- #1

Benny

- 584

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Hi, can someone help me out with the following question?

Q. Show that the Fourier series for the function y(x) = |x| in the range -pi <= x < pi is

[tex]

y\left( x \right) = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\cos \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^2 }}}

[/tex]

By integrating term by term from 0 to x, find the function g(x) whose Fourier series is

[tex]

\frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}

[/tex]

Deduce the sum S of the series: [tex]1 - \frac{1}{{3^3 }} + \frac{1}{{5^3 }} - \frac{1}{{7^3 }} + ...[/tex]

I took the Fourier series for y(x) and I integrated it as follows.

[tex]

\int\limits_0^x {\left( {\frac{\pi }{2}} \right)} dt - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\left( {\int\limits_0^x {\frac{{\cos \left( {2m + 1} \right)t}}{{\left( {2m + 1} \right)^2 }}dt} } \right)}

[/tex]

[tex]

= \frac{{\pi x}}{2} - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}

[/tex]

I don't know what to do with it to find the function whose Fourier series is

[tex]

\frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}

[/tex]

Can someone help me get started? I'm not sure what to do.

Edit: Do I just equate the integral I evaluated to the integral of |x|(considering x positive and negative separately) and solve the equation for the sine series?

Q. Show that the Fourier series for the function y(x) = |x| in the range -pi <= x < pi is

[tex]

y\left( x \right) = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\cos \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^2 }}}

[/tex]

By integrating term by term from 0 to x, find the function g(x) whose Fourier series is

[tex]

\frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}

[/tex]

Deduce the sum S of the series: [tex]1 - \frac{1}{{3^3 }} + \frac{1}{{5^3 }} - \frac{1}{{7^3 }} + ...[/tex]

I took the Fourier series for y(x) and I integrated it as follows.

[tex]

\int\limits_0^x {\left( {\frac{\pi }{2}} \right)} dt - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\left( {\int\limits_0^x {\frac{{\cos \left( {2m + 1} \right)t}}{{\left( {2m + 1} \right)^2 }}dt} } \right)}

[/tex]

[tex]

= \frac{{\pi x}}{2} - \frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}

[/tex]

I don't know what to do with it to find the function whose Fourier series is

[tex]

\frac{4}{\pi }\sum\limits_{m = 0}^\infty {\frac{{\sin \left( {2m + 1} \right)x}}{{\left( {2m + 1} \right)^3 }}}

[/tex]

Can someone help me get started? I'm not sure what to do.

Edit: Do I just equate the integral I evaluated to the integral of |x|(considering x positive and negative separately) and solve the equation for the sine series?

Last edited: