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The solution to an exercise I've done approximates $ f(x) = x(1-x)$ as a Fourier series, but does not mention how I can prove that $f(x)$ is indeed equal to the solution series.

What I've done is :

$$g(n) = \int_0^1 f(x)e^{-2{\pi}inx}dx = \int_0^1 (x-x^2)e^{-2{\pi}inx}dx = \int_0^1 (x-x^2)\frac d{dx}\Big(\frac {e^{-2{\pi}inx}}{-2{\pi}in} \Big)dx $$

Integrating by parts two times, I get that the fourier coefficients are $c_n =\frac {-1}{2\pi^2n^2}$, from there (and we can easily calculate that $c_0 = 1/6$) I am able to conclude that

$$f(x) = \frac{1}{6} - \sum_1^\infty \frac{\cos(2\pi nx)}{\pi^2n^2} $$.

The problem is, while doing this, we have never proved that $f(x)$ indeed does converge to the fourier series that we have calculated. What is the argument I can make to conclude that $f(x) = x(1-x)$ does indeed converge to that series? Thank you

mrf
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aga7689
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  • What kind of convergence theorems for Fourier series do you know? I'm guessing you know at least one that covers your function. – mrf Apr 18 '16 at 08:26
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    The correct thing to say is that the series converges to $f(x)$, not the other way around. $f(x)$ is just a function - it doesn't converge. – Newbie Apr 18 '16 at 08:36
  • That's true, typed it too fast. – aga7689 Apr 18 '16 at 08:37

1 Answers1

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It depends on what kind of convergence do you speak.

The usual criterion is the following

If $f \in L^2[a,b]$ then the Fourier series of $f$ does converge to $f$ in $L^2[a,b]$.

Of course in your case, it is clear that $x(1-x)$ is square integrable on $[a,b]$ so you have the $L^2$-convergence. It is important to note that you don't have a priori the ponctual convergence. Actually in your case, since your function is continuous, you can use the Dirichlet's theorem (https://en.wikipedia.org/wiki/Dirichlet_conditions) to know that your convergence is also ponctual.

C. Dubussy
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