3( x − a)2 ( y − b) f (a, b) + 3xx(yx − a )( y − b)2 f
(x − a )3 f
(a , b )
+ ( y − b)3 f yyy ( a, b) } + ...............
xyz
(a ,b x)xx
.......(1)
Given a = 1, b = -2
f ( x, y) = x2 y + 3 y − 2 ⇒ f (1 , −2) = −10 f x ( x, y) = 2 xy ⇒ fx (1, −2) 2 ) = 2(− 1) 1)(− 2) 2) = − 4 f y ( x, y) = x2 + 3 ⇒ fy ( 1, −2) = +4 f ( x, yx)x = 2 y ⇒ f (1, −2xx) = −4 f ( x, yx)y = 2 y ⇒ f (1, −2xy) = 2 f ( x, yyy) = 0 ⇒ f (1, −y2) 2y ) = 0 f ( x, xyx)x = 0 ⇒ f (1, −x2xx) = 0 f ( x, xyx)y = 2 ⇒ f (1, −x2) 2xy) = 2 f ( x, xyy)y = 0 ⇒ f (1, −x2yy) = 0 f
( x, yyy)y = 0 ⇒ f
( x, yyy)y = 0
Putting a = 1, b = -2 and values in (1), we get
x2 y+ 3 y− 2 = 10[( x− 1)(− 4) + ( y+ 2)( 4)] +
1 2!
[ x− 1)2 (− 4)
+ 2( x− 1)( y+ 2)(2) + ( y+ 2)2 (0 ( 0)] +
1 3!
[( x− 1)3 (0 ( 0)
+ 3( x− 1)2 ( y+ 2)( 2) + 3( x− 1)( y+ 2)2 (0) + ( y+ 2)3 (0 ( 0)]. or x 2 y + 3 y − 2 = −10 − 4( x − 1) + 4( y + 2) − 2( x− 1)2 + 2( x− 1)( y+ 2) + ( x− 1)2 ( y+ 2). ax
Example: Expand e by Maclaurin’s theorem. ax
0
Solution: y = e , them (y)0 = e = 1 (by putting x = 0),
y = aeax , ⇒
1
( 1y)0 = ae0 = a,
y2 = a2 eax ⇒ ( y2 )0 = a2 e0 = a2
∴
...............
....................
...............
.....................
yn = an eax
⇒ ( yn )0 = an e0 = an .
By Maclaurin’s theorem, we get n 3 x x x ( y2 ) 0 + ( y3 ) 0 + ....... + ( yn ) 0 + ............. y = ( y) 0 + x( y1 ) 0 + 2! 3! n! 2
2
⇒ e = 1 + xa +
x
ax
2!
n
3
a +
x
2
3!
a + ...... . + 3
x
n!
a n + ....... ....
Note: If a = 1, then
x2
e = 1+ x + x
2!
+
x3 3!
+ .........
xn n!
+ ....
This is known as Exponential series. x
Example: Expand e sex x by Maclaurin’s theorem. x
0
Solution : Let y = e sec x then (y)0 = e sec (0) = 1
y1 = e x sec x + ex sec xtan x
= e x sec x(1 + tan x) = y (1 + tan x ), ( y1 )0 = ( y )0 [1 [1 + tan(0)] = 1 y2 = ysec2 x + y1 (1 + tan x), ( y2 )0 = ( y)0 sec2 (0 (0) + ( y1 ) 0 (1 (1 + tan 0) = 1+ 1 = 2, y3 = y1 sex2 x + 2 ysec 2 xtan x+ y2 (1 + tan x) + y1 sec2 x = 2 y1 sec 2 x+ 2 ysec 2 xtan x+ y2 (1 + tan x) ( y3 )0 = 2( y1 )0 1 + 2( y) 0 (0) + ( y2 )0 .1 = 4
∴ By Maclaurin’s theorem, we get e sec x = ( y )0 + x ( y1 ) 0 + x
= 1 + x.1 +
x 2
2!
x3
( y2 ) 0 +
x2 3!
