Finding the Fourier cosine series for ##f(x)=x^2##

[chwala]

Homework Statement

See attached.

Relevant Equations

Fourier cosine series

I was just going through my old notes on this i.e

The concept is straight forward- only challenge phew is the integration bit...took me round and round a little bit... that is for $A_n$ part.

My working pretty ok i.e we shall realize the text solution. Kindly find my own working below.

Now to my question supposing we have to find say Fourier cosine series for $f(x)= x^7$. The integration by parts here will take like forever to do. Do we have software for this? Will Wolfram help?

 

[BvU]

Will Wolfram help?

Pretty easy to check for yourself . Let us know !

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[chwala]

Pretty easy to check for yourself . Let us know !

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Will do check @BvU ... however, if one was to do this by hand ... No technology...how long would it take to work to solution?

 

[Hill]

The $f(x)= x^7$ is odd. Perhaps Fourier sine series rather than cosine?

 

[pasmith]

Now to my question supposing we have to find say Fourier cosine series for $f(x)= x^7$. The integration by parts here will take like forever to do. Do we have software for this? Will Wolfram help?

Set I_n = \int_0^{2\pi} x^{2n+1} \sin kx\,dx, \qquad k = 1, 2, \dots. so that integrating by parts twice, <br /> I_n = - \frac{(2n+1)(2n)}{k^2}I_{n-1} - \frac{(2\pi)^{2n+1}}{k}. This recurrence relation can be solved subject to the initial condition I_0 = -\frac{2\pi}{k} to obtain <br /> I_N = \frac{(2N + 1)!(-1)^N}{k^{2N}} \frac{2\pi}{k}\sum_{n=0}^N \frac{(-1)^{n+1}(2\pi k)^{2n}}{(2n+1)!}.