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As I understand it:

$\cos^2(t)$ is even because it is a product of two even functions $\cos(t)$.

The Fourier series and Fourier cosine series of an even function is the same link.

So in the fourier series expansion $\cos^2(t)={a_0 \over 2}+\sum_{n=1}^\infty(a_n\cos(n \omega t) + b_n \sin(n \omega t))$, I expect $a_n\neq 0$ and $b_n=0$.

I try to get the the coefficients with wolfram alpha like this:

$a_n$ with FourierCosCoefficient[$\cos(t)^2,t,n$] gives zero link

$b_n$ with FourierSinCoefficient[$\cos(t)^2,t,n$] gives non-zero link

Which of my assumptions are wrong?

(I also got $a_n=0$ when calculating by hand so the question is not primarily about wolfram alpha but about the relation between Fourier series, Fourier cosine series, and even functions)

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  • $\begingroup$ Looks to me that already $a_0 = \frac{1}{2} \ne 0$... Can you detail your computations ? $\endgroup$ Commented Mar 9, 2019 at 11:30
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    $\begingroup$ I'm not familiar enough with WolframAlpha to say, but it seems to me this is likely to be a misunderstanding of what the function does. For example, where's the $\omega$ in that call? Maybe it's trying to project the function on a different period? As it stands, $cos^2(t) = 1/2+1/2cos(2t)$, so that's all your Fourier series. $\endgroup$
    – Okarin
    Commented Mar 9, 2019 at 11:41
  • $\begingroup$ The correct answer is $a_0=a_2=\frac 1 2$ and all other coefficients $0$. $\endgroup$ Commented Mar 9, 2019 at 11:41
  • $\begingroup$ Surprisingly Mathematica gives the same answer. OTOH to the integral FourierCosCoefficient[(1+Cos[2t])/2,t,n]it gives the expected (DiscreteDelta[n-2]+2 DiscreteDelta[n])/2 $\endgroup$ Commented Mar 9, 2019 at 11:50
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    $\begingroup$ FWIW, I asked this at Mathematica.SE. $\endgroup$ Commented Mar 9, 2019 at 12:06

3 Answers 3

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The simplest way to find the Fourier series of this function is to write it as $\frac {1+\cos(2t)} 2$. This expression is in fact the Fourier series!.

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    $\begingroup$ I'm sure we all know this. The question is why WolframAlpha gives a different answer. $\endgroup$ Commented Mar 9, 2019 at 11:46
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Just read the documentation: https://reference.wolfram.com/language/ref/FourierSinCoefficient.html

The command FourierSinCoefficient does not compute the coefficients $b_n$ in the full Fourier series, but the coefficients in the Fourier sine series (= the full Fourier series of the odd extension of the function in question).

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  • $\begingroup$ Thanks, this shows why I was wrong to assume I could calculate $b_n$ with FourierSinCoefficient. I still think I should be able to calculate $a_n$ with FourierCosCoefficient. $\endgroup$
    – user652290
    Commented Mar 9, 2019 at 14:05
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It is clear that, for each $n\in\mathbb N$,$$\int_{-\pi}^\pi\cos^2(t)\sin(nt)\,\mathrm dt=0,$$since $t\mapsto\cos^2(t)\sin(nt)$ is an odd function.

On the other hand,$$a_0=\frac1\pi\int_{-\pi}^\pi\cos^2(t)\,\mathrm dt=1\neq0.$$It turns out that $a_1=0$, but $a_2=\frac12\neq0$.

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  • $\begingroup$ I'm sure we all know this. The question is why WolframAlpha gives a different answer. $\endgroup$ Commented Mar 9, 2019 at 11:48
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    $\begingroup$ No, we don't all know this, since the OP claimed to have obtained $a_n=0$ while calculating by hand. $\endgroup$ Commented Mar 9, 2019 at 11:49

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