.4 + ........ 3! 2 ⇒ e x sec x = 1 + x + x 2 + x 3 + .......... 3 2!
.2 +
x 2
( y3 ) 0 + ............
Example: Use Maclaurin’s theorem to find the expansion of log(l + e ) in ascending powers of x to x
the containing x 4 .
= log(l + e 0 ) = log 2
Solution: Let y = log(1 + e ). then ( y ) 0 x
y1 = y2 =
e x 1 + e x
,
(1 + e x )e x − e x .e x (l + e )x2
( y1 ) 0 =
=
e0
1
=
l + e0
2
e x[1 + e x − e x ] (1 + e )x2
=
ex
1 . (l + e )x (1 + e )x
e x = y1 = y1 1 − x x 1+ e 1+ e 1
⇒ y 2 = y1 (1 − y1 ),
( y2 ) 0 = ( y1 ) 0 [1 − ( y1 )0 ] =
y3 = y2 (1 − y1 ) + y1 (− y2 ),
( y3 ) 0 = ( y2 ) 0 − 2( y1 ) 0 ( y2 ) 0
= y2 − 2 y1 y2
=
y4 = y3 − 2 y3 y1 − 2 y22 ,
1 2
1 1
− 2. . = 0 2 4
( y4 ) 0 = ( y3 ) 0 − 2( y3 ) 0 ( y1 ) 0 − 2( y22 ) 0
= 0 − 0 − 2. ∴ By Maclaurin’s theorem, we get log(1 + e ) = ( y ) 0 + x ( y1 ) 0 + x
1 1 1 1− = 2 2 2
x 2
16
=−
1 8
x 2
( y 3 ) 0 + ........ 3! x 4 1 1 x 2 1 x 2 = log 2 + x. + . + .0 + . − + ............ 2 2! 4 3! 4! 8 2!
( y 2 ) 0 +
1
1
1
1
2
8
192
⇒ log(1 + e x ) = log 2 + x + x 2 −
x 4 + ..........
Example: Expand log(1 + sin x ) by Maclaurin’s theorem Solution: Let y
= log(1 + sin x), then ( y ) 0 = log1 = 0.
….(1)
1 1 cos 2 x − sin 2 x 1 cos x 2 2 y1 = = .cos x = 1 1 1 1 + si sin x 1 + si sin x 2 1 + sin 2 x + 2 sin x cos x cos 2 2 2 2
1 1 1 1 1 cos 2 x − sin 2 x cos x − sin x 1 − tan x 2 2 = 2 2 = 2 = 2 1 1 1 1 1 cos sin 1 tan x x x + + cos sin x x + 2 2 2 2 2 (by dividing Num. and Den. By cos
1 1 ⇒ y1 = tan π − x 2 4 1 1 1 y 2 = sec 2 π − x . − , 4 2 4
1 2
1
∴ ( y1 ) 0 = tan π = 1 4
1
1
2
4
∴( y 2 ) 0 = − sec 2 π = −1
x)
……(2)
……(3)
1 1 1 1 1 1 2 sec 2 π − x − × tan π − x − 2 2 2 2 2 4 4 1 1 1 1 1 = − − sec 2 π − x × tan π − x 2 2 4 4 2
y3 = −
=
1
∴ ( y3 )0 = −( y2 )0 ( y1 ) 0 = +1
− y 2 , y1 ,
…….(4)
∴ ( y 4 ) 0 = −( y 3 ) 0 ( y1 ) − ( y 2 ) 0 = −1 − 1 = −2 …..(5) 2
y4 = −[ y4 y1 + y2 2 ], …. …. ….. ….. …. …. ….. …..
∴
By Maclaurin’s theorem, we get.
log(1 + sin x ) = ( y ) 0 + x ( y1 ) 0 +
= 0 + x −
= x −
x 2 2
x 2 2!
