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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