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Problem statement:

Let us assume that the Fourier series of an unknown function $f(x)$ is

$$ \sum ^{\infty} _{n=1} \frac{\sin(nx)}{n}. $$

Find the function $f(x)$ to which it belongs.

Attempt at solution:

The first thing I did was plot for $x \in (-\pi, \pi)$ for $n = \{1, 2, \ldots, 10\}$.

enter image description here

From that we can see there is a dip around $x = -0.5$ and then an increase around $x = -0.2$ or something.

As we are given a Fourier sine series we know that the function $f(x)$ is an odd function and therefore $a_n = 0$. Therefore, we know that $f(x)$ can be written as

$$ f(t) = \sum ^{\infty}_{n=1} b_n\sin \left(\frac{n\pi t}{L}\right). $$

By comparing the equation above to the series given at the start of the problem we immediately see that $L = \pi$ and $b_n = \frac1n$. Therefore, we are searching for a function $f(x)$ which makes the following hold:

$$ b_n = \frac1L \int^L _{-L}f(t)\sin \left(\frac{n \pi t} {L} \right)dt = \frac1n. $$ Substituting $L = \pi$ we get

$$ \frac{1}{\pi} \left( \int ^{0} _{-\pi} f(t) \sin (nt)dt + \int ^{\pi} _0 f(t) \sin (nt)dt\right) = \frac1n $$

Our function $f(x)$ must therefore fulfill the following integral:

$$ \int ^{0} _{-\pi} f(t) \sin (nt)dt + \int ^{\pi} _0 f(t) \sin (nt)dt = \frac{\pi}{n}. $$

This is where I get stuck as I'm unsure of the validity of my prior steps. I would appreciate a nudge in the right direction.

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jommi
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    you could think of sum as a lebesgue integral with respect to the counting measure – vidyarthi Feb 24 '19 at 13:22
  • or you could also use the fourier inversion theorem – vidyarthi Feb 24 '19 at 13:24
  • This is from an introductory course in Fourier analysis. We haven't covered the inversion theorem and there is no mention of the Lebesgue integral. Considering the depth of this course (or lack thereof) there has to be "simpler" way. – jommi Feb 24 '19 at 13:29

1 Answers1

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Your graph is an approximation to the function you are after. It seems that in the interval $[0,\pi]$ the function $f$ is decreasing linearly from some positive starting value to $0$. I therefore suggest you try the function $$g(x):=\pi-x\qquad(0< x\leq\pi)\ ,$$ extend it odd and $2\pi$-periodically to all of ${\mathbb R}$, and see what happens.

Christian Blatter
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