+
+
x 3 3
x 2 2!
x 3 3!
−
− x4
12
( y 2 ) 0 +
x4 4!
x 3 3!
( y 3 ) 0 +
x 4 4!
( y 4 ) 0 + .......
(−2) + ..........
+ .......
−1
Example: Expand sin x by Maclaurin’s theorem. Solution: Let y =
sin −1 x
Differentiating it with respect to x, we get ge t
…..(1)
y1 =
1
⇒ y1 (1 − x 2 ) = 1 2
(1 − x ) 2
…..(2)
Differentiating it again, we have
y1 (−2 x) + 2 y1 y 2 (1 − x 2 ) = 0 ⇒ − xy1 + y 2 (1 − x 2 ) = 0 2
⇒ y 2 (1 − x 2 ) − xy1 = 0
……(3)
Differentiating (3) n times by Leibnitz’s theorem, we have
yn+ 2 (1 − x2 ) + n c1 yn+1 (−2 x) + n c2 yn (−2) − yn +1 . x + n c1 yn .1 = 0 ⇒ y n + 2 (1 − x 2 ) − ny n +1 (−2 x) +
n(n − 1) 2!
y n (−2) − y n +1 . x − ny n = 0
⇒ (1 − x 2 ) y n+2 − (2n + 1) xyn+1 − n 2 y n = 0
…..(4)
Putting x = 0 in (1), (2), (3), we get n+2
(y
2
) 0 - n ( yn)0 = 0
Putting n = 1, 2, 3…… , we have
( y3 ) 0 − 1( y1 ) 0 = 0 ⇒ ( y3 ) 0 = ( y1 ) 0 ⇒ ( y3 ) 0 = 1 ( y 4 ) 0 − 4( y 2 ) 0 = 0 ⇒ ( y 4 ) 0 = 4( y 2 ) 0 ⇒ ( y 4 ) 0 = 0. ( y 5 ) 0 − 9( y 3 ) 0 = 0 ⇒ ( y 5 ) 0 = 9( y3 ) 0 ⇒ ( y5 ) 0 = 9 = 3 2 ( y 6 ) 0 − 16( y 4 ) 0 = 0 ⇒ ( y 6 ) 0 = 16( y 4 ) 0 ⇒ ( y 6 ) 0 = 0
( y 7 ) 0 − 25( y 5 ) 0 = 0 ⇒ ( y 7 ) 0 = 25( y 5 ) 0 ⇒ ( y 7 ) 0 = 25 × 9 = 3 2.5 2 Hence by Malaria’s theorem, we have −1
sin x = ( y ) 0 + x( y1 ) 0 +
= 0 + x.1 + = x −
x
3
3!
+
x 2 2!
.0 +
2
3 . x 5!
5
x 2 2! x 3
+
3! 2
( y 2 ) 0 + .1 + 2
x 4
3 .5 . x 7!
4! 7
x3 3!
.0 +
( y 3 ) 0 + ....... x 5 5!
(3) +
. + .............
2
x 6 6!
.0 +
x 7 7!
.(3 2 5 2 ) + .........
1 1
1 3 1
1 3 5 1
2 3
2 4 5
2 4 2 7
⇒ sin −1 x = x + . x 3 + . . x 5 + . . . x 7 + ....... −1
2
Example: Expand (sin x) in ascending powers of x. Solution: Let y
= (sin −1 ) 2
y1 = 2 ( sin −1 x )
(y1) 0 = 0
1
(1 − ) 2
⇒ (1 − x 2 ) y1 = 4(sin −1 x) 2 = 4 y 2
⇒ (1 − x 2 ) y1 = 4 y
∴ ( y1 ) 0 = 0
2
⇒ (1 − x 2 )2 y1 y 2 − 2 xy1 = 4 y1 2
⇒ (1 − x 2 ) y 2 − xy1 − 2 = 0
∴ ( y 2 ) 0 = 2
Differentiating n times by Leibnitz’s theorem, we get
n(n − 1) 2 − + − + y ( 1 x ) ny ( 2 x ) y ( 2 ) + + n 2 n 1 n − [ y n +1 x + ny n ] = 0 2!
⇒ (1 − x 2 ) y n + 2 − (2n + 1) xy n+1 − n 2 y n = 0
∴ ( y n+ 2 ) = n2 ( y n ) 0
…(1)
Putting n = 1,2,3,….., in (1), we get
( 3y)0 = 1. ( 1y) 0 = 0 ( 5y)0 = 32. ( 3y)0 = 0
( 4y) 0 = 22.( 2y) 0 = 2 2.2; ( 6y) 0 = 4 2.( 4y) 0 = 2 2.2 4.2;
Hence
(sin
−1
x) = ( y0 ) + x( y1 ) 0 + 2
=
2. 2x 2!
+
22.2. 4x 4!
+
x 2 2!
( y2 ) 0 +
42.22 .2. 6x 6!
x3 3!
( y3 ) 0 + ......
+ ................
Deductions:
1. If we put x = sin sin θ in the the above above resu result lt,, we we get. get.
......(2)
θ = 2
2 sin 2 θ 2!
+ 2 .2 2
sin 4 θ 4!
+4 . 2
2 2.2 sin 6 θ 6!
+ .......
2. If we differentiate both sides sides of (2) w.r.t x, we get
2 sin −1 x
= 2 x + 2 .2 2
(1 − x ) 2
⇒
sin −1 x
(1 − x ) 2
x 3 3!
+ 4 .2 .2 2
2
x 5 5!
+ .........
2 3 2.4. x 5 = x + x + + ......... 3 3.5 -1
3
Example: Expand sin (x + h) in power of x as far as the term in x . -1
Solution: First we observe that we are to expand sin (x + h) in ascending powers of x. so let -1
f(h) = sin h. Then -1 f(h + x) = sin ( h + x) Thus we are to expand f (h + x) in power of o f x. So by Taylor’s theorem, we have 2 x x 3 ’(h)+ f " (h) + f " ' (h) + .... f( h + x) = f (h) + xf ’(h)+ …….(1) 2! 3! –1
Now f(h) = sin h.. Therefore
1
f ’(h) =
1 − h2
f ’’ (h) =
= (1 − h2 ) −1 / 2
h(1 − h2 )−3 / 2
f '''( h) = (1 − h2 ) −3 / 2 + h(−3/ 3 / 2)(1 − h2 ) −5 / 2 (−2 h)
= (1 − h2 )−3 / 2 + 3h2 (1 − h 2 ) −5 / 2 = (1 − h 2 ) −5 / 2[(1 − h 2 ) + 3h 2 ) =(1 − h2 )−5 / 2 (1 + 2h 2 ), etc.
Substituting these values in (1), we have -1
-1
2 1/2
Sin (h + x ) = Sin h + (1-h )
2
2 -3/2
x + (x /2!) h (1-h )
3
2 -5/2
+ (x /3!) (1-h )
2
(1-2h )+…
Example: Use Taylor’s theorem to prove that sin θ sin 2θ sin 3θ -1 -1 tan (x + h) = tan x +(hsinθ +(hsin θ )) − (h sin θ )2 + ( h sin θ )3 − ..... 1 2 3 n-1 n sin nθ -1 + ....., ….+(-1) (hsinθ (hsin θ )) where 0 = cot x n -1
Solution: Let f(x) = tan x. then -1
f (x + h) = tan (x + h ). Expanding f (x + h) in power of h by Taylor’s theorem we have f ( x + h) = f (x) +
h 1!
f '( x) +
h2 2!
f ''( x) + ..... +
-1
n
hn n!
f ''( x) + ....
