I'm working on the following problem for my PDE's class as study for a test.

Sketch the Fourier series of $$ f(x) = 2x^2$$ On the interval of $[-1,1]$.

My professors answer key states that the Fourier series for this function is $2x^2$ repeated over the interval $[-1,1]$ (I'd post the graph, but I'm not sure how.)

My understanding of the Fourier series is that for any function $f(x)$ is given by

$$ A_0 + \sum_{n=1}^\infty A_n \cos\bigg(\frac{n \pi x}{L} \bigg ) + \sum_{n=1}^\infty B_n \sin\bigg(\frac{n \pi x}{L} \bigg ) \tag{1}\label{1} $$

Where $$ A_0 = \frac{1}{2L} \int_{-L}^L f(x) dx$$ $$ A_n = \frac{1}{L} \int_{-L}^L f(x) \cos\bigg(\frac{n \pi x}{L} \bigg )dx$$ $$ B_n = \frac{1}{L} \int_{-L}^L f(x) \sin\bigg(\frac{n \pi x}{L} \bigg )dx$$

There are easier techniques to graphing the Fourier series of some function, but in attempting to solve this problem the hard way I find that

$$ A_0 = \frac{2}{3} $$ $$ A_n = \frac{8}{(n \pi)^2} (-1)^n $$ $$ B_n = 0 $$

When you plug these coefficients into $(\ref{1})$, you do not get $2x^2$ as my professor states.

I've worked this problem three times now and cannot seem to find my mistake. So my question, what are the correct coefficients for $f(x)$ along the interval $[-1,1]$?

  • $\begingroup$ I think that your professor means that the sum of the fourier series is equal to the value of the function $f(x)=2x^2$ repeated over [-1,1] for every value of $x$. This because the series converges, and it's not interesting what its fourier coefficients look like. Remember, you should only sketch the graph here. $\endgroup$ – Scounged Nov 18 '16 at 0:27
  • $\begingroup$ @Scounged, even then, $\frac{2}{3}$ + the sum for $A_n$ doesn't converge to $2x^2$. $\endgroup$ – Kosta Nov 18 '16 at 0:30
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    $\begingroup$ Why not? ${}{}$ $\endgroup$ – Qiaochu Yuan Nov 18 '16 at 0:35
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    $\begingroup$ Don't understand what problem you have: the coefficients you found are correct, and inserted in the formula do approach $2*x^2$ ! $\endgroup$ – G Cab Nov 18 '16 at 0:40
  • $\begingroup$ @QiaochuYuan The sum of $A_n \cos(n \pi x)$ converges to 0. Since $B_n = 0 $, the Fourier series is 2/3. I know this is wrong, but I don't know why. $\endgroup$ – Kosta Nov 18 '16 at 0:44

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