UNIVERSIDAD NACIONAL
DEL
CALLAO
FACULTAD DE INGENIERIA MECANICA ENERGIA
INTEGRANTE :
TRABAJO
: DIRIGIDO (RESOLUCION PROBLEMAS)
CURSO DIRIGIDO: METODOS NUMERICOS PROFESOR
:
LIC. COLLANTES GRUPO HORARIO :
02M
1
2006
SOLUCIONARIO DE PROBLEMAS DE METODOS NUMERICOS
1. determina determinarr interval! interval! "#e $nten% $nten%an an !l#$ine !l#$ine! ! a la! !i%#iente! !i%#iente! e$#a$ine!& 'a ', '$ 'd
()*)( + 0 -(2) e( + 0 (2)2(2)-(*+ 0 (*-.001(2-.002(1.001+0
Problema 1: Nos auxiliamos de un software gráfico como el Matlab, de esta manera podemos determinar un intervalo apropiado donde buscar la solución de las ecuaciones pedidas. Para cada caso mostramos las gráficas obtenidas por el matlab. a)x!"x las gráficas #ue se obtienen para la intersección de $x e $ x!"x muestran #ue un intervalo donde %a$ una solución es: & '(.,(.*+.
2
2006
SOLUCIONARIO DE PROBLEMAS DE METODOS NUMERICOS
1. determina determinarr interval! interval! "#e $nten% $nten%an an !l#$ine !l#$ine! ! a la! !i%#iente! !i%#iente! e$#a$ine!& 'a ', '$ 'd
()*)( + 0 -(2) e( + 0 (2)2(2)-(*+ 0 (*-.001(2-.002(1.001+0
Problema 1: Nos auxiliamos de un software gráfico como el Matlab, de esta manera podemos determinar un intervalo apropiado donde buscar la solución de las ecuaciones pedidas. Para cada caso mostramos las gráficas obtenidas por el matlab. a)x!"x las gráficas #ue se obtienen para la intersección de $x e $ x!"x muestran #ue un intervalo donde %a$ una solución es: & '(.,(.*+.
2
3
2 . 5
2
1 . 5
1
0 . 5
0
0 . 5
1 1
0 . 8
0 . 6
0 . 4
0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1
b) x -ex) las gráficas #ue se obtienen para la intersección de $x e $ -ex) muestran #ue un intervalo donde donde %a$ una solución es: & '(./,(.*+
1 0 . 8 0 . 6 0 . 4 0 . 2
0 0 . 2 0 . 4 0 . 6 0 . 8 1 1
0 . 8
0 . 6
0 . 4
0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1
3
c) x -x! 0 x 2 !) las gráficas se muestran a continuación. 1 0 . 8 0 . 6 0 . 4 0 . 2
0 0 . 2 0 . 4 0 . 6 0 . 8 1 1
0 . 8
0 . 6
0 . 4
0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1
el intervalo donde se encuentra la solución es: & '(.,(.*+. d)x-x3! 2 .((1x3 2 1.1(1).(( las gráficas se muestran a continuación: 5
0
5
1 0
1 5
2 0 6
5 . 5
5
4 . 5
4
3 . 5
3
2 . 5
2
podemos considerar entonces el intervalo & '"4.4, ".4+. 5entro de este intervalo podemos encontrar una ra67 real.
4
2. !ean /(' + 2( $! 2(' ()2'2
(0 + 0
a' determine el ter$er 3linmi de Talr P*(' 4!el 3ara a3r(imar / 0.-' ,' #!e la /rm#la del errr en el terema de Talr determine $n ella #na $ta !#3erir 3ara el errr 5 / 0.-') P*0.-' 5 . $al$#le el errr real.
= f ( x0 ) +
P3 ( x )
f ' ( x0 ) 1!
( x − x0 ) +
f " ( x0 ) 2!
2
( x − x0 ) +
f "' ( x0 ) 3!
( x − x0 )
3
8lrededor de x0 = 0 f ( x )
= 2 x cos ( 2 x ) − ( x − 2 ) 2
f ' ( x )
= 2 cos ( 2 x ) + 2 x 2 (− sen 2 x − ) 2 ( −x = 2c0s ( 2 x ) − 4 xsen 2 x − 2 ( x − 2 )
f " ( x )
= −4sen ( 2 x ) − 4sen2 x+ 4 x ( 2cos 2x − ) = −8 sen ( 2 x ) − 8x cos 2 x − 2
f "' ( x )
= −16 cos ( 2 x ) − 8[ cos 2 x − 2 xsen 2x] =− 24 cos ( 2 x ) + 16 xsen ( 2 x )
f ( 0 )
= −4
f ' ( 0 )
=6
f " ( 0 )
= −2
f "' ( 0 )
= −24
f ( 0.4 )
2)
2
2
24
2
6
≈ P 3 ( 0.4 ) = −4 + 6 ( 0.4 − 0 ) − ( 0.4 − 0 ) 2 −
( 0.4 − 0 )
3
9uego : P 3 ( 0.4 ) = −4 + 2.4 − 0.16 − 0.256 P 3 ( 0.4 )
= 2.016 5
f 3 ( 0.4 )
= 2( 0.4 ) cos ( 0.8) − ( 0.4 − 2) 2
f 3 (0.4)
−
P 3 (0.4)
=
0.013365367
*. U!e la aritmti$a de rednde de tre! $i/ra! !i%ni/i$ativa! 3ara l! !i%#iente! $7l$#l!. Cal$#le el errr a,!l#t errr relativ $n el valr e(a$t determinad a 3r l men! 8 $i/ra!. a' ,' $' d' e' /' %'
1** 0.921 1** ) 0.-99 121 0.*2:' 119 121)119' 0.*2: 1*;1- ) 6;: ' ; 2 e )8.-' )10 6 e )*;62 ) 22;:' ; 1;1:'
8 ! cifras significativas a) 133 + 0.921 = 133.9210 ≈ 134 b) 133 + 0.499 = 132.5010 ≈ 133
c)
( 121 − 0.327 ) − 119 = ( 121) − 119 = 2 1 44 2 4 4 3 120.6730
d)
13 14
= 0.929 6
= 0.857 7 2e = 5.44 9uego :
0.929 − 0.857 5.44 − 5.40
=
0.07 0.04
= 1.75
f) −10π = −31.4 6e = 16.3
−
3 62
= −0.05
6
9uego : −10π + 6e −
3
= −15.15
62
%) π = 3.14 22 7 1 17
− 22
=
3.14
=
0.06
π
7
1
=0
17
-. determine la ra3ide< de $nver%en$ia de la! !i%#iente! !#$e!ine! a' lim
!en
1
+ 0
n
n →∞
,' Lim
!en
1 n
2
+ 0
n → ∞
$' lim
!en
1 n
'2 + 0
n → ∞
d' lim Ln n1' Ln n ' + 0 n → ∞
a)
1 lim sen n→∞ n Pn
