This document describes using the Lagrange polynomial interpolation method to find a polynomial of degree 3 that passes through 4 given points. It gives the 4 points as (0, -5), (1, 1), (3, 25), (4, 55). It then shows the calculations to find the Lagrange basis polynomials L0, L1, L2, L3 and combines them to obtain the Lagrange interpolation polynomial P4(x) = x3 - 2x2 + 7x - 5.

1. 1.- Lee el documentoPOLINOMIODELA GRANGE.pdf y determinalasoluciónde lossiguientes
sistemasde ecuacionesutilizandoel métodoantesmencionado
a) Determinarel polinomiode interpolaciónde LaGrange de grado n = 3 que pasa por lospuntos
indicadosenlatabla,siguiendoel ordenindicado:
X0 = 0 f0=-5
X1=1 f1=1
X2=3 f2=25
X3=4 f3=55
L0= (x-x1)(x-x2)(x-x3) /(x0-x1)(x0-x2)(x0-x3)= -1/12 [ (-4+x) (-3+x)(-1+x)]
L1= (x-x0)(x-x2)(x-x3)/(x1-x0)(x1-x2)(x1-x3) =1/6 [(0+x)(-3+x)(-4+x)]
L2=(x-x0)(x-x1)(x-x3)/(x2-x0)(x2-x1)(x2-x3)= -1/6[ (-4+x)(-1+x)(0+x)]
L3= (x-x0)(x-x1)(x-x2)/(x3-x0)(x3-x1)(x3-x2)= 1/12 [(0+x)(-1+x)(-3+x)]
P4=f0L0[X]+f1L1[x]+f2L2[x]+f3L3[x] =5/12 [ (-4+x)(-3+x)(-1+x) ] +1/6 [ (0+x)(-3+x)(-4+x)] -25/16[ (-
4+x)(-1+x)(0+x)] +55/12 [(0+x)(-1+x)(-3+x)]
=(5(-4+x)(-3+x)(-1+x))/12 +( x(x-3)(x-4)) /6– (25x(x-4)(x-1)) /6+( 55x(x-1)(x-3))/12
=(5(-4+x)(-3+x)(-1+x))/12 +( 4x(x-4)(-12x+11)) /12 +( 55x(x-1)(x-3))/12
=(5(-4+x)(-3+x)(-1+x) +4x(x-4)(-12x+11) + 55x(x-1)(x-3) )/12
= (12 (𝑋3 − 2𝑋2 + 7𝑋 − 5)) /12
= 𝑿𝟑 − 𝟐𝑿𝟐 + 𝟕𝑿 − 𝟓

2. (0,5)
(1,1)

3. (3,25)
(4,55)