Taller Resuelto de Analisis Numerico

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Taller Resuelto de Analisis Numerico Uploaded by Andrea Ospino García

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Andres Ramirez Ejercicios 3 y 8

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TALLER PRIMER SEGUIMIENTO

Autores: ZULY MILENA ALTUVE BERDUGO – 20111170! MARIA "AMILA ES"ALANTE "ARRAS"AL # 2011117002 MANUEL ANDRES $UENTES "UADRADO – 201111700% VANESSA VEGA &ERNANDEZ # 20111170%'

Do(e)te Sign up to vote on this title

Es*+ LEIDER SAL"EDOGAR"IA Useful  Not useful

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Ejericios resueltos de analisis numerico guia03

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TALLER PRIMER SEGUIMIENTO n

| xi| es u) )or+ 1+ Prue.e /ue ‖ x‖1=∑ i =1 Sugerencia: Debe probar que

‖ x‖1

cumpla con los cuatros axiomas de

norma vectorial que son la no negatividad, la nulidad, la homogeneidad y l desigualdad triangular.

Sou(34): n x =⟨ x , x , … , x n ⟩ ∈ R 1

2

y = ⟨ y 1 , y 2 ,… , y n ⟩ ∈ R

1) Primer axioma

n

‖ x‖1 ≥ 0

| x|i ≥ 0 para i =1,2, … ,n

Se sabe que

You're Reading a Preview

De esta manera

| x|1 +| x|2 +…Unlock +| x|n ≥full0 access with a free trial. Download With Free Trial

n

Luego

| x |≥ 0 ∑ = i

i

Por tanto

1

‖ x‖1 ≥ 0

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 2) Segundo axioma

‖ x‖1= 0 ↔ x =0⃗

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‖kx‖1=|k |‖ x‖1

3) Tercer axioma

‖kx‖1=‖k ( x 1 , x 2, … x , n )‖1=‖( kx 1 , kx 2, … , kx n )‖1

Tenemos que

n

n

i=1

i=1

|kxi|=∑ |k || x i|=¿|k |∑ | xi|=|k |‖ x‖

1

=

n

¿ ∑ = i 1

4) Cuarto axioma

‖ x + y‖1 ≤‖ x‖1+‖ y‖1

‖ x + y‖1 ≤‖ x‖1+‖ y‖1

Para probar que

, se ace uso de !a desigua!dad de

"in#o$s#%, !a cua! estab!ece que

[∑ [ ] ] [ ∑ [ ] ] [ ∑ [ ] ] 1

n

p p

i =1

n

1

n

p p

x i + y i ≤ xi y i + You're Reading a Preview i =1 i =1

1

p p

Unlock full access with a free trial.

Tenemos que p=1 seg&n !a desigua!dad de "in#o$s#% nos queda' 1 1 Download With Free Trial n n

‖ x + y‖1 ≤

[ ][ [ x ] ∑ = i

1

+

i 1

Por tanto'

[ y ] ∑ = i

i 1

1

]

‖ x + y‖1 ≤‖ x‖1+‖ y‖1 Sign up to vote on this title

La norma

‖ x‖

1



Useful

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cump!e con !os cuatro axiomas 1, 2, 3, 4 entonces

una norma (ectoria!

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es

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Pa +)= det + A − λI )

[

A − λI =

[

][ ]

−1 −1 d − 1 − λ −1 d

d

−1 −1

0 1 0

0 0 1

][ ]

−1 − 1 λ 0 0 d −1 − λ 0 λ 0 0 0 λ −1 d

d

A − λI = −1

−1

[

d − λ

A − λI = −1

Sea

1 0 0

−1

− 1 −1 d − λ −1 −1 d − λ

]

n = d − λ , entonces You're Reading a Preview n −1

[

( A − λI )= DET −1trial.n P ( n ( λ ) )= DET Unlock full access with a free

−1 −1

−1 −1 n

]