=0
1 = sen n
lim Pn
= P = 0
Para determinar la rapide7 de convergencia, determinaremos el l6mite par algun lim
P n
1 −
+
P n
−
P
P
α
7
n 1 1 n + 1 sen ( ) n + 1 n + 1 = lim n + 1 1 1 n sen n sen n n
n sen lim
n→∞
sen 1 1 n 1 + 1 1 1 + n ( n + 1) = 1 = lim n→∞ sen 1 n 1 n ∴ converge linealmente b)
1 = 0 n2
lim sen
n→∞
1 = sen 2 n P 0 = 0 Pn
1 Sen 2 Pn +1 − P ( n + 1) = P − P 1 sen 2 n
1 Sen 2 2 ( n + 1) ( n + 1) 1 n 2sen 2 n
n 2 ( n + 1)
⇒ lim
n→∞
2
1 sen 2 ( n + 1) 1 2 2 1 ( n + 1) = lim =1 1 n→∞ 1 + sen 1 2 n n 1 n2
8
∴ converge linealmente e
1 c) lim sen = 0 n→∞ n
1 Pn = sen n
e
e
sen 1 n + 1 Pn+1 − P = 1 lim = lim e Pn − P 1 sen n ∴ converge linealmente lim Ln ( n+ 1)− Ln ( n )=
d)
n→∞
Pn
0
= Ln ( n + 1) − Ln ( n )
Pn+1 = Ln ( n + 2 )
− Ln ( n + 1)
8. A3li"#e el mtd de la ,i!e$$i=n 3ara en$ntrar !l#$ine! e(a$ta! dentr de 10)8 3ara l! !i%#iente! 3r,lema! a' ( e( + 0 ,' e( ) (2*()2 + 0
3ara 3ara
≤ x ≤1 0 ≤ x ≤1
0
;eguimos los siguientes pasos: &ntervalo inicial [ a , b] a1 = a b1 = b
a1 + b1 2
;i : f ( a1 ) f ( c1 ) < 0 entonces b1 = c1 ;ino ;i f ( c1 ) f ( a1 ) < 0 entonces a1 = C 1 ;ino No es posible %allar una solución en el intervalo inicial dado
9
=epetimos este proceso %asta una tolerancia de 10−5 >eamos el primer caso: a)
x − e − x
?
=0
0
≤ x ≤ 1
i = 1 f ( x )
a1 = 0 f ( a1 )
= x − e− x
⇒
a=0 b =1
i = 2 ? como f ( b1 ) f ( c1 )
= −1 b1 = 1 f ( b1 ) = 0.632121 c1 = 0.5 f ( c1 ) = −0.106531 <0
⇒ a2 = c1 = 0.5 ⇒ f ( a2 ) = −0.106531 = b = 1 ⇒ f ( b2 ) = 0.632121 c2 = 0.75 ⇒ f ( c2 ) = 0.277633
b2
i = 3 ? como f ( a2 ) f ( c2 )
<0
⇒ a3 = a2 = 0.5 = c2 = 0.75 a + b3 c3 = 3 = 0.625
b3
2
PROBLEMA 8
;eguimos los siguientes pasos: &ntervalo inicial [ a , b] a1 = a b1 = b
a1 + b1 2
;i : f ( a1 ) f ( c1 ) < 0 entonces b1 = c1 ;ino ;i f ( c1 ) f ( a1 ) < 0 entonces a1 = C 1
10
;ino No es posible %allar una solución en el intervalo inicial dado =epetimos este proceso %asta una tolerancia de 10−5 >eamos el primer caso: b)
x − e
− x
=0
;
0
≤ x ≤ 1
i = 1 f ( x )
a1 = 0 f ( a1 )
= −1 b1 = 1 f ( b1 ) = 0.632121 c1 = 0.5 f ( c1 ) = −0.106531
= x − e− x
⇒
a=0 b =1
i = 2 ? como f ( b1 ) f ( c1 )
<0
⇒ a2 = c1 = 0.5 ⇒ f ( a2 ) = −0.106531 = b = 1 ⇒ f ( b2 ) = 0.632121 c2 = 0.75 ⇒ f ( c2 ) = 0.277633
b2
i = 3 ? como f ( a2 ) f ( c2 )
<0
⇒ a3 = a2 = 0.5 = c2 = 0.75 a + b3 c3 = 3 = 0.625
b3
2
@ en 1A iteraciones se obtiene : c17 = 0.567142 a1 = 1.2 b1 = 1.3 c1 = 1.25
= 0.1548 f ( b1 ) = −0.1323 f ( c1 ) = 0.01915 f ( a1 )
⇒ a1 = c1 9uego :
11
= c1 = 1.25 b2 = b1 = 1.3 c2 = 1.275 a2
c)
f ( x )
= 0.0915 f ( b2 ) = −0.1322 f ( c2 ) = −0.0548 f ( a2 )
= e x − x 2 + 3x − 2
,
0≤x
≤1
i = 1 a1 = a = 0 b1 = b = 1 c1 =
a1 + b1 2
= −1 f ( b1 ) = 2.718282 f ( c1 ) = 0.898721 f ( a1 )
= 0.5
= 2 a2 = a1 = 0 b2 = c1 = 0.5 a + b2 c2 = 2 = 0.25 i
2
= −1 f ( b2 ) = 0.898721 f ( c2 ) = −0.028475 f ( a2 )
= c2 = 0.25 b3 = b2 = 0.5 a + b3 c3 = 3 = 0.375 a3
2
= −0.028475 f ( b3 ) = 0.898721 f ( c3 ) = 0.439366 f ( a3 )
@ en 1A iteraciones se obtiene : c17 = 0.257534 &ntervalo de extremos 'aBi,bBi+ con punto medio cBi a) i aBi bBi
cBi
( ( 1.(((((((((((((( (.4((((((((((((( 1.(((((((((((((( (.4((((((((((((( 1.(((((((((((((( (.A4((((((((((((
12
.(((((((((((((( (.4((((((((((((( (./4((((((((((( !.(((((((((((((( (.4((((((((((((( (.4/4(((((((((( .(((((((((((((( (.4/4(((((((((( (.4C!A4((((((((( 4.(((((((((((((( (.4/4(((((((((( (.4A*14(((((((( /.(((((((((((((( (.4/4(((((((((( (.4A(!14((((((( A.(((((((((((((( (.4/4(((((((((( (.4//(/4(((((( *.(((((((((((((( (.4//(/4(((((( (.4/*!4C!A4((((( C.(((((((((((((( (.4//(/4(((((( (.4/A!**14(((( 1(.(((((((((((((( (.4//(/4(((((( (.4//*C4!14((( 11.(((((((((((((( (.4//*C4!14((( (.4/A1!*/A1*A4(( 1.(((((((((((((( (.4/A1!*/A1*A4(( (.4/A/(A1*A4( 1!.(((((((((((((( (.4/A1!*/A1*A4(( (.4/A1CCA(A(!14 1.(((((((((((((( (.4/A1!*/A1*A4(( (.4/A1/C1*C4!1! 14.(((((((((((((( (.4/A1!*/A1*A4(( (.4/A14!C!(//(/ 1/.(((((((((((((( (.4/A1!*/A1*A4(( (.4/A1/!(1/C4! 1A.(((((((((((((( (.4/A1!*/A1*A4(( (.4/A1*/4AA
(.A4(((((((((((( (./4((((((((((( (./4((((((((((( (.4C!A4((((((((( (.4A*14(((((((( (.4A(!14((((((( (.4A(!14((((((( (.4/*!4C!A4((((( (.4/A!**14(((( (.4/A!**14(((( (.4/A!**14(((( (.4/A/(A1*A4( (.4/A1CCA(A(!14 (.4/A1/C1*C4!1! (.4/A14!C!(//(/ (.4/A1/!(1/C4!
@ en 1A iteraciones se obtiene : c17 = 0.567142 a1 = 1.2 b1 = 1.3 c1 = 1.25
= 0.1548 f ( b1 ) = −0.1323 f ( c1 ) = 0.01915 f ( a1 )
⇒ a1 = c1 9uego :
13
= c1 = 1.25 b2 = b1 = 1.3 c2 = 1.275 a2
d)
f ( x )
= 0.0915 f ( b2 ) = −0.1322 f ( c2 ) = −0.0548 f ( a2 )
= e x − x 2 + 3x − 2
,
0≤x
≤1
i = 1 a1 = a = 0 b1 = b = 1 c1 =
a1 + b1 2
= −1 f ( b1 ) = 2.718282 f ( c1 ) = 0.898721 f ( a1 )
= 0.5
= 2 a2 = a1 = 0 b2 = c1 = 0.5 a + b2 c2 = 2 = 0.25 i
2
= −1 f ( b2 ) = 0.898721 f ( c2 ) = −0.028475 f ( a2 )
= c2 = 0.25 b3 = b2 = 0.5 a + b3 c3 = 3 = 0.375 a3
2
= −0.028475 f ( b3 ) = 0.898721 f ( c3 ) = 0.439366 f ( a3 )
@ en 1A iteraciones se obtiene : c17 = 0.257534
14
:. en $ada #na de la! !i%#iente! determine #n interval a>, en "#e $nver%e la Itera$i=n de #n 3#nt /i?. E!time la $antidad de itera$ine! ne$e!aria! 3ara ,tener la a3r(ima$i=n $n #na e(a$tit#d de 10 )8 reali$e l! $7l$#l!. a' ( +
2
− e + x x
2
3
,' ( +
e
x
'@
3
$' ( + 6)( d' ( +
5 x
'2
2
e' ( + 8)( /' ( + 0.8 !en ( $! (' !l#$in
a)
x =
2 − e x
+ x2
3
= g ( x)
&niciali7ando en x0 generamos xn de la manera siguiente xn+1 = g ( xn ) x1 = g ( x0 )
b)
∀n ≥ 0
= 0.200426
x2
= g ( x1 ) = 0.272749
x3
= g ( x2 ) = 0.253607
x9
= g ( x8 ) = g ( 0.257534) = 0.257529
g ( x ) x0
=
5 x
2
+2= x
= 2.8
x1 = g ( x0 )
= 2.6377
15
x2
= g ( x1 ) = 2.7186
x3
= g ( x2 ) = 2.6765
x16
= g ( x15 ) = 2.6906
>eamos a%ora la ( f i ) f) g ( x ) = 0.5 ( senx + cos x ) = x x0
= 0.8
x1 = g ( x0 )
= 0.707031
x2
= g ( x1 ) = 0.704936
x3
= g ( x2 ) = 0.704819
x4
= g ( x3 ) = 0.7048124
x5
= g ( x4 ) = 0.7048122
;ólo son necesarios iteraciones para lograr una tolerancia de 10−5 a)
I
= [ 0.1]
b)
I
= [ 2.5 , 3.0] ,
c)
I
= [ 0.25 ,1] ,
x0
d)
I
= [ 0.3 , 0.7]
, x0
e)
I
= [ 0.3 , 0.6] ,
f)
I
= [ 0,1] ,
, x0
x0
= 0.5
x9
= 0.25753
= 2.8
x16
= 0.69065
= 0.28
x14
= 0.90999
= 0.5
x32
= 0.469626
= 0.5
x43
= 0.44806
x0
x0
= 0.8
x4
= 0.704812
Problema A: 8plicando el mDtodo de punto fiEo dando un punto inicial x( para luego generar una sucesiFn de puntos mediante la siguiente relación: P-n21)g-P-n)) para todo nG( Hasta una tolerancia de error de (.((((1 para cada caso. a)x( (.4 i xBi g-xBi) ( (.4((((((((((((( (.((/!(CCC/ 1 (.((/!(CCC/ (.AAC(/4(C*!A
16
(.AAC(/4(C*!A (.4!/(A14/4*1! ! (.4!/(A14/4*1! (.4*44(!A//C (.4*44(!A//C (.4A/4/!/!!4(C 4 (.4A/4/!/!!4(C (.4A4C*C*41/1C / (.4A4C*C*41/1C (.4A41441*! A (.4A41441*! (.4A4!C1!/144 * (.4A4!C1!/144 (.4A4C(*1/AC/ C (.4A4C(*1/AC/ (.4A4!(4CA!*!! se %an reali7ado C iteraciones. b) x( .* i xBi ( 1 ! 4 / A * C 1( 11 1 1! 1 14 1/
.*((((((((((((( ./!AA441(((* .A1*/C1!AAC* ./A/4(//!1A/(/* ./CAC/4!/4(4 ./*/C(/!C(A/! ./C4A(CA14 ./*C//(*C(A/ ./C114(1(//* ./C(!*A/A/(/44 ./C(A*1(!4C4*** ./C(4A**A!!*1 ./C(/*/4!4!A ./C(/C!AC/A ./C(/4/A/A!! ./C(//**C(*A ./C(/C*C!*(
g-xBi) ./!AA441(((* .A1*/C1!AAC* ./A/4(//!1A/(/* ./CAC/4!/4(4 ./*/C(/!C(A/! ./C4A(CA14 ./*C//(*C(A/ ./C114(1(//* ./C(!*A/A/(/44 ./C(A*1(!4C4*** ./C(4A**A!!*1 ./C(/*/4!4!A ./C(/C!AC/A ./C(/4/A/A!! ./C(//**C(*A ./C(/C*C!*( ./C(//1CA1!(
c) x( (.* i
( 1 ! 4 / A * C 1( 11 1 1! 1
