FourierSeries command , for arbitrary period T?

Question

ExptoTrig[FourierSeries[Piecewise[{{-Pi,-2Pi<x<=0},{Pi,0<x<=2Pi}}],x,3]]

You can see, the function goes from $(-2\pi,2\pi]$

But Wolfram Alpha gives a wrong answer to this, because she computes it as if the function would go from $-\pi$ to $\pi$.

(The right answer should be $4\sin(\frac{x}{2}) + ..$.)

So, how can I use this command when the function's period T is not 2$\pi$ exactly?

Why not rescale your function first so that the function applies, and scale back after the operation? — J. M.'s missing motivation, May 19 '16 at 11:56
Oh we use one of the magic words of math for that: let $x=2u$… — J. M.'s missing motivation, May 19 '16 at 12:13
I think FourierParameters is what you should go for, but what settings would match what you want, I do not know. Just try it out. — Lotus, May 19 '16 at 12:38
@Lotus Thank you, I don't know how to use it , but I will read the docs. — Filkor, May 19 '16 at 12:39

score 4 · Accepted Answer · answered May 19 '16 at 12:44

4

f[x_] = Piecewise[{{-Pi, -2 Pi < x <= 0}, {Pi, 0 < x <= 2 Pi}}];
T = 4 \[Pi];
fr = FourierTrigSeries[f[x], x, 3, FourierParameters -> {1, 2 \[Pi]/T}]
(* 4 Sin[x/2] + 4/3 Sin[(3 x)/2] *)

answered May 19 '16 at 12:44

Wow , Jesus, nice! – Filkor May 19 '16 at 12:48

score 1 · Answer 2 · answered May 19 '16 at 12:56

Thanks @rewi. I just write down the Wolfram Alpha version, (based on his answer). So I can remember.

FourierTrigSeries[Piecewise[{{-Pi, -2 Pi < x <= 0}, {Pi, 0 < x <= 2 Pi}}], x, 3, FourierParameters -> {1, 1/2}]

where FourierParameters' second parameter is $\omega = \frac{2\pi}{T}$

FourierSeries command , for arbitrary period T?

2 Answers2

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