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Ex. 3.9. Develop Fourier-Legendre series for f(x) = sin(21x). Graph on common axes the partial sums Sm, m = 1,2, ...,M up to
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Answer #1

SOLUTION:

Given That

f(x)=sin(2πx) with m=1,2,.......M

So

The Legendre polynomials form

a complete Orthogonal system over the interval [-1,1]

with respect to the weight function==> WI) = 1,

The function f(x) may be expanded interms of them as

                           (1) - Σα, ΡΑ(α)

To obtain, the coefficient a_n in the expansion multiply both sides by Pm (2) and integrate,

                  1 pcm)f(x)dır Əa | Pa(w)Pm(a)dr n=0 J-1

where the coefficient

an = + | Pn(z)f(x)dx J-1

Here   

  f(1) = sin(272)

To find the coefficients do, d1, 02, ...:

The Legendre's Polynomial is as follows:

Pol.2) = 1

P1(x) = 2

P2(x) = 5(3.- 1)

P_3(x)=\frac{1}{2}(5x^3-3x)

P_4(x)=\frac{1}{8}\left ( 35x^4-30x^2+3 \right )

P5(x) == (6325 – 70x3 + 152)

To find a_0:

       n=0\Rightarrow P_0=1\Rightarrow a_0=\frac{1}{2}\int_{-1}^{1}1sin(2\pi x)dx

                                                    a_0=0

To find a_1:

      n=1\Rightarrow P_1=x\Rightarrow a_1=\frac{3}{2}\int_{-1}^{1}xsin\left ( 2\pi x \right )dx

                                                    a_1=\frac{-3}{2\pi }

To find a_2:

     n=2\Rightarrow P_2=\frac{1}{2}(3x^2-1)

                   \Rightarrow a_2=\frac{5}{2}\int_{-1}^{1}\frac{1}{2}(3x^2-1)sin(2\pi x)dx=0

                     a_2=0

To find a_3:

    n=3\Rightarrow P_3(x)=\frac{1}{2}(5x^3-3x)

   a_3=\frac{7}{2}\int_{-1}^{1}\frac{1}{2}(5x^3-3x)sin(2\pi x)dx

    a_3=\frac{7}{4}\left [ \frac{15-4\pi^2}{2\pi^3} \right ]

To find a_4:

     n=4\Rightarrow P_4=\frac{1}{8}\left ( 35x^4-30x^2+3 \right )

     a_4=\frac{9}{2}\int_{-1}^{1}\frac{1}{8}\left ( 35x^4-30x^2+3 \right )sin(2\pi x)dx

       a_4=0

To find a_5:

n = 5= P3(x) = (63x5 – 70x} + 152)
a_5=-\frac{(945-480\pi^2+16\pi^4)}{16\pi^5}

From the above calculations,

we observe that a_0=a_2=a_4=....=0 ,

That is , all even a_n vanish. sin(2\pi x) is an odd function.

   a_1=\frac{-3}{2\pi } ,        a_3=\frac{7}{4}\left [ \frac{15-4\pi^2}{2\pi^3} \right ] ,          a_5=-\frac{(945-480\pi^2+16\pi^4)}{16\pi^5}

         

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