xBi (.*(((((((((((( (.//11(**A11CC (.*(A*1C*1CC (.*/!!(1(*// (.**C(141(A (.C((4(*/1(((! (.C(4/C4AA(1*(1/ (.C(*(AACC1(!A (.C(C11/!(4(4C/* (.C(C/(1!AAC/ (.C(C*!114*!1* (.C(CC!/A!1!41 (.C(CC/C!*A(1A (.C(CCC(1C*1(4/ (.C(CCCC//A11CA4
g-xBi) (.//11(**A11CC (.*(A*1C*1CC (.*/!!(1(*// (.**C(141(A (.C((4(*/1(((! (.C(4/C4AA(1*(1/ (.C(*(AACC1(!A (.C(C11/!(4(4C/* (.C(C/(1!AAC/ (.C(C*!114*!1* (.C(CC!/A!1!41 (.C(CC/C!*A(1A (.C(CCC(1C*1(4/ (.C(CCCC//A11CA4 (.C1(((!CA44!(1
17
d)x( (.4 i
xBi
g-xBi)
(.4((((((((((((( (.A1!4C4CCC/ 1 (.A1!4C4CCC/ (.*/*/A*/4AAA (.*/*/A*/4AAA (.4/A//(/A*(/A ! (.4/A//(/A*(/A (.AC!C*CC!! (.AC!C*CC!! (./4C4C11A1C 4 (./4C4C11A1C (.A41C/A!4/A1 / (.A41C/A!4/A1 (./4(CCC1A1AA A (./4(CCC1A1AA (.A*1/!(4( * (.A*1/!(4( (./A1!AA!/ACC C (./A1!AA!/ACC (.A144CCA/44 1( (.A144CCA/44 (./*444C(!CC 11 (./*444C(!CC (.A(//!!/C41! 1 (.A(//!!/C41! (./**!4*4(( 1! (./**!4*4(( (.A(1/A4!CC( 1 (.A(1/A4!CC( (./C1A4C(*4 14 (./C1A4C(*4 (./CC/1/C*/*14 1/ (./CC/1/C*/*14 (./C!/41**(* 1A (./C!/41**(* (./C*1/(1!1/!4( 1* (./C*1/(1!1/!4( (./CA4A1A/ 1C (./CA4A1A/ (./CA!ACA4C(A/ ( (./CA!ACA4C(A/ (./C4!*1*!1ACA 1 (./C4!*1*!1ACA (./C/*4/1/*!C (./C/*4/1/*!C (./C4A(4!*1/4 ! (./C4A(4!*1/4 (./C/4*1(/(!1 (./C/4*1(/(!1 (./C4C4A4!(1 4 (./C4C4A4!(1 (./C/4C!!*A / (./C/4C!!*A (./C/(/CCC(/** A (./C/(/CCC(/** (./C/!!A!1!C/1 * (./C/!!A!1!C/1 (./C/1CCA*/!* C (./C/1CCA*/!* (./C/*//*ACA/! !( (./C/*//*ACA/! (./C/1/*4/! !1 (./C/1/*4/! (./C/4AA//A* ! (./C/4AA//A* (./C/1C(1(1*4(1 se %an reali7ado ! iteraciones. e)x( (.4 i xBi
( 1 ! 4 /
(.4((((((((((((( (.(**C(/!*/ (.*11CCA!A!( (.!*(A*14/ (./C*C*AA!C/ (.!1!A(A!C1!1 (./1/*/(!14C*!
g-xBi) (.(**C(/!*/ (.*11CCA!A!( (.!*(A*14/ (./C*C*AA!C/ (.!1!A(A!C1!1 (./1/*/(!14C*! (.!A4*/AA*1*1*
18
A (.!A4*/AA*1*1* (.4/*14*1/(C * (.4/*14*1/(C (.1(*/*!*C1C C (.1(*/*!*C1C (.4!/CC14*4(4 1( (.4!/CC14*4(4 (.!4/1(!/CAA 11 (.!4/1(!/CAA (.41/C(*4/*A1 1 (.41/C(*4/*A1 (.414C1*1/1A 1! (.414C1*1/1A (.4(((4(!C*(( 1 (.4(((4(!C*(( (./1C(/!A/1 14 (./1C(/!A/1 (.C4/*CA4(A 1/ (.C4/*CA4(A (./*44A!C!/C!/ 1A (./*44A!C!/C!/ (.C(!!(1*!1A 1* (.C(!!(1*!1A (.A*A44AA 1C (.A*A44AA (.*/**!/411/C ( (.*/**!/411/C (.A4/1!/*AA1 1 (.A4/1!/*AA1 (.*//(A1* (.*//(A1* (.AA!C/*/A*A ! (.AA!C/*/A*A (.*!AAC/(A1 (.*!AAC/(A1 (.A*4/!((C!(( 4 (.A*4/!((C!(( (.*!(4!4! / (.*!(4!4! (.AC*A!!1A!( A (.AC*A!!1A!( (.*1A(C41!!C(( * (.*1A(C41!!C(( (.ACA/*(C1*/* C (.ACA/*(C1*/* (.*1!/(C!11* !( (.*1!/(C!11* (.*((A/C!1 !1 (.*((A/C!1 (.*1(A**A!1!* ! (.*1(A**A!1!* (.*(A1(!1!!A !! (.*(A1(!1!!A (.*(C1C4A*A*(( ! (.*(C1C4A*A*(( (.*(!C*C(AAC( !4 (.*(!C*C(AAC( (.*(*1/C11!C*/ !/ (.*(*1/C11!C*/ (.*(*1!A*!4 !A (.*(*1!A*!4 (.*(A4(A(4A( !* (.*(A4(A(4A( (.*(4!CA! !C (.*(4!CA! (.*(A(*(CA* ( (.*(A(*(CA* (.*(4/*/**1(C 1 (.*(4/*/**1(C (.*(/*(/(4// (.*(/*(/(4// (.*(4C(A44444C ! (.*(4C(A44444C (.*(//***1*4 se %an reali7ado ! iteraciones. f) x( (.* i xBi
g-xBi)
( (.*((((((((((((( (.A(A(!1((1!! 1 (.A(A(!1((1!! (.A(C!/4CA1!4* (.A(C!/4CA1!4* (.A(*1C(**//(1 ! (.A(*1C(**//(1 (.A(*1(4!/4C( (.A(*1(4!/4C( (.A(*1(C/1!( se %an reali7ado iteraciones.
19
. en $ada #na de la! !i%#iente! e$#a$ine! determine #na /#n$i=n #n interval a>, dnde la! intera$$i=n de 3#nt /i? $nver%en en #na !l#$i=n 3!itiva de La e$#a$i=n> ,ten%a la! !l#$ine! $n #na e(a$tit#d de 10 )8 a' *(2 e( + 0 ,' () $(+ 0
solucion: a)
3 x 2 − e x f ( x )
=0
= 3x 2 − e x
e x
1/ 2
Para el problema Ac, podemos buscar una solución en el intervalo
[ 0.25 , 1] por el mDtodo de punto fiEo, iniciali7ando en
x0
= 0.28 . ;e
obtiene
= 0.28 x1 = g ( x0 ) = 0.664110 x2 = g ( x1 ) ?0.804728 x0
x14
b)
= g ( x13 ) = g ( 0.90999) = 0.9099996
x = cos x = g ( x )
&niciali7amos en x0 = 0.7 x1 = g ( x0 )
= 0.764842
x2
= g ( x1 ) = 0.721491
x3
= g ( x2 ) = 0.750821
x24
= g ( x23 ) = g ( 0.739089) = 0.739082
20
Ac) x( (.* i ( 1 ! 4 / A * C 1( 11 1 1! 1
xBi
g-xBi)
(.*(((((((((((( (.//11(**A11CC (.*(A*1C*1CC (.*/!!(1(*// (.**C(141(A (.C((4(*/1(((! (.C(4/C4AA(1*(1/ (.C(*(AACC1(!A (.C(C11/!(4(4C/* (.C(C/(1!AAC/ (.C(C*!114*!1* (.C(CC!/A!1!41 (.C(CC/C!*A(1A (.C(CCC(1C*1(4/ (.C(CCCC//A11CA4
c)
x = cos x = g ( x )
(.//11(**A11CC (.*(A*1C*1CC (.*/!!(1(*// (.**C(141(A (.C((4(*/1(((! (.C(4/C4AA(1*(1/ (.C(*(AACC1(!A (.C(C11/!(4(4C/* (.C(C/(1!AAC/ (.C(C*!114*!1* (.C(CC!/A!1!41 (.C(CC/C!*A(1A (.C(CCC(1C*1(4/ (.C(CCCC//A11CA4 (.C1(((!CA44!(1
&niciali7amos en x0 = 0.7 x1 = g ( x0 )
= 0.764842
x2
= g ( x1 ) = 0.721491
x3
= g ( x2 ) = 0.750821
x24
= g ( x23 ) = g ( 0.739089) = 0.739082
b)&niciali7ando en x( (.A se obtiene los siguientes resultados al aplicar el mDtodo de punto fiEo dado por: x-n21) g-x-n)) para todo n G (, la tabla se muestra a continuación. & xBi g-xBi) cos-xBi) ( (.A((((((((((((( (.A/*1*A*C 1 (.A/*1*A*C (.A1C1/!C4CA4! (.A1C1/!C4CA4! (.A4(*1!**!C4 ! (.A4(*1!**!C4 (.A!11*AA4A!!/ (.A!11*AA4A!!/ (.A11*!/A1/ 4 (.A11*!/A1/ (.A!4*((((4CC / (.A!4*((((4CC (.A14(*/41//(( A (.A14(*/41//(( (.A!A4(4!14(1* * (.A!A4(4!14(1* (.A(1*4*4!C/A/ C (.A(1*4*4!C/A/ (.A!*!!/1(!41(( 1( (.A!*!!/1(!41(( (.A!C4**/C4!4 11 (.A!C4**/C4!4 (.A!*A*A(C//(C 21
1 (.A!*A*A(C//(C (.A!C!11A1(!!*(1 1! (.A!C!11A1(!!*(1 (.A!*C!*C1/CA( 1 (.A!*C!*C1/CA( (.A!C1*ACA/C44 14 (.A!C1*ACA/C44 (.A!C(14*A!C(1 1/ (.A!C(14*A!C(1 (.A!C1!1A*/!/A11 1A (.A!C1!1A*/!/A11 (.A!C(4!A(/*/4( 1* (.A!C(4!A(/*/4( (.A!C1(/!((A!/ 1C (.A!C1(/!((A!/ (.A!C(A(*A!A( ( (.A!C(A(*A!A( (.A!C(CA!**!C44 1 (.A!C(CA!**!C44 (.A!C(A*//A1/C! (.A!C(A*//A1/C! (.A!C(*CC1*(41! ! (.A!C(*CC1*(41! (.A!C(*1CA(C4( (.A!C(*1CA(C4( (.A!C(*A11(C(A( se %an reali7ado iteraciones.
9. a3li"#e el mtd de Netn 3ara ,tener !l#$ine! $n #na e(a$tit#d de 10)8 3ara l! !i%#iente! 3r,lema! a' e( e)( 2 $! ( )6 + 0
3ara
,' Ln ()1' $! ()1' + 0
1
≤ x≤2
3ara
1 .3
≤x≤2
$' 2( $! 2( ( )2'2 + 0
3ara
2
≤ x≤3
3
≤ x≤4
d' e( * (2 + 0
3ara
0
≤ x ≤1
3
≤ x≤5
e' ()2'2 Ln ( + 0
3ara
1
≤x≤2
e
≤x≤4
1
≤ x ≤ 2
;olucion: a)
f ( x )
= e x + e − x + 2Cos ( x ) − 6
f ' ( x )
= e x .e x − 2Senx
,
&niciali7amos en x0 = 1.5 Itili7amos la relación: xi +1 − xi
b)
−
f ( xi ) f ' ( xi )
Ln ( x − 1) + Cos ( x − 1) f ( x )
?
∀i = 0,1,2,....