Download With Free Trial

|

n

P ( n ( λ ))= DET ( A − λI )= −1

−1

|

− 1 −1 n −1 −1 n


P ( n ( λ ))= n

−1 −(−1 ) −1 −1 +(−1 ) −1 n  n −1 n −1 −1

|− | | n

1

| |

|

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1 0− 3 −2|2 242

1210

Se obtiene que' P ( n ( λ ))= n −3 n −2=( n −2 ) ( n + 2 n +1 ) 3

2

2

-actori.ando a

n + 2 n +1

P ( n ( λ ))= n −3 n −2=( n −2 ) ( n + 1 )( n + 1 ) 3

Para a!!ar !os (a!ores caractersticos acemos P ( n ( λ ) )= 0 P ( n ( λ ))= ( n−2 ) ( n + 1 ) ( n + 1 )= 0 You're Reading a Preview

ntonces tenemos

Pero

n = d − λ , entonces

Unlock full access n −2with =0anfree + 1trial. =0


•

d − λ −2 =0 Sign up to vote on this title

λ 1= d −2

 •

d − λ + 1=0

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ρ ( A )=|d + 1|

n∗n %+ Se A ∊ R u) tr36 3)ert3.e- 5e>3)3os e )?ero 5e (o)53(34

−1 5e A (oo: Con ( A) =‖ A‖‖ A ‖ −1 < Prue.e /ue Con ( A) =Con ( A ) 9 Con ( A ) ≥ 1

Su@ere)(3: Prt 5e e(o /ue 1 ‖ I ‖ + Sou(34): a) ti!i.ando !a deinici/n de n&mero de condici/n se tiene que A

−1

¿ Con ( A ) =‖ A‖‖ A ‖ % ¿ 1 − −1 ‖ Aaccess ‖¿ with a free trial. Con ( Unlock A ) =full You're ¿ Reading a Preview

−1

Download With Free Trial A

Pero se sabe qu0'

−1

¿ ¿ ¿

De esta manera se tiene que'

Con ( A

−1


) =‖ A− ‖‖ A ‖ Useful 1

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A



−1

¿ ¿ ¿ −1 ‖ A‖‖ A ‖=‖ A−1‖¿ Demostrando que' Con ( A) =Con ( A

b) De

acuerdo

‖ A‖‖B‖≥‖ AB‖

−1

con

) !a

proposici/n

,se sabe que'

‖ A‖‖ A−1‖≥‖ A A−1‖ Pero tambi0n es conocido e! eco que' You're Reading a Preview −1 A A = I  Por !o que' ‖ A A ‖=‖ I ‖ −1

Unlock full access with a free trial.

ntonces'


‖ A‖‖ A− 1‖≥‖ I ‖ Se p!antea en e! enunciado que

‖ I ‖=1.


n ese sentido se airma que'

‖ A‖‖ A−1‖≥ 1



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'+ D5  tr36


 

A =

 Search document

[ − ] - 5o)5e 1

ε

1

Slides MetNum u2

1 +ε 1

( )

2 +ε Con ( A ) = )or 3)>3)3t /ue ε

ε ˃0.



Prue.e us)5

2

+

Sou(34):

‖ A‖∞

5nicia!mente procedemos a a!!ar 2

‖ A‖∞= max ∑ |anj|=max j =1

{∑ | 2

j =1

2

a1 j|,

}

|a | ∑ = 2 j

j 1

¿ max {|a11|+|a12| ,|a21|+|a22|} ¿ max {|1|+|1+ ε|,|1− ε|+|1|} 1−ε + 1 }= ¿ max { 1You're + 1 + ε ,Reading max { 2 + ε , 2− ε } a Preview Unlock full access ε a free trial. ¿ 2 +with

Download With Free Trial ‖ A‖∞=2 + ε

Posteriormente a!!amos !a in(ersa de !a matri.' −1

A =

1

de ( A

[) − − ]= −( + ) ( − ) [ d #

" a

1

1

1 ε 1

ε

 Sign up to vote on this title 1  Useful −1−ε =  12Not 1useful−1−ε 1 1 ε −1 ε ε −1

