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TALLER ANALISIS NUMERICO

INTERPOLACION POR EL METODO DE LAGRANGE

ANA MARIA DELGADO CUADRADO

DOCENTE

GERMAN ISAAC SOSA MONTENEGRO

UNIVERSIDAD POPULAR DEL CESAR

VALLEDUPAR CESAR

2016

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1. Obtener el polinoio !"#n$o l# %&r!l# $e interpol#'i&n $e L#Gr#n(e 'on lo" "i(!iente"

p#re" $e )#lore" (7,6 ) , (30,22) * e interpol#r en el p!nto x=4

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

L0(x )=

xx1

x0x1=x30730

=123

x+30

23

L1(x )=

xx0

x1x0=

x7307

= 1

23x

7

23

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

p (x )=6 (L0(x ))22(L1(x ))

p (x )=6 (123 x+ 3023 )22( 123 x 723 )p (x )=

1623

x26

23

Interpol#'i&n en el p!nto x=4

L0(4 )=4x1

x0x1=430

730=1.4783

L1(4 )=

xx0

x1x0=47307

=0.4783

Polinoio $e interpol#'i&n+

p (x )=6 (L0(4 ))22(L1(4 ))

p (x )=6 (1.4783 )22(0.4783 )

p (x )=1.6528

2. Obtener el polinoio $e interpol#'i&n !"#n$o l# %&r!l# $e interpol#'i&n $e l#(r#n(e 'on

lo" "i(!iente" )#lore" (1,10 ) , (4,10 ) , (7,34 ) * e interpol#r en el p!nto x=3

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El polinoio $e interpol#'i&n $e l#(r#n(e e"+

x

(0x1)(x0x2)=(x+4)(x+7)(1+4)(1+7)

= 1

40x

2+11

40+ 7

10

L0 (x )=(xx

1)(xx2)

x

(1x0)(x1x2)= (x1)(x+7)(41)(4+7)

=1

15x

22

5x+

7

15

L1(x )=

(xx0)(xx

2)

x

(2x0)(x2x1)= (x1)(x+4)(71)(7+4)

= 2

24x

2+1

8x

1

6

L2(x )=

(xx0)(xx

1)

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

p (x )=10(L0 (x ))+10 (L1 (x ))+34 (L2 (x ) )

p (x )=10

( 1

40 x

2

+

11

40 +

7

10

)+10

( 1

15 x

2

2

5 x+

7

15

)+34

(

2

24 x

2

+

1

8 x

1

6 )

p (x )=x2+3x+6

Interpol#'i&n en el p!nto x=3

x

(0x1)(x0x2)=(3+4)(3+7)(1+4)(1+7)

=6.4000

L0(3 )=(xx1)(xx2)

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x

(1x0)(x1x2)=(31)(3+7)(41)(4+7)

=9.6000

L1(3 )=

(xx0)(xx

2)

x

(2x0)(x2x1)=(31)(3+4)(71)(7+4)

=1.5000

L2(3)=

(xx0)(xx

1)

Polinoio $e interpol#'i&n+

p (x )=10

(L

0(3 )

)+10

(L

1(3 )

)+34

(L

2(3)

)p (x )=10 (6.4000 )+10 (9.6000 )+34 (1.5000 )

p (x )=109

3. Obtener el polinoio $e interpol#'i&n !"#n$o l# %&r!l# $e interpol#'i&n $e l#(r#n(e 'on l#

"i(!iente t#bl# $e )#lore", E interpol#r en el p!nto x=1

-. / / 2 /6. /16 /3 /10 /30

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

x

(0x1)(x0x2)(x0x3)= (x+3 ) (x2 ) (x+6)

(4+3 ) (42) (4+6 )=12x3+84x2432

L0(x )=

(xx1 )(xx2 ) (xx3 )

x

(1x0)(x1x2)(x1x3)= (x+4 )(x2)(x+6)(3+4)(32)(3+6)

=15x3120x260x+720

L1(x )=

(xx0)(xx

2)(xx

3)

