For on f (x) = ln (x + 1) and let x_0 = 0: x_1 = 0.6; x_2 = 0.9 construct the Lagrange polynomial of degree one and the Lagrange polynomial of degree two to approximate the value of f (0.45) use that of one to find the absolute error
For on f (x) = ln (x + 1) and let x_0 = 0: x_1 = 0.6; x_2 = 0.9 construct the Lagrange polynomial of degree one and the Lagrange polynomial of degree two to approximate the value of f (0.45) use that of one to find the absolute error
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
For on f (x) = ln (x + 1) and let x_0 = 0: x_1 = 0.6; x_2 = 0.9 construct the Lagrange polynomial of degree one and the Lagrange polynomial of degree two to approximate the value of f (0.45) use that of one to find the absolute error
Transcribed Image Text:Para la funcion f(x)=In(x+1) y sea Xo=0;x,=0.6; x2=0.9 construya el polinomio
de Lagrange de grado uno y el polinomio de Lagrange de grado dos. para aproximar
el valor de f(0.45) utilice el polinomio de grado uno, para encontrar el error
absoluto.
Seleccione una:
a. P1(x) =0.874548x; P2(x) = -0.268961x2 +0.955236x; P1(0.45) =0.5939;
|f(0.45)-P1(0.45)|=0.0212983
Ob.
1. P1(x) =0.874548x; P2(x) = -x+0.955236x; P1(0.45) =0.393546;
|f(0.45)-P1(0.45)|=0.0212983
O c.C. P1(x) =0.874548x; P2(x) = -0.268961x2+0.955236x; P1(0.45)
=0.393546; |f(0.45)-P1(0.45)|=0.2
o d. P1(x) =0.874548; P2(x) = -0.268961x2 +0.955236x; P1(0.45) =0.393546;
|f(0.45)-P1(0.45)|=0.0212983
O e. P1(x) =0.874548x; P2(x) = -0.268961x2+0.955236x; P1(0.45)
=0.393546; |f(0.45)-P1(0.45)I=0.0212983
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