] [

]

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2

‖ A ‖∞= max ∑ |anj|= max −1

Slides MetNum u2

j = 1

{∑ | 2

j = 1

2

a1 j|,

∑= |a



}

|

2 j

j 1

¿ max {|a11|+|a12| ,|a21|+|a22|} ¿ max ¿ max

¿ max

{ + { 1

ε

2

+

+ε

1

ε

,

1

2 ε 2−ε

ε

2

,

ε

2

−ε ε

}

2

+

1

ε

2

{| | | | | | | |} 1

ε

} { =max

1

+

2

−1− ε ε −1

2

,

1

− ε +1

ε

+ε + 1 ε

2

,

ε

ε

2

2

+

1

ε

2

}

= 2 +2ε ε

‖ A− ‖∞= 2 +ε 1

ε

2


-ina!mente se a!!a e! n&meroUnlock de condici/n' full access with a free trial.

( )

( )

( 2+ε ) 2 + ε Trial Con ( A ) =‖ A‖Download ‖ A−1‖=( 2 +With ε ) Free = = ε2 ε2

Demostrando que'

( )

Con ( A) =

2 +ε

2

+ε

2

ε

2

ε


 n

de ( A )

2

∏ λ

λ λ , … , λ

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

So!uci/n' Se tiene que'

(¿ λ i− λ ) n

P A ( λ ) = de ( A − λI )=

¿ ∏ = i 1

Cuando λ = 0 '

(¿ λ i− 0 ) n

P A ( 0 )= de ( A − ( 0 ) ( I ) )=

¿ ∏ = i 1

(¿ λi ) n

P A ( 0 )= de ( A ) =

¿ ∏ = i

1


Demostrando que'

Unlock full access with a free trial. n

∏ =

Download With Trial ( A )= λ de 1 , λ 2 , … , λn i = λFree i 1

+ Su*o)@ /ue os )?eros 1-  2-F- ) so) *ro3(3o)es  1- 2 ) 9 /ue (5 u)o es e H3o error *os3.e 5e E+ Prue.e /ue H3o error *os3.e e)  su  1  2  F  ) es )E+  Sign up to vote on this title



Se sabe que' E

´ ¿ % %

´ ∈ R %

n

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7+ A*3/ue e Jto5o 5e *u)to >3Ko *r *ro3r  r86



& ( x )=√ x −

Use e or 3)3(3  0  0+C e 3tere st /ue E R ≤ 0.001 + "o)s3@)e o

resut5os o.te)35os e) (5 3ter(34) e) u) t.+ Tr.Ke (o) se 58@3tos 5e *re(3s34)+ Para ap!icar e! m0todo de punto i
√ x − x

1 ¿ x = √ x 2 ¿ x

 Para ta! eecto tenemos !as siguientes opciones

=( √ x )2−− ' x = x 2

2

Para dico proceso se!eccionamos g +x) % p!anteamos ormu!a de iteraci/n ( ( x ) =√ x x k = ( ( x k −1) x k =√ x k −1

k

x k = ( ( x k − 1 )

1 2 3 4 B ? > @ A 1 11

>>1? @4@AB A1>3 AB>?2 A>@B>1 A@A22> AA4BA@ AA>2AB AA@?4? AAA322 AAA??

¿ & You're ( x k ) ∨¿Reading a EPreview A =| x k − x k −1|

E R=

E A

Unlock full access with a free trial. 133>@A 2>1? 2A2@A2 >?1@ 133>@A 1BA13 Download With Free Trial 4BAA >?1@ @2AA? 2A?A 4BAA 423A> 1?B? 2A?A 2142@ B3>1 1?B? 1>>2 2?A> B3>1 B41  13B1 2?A> 2>4 Sign up to vote on this title ?>? 13B1  Useful  Not useful13B3 33@ ?>? ?>> 1?A 33@ 33@