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x

(2x0)(x2x1)(x2x3)=(x+4)(x+3)(x+6)(2+4)(2+3)(2+6)

=20

3x

3+260

3x

2+360x+480

L2(x )=

(xx0)(xx

1)(xx

3)

x

(3x0)(x3x1)(x3x2)= (x+4)(x+3)(x2)(6+4)(6+3)(62)

=12x360x2+24x+288

L3(x )=

(xx0)(xx

1)(xx

2)

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

p (x )=16 (L0(x ))5 (L1(x))10 (L2 (x ))50(L3 (x ))

p (x )=16 (12x3+84x2432 )5 (15x3120x260x+720 )

10 (203 x3+260

3x

2+360x+480)50 (12x360x2+24x+288)p (x )=416.33x3+1389.33x24500x15888

Interpol#'i&n en el p!nto x=1

x

(0x1)(x0x2)(x0x3)= (1+3) (12 ) (1+6 )

(4+3 ) (42) (4+6 )=336

L0(x )=

(xx1 )(xx2 ) (xx3 )

x

(1x0)(x1x2)(x1x3)= (1+4)(12)(1+6)

(3

+4

)(3

2

)(3

+6

)

=525

L1(x )=

(xx0)(xx

2)(xx

3)

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x

(2x0)(x2x1)(x2x3)=(1+4)(1+3)(1+6)(2+4)(2+3)(2+6)

=933.33

L2(x)=

(xx0)(xx

1)(xx

3)

x

(3x0)(x3x1)(x3x2)= (1+4)(1+3)(12)(6+4)(6+3)(62)

=240

L3(x )=

(xx0)(xx

1)(xx

2)

Polinoio $e interpol#'i&n+

p (x )=16 (L0(1 ))5 (L1(1))10 (L2(1))50 (L2(1))

p (x )=16 (336)5 (525)10 (933.33 )50 (240)

p (x )=18582.3

4. C#l'!l# el polinoio interpol#$or $e l#(r#n(e $e l# %!n'i&nf(x )=

1

x * en lo" no$o"

x0=1,x

1=2,x

2=3

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

x

(0x1)(x0x2)=(x2)(x3)(12)(13)

=2x210x+12

L0(x )=

(xx1)(xx

2)

x

(1x0)(x1x2)=(x1)(x3)

(2

1

)(2

3

)

=x2+4x3

L1(x )=

(xx0)(xx

2)

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x

(2x0)(x2x1)=(x1)(x2)(31)(32)

=1

2x

23

2x+1

L2(x )=

(xx0)(xx

1)

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

p (x )=f(1)(L0 (x ))+ f(2)(L1(x ))+ f(3) (L2(x ))

p (x )=1(2x210x+12)+0.5(x2+4x3)+0.33( 12 x232 x+1)p (x )=1.665x28.495x+10.83

5. Interpol#r l# %!n'i&n f(x )=xsen(x) por !n polinoio "e "e(!n$o (r#$o 4!e p#"e por lo"

p!nto"x

0=2,x

1=2.5,x

2=3

* '#l'!l#r el )#lor interpol#$o $e f(2.1)

x

(0x1)(x0x2)=(x2.5)(x3)(22.5)(23)

=2x211x+15

L0(x )=

(xx1)(xx

2)

x

(1x0)(x1x2)= (x2)(x3)(2 .52)(2 .53)

=x2+5x6

L1(x )=

(xx0)(xx

2)

x

(2x0)(x2x1)=(x2)(x2 .5)(32)(32.5 )

=1

2x

29

4x+

5

2

L2(x )=(xx0)(xx1)

El polinoio $e interpol#'i&n $e l#(r#n(e e"+

p (x )=f(1)(L0 (x ))+ f(2)(L1(x ))+ f(3) (L2(x ))

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p (x )=0.069 (2x211x+15)+0.109 (x2+5x6 )+0.157( 12 x294 x+ 52 )p (x )=0.1075x20.5672x+0.7735

V#lor interpol#$o $e f(2.1)

f(2.1)=p (2.1)=0.0563