…….(1)
-1
Now f ( x ) = tan x. Therefore f ’’(x) = D tan x = (-1)
n-1
n
-1
(n-1)! sin θ sin nθ n θ .. Where θ = cot x
Putting n= 1, 2, 3, ……..in it , we get 2
3
f ’(x)= sinθ sin θ sin sinθ θ , f ’’(x) = -1! sin θ sin2θ sin2 θ f ’’’(x) = 2! Sin θ sin3θ sin3 θ .etc. .etc. Substituting these values in (1), we have -1
-1
tan (x + y) = tan x +hsinθ +hsinθ sinθ sin θ -
h2 2!
+ -1
= tan x + hsinθ hsin θ
sin θ 1
2 − ( h si s in θ )
x
sin 2θ 2
sin θ sin 2θ + 2
h
n
n!
h3 3!
2 !sin 3 θ sin 3θ − .. .....
(−1)n −1 ( n − 1) !sin n θ sin nθ + ....
+ (h ssiin θ )3
sin 3θ 3
..... + ( − 1) 1)
n −1
( h ssiin θ )
n
sin nθ
n
+ .......
π by Taylor’s Theorem. 4
Example: Expand e cosy near the point 1, Solution: By Taylor’s theorem
2
3
∂ ∂ ∂ ∂ 1 ∂ 1 ∂ F(x + h, y + k) = F (x, y) + h + k F + h + k F + h + k F + .....(1) 2! ∂ x 3! ∂ x ∂ y ∂ y ∂ y ∂ x x
Again e cosy = F(x, y)
=F 1 + ( x − 1).
π π π = F 1 + h + k + y − , where h = x -1, k = y 4 4 4 4
π
π e = 4 2 ∂ F π e 1. = ⇒ ∂ y 4 2
⇒ F 1.
x
F(x, y) = e cosy
∂ F = e x cos y ∂ y ∂ F = −e x sin y ∂ x
e ∂ 2 F π = 1 . ⇒ ∂ x 2 4 2
∂ 2 F = e x cos y 2 ∂ x
∂ 2 F = −e x cos y 2 ∂ y
∂ 2 F = −e x cos y ∂ x∂ y
e ∂ 2 F π 1 . = ⇒ ∂ x 2 4 2 ∂ 2 F π − e 1. = ⇒ ∂ y 2 4 2
∂ 2 F π − e ⇒ 1. = ∂ x∂ y 4 2
Substituting these values in Taylor’s theorem, we get x
e cosy =
+
e π − e + ( x − 1) + y − 2 4 2 2 e
1
( x− 1)
2 !
2
π −e π − e y + 2( x− 1) y− + − + .......... 4 2 4 2 2
e
2
ADDITIONAL PROBLEMS:
1. Show that by Maclaurin’s theorem,
log(1 + x) = x −
x2 2
+
x3 3
−
x4 4
+ ........... + (− 1) 1)
n −1
xn n!
2. By Maclaurin’s theorem, prove that
e ax sin bx = bx + abx 2 +
3a b − b 2
3
3!
x 3 + .. .. .. ... . +
3. Apply Maclaurin’s Maclaurin’s theorem theorem to obtain the term upto 4. If y
= e a sin
−1
x
(a
2
+ b2 ) n!
4
n/2
x n sin(n tan −1
, show that (1 − x2 ) yn + 2 − (2 n+ 1) xyn +1 − ( n2 + a2 ) yn = 0
e
= 1 + ax +
Also deduce that
a2 x2 2!
+
a (12 + a 2 ) 3!
eθ = 1 + sin θ +
x 3 + ........ 1 2!
sin 2 θ +
1 3!
sin 3 θ + .....
5. By Maclaurin’s theorem, show that
e
a cos−1 x
) + .. .
a in the expansion of log(1 + sin 2 x)
Hence by Maclaurin’s theorem show that a sin −1 x
b
a2π a 2 x 2 a (1 + a 2 ) 3 a(22 + a 2 ) 4 = 1 − ax + − x + x ........ e 2 ! 3 ! 4 !
