Métodos Numéricos Colaborativo Dos

4. Se tienen los siguientes datos para que halle un polinomio P(x) de grado desconocido, con el método de Diferencias divididas de Newton:

X 0 1 2 3

F(x) 1 6 8 12

Y con la ecuación o polinomio que logre aproxime el valor de P (1.7).

Según la tabla de datos 〖(x〗_0 ,x_1,x_2,x_3 ) el polinomio p(x) será de grado 3 (n=3)b

p(x)=b_0+(x-x_0 ) b_1+(x-x_0 )(x-x_1 ) b_2+(x-x_0 )(x-x_1 )(x-x_2 ) b_3

Calculamos 〖b_0,b〗_1,b_2,b_3

b_0=f(x_0 )=f(0)=1-------→ b_0=1

b_1=f(x_1,x_0 )=((x_1 )-f(x_0 ))/(x_1-x_0 ) =(f(1)-f(0))/(1-0)

b_1=(6-1)/(1-0) =5/1 =5------→b_1=5

b_2=f(x_2,x_1,x_0 )=((f(x_2 )-f〖(x〗_1))/(x_2-x_1 )-(f(x_1 )-f〖(x〗_0))/(x_1-x_0 ))/(x_2-x_0 )

b_2=((f(2)-f(1))/(2-1)-(f(1)-f(0))/(1-0))/(2-0)=((8-6)/(2-1)-(6-1)/(1-0))/(2-0)

b_2=(2/1-5/1)/2 =(2-5)/2= -3/2 -------→ b_2=-3/2

b_3=f(x_3,x_2,x_1,x_0 )=(f(x_3,x_2,x_1,)-f(x_2,x_1,x_0 )_ )/(x_3-x_0 )

b_3=((f(x_3,x_2 )-f〖(x〗_2,x_1))/(x_3-x_1 )-(f(x_(2,) x_1 )-f〖(x〗_1,x_0))/(x_2-x_0 ))/(x_3-x_0 )

b_3=(((f〖(x〗_3)-f〖(x〗_2) )/(x_3-x_2 ))-((f(x_2 )-f〖(x〗_1))/(x_2-x_1 )))/((x_3-x_1)/(〖 x〗_3-x_0 ))-(((f〖(x〗_2)-f〖(x〗_1) )/(x_2-x_1 ))-((f(x_1 )-f〖(x〗_0))/(x_1-x_0 )))/((x_2-x_0)/)

b_3=(((12-8 )/(3- 2 ))-((8-6)/(2-1)))/((3-1)/( 3-0 ))-(((8-6)/(2-1))-((6-1)/(1-0)))/((2-0)/)

b_3=(((4 )/(1 ))-(2/1))/(2/( 3 ))-((2/1)-(5/1))/(2/)

b_3=2/( 2/( 3 ))-(-3)/2=(2+3)/(2/3)= 5/(2/3) = 5/6

b_3= 5/6

El polinomio p(x) será:

p(x)=b_0+(x-x_0 ) b_1+(x-x_0 )(x-x_1 ) b_2+(x-x_0 )(x-x_1 )(x-x_2 ) b_3

p(x)=1+(x-0)(5)+(x-0)(x-1)(-3/2)(x-0)(x-1)(x-2)(5/6)

p(x)=1+5x-3/2 x(x-1)+5/6 (x)(x-1)(x-2)

p(x)=1+5x-( 3/2 〖 x〗^2 + 3/2 x) +(5/6 〖 x〗^2-5/6 x)(x-2)

p(x)=1+5x- 3/2 〖 x〗^2 + 3/2 x +5/6 〖 x〗^3-10/6 x^2-5/6 〖 x〗^2+10/6 x

p(x)=1+(5x+ 3/2 x + 10/6 x )+((-3)/2 〖 x〗^2-10/6 x^2-5/6 〖 x〗^2 )+5/6 x^3

p(x)=1+((30x+9x+10x)/6)+((-9x^2-10x^2-5x^2)/6)+5/6 x^3

p(x)=1+49/6 x-24/6 x^2+5/6 x^3

p(x)=1+49/6 x-4x^2+5/6 x^3

p(1.7)=1+49/6 (1.7)-4(〖1.7)〗^2+5/6 〖(1.7)〗^3

p(1.7)=1+13.883333-11.56+4.094167

p(1.7)=7.4175

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