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Suppose that $$\sum_{n \in \mathbb{Z}-\{0\}} n^{-1}e^{inx}$$ is the Fourier series of some function $g(x) \in L^{2}[-\pi, \pi]$. Find a closed form for $g(x)$.

I have been stuck on this all day. My only two observations that seemed might be useful were that we must have $$ \hat{g}(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi}g(x) \, e^{ikx} dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}g(x) \, \cos(nx) dx + \frac{i}{2\pi}\int_{-\pi}^{\pi}g(x) \, \sin(nx) dx = n^{-1} $$

for all $n \in \mathbb{Z}, n \neq 0$ and the observation that the original Fourier series, under the substitution $z = e^{ix}$ is basically the power series for $\ln(1-x)$. I can't seem to get anywhere on this, I also can't seem to see why it matters that $g \in L^{2}$ other than the fact that it means I don't have to find a continuous function. I tried exploring both observations at length and never got anything.

Prince M
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  • I assume $k=n$. If it were $\sum_n e^{inx}$ it is just a geometric series. Observe that taking derivatives of $n^{-1}ie^{inx}$ gives you $ie^{inx}$. – orole Feb 11 '18 at 04:22
  • Oops, yes. Will fix now – Prince M Feb 11 '18 at 04:22
  • @orole there might be a slight typo in what you wrote? – Prince M Feb 11 '18 at 04:29
  • If $\sum_{n \in \mathbb{Z}-{0}} n^{-1}e^{inx}$, then $\hat{g}(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi}g(x) , e^{-inx} dx$ (Look at the sign of $n$ in the second equation). Please refer to equation (20) here for more info: http://mathworld.wolfram.com/FourierSeries.html – Mehrdad Zandigohar Feb 11 '18 at 06:13

1 Answers1

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Two facts: (a) Differentiation of a function multiplies its Fourier coefficients by $in$, and integration divides them by $in$;

(b) The Dirac Delta function at 0, denoted $\delta_0$, has all Fourier coefficients equal to $1/(2\pi)$ because $$ c_n = \frac{1}{2\pi}\int_{-\pi}^\pi \delta_0(x) e^{-i n x}\,dx = \frac{1}{2\pi} $$ for all $n$.

Combining these, we see that we need to adjust Dirac Delta so that its 0th coefficient is 0, and then integrate it. The adjusted form is $2\pi \delta_0 - 1$; this has Fourier coefficients $$ c_n = \begin{cases} 1, \quad n\ne 0 \\ 0, \quad n=0\end{cases} $$ Integration results in a function that jumps up by $2\pi$ at $0$, and is decreasing with slope $-1$ the rest of the way. Such a function can be written as $$ f(x) = \begin{cases} \pi-x, \quad &0<x<\pi \\ -\pi-x, \quad & -\pi < x < 0\end{cases} $$ Now you can ignore the argument about Dirac Delta, and just check that this $f$ has Fourier coefficients $$ c_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x) e^{-i n x}\,dx = \frac{i}{n} $$ for all $n$, except $c_0=0$.

Conclusion: the function you want, with coefficients $n^{-1}$, is $-i f$.

  • Yes, very helpful. I asked another student what he got and it was the same function for g, but he had no explanation other than repeated trial and error with informed guesses. – Prince M Feb 11 '18 at 19:32
  • Out of curiosity can you think of a reason why asking a question like this would be constructive? Other than the fact that it’s able to be solved is trying to recover the original function a natural question in Fourier analysis? I don’t do much analysis – Prince M Feb 11 '18 at 19:48