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?
Best Answer
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).