What are the fourier series for: $\sin(\pi*x)+\cos(3\pi x)$ and $\sin(3x)$

I was taught that the purpose of the Fourier series was to describe periodic functions in the form of an infinite sum of cosines and sines. But if the function is already in the form of sin and cos such as above, then wouldn't it be redundant to express it in the form it is already in? I can see that the functions above are not in the form of a series, but how can you put sin and cosine into a series of itself? Is the question a trick question and the answer just the function provided?

  • $\begingroup$ If you denote $\mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $\mathcal{F}(\sin)(x) = \sin(x)$, $\mathcal{F}(\cos)(x) = \cos(x)$, the series are themselves. That's it. $\endgroup$ – Klaramun Nov 27 '15 at 8:55
  • $\begingroup$ @Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'? $\endgroup$ – VanGo Nov 27 '15 at 8:59
  • $\begingroup$ Ok now I have seen the edit. The second function $\sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n \sin\Big(\frac{2 \pi n x}{T} \Big) + b_n \cos\Big(\frac{2 \pi n x}{T} \Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines. $\endgroup$ – Klaramun Nov 27 '15 at 9:03
  • $\begingroup$ I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps. $\endgroup$ – Klaramun Nov 27 '15 at 9:17

The Fourier series express a periodic function in terms of sines and cosines, but with a precise relation between their periods. Let me explain.

If $f$ is a periodic function with period $T$ (i.e. $f(x+T) = f(x)$), then roughly speaking (I'll not worry about convergence issues) its Fourier series will be

\begin{equation*} \begin{split} \mathcal{F}(f)(x) & = \sum_{n=1}^{+\infty} a_n \sin\Big( \frac{2\pi n}{T}x\Big) + b_n \cos\Big( \frac{2\pi n}{T}x\Big) = \\ & = \underbrace{\Big( a_1 \sin\Big( \frac{2\pi}{T}x\Big) + b_1 \cos\Big( \frac{2\pi}{T}x\Big) \Big)}_{\text{First term, period } T_0} + \underbrace{\Big( a_2 \sin\Big( \frac{2\pi·2}{T}x\Big) + b_2 \cos\Big( \frac{2\pi· 2}{T}x\Big) \Big)}_{\text{Second term, period } 1/2·T_0} + \cdots \end{split} \end{equation*}

So the relation between the periods of each term of the series is that they decrease in $\frac{1}{n}$ at the $n$-th term with respect to the first one, the periods are related in a rational way.

In the case of the function $\cos(\pi x) + \sin(3 x)$, the periods of the terms are not related in a rational way (in fact the first one is rational, but not the second one), so you must compute its Fourier series, which will not be the same as the function itself (first you need to determine the period $T$ of such a function).

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  • $\begingroup$ I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here? $\endgroup$ – VanGo Nov 27 '15 at 9:59
  • $\begingroup$ You re-edited your function, it ends up being $\sin(\pi x) + \cos(3 \pi x)$; it is its Fourier series itlsef because of what I just told you. $\endgroup$ – Klaramun Nov 27 '15 at 10:04

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