=0
,
1.3
JJJJJJJ-K)
≤2≤2
= Ln ( x − 1) + Cos ( x − 1)
22
f ' ( x )
=
1 x − 1
− Sen ( x − 1)
&niciali7amos en x0 = 1.3 x @ luego usamos -K) c)
f ( x )
= 2 xCos ( 2 x ) − ( x − 2 ) 2
f ' ( x )
= 2Cos ( 2 x ) − 4 xSen ( 2 x ) − 2 ( x − 2 )
&niciali7amos en x0 = 2.5 $ usamos -K) &niciali7amos en x0 = 3.5 $ usamos -K) d)
f ( x )
= e x − 3x 2
f ' ( x )
= e x − 6 x
d.1 ) &niciali7amos x0 = 0.5 $ usamos -K) d. ) &niciali7amos x0 = 4 $ usamos -K) e)
f ( x )
= ( x − 2 ) 2 − Ln ( x )
f ' ( x )
= 2 ( x − 2) −
1 x
e.1 ) &niciali7amos x0 = 1.5 $ usamos -K) e. ) &niciali7amos x0 = 3.8 $ usamos -K)
a)
x1 = x0 −
x2
= x1 −
x5
= x4 −
f ( x0 ) f ' ( x0 ) f ( x1 ) f ' ( x1 )
f ( x4 ) f ' ( x4 )
= 1.5 −
f ( 1.5 ) f ' ( 1.5 )
= 2.0097 −
= 1.8579 −
= 1.5 −
f ( 2.0097 ) f ' ( 2.0097 )
f ( 1.8579 ) f ' ( 1.8579 )
( −1.1537 ) 2.26357
= 2.0097
= 2.0097 −
0.7451 2.2635
= 1.8746
= 1.85792
23
x1 = x0 −
b)
x4
= x3 −
x1 = x0 −
c)
f ( x0 ) f ' ( x0 )
f ( x3 ) f ' ( x3 )
f ( x0 ) f ' ( x0 )
= 1.3 −
f ( 1.3 ) f ' ( 1.3)
= 1.3977 −
= 2.5 −
= 1.3 −
( −0.2486)
f ( 1.3977 ) f ' ( 1.3977 )
f ( 2.5 ) f ' ( 2.5 )
= 2.5 −
1.4696
= 1.3818
= 1.3977 −
1.1683
( −0.3308)
( −0.0000006) 2.1299
= 1.397748
= 2.3724
c.1) x3
= x2 −
f ( x2 ) f ' ( x2 )
c.) x1 = x0
x4
= x3 −
−
= 2.3706 −
f ( x0 ) f ' ( x0 )
f ( x3 ) f ' ( x3 )
= 3.5 −
= 3.7221 −
0.000007 8.80466 f ( 3.5 ) f ' ( 3.5 )
= 2.370686
= 3.5 −
f ( 3.7221) f ' ( 3.7221)
3.027316
( −10.69000 )
= 3.7221 −
== 3.7831
( −0.00004) = 3.722112 ( −16.3468)
5e manera análoga para las funciones restantes
Problem 9 ) es#l$%os ob$e&i%os i i
(i)
0 1.50000000000000 *1.15370636617810 1 2.00968466094811 0.74512833500593 2 1.87461161044704 0.07337568063074 3 1.85814662637326 0.00097930038952 4 1.85792087107424 0.00000018179411 5 1.85792082915020 0.00000000000001 +& si%o &ecesris 5 i$ercio&es b) i
0 1 2 3
i 1.30000000000000 1.38184713964704 1.39732073293914 1.39774816447362
(i) *0.24863631520033 *0.03475699865755 *0.00091039034054 *0.00000066246944
24
4 1.39774847595858 *0.00000000000035 +& si%o &ecesris 4 i$ercio&es c) c.1) i
i
(i)
0 2.50000000000000 1.16831092731613 1 2.37240732118090 0.01513958337497 2 2.37068782574746 0.00000798693159 3 2.37068691766252 0.00000000000225 +& si%o &ecesris 3 i$ercio&es c.2) i
i
(i)
0 3.50000000000000 3.02731578040313 1 3.78319116470837 *1.03355918205604 2 3.72416540142925 *0.03350943779715 3 3.72211549727338 *0.00004441336485 4 3.72211277310661 *0.00000000007867 +& si%o &ecesris 3 i$ercio&es obsr-ese #e ls rces ob$e&i%s so& %iere&$es. %) %.1) i
i
(i)
0.50000000000000 0.89872127070013 1 1.16508948243844 *0.86609073643071 2 0.93622693756065 *0.07922198819621 3 0.91039666487202 *0.00115808989446 4 0.91000766186313 *0.00000026595184 5 0.91000757248871 *0.00000000000001 se +& reli%o 5 i$ercio&es %.2) i
i (i) 4.00000000000000 6.59815003314424 1 3.78436114516737 1.04337931099471 2 3.73537937507954 0.04474262358146 3 3.73308389787410 0.00009450832188 4 3.73307902865469 0.00000000042450 se +& reli%o 4 i$ercio&es obsr-ese #e ls rces ob$e&i%s so& %iere&$es. e) e.1)
25
i i (i) 0 1.50000000000000 *0.15546510810816 1 1.40672093513510 0.01071863069140 2 1.41236995725119 0.00003995300368 3 1.41239117172501 0.00000000056286 4 1.41239117202388 0.00000000000000 se +& reli%o 4 i$ercio&es
e.2) i
i
(i)
0 3.80000000000000 1.90499893326766 1 3.22910126605543 0.33848606950532 2 3.07155735110681 0.02605044324437 3 3.05722460270714 0.00021634875099 4 3.05710355863191 0.00000001543548 5 3.05710354999474 0.00000000000000 se +& reli%o 5 i$ercio&es obsr-ese &#e-me&$e #e ls rces %iiere& r c% cso.
10. en$#entre #na a3r(ima$i=n de la e$#a$i=n de la 3,la$i=n 1 86->000 + 1 000>000 e
435,000 λ
$n #na e(a$tit#d de 10)- 3ara
e )1 '
U!e e!te valr 3ara 3rede$ir la 3,la$i=n "#e a,r7 al /inal del !e%#nd aF> !#3niend "#e la ta!a de inmi%ra$i=n d#rante e!te aF !e mantiene en -*8>000 3r aF. Nta el $re$imient de #na 3,la$i=n n#mer!a 3#ede mdelar!e d#rante 3erid! ,reve! dN (t ) dt
+ N t'GGGGGGG1'
indi$e $n!tante de natalidad . N t' $antidad de a,itante! en el tiem3 t La !l#$in de 1' e! N t' + N0 e t > n e! la 3,la$i=n ini$ial. La e$#a$i=n de 1' e! valida !i n a inmi%ra$i=n del e(terir> !i !e 3ermite Inmi%ra$i=n $n #na ta!a $n!tante H la E.D. !era
26
dN (t ) dt
+ N t' HGGGG......2'
Sl#$in de 2' N t' + N0 e
t
V λ
e )1 '
