Trabajo Colaborativo No 3

TRAB TRABAJ AJO O COLAB OLABOR ORA ATIVO IVO No 3

1

Trabajo Colaborativo No 3 Evaluación unidad 3 “Unidad 3. Difereniai!n e Inte"rai!n N#$%ria & '#aione( Difereniale( Ordinaria(.)

Jo(% *ernando Bola+o( Ra$,reIn"enier,a en 'letr!nia %todo( n#$%rio(/C#r(o 0

a&o 1 de 2145 6antia"o de Cali T#tor Je(7( O$ar var"a(

%todo( n#$%rio(

2 Introducción

'n e(te trabajo olaborativo (e de(arrollaran lo( ejeriio( ontenido( en la "#,a de atividade( de la #nidad 38 Difereniai!n N#$%ria5 Re"la del Tra9eio5 Re"la de 6i$9(on5 Inte"rai!n de Ro$ber"5 Inte"rale( #lti9le(5 %todo de '#ler5 %todo( de Ta&lor & %todo de R#n"e :#tta. Lo( re(#ltado( (e 9re(entaran en lo 9o(ible en for$a de odi"o en ;&t

3 'val#ai!n #nidad 3

“ Diferenciación e Integración Numérica y Ecuaciones Diferenciales Ordinarias.”

Punto 1.

;lantee & (ol#ione do( ejeriio( (obre Difereniai!n N#$%ria e=9liando 9a(o a 9a(o el 9roedi$iento #tili-ado. Eem!lo no 1

Dado( lo( 9#nto( de dato( #nifor$e$ente e(9aiado(8 =



.1

.2

.3

.0

f>=?

.

.1@

.1301

.10

.14@4

Cal#lar f>=? & f>=? en =   & = .2 #(ando a9ro=i$aione( de diferenia( finita( de ><2 ?.

"olución#


0

f ' ' ( 0 )=

2 f ( 0 )−5 f ( 0.1 )+ 4 f ( 0.2 )− f ( 0.3 )

(0.1 ) ²

2 ( 0 )−5 ( 0.0819 )+ 4 ( 0.1341 )−0.1646

( 0.1 ) ²

= − 3.77

De la (i"#iente tabla de oefiiente( de la( a9ro=i$aione( entrale( de la( diferenia( finita( de ><2? f>=H2
f>=H
f>=?

f>=F
2=?

H1



1

<2f>=?

1

H2

1

f>=F2
2<3f>=?

H1

2



H2

1

0f >0?>=?

1

H0



H0

1

f ' ( 0.2 )=

−f ( 0 )+ f ( 0.3) −0.0819 + 0.1646 = = 0.4135 2 ( 0.1 ) 0.2


G Eem!lo no $

U(ando lo( dato( del eje$9lo anterior al#le f>? tan 9rei(a$ente o$o (ea 9o(ible

"olución#

A9lia$o( la

e=tra9olai!n de Ri
a9ro=i$aione( de diferenia( finita(.

Iniia$o( on do( a9ro=i$aione( de diferenia <? 9ara f>?5 la 9ri$era #(ando <  .2 & la (i"#iente #(ando <  .1. U(ando la (i"#iente tabla tene$o( E#e8

f>=?

f>=F
f>=F2
2=?

H3

0

H1

<2f>=?

2

HG

0

H1

2<3f>=?

HG

1

H20

10

H3

0f >0?>=?

3

H10

2

H20

11

g ( 0.2 )=

−3 f ( 0 )+ 4 f (0.2 )− f ( 0.4 ) = 2 ( 0.2 )

f>=F3
f>=F0
3 ( 0 )+ 4 ( 0.1341 )−0.1797 0.4

=0.8918

f>=FG
H2


 p

G=

2

g ( h1 / 2 )− g ( h1) p

2

−1

Con 9  2 no(otro( obtene$o( E#e

f ' ( 0 )≈G =

2² g ( 0.1 )− g ( 0.2 ) 2² −1

=

4 ( 0.9675 )−0.8918 3

= 0.9927

K#e e( la a9ro=$ai!n de diferenia finita de >< ⁴?

P u n t oNo2 .

