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Métodos Numéricos Fase 2

The document contains two numerical exercises: 1) It defines a 5x5 matrix A where the element aij is defined by (i+j-1/k)-1, with k=1. It then calculates each element of the matrix. 2) It defines a polynomial interpolation problem to find a function P(x) that passes through 5 given x-y points. It calculates the Lagrange basis polynomials L0-L4 and defines P(x) as a linear combination of the basis polynomials weighted by the y-values.

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Julieth Daniela
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33 views4 pages

Métodos Numéricos Fase 2

The document contains two numerical exercises: 1) It defines a 5x5 matrix A where the element aij is defined by (i+j-1/k)-1, with k=1. It then calculates each element of the matrix. 2) It defines a polynomial interpolation problem to find a function P(x) that passes through 5 given x-y points. It calculates the Lagrange basis polynomials L0-L4 and defines P(x) as a linear combination of the basis polynomials weighted by the y-values.

Uploaded by

Julieth Daniela
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Métodos numéricos

Fase 2

Julieth Daniela bolívar

Tutor
Francisco Javier castellano

Universidad nacional abierta y a distancia


Cead
Ingeniería de alimentos
ibague 2019
INTRODUCCION

Ejercicio 1.
A=(a,i,j)5x5
aij(i+j-c)-1, con c= 1/k; k=0 1
aij =(i+j- 1/k)-1

1 2 3 4 5
1 1 0 0 0 0
2
i= 0 1 0 0 0
3 0 0 1 0 0
4 0 0 0 1 0
5 0 0 0 0 1

1
2
𝑏𝑖 = ||3||
4
5

1 1 1 1 1
3 5 7 9
1 1 1 1 1
A= 2 3 4 5 6
1 1 1 1 1
3 4 5 6 7
1 1 1 1 1
4 5 6 7 8
1 1 1 1 1
5 6 7 8 9
1
a11=( 1+1-1)-1=(1)-1, c = =1
1
1
a22=( 2+2-1)-1=(3)-1, c = =1
3
1
a33=( 3+3-1)-1=(5)-1, c = =1
5
1
a44=( 4+4-1)-1=(7)-1=7
1
a55=( 5+5-1)-1=(9)-1=9
1
a21=( 2+1-1)-1=(2)-1=2
1
a22=( 2+2-1)-1=(3)-1=3
1
a23=( 2+3-1)-1=(4)-1=4
1
a24=( 2+4-1)-1=(5)-1=5
1
a25=( 2+5-1)-1=(6)-1=6
1
a31=( 3+1-1)-1=(3)-1=3
1
a32=( 3+2-1)-1=(4)-1=4
1
a33=( 3+3-1)-1=(5)-1=5
1
a34=( 3+4-1)-1=(6)-1=6
1
a35=( 3+5-1)-1=(7)-1=7
1
a41=( 4+1-1)-1=(4)-1=4
1
a42=( 4+2-1)-1=(5)-1=5
1
a43=( 4+3-1)-1=(6)-1=6
1
a44=( 4+4-1)-1=(7)-1=7
1
a45=( 4+5-1)-1=(8)-1=8

Ejercicio 2.
x -6.5 -4 -1.5 1 1.5
y 10 2.3 -1.2 -8.5 -2.7

P(x)=Y0L0+ Y1L1+… YnLn


(𝑥−3.5)(𝑥−1)(𝑥−1.5)(𝑥+4) (𝑥−3.5)(𝑥−1)(𝑥+1.5)(𝑥+4)
L0(x)=(−6.5−3.5)(−6.5−1)(−6.5+1.5)(−6.5+4) =
937.5

(𝑥−6.5)(𝑥+1.5)(𝑥−1)(𝑥−3.5) (𝑥+6.5)(𝑥+1.5)(𝑥−1)(𝑥−3.5)
L1(x)=(−4+3.5)(−4+1.5)(−4−1)(−4−3.5) =
234.375

(𝑥+6.5)(𝑥+4)(𝑥−1)(𝑥−3.5) (𝑥+6.5)(𝑥+4)(𝑥−1)(𝑥−3.5)
L2(x)=(−1.5+6.5)(−1.5+4)(−1.5−1)(−1.5−3.5) =
156.25

(𝑥+6.5)(𝑥+4)(𝑥+1.5)(𝑥−3.5) (𝑥+6.5)(𝑥+4)(𝑥+1.5)(𝑥−3.5)
L3(x)= (1+6.5)(1+4)(1+1.5)(1−3.5) =
234.375

(𝑥+6.5)(𝑥+4)(𝑥+1.5)(𝑥−1) (𝑥+6.5)(𝑥+4)(𝑥+1.5)(𝑥−3.5)
L4(x)=(3.5+6.5)(3.5+4)(3.5+1.5)(1−3.5) =
937.5

(𝑥−3.5)(𝑥−1)(𝑥+1.5)(𝑥+4) (𝑥+6.5)(𝑥+1.5)(𝑥−1)(𝑥−3.5)
P(x)=10 ( ) + 2.3 ( )+
937.5 234.375
(𝑥+6.5)(𝑥+4)(𝑥−1)(𝑥−3.5) (𝑥+6.5)(𝑥+4)(𝑥+1.5)(𝑥−3.5)
(−1.2) ( ) + (−8.5) ( )+
156.25 234.375
(𝑥+6.5)(𝑥+4)(𝑥+1.5)(𝑥−3.5)
(−27) ( )
937.5

P(x)=0.000639x4-0.078293x3-0.66896x2-3.11642666x-4.63696.

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