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Chapter 12: Q13P (page 582)

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) = 鈭�l=0鈭�clPl(x)

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

And

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

Hence,

cm摆2/2尘+1闭=鈭�-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= 鈭�-11 f(x) P0(x)dx

2c0= 鈭�-11 x4dx

c0=1/5

Now calculating for m=1:

2/3 c1= 鈭�-11 f(x) P1(x)dx

2/3 c1= 鈭�-11 x4(x) dx

Since Integrand is old,

c1=0

Now calculating for m=2:

2/5 c2= 鈭�-11 f(x) P2(x)dx

2/5 c2= 1/2 鈭�-11 x4(3x2-1) dx

2/5 c2= 1/2 鈭�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脳 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

Expand each of the following polynomials in a Legendre series. You should get the same results that you got by a different method in the corresponding problems in Section 5.

7x4-3x+1

Verify the formula stated for j0,j1and j2in terms of sinxand cosxand also find the value of the, J3/2(x)and J5/2(x) .

Prove that the functions Ln(x)are orthogonal on(0,鈭�)with respect to the weight functione鈭�x

Hint: Write the differential equation(22.21)asexddxxe鈭�xy'+ny=0, and see Sections7and19 .

Find the norm of each of the following functions on the given interval and state the normalized functionP2(x)on(-1,1)

Multiply(5.8e)by {P1}(x)and integrate from -1 to 1. To evaluate the middle term, integrate by parts.
See all solutions

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