S o l u c i o n el o ss i g u i e n t e se j e r c i c i o su t i l i z a n d ol aRe g l ad e l T r a p e c i o . ( n =4 )

2

∫ 0

P r e s e n t a n d os o l u c i ó nd e :

x

2

3

1+ √ ( x )

dx

∫ 1

3

√ x e dx x


4

Us a n d ol af ó r mu l ag e n e r a ld el aRe g l ad el o st r a p e c i o s :

b

∫ f ( x) dx≈ h2 [ f ( a )+ 2 f ( a + h )+ 2 f ( a + 2 h)+ .... + f (b )] a

h=

b −a n

En c o n t r a mo sh :

h=

2− 0 4

=

2 4

=

1 2

Su s t i t u y e n d oe nl af ó r mu l ag e n e r a lq u e d a :

2

∫ 0

x

3

1 + √ ( x )

dx≈

1 1

[

1

0

( 0+ )

3

2 2 1 + √ ( 0 )

3

( 0 + 2∗

2

+2 1+

√

(0 + 1 ) 2

+2 1+

√

1 2

(0 +2∗

3

)

( 0 + 3∗ 1 2

+2 )

1+

√

1 2

( 0 +3∗

3

) 1 2

+ )

2

3

1 + √ ( 2 )

]

= 7.160155

1 4




P r e s e n t a n d os o l u c i ó nd e : 2

∫

3

√ x e dx x

1

1 / 3 x x e

Us a n d ol af ó r mu l ag e n e r a ld el aRe g l ad el o st r a p e c i o s :

b

∫ f ( x) dx ≈ h2 [ f ( a )+ 2 f (a + h)+ 2 f (a +2 h )+ .... +f ( b )] a


@ 2

∫

3

√ x e dx ≈ 5.45968 x

1

P u n t o3 .

S o l u c i o n e sl o ss i g u i e n t e se j e r c i c i o su t i l i z a n d ol aRe g l ad eSi mp s o n1 / 3y3 / 8 . ( n =4 ) 3

∫ 1

4

x

e dx x

∫ 2


∫ 1

x

e dx x

x

e

ln ( x ) dx


1

Su s t i t u i mo se nl af ó r mu l ag e n e r a l 3

∫ 1

x

3−1

e dx = x

6

(

f ( 1 )+ 4 f

( + ) + ( )) 1 3 2

f

3

Re e mp l a z a mo syr e s o l v e mo sp a r af ( x ) 3

∫ 1

x

e dx = x

3

∫ 1

1 3

(

1

e

1

e

2

+ 4 + 2

e

3

3

x

)=

e dx = 8.0639 x


∫ 2

x

e

ln ( x ) dx

8.0639


11

Us a n d ol af ó r mu l ag e n e r a l d e l mé t o d od eS i mp s o n b

∫ f ( x) dx =

b −a 6

a

(

f ( a )+ 4 f

( + ) + ( )) a b

f b

2

Su s t i t u i mo se nl af ó r mu l ag e n e r a l 4

∫

x

e

ln ( x ) dx =

2

4 −2 6

(

f ( 2)+ 4 f

( )+ ( )) 2+ 4 2

f

4

Re e mp l a z a mo syr e s o l v e mo sp a r af ( x ) 2

∫ 1

x

e

ln ( x ) dx =

1

(e ln ( 2 )+ 4 e ln (3 )+ e 3 2

3

4

2

∫ 1

x

e

ln ( x ) dx = 56.359

ln ( 4 )

) = 56.359

%todo( n#$%rio( Punto %. Ap l i c a re l mé t o d od eT a y l o rd eor d e ndo sal ae c u a c i ó ny ’=Co s ( x y ) , c o nl ac o n di c i ó ni n i c i a l :y ( 0 )= 1 .

P a r ar e s o l v e re s t ee j e r c i c i ou t i l i z a r e mo sl as e r i ed et a y l o rq ued i c eq u e : ( k )

P n ( x )=f ( a )+ f ´ ( a )( x −a )+

Yt o ma r e mo sah=0 . 5 L u e g o :

y ' ( x n)= f ( x , y n)= cos ( xn y n )

y ' ' =

y ' ' =

(n )

f ´´ ( a ) f ( a ) f ( a ) ( x −a )2 + . .. + ( x −a )k + .. . + ( x −a )n . k! n! 2!

d cos( xy ) = −sen ( xy ) y − sen ( xy ) y ' dx

d cos( xy ) = −sen ( xy )( y + cos ( xy )) dx

i t e r a mo si n i c i a n doc o nh =0 . 5

x 1= x 0+ h=0.5

%todo( n#$%rio( 2

y 1= y 0 + h cos ( x0 y 0 )+

h

2

2

0.5 ( −sen ( x y )( y + cos( x y )))=1 + 0.5cos0 + ( −sen 0 ( 1+ cos(0 )))=1.5 0

0

0

0

0

2

x 2= x 1+ h =1

2

2

h 0.5 y2= y 1 +h cos( x 1 y1 )+ ( −sen ( x1 y 1)( y1 + cos( x 1 y1 )))=1.5+ 0.5cos (0.51.5)+ ( −sen (0.51.5)( 1.5+ cos (( 0.51.5))))=1.6757 2

2

Trabajo Colaborativo No 3

Recommend Documents