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I've been struggling with finding Fourier series of the given function for a while now enter image description here

I've calculated my coefficients using formulas :

Though my results does approximate function sufficiently enough (see the picture), I'm pretty sure that I did something wrong, as online calculators give me different values for coefficients, and the sum of some series(like (-1)^n/n^2) does not match expected values(integrals are calculated correctly). Is there something wrong with how I approach solving those integrals for coefficients or maybe there are some other pitfalls that I could have encountered? I would really appreciate it if you could help me out at least with calculating . Thank you in advance.

enter image description here

Resulting Furier series

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JimmyNeutron
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    @Btzzzz are you sure about that? – Doug M Mar 29 '18 at 00:52
  • Yes, one of the tasks is to find sum of the series like (-1)^2n/n^2 using found Fourier series, and my calculations don't converge with what wolfram gives me. It's like there's a linear coefficient missing or something. – JimmyNeutron Mar 29 '18 at 00:55
  • You don't show what results you arrive at after integration, but what you show so far looks correct. – Doug M Mar 29 '18 at 00:56
  • I added my resulting series. You can try to find the sum of (-1)^(2n)/n^2 based on that, considering f(2)=2, and the result won't give you the right answer – JimmyNeutron Mar 29 '18 at 01:01

1 Answers1

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Here are my calculations. I haven't thoroughly checked them, but hopefully they will help.

$a_n = \frac {1}{2}[\int_{-2}^1 (1-x)\cos \frac {n\pi}{2} x - \int_1^2 (1-x)\cos \frac {n\pi}{2} x]\\ \frac 12[\frac {2(x-1)}{n\pi}\sin\frac {n\pi}{2} x + \frac {4}{n^2\pi^2} \cos\frac {n\pi}{2} x|_1^2 + \frac {2(x-1)}{n\pi}\sin\frac {n\pi}{2} x + \frac {4}{n^2\pi^2} \cos\frac {n\pi}{2} x|_1^{-2}]$

Evaluated at $x = 2,$ and $x = -2,$

$\sin\frac {n\pi}{2} x = 0\\ \cos\frac {n\pi}{2} x = (-1)^n$

at $x = 1, (x-1) = 0$

$\cos\frac {n\pi}{2} x = 0$ if $n$ is odd, $-1$ if $n\equiv 2 \pmod 4$, $1$ if $n\equiv4\pmod 4$

$a_n = -\frac {4}{n^2\pi^2}, \frac {8}{n^2\pi^2},-\frac {4}{n^2\pi^2},0 $ when $n \equiv 1,2,3,4$ respectively.

$b_n = \frac {1}{2}[\int_{-2}^1 (1-x)\sin \frac {n\pi}{2} x - \int_1^2 (1-x)\sin \frac {n\pi}{2} x]\\ \frac12[\frac {4}{n^2\pi^2} \sin\frac {n\pi}{2} x - \frac {2(x-1)}{n\pi}\cos\frac {n\pi}{2} x]$

Evaluated at:

$x = 2\\ \frac {2}{n\pi}\cos n\pi\\(-1)^n\frac {2}{n\pi}$

$x = -2\\ \frac {6}{n\pi}\cos n\pi\\(-1)^n\frac {6}{n\pi}$

$x = 1\\ \frac {4}{n^2\pi^2} \sin\frac {n\pi}{2}\\ \frac {4}{n^2\pi^2}, 0 ,-\frac {4}{n^2\pi^2}, 0$

$b_n = -\frac{4}{n\pi} - \frac {4}{n^2\pi},\frac{4}{n\pi},-\frac{4}{n\pi} + \frac {4}{n^2\pi}, \frac{4}{n\pi}$

Doug M
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