91Ó°ÊÓ

The general expression for the function according to Legendre polynomial is given by, f(x) = ∑_l=0^∞c_lP_l(x) ..… (1)

To find the coefficients c_lyou multiply with P_m(x)and integrate. Because Legendre polynomials are orthogonal, all the integrals on right are 0 except the one contains c_mand you can evaluate it as follows:

∫_-1¹ [P_m(x)]² dx= 2/2m+1

So you get:

∫_-1¹ f(x)P_m(³æ)»å³æ=∑_l=0^∞c_l∫_-1¹P_l(x)P_m(x) dx

∫_-1¹ f(x)P_m(x)dx=c_m∫_-1¹[P_m(x)]² dx ..... (2)

∫_-1¹ f(x)P_m(x)dx=c_m 2/2m+1

You know:

Degree of polynomial f(x)=[x-x³] is 3.

So now

3∫_-1¹f(x)P_m(x) dx=0 …… (3)

When 3<m or m>3.

For m>3, using equation (2):

3∫_-1¹f(x)P_m(x) dx=c_m [2/2m+1]

0=c_m [2/2m+1]

c_m=0

You get,

c₄=c₅=0 ..... (a)

Use the result of equation (2), putting m=0 P₀(x)=1 in equation (2).

∫_-1¹ f(x)P₀(x)dx=c₀ [2/2(0)+1]

∫_-1¹ f(x)dx=2 c₀

Using given, ∫f(x)dx= [x-x³].

By integrating:

[x²/2 - x⁴/4]_-1¹=2c₀

[(1)²/2 - (1)⁴/4][(-1)²/2 - (-1)⁴/4]=2c₀ ..... (b)

0=2c₀

c₀=0

Use the result (2), putting m=1 and P₁(x)=x in equation (2).

∫_-1¹ f(x)P₁(x)dx=c₁ [2/2(1)+1]

∫_-1¹ f(x)dx=2/3 c₁

Using given:

∫f(x)dx= [x-x³] ∫_-1¹[x²-x⁴] dx =2/3 c₁

[x³/3 - x⁵/5]_-1¹ =2/3 c₁

[1³/3 - 1⁵/5] [(-1)³/3 - (-1)⁵/5]=2/3 c₁

4/15=2/3 c₁

So,

c₁= 2/5 ..... (c)

Use the result (2), putting m=2 and P(x)=3/2 x²-1/2 in equation (2).

∫_-1¹f(x) P₂(x)dx=c₂[2/2(2)+1]

∫_-1¹(x-x³) (3/2 x²-1/2) dx=c₂[2/4+1]

And ∫x=[x-x³]

∫_-1¹[- 3(x)⁵/2+2(x)³-x/2] dx=2/5 c₂

[- (1)⁶/4+(1)⁴/2-(1)²/4] -[- (-1)⁶/4+(-1)⁴/2-(-1)²/4] = 2/5 c₂ ..... (d)

0=2/5 c₂

c₂=0

Using result (2), put m=3.

∫_-1¹f(x) P₃(x) dx=c₃ [2/2(3)+1]

∫_-1¹(x-x³) 1/2 (5(x)³-3x) dx = c₃ [2/6+1]

P₃(x) = 1/2 (5(x)³-3x)

f(x) = [x-x³]

So,

∫_-1¹ [- 5 (x)⁶/2+4(x)⁴-3(x)²/2] dx=2/7 c₃

Simplify further as follows:

[- 5(x)⁷/14+4(x)⁵/5-x³/2]_-1¹=2/7 c₃

[- 5(1)⁷/14+4(1)⁵/5-1³/2] [- 5(-1)⁷/14+4(-1)⁵/5-(-1)³/2]=2/7 c₃

-4/35 =2/7 c₃

c₃= -2/5 ..... (e)

Putting the all values & e in given function:

f(x) = ∑_l=0^∞c_lP_l(x)

f(x)= (0) P₀(x)+(2/5) P₁ +(0)P₂(x)-(2/5) P₃(x)+0

f(x) = 2/5 P₁(x) - 2/5 P₃(x)+0

f(x) = 2/5 P₁(x) - 2/5 P₃(x)

Hence, this is the solution.