Find the best (in the least squares sense) second-degree polynomial approximation to each of the given functions over the interval -1<x<1.

x4

Short Answer

Expert verified

The second-degree polynomial approximation for the function f(x) = x4 is f(x)=1/5+5/7 (3x2-1).

Step by step solution

01

Concept of Legendre series:

Legendre series of the function f(x):

f(x) = ∑l=0clPl(x)

-11 f(x)pm(x) dx=∑l=0cl-11 Pm(x) Pl(x)

And

l=0cl-11 Pm(x) Pl(x) =cm [2/2m+1]

Hence,

cm[2/2m+1]=∫-11 f(x)pm(x) dx

02

Obtain the value of different coefficients of the series:

Given function is, f(x) = x4.

Now calculating for m=0:

2c0= ∫-11 f(x) P0(x)dx

2c0= ∫-11 x4dx

c0=1/5

Now calculating for m=1:

2/3 c1= ∫-11 f(x) P1(x)dx

2/3 c1= ∫-11 x4(x) dx

Since Integrand is old,

c1=0

Now calculating for m=2:

2/5 c2= ∫-11 f(x) P2(x)dx

2/5 c2= 1/2 ∫-11 x4(3x2-1) dx

2/5 c2= 1/2 ∫01 x4(3x2-1) dx

c2= (3/7-1/5)

Solving further,

2/5 c2 = 8/35

2/5 c2 = 4/7

c2 = 4/7× 5/2

c2= 10/7

03

Put the value of coefficients in the function and observe the best second degree polynomial:

After putting the coefficients in function, f(x) = 1/5 P0(x)+10/7 P2(x).

The best second-degree polynomial is f(x) = 1/5 +5/7 (3x2-1).

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Most popular questions from this chapter

Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let f(x) be the given function to be approximated. Let the functions pl(x)be the normalized Legendre polynomials, that is, pl(x) = √(2l+1)/2 Pl(x) , so that

-11[pl(x)"]"2dx=1.

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To show the first few terms of J0(x),J1(x),J-1(x),J2(x),J-2(x). Show thatJ-1(x)=-J1(x)andJ-2(x)=J2(x).

Find the best (in the least squares sense) second-degree polynomial approximation to each of the given functions over the interval -1<x<1.

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