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resut5os o.te)35os 5e (5 3ter(34) e) u) t.+ Tr.Ke (utro 58@3tos 5e *re(3s34)+ Tenemos que a = 1>B % b  = 2>B

# o=

a 0 + "0 2

=

1.75 + 2.75 =2.25 2

3 3 3 3 3 3 3 8= x ( √ 8 ) = x 8= x 0= x − 8 & ( x ) = x − 8 √

E A =|# k − # k −1|

k  1 2

ak

# k

"k

& ( ak )

& ( #k )

& ( " k )

|#

1>B 1>B 1>B

22B 2 2

2>B 22B 22B

2?4? 2?4? 2?4?

33A?  

12>A?@ 33A? 33A?

 

Luego de 3 iteraciones +# = 2)You're e! proceso se detiene %a que Readingiterati(o a Preview

E A < 10

De esta manera podemos decir que 2 cero o ra. aproximada de Unlock full access with aes freeun trial. unci/n dada Download With Free Trial

!+ L eo(355 (3 rr3. 5e u) (oete se *ue5e ((ur us)5o s3@u3e)te >oru rr3.-

*

) = * ln

(

m0 m 0− +

)−¿ Do)5e

)

es  eo(355 (




es  eo(355 (o)  /ue eUseful (o.ust3.e se ret3 Not useful 



(oete- m es  s 3)3(3 5e (oete e) e t3e*o  =0 - + 0

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(o)  0=26 - tr.Ke (o) o(o < 58@3tos 5e *re(3s34) 9 ter3)e 5 ) (  )=100

3terr st (u)5o

) (  )= * ln

(

m0 m0− +

) (  )=2200 ln

) (  )= 2200 /

(

m .

+

)−¿

|160000| |160000 −2680  |

−9,8 

)

428800000

( 160000− 2680  )2

− 9, 8

|160000| |160000− 2680  |

You're Reading a Preview /

) (  )= 2200

(

)−

2680 160000−2680 

Unlock 9, 8 full access with a free trial.

Download With Free Trial /

) (  )=

5896000 −9, 8 160000 −2680 

ntonces !a /rmu!a de interacci/n es' ( (  k −1 ) = x −

) (  ) =  k /


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k

 k

|) (  k − )|

a =¿| k −  k − 1| E ¿

1 2

1>A1>?1B2 >14>@11B

23B41A3411 1A4A3>>>3A

1@@23@4@ >22@34B

1

De acuerdo con !a condici/n inicia!, se detiene e! proceso de iteraci/n en #=2

Con !a anterior tab!a podemos apreciar que e! tiempo en e! cua! !a (e!ocidad ac arriba de! coete supera !os mi! metros es 2? s, tambi0n e! tiempo en que (e!ocidad aciende a 1A4A3>>>3A ms es de >14>@11B s

10+ A*3/ue e Jto5o 5e  se()te *r resoer  e(u(34) x = (o3e)(e (o) x 0=0

x 1=0.1

9

+ Ter3)e (u)5o

Er ≤ 10

"o)s3@)e os resut5os o.te)35os 5e (5 3ter(34) e) u) t. Tr.Ke (o) (utro 58@3tos 5e *re(3s34)+ − x 2

x =e

You're Reading a Preview Unlock full access with a free trial. − x 2

0 = x −e

Download With Free Trial − x2

& ( x )= x − e x

(¿ ¿ k + 1 −e− x ) & ( k + 1 ) =¿ Sign up to vote on this title 2

k + 1

 ntonces !a /rmu!a de interacci/n es'

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x k +1

|& (k + 1 )|

Ea

Er

1 2 3 4 B

AAB 2?@A BAA> ?B3 ?B31

4>22 444 A@ 3 4

@AB 2@? 2A2 B33 1

@A 143B 4?@ @1? 1

La ra. de !a ecuaci/n con !a condici/n inicia! es x = ?B31

You're Reading a Preview Unlock full access with a free trial.



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Taller Resuelto de Analisis Numerico

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