# Recover general formula for fourier series?

Hi,

Is there a way to recover the general formula for a fourier series? That is, f.fourier_series_sine_coefficient(9,pi) will give me the 9th sine coefficient, but I'd like to know more generally how I can construct the Nth sine coefficient.

I'm doing some complicated (to me) fourier series, and I'm having trouble figuring out the formula for the Nth.

Thanks for any help.

Andrew

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Can you explain what you mean by that? Generally the coefficients of the fourier series are not related to one another. One can write a expression for the nth coefficient only in special cases, and I am not sure there is a way of identifying whether a given formula falls in that category.

It may be a poor question then .. my frame of reference is the typical fourier series for periodic functions found on-line (square, triangle, saw), which all seem to reduce to nice clean equations for the nth coefficient. I didn't realize most would not behave so nicely .. though, I don't feel so bad now for scratching my head about the more complex function for which I was trying to generate a series .. smile. Thanks for the input/answer.

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In simple cases you can try:

sage: maxima('totalfourier(abs(x),x,%pi)').sage()

1/2pi + 2sum(((-1)^n - 1)cos(nx)/n^2, n, 1, +Infinity)/pi

sage: maxima('totalfourier(x^3,x,%pi)').sage()

-2sum((pi^2n^2 - 6)(-1)^nsin(n*x)/n^3, n, 1, +Infinity)

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I would recommend also the class notes by Prof. David Joyner:

http://www.usna.edu/Users/math/wdj/te...

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Also:

sage: maxima('foursimp(fourier(x^3,x,%pi))')

[%t26,%t27,%t28]

sage: maxima('[%t26,%t27,%t28]')

[a=0,a[n]=0,b[n]=-2(%pi^2n^2-6)*(-1)^n/n^3]

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Thanks very much for these .. I will study/play with them, and hopefully my function falls under 'simple case' .. thanks again.