!#3n%a "#e $ierta 3,la$i=n tiene ini$ialmente 1 000 000 de a,itante!> "#e -*8 000 de ell! inmi%ran a!ta la $m#nidad el 3rimer aF "#e 1 86- 000 Se en$#entran en ella la /inal del aF 1. Si "#erem! determinar la natalidad de e!ta 3,la$i=n de,em! determinar 1 86->000 + 1 000>000 e
435,000 λ
e )1 '
5efiniendo f ( x )
= 1000000e x +
435000 x
( e x −1) −1564000
Isando el mDtodo de Newton con punto inicial x0 = 2 $ aplicando -K) resulta #ue: λ
= 0.0000001 x10 6 = ( 0.1 x10 −6 ) x10 6 = 0.1009
@ reempla7ando en : N ( t )
= N 0e λt +
V
e t − 1) ( λ λ
Para los valores: t = 2
= 0.1009 V = 425000 N 0 = 1000000 λ
;e obtiene: N ( t ) = 2165406 al final del do aLo Problem 10 lc#los reli%os
i 0 1 2
i 106 (i)106 0.00000200000000 7.21467580044807 0.00000113090498 2.34162681259594 0.00000047618624 0.60309228052234
27
3 4 5 6
0.00000016051940 0.00000010263928 0.00000010099920 0.00000010099793
0.08197903100495 0.00219945011100 0.00000170006609 0.00000000000102
11. a' Dem#e!tre "#e 3ara $#al"#ier enter 3!itiv la !#$e!in de/inida 3r Pn+
1 n
k
$nver%e linealmente en P + 0
,' Para $ada 3ar de enter ! m> determine #n n#mer N 3ara la $#al 1 N
a) P n = P 0
k
J 10)m
1 n k
=0 1
Pn +1 − P Pn − P
n ( =
k
+ 1)
k
−0
1 n k
k 1÷ n = = 1÷ k + n 1 ( ) 1+ ÷ n
k
k
P − P 1 ÷ ⇒ lim n+1 = lim ÷ =1 1 n→∞ Pn +1 − P n→∞ 1+ ÷ n ∴ P n converge linealmente b)
;ean k , m > 0 entonces ⇒ 10m ∈ N
⇒ ∃ N ∈ ¥ /10m < N k 28
⇒ N − < 10 − K
⇒
1 K
N
< 10 −
m
m
12. n
a' Dem#e!tre "#e la !#$e!i=n Pn + 10 2 $.v $#adrati$amnte en $er ,' Dem#e!tre "#e la !#$e!i=n Pn + 10 n n $.v $#adrati$amnte a $er !in im3rtar el tamaF del e(3nente K1. −
−
k
Sl
a) P n = 10−2
n
= 0
P
( n+1)
P n+1 = 10−2 Pn+1 − P Pn
− P
2
=
( n+1) − 2 10
10−2n ÷
=
( n +1) − 2 10
−2( n+1) 10
=1
∴ P n converge cuadráticamente b) ;upongamos #ue P n = 10− n
k
converge cuadráticamente a cero -P ()
9uego : Pn+1 Pn
− P
− P 2
=
10
−( n+1) k
10− nk ÷
2
= 102 n
k
−( n+1) k
;i P n converge cuadráticamente , entonces lim
n→∞
Pn+1 − P n Pn − P
2
2
= λ < ∞ -es finito)
;ea k cual#uiera pero fiEo, se cumple #ue : 2n k − ( n + 1) k ≥ 0 ∀n ≥ 4 k k 9uego : lim 102n −( n +1) → ∞ n→∞
29
∴ P n no converge cuadráticamente
1*. dem#e!tre "#e el mtd de la ,i!e$$i=n de #na !#$e!i=n $n #na $ta de errr "#e $nver%e linealmente a $er !l
9a cota del error de cada iteración por el mDtodo de la bisección viene dado por: C n
=
b−a 2n
@ Dsta converge linealmente a cero, pues : C n+1 − 0 lim n→∞ C n − 0
=
b−a 2n +1
lim n→∞ b − a
1
= <∞ 2
2n
1-. la! !i%#iente! !#$e!ine! !n linealmente $nver%ente! %enere 8 3rimer! trmin! de la !#$e!in Pn 3r medi del mtd 2 de Aiten.
a) b) c) d)
Pn - 0 e Pn"1 2 Pn"1)! Pn - e Pn"1 !)1 Pn ! Pn"1 Pn cos Pn"1
P( (.4 P( (.A4 P( (.4 P( (.4
n ≥1 n ≥1 n ≥1 n ≥1
;ol: a) P 0 = 0.5 P 1 =
P 2
P 3
= =
P 0
2−e
+ P 02
3 2 − e P 1 + P 12 3 2 − e P 2 3
+ P 22
= = =
2 − e0.5 + ( 0.5 ) 3 2 − e0.2004
2
= 0.2004
+ ( 0.2004 ) 2 3
2 − e0.2727
+ ( 0.2727 ) 2 3
= 0.2727 = 0.2536
30
P n
P 4
P 5
P 6
P 7
=
= = = =
P 2 − e n−1 + P n2−1
3
2−e
( 0.2536 )
+ ( 0.2536 ) 2 3
2−e
( 0.2586 )
+ ( 0.2586 ) 2 3
2−e
( 0.2573 )
+ ( 0.2573 ) 2 3
2−e
( 0.2576 )
+ ( 0.2576 ) 2 3
= 0.2586 = 0.2773 = 0.2576 = 0.2575
31
Por el mDtodo de 8tinen se obtiene: Pn
= P n −
( Pn+1 − P n )
2
( Pn+ 2 − 2Pn +1 + P n )
ntonces: 2
2
[ 0.2004 − 0.5 ] = 0.5 − = 0.2576 P0 = P 0 − 0.2727 − 2 ( 0.2004 ) + 0.5 ( P2 − 2 P1 + P 0 ) ( P1 − P 0 )
( P2 − P 1 )
2
2
[ 0.2727 − 0.2004 ] = ( 0.2004 ) − = 0.2575 P1 = P 1 − 0.2536 − 2 ( 0.2727 ) + 0.2004 ( P3 − 2P2 + P 1 ) P2
P3
P4
P5
= P 2 − = P 3 − = P 4 − = P 5 −
( P3 − P 2 )
2
( P4 − 2P3 + P 2 ) ( P4 − P 3 )
2
( P5 − 2P4 + P 3 ) ( P5 − P 4 )
= 0.2575
2
( P6 − 2P5 + P 4 ) ( P6 − P 5 )
= 0.2575
= 0.2575
2
( P7 − 2 P6 + P 5 )
= 0.2575
b) P 0 = 0.75
e P 0 P 1 = ÷ 3÷
1/ 2
e P 1 P 2 = ÷ 3÷
1/ 2
e P 2 P 3 = ÷ 3÷
1/ 2
eP n−1 P n = 3 ÷÷
e0.75 = ÷÷ 3
1/ 2
e0.84 = ÷÷ 3
1/ 2
= 0.8400
e0.8787 = ÷÷ 3
= 0.8787 1/ 2
= 0.8959
1/ 2
1/ 2 e P 3 e0.8959 P 4 = ÷ 1/ 2 = 3 ÷÷ = 0.9036 3÷
32
e P 4 1/ 2 e0.9036 1/ 2 P 5 = ÷ ÷ = 0.9071 3 ÷ = 3 ÷ e0.9071 P 6 = 3 ÷÷
1/ 2
e0.9087 P 7 = 3 ÷÷
= 0.9087 1/ 2
= 0.9094
@ los tDrminos por el mDtodo de 8itNem son: ( P1 − P 0 )
2
2
[ 0.84 − 0.75 ] P0 = P 0 − = 0.75 − = 0.9096 − + P 2 P P − + 0.8787 2 0.84 0.75 ( 2 ( ) 1 0) ( P2 − P 1 )
2
2
[ 0.8787 − 0.84 ] = 0.84 − = 0.9099 P1 = P 1 − 0.8959 − 2 ( 0.8787 ) + 0.84 ( P3 − 2P2 + P 1 ) ( P3 − P 2 )
2
( P4 − P 3 )
2
2
[ 0.8959 − 0.8787 ] = 0.8787 − = 0.9100 P2 = P 2 − 0.9036 − 2 ( 0.8959 ) + 0.8787 ( P4 − 2P3 + P 2 ) P3
P4
P5
= P 3 − = P 4 − = P 5 −
( P5 − 2P4 + P 3 ) ( P5 − P 4 )
2
( P6 − 2P5 + P 4 ) ( P6 − P 5 )
= 0.9100 = 0.9100
2
( P7 − 2 P6 + P 5 )
= 0.9100
Problem 14 & los roblems si#ie&$es +emos %e #$ilir l rm#l %e :i$e&, %% or P(&) < (&) = ((&>1) = (&))2/((&>2) = 2(&>1) > (&)) o&%e los (&) rerese&$& los eleme&$os %e l s#cesi& orii&l @ los P(&) los ob$e&i%os or el m$o%o %e :i$e& %%os e& l rimer @ se#&% col#m& resec$i-me&$e r c% cso ) (&) P(&)
0.20042624309996 0.27274906509837 0.25360715658413 0.25855037626494 0.25726563633509
0.25761321071575 0.25753583232667 0.25753066000103 0.25753031065960 0.25753028713916
33
b)
(&) 0.84003968489841 0.87872234966328 0.89588343485123 0.90360367538226 0.90709843499833
c)
0.90956750686718 0.90991689374599 0.90998883848610 0.91000369764865 0.91000677062656
(&)
P(&)
0.57735026918963 0.53031500464853 0.55843861277518 0.54144839213746 0.55164979836866
%)
P(&)
0.54791507369074 0.54784704206864 0.54782256546363 0.54781366575393 0.54781044996893
(&)
P(&)
0.87758256189037 0.63901249416526 0.80268510068233 0.69477802678801 0.76819583128202
0.73608669171302 0.73765287139640 0.73846922087626 0.73879806517359 0.73895771094142
18. $n!idere la /#n$i=n /('+ e6( * Ln2'2 e2( Ln' e-( Ln2'* a3li"#e el metd de Netn $n P 0 + 0 3ara a3r(imar #na rai< de / %enere termin! a!ta "#e 5Pn1)Pn5 J 0.0002. $n!tr#a la !#$e!i=n. Pn me?r la $nver%en$ia Sl
f ( x )
= e6 x + 3 ( Ln 2 ) 2 e2 x − ( Ln8) e4 x − ( Ln 2 ) 3
f ' ( x )
= 6e6 x + 6 ( Ln 2 ) 2 e2 x − 4 ( Ln8 ) e4 x
l mDtodo de Newton con x0 = 0 nos da: xi +1 = xi
−
f ( xi ) f ' ( xi )
⇒ x1 = x0 − x2
f ( x0 ) f ' ( x0 )
= x1 −
f ( x1 ) f ' ( x1 )
∀i ≥ 0
=0−
0.289 0.5650
= −0.0511 −
= −0.0511 0.0092 0.5650
= −0.0898 34
x3
x17
= x2 −
f ( x2 )
= −0.0898 −
f ' ( x2 )
= x16 −
f ( x16 )
0.0029 0.2380
= −0.1827 −
f ' ( x16 )
= −0.1182
0.000000004 0.000005
= −0.1828
8l aplicar el mDtodo de 8it Nen se obtiene: n x
( xn+1 − xn )
=x − n
2
( xn+ 2 − 2 xn+1 + xn )
9uego :
= x0 −
x0
1 x
=x −
15 x
1
( x1 − x0 )
2
x3
=x − 15
= −0.21022
( x2 − 2 x1 + x0 ) ( x2 − x1 )
(
2
− 2x + x ) 2
1
( x16 − x15 )
= −0.19657
2
( x17 − 2 x16 + x15 )
= −0.1828
Problem 15 Aos res#l$%os ob$e&i%os emle&%o el m$o%o %e BeC$o& i&ici&%o e& 0<0 co& #& $oler&ci %e error %e 0.0002 se m#es$r& e& l si#ie&$e $bl i i (i) 0 1 2 3 4 5 6 7 8 9 10 11
0 *0.05114213657334 *0.08984247581503 *0.11824473602582 *0.13856559584001 *0.15281619296930 *0.16266025259029 *0.16938617562231 *0.17394606447475 *0.17702081377114 *0.17908645522777 *0.18047067668126
0.02889284808584 0.00921156839215 0.00288671499445 0.00089158298625 0.00027219697652 0.00008237076153 0.00002476593507 0.00000741200380 0.00000221115908 0.00000065817836 0.00000019561985 0.00000005808176
35
12 *0.18139668915355 0.00000001723331 13 *0.18201546138155 0.00000000511090 14 *0.18242861474714 0.00000000151528 15 *0.18270433495676 0.00000000044916 16 *0.18288827515368 0.00000000013312 se +& $e&i%o #e eec$#r 16 i$ercio&es r ob$e&er l $oler&ci e%i%. :l licr el m$o%o %e :i$e&, se ob$ie&e l si#ie&$e s#cesi& *0.21022028116731 *0.19657860612537 *0.18966284672758 *0.18627100853214 *0.18465155747288 *0.18389421250327 *0.18354544501206 *0.18338660154990 *0.18331481859909 *0.18328255338501 *0.18326810401886 *0.18326164926631 *0.18325877060686 *0.18325748847678 *0.18325691746531 *0.18270433495676 *0.18288827515368 #e se ob$ie&e e& 15 sos, &$ese #e e& me&os sos, l s#cesi& Dl$im co&-ere ms ri%o #e l #e ob$#-imos rimero. ec#r%ese #e el m$o%o %e :i$e& -ie&e %%o or P(&) <(&) = ((&>1) = (&))2/((&>2) = 2(&>1) > (&))
16. la! !#$e!ine! $nver%en a 0. #!e el mtd Pn a!ta "#e 5Pn5 8(10)2 ) P n + b) P n +
1
2
de Aiten %enerar
n ≥1
n 1 n
2
n ≥1
Sl
a) Pn = 1/ n
∀n ≥ 1
P 1 = 1 P 2
=
1 2
36
P 3
=
1
P 4
=
1
P 9
= 0.1111
3 4
@ los tDrminos #ue se obtienen por el mDtodo de 8tinen son: P n
= P − n
P1 = P 1 −
( Pn+1 − P n )
( Pn+ 2 − 2Pn+1 + P n ) ( P2 − P 1 )
( P3 − P 2 )
= P −
P 3
= 0.1250
2
= P 9 −
b) Pn = P 1 = P 2
=
P 3
=
P 4
=
2
( P10 − P 9 ) ( P11 − 2 P10
= 0.1666
2
+ P 9 )
= 0.05 ≤ 0.05
∀n ≥1
n2 1
= 0.2500
( P4 − 2P3 + P 2 )
1
1
2
( P3 − 2 P2 + P 1 )
P 2
P9
2
=1
1 22 1 32 1 42
1
= = 0.25 4
1
= = 0.1111 9
=
1 16
= 0.1625
Por el mDtodo de 8it nen se obtiene: Pn
= P n −
( Pn+1 − P n )
2
( Pn + 2 − 2 Pn+1 + P n )
37
