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Métodos numéricos iterativos .pdf

ex 35)

Newton de grau 3
p3(x)= a0+a1(x-x0)+a2(x-x0)(x-xn)+a3(x-x0)(x-x1)(x-x2)
a0=f[x0]
a1=f(x0,

f[x0,x1]=f(x1)-f(x0)/x1-x0=0-1/0.1-0.0=-1/(1/10)=-10 = a1

f[x1,x2]=f(x1)-f(x2)/x2-x1=-1-0/0.3-0.1=-1/(2/10)=-5

f[x2,x3]=f(x3)-f(x2)/x3-x2=0-(-1)/0.4-0.3=1/0.1=10

f[x0,x1,x2]=f[x1,x2]-f[x0,x1]/x2-x0=-5-(-10)/0.3-0.0=-5+10/(3/10)=50/3 =a2

f[x1,x2,x3]=f[x2,x3]-f[x1,x2]/x3-x1=10-(-5)/0.4-0.1=50

f[x0,x1,x2,x3]= 50-(50/3)/0.4=150-50/3/(4/10)= 100*10/(3*4)=250/3 =a3

p3(0.2) = 48.6979

(0.95000)