Find the Fourier series of $f(x) = \frac{a_0}{2} + \sum_{k = 1}^{n}(a_k\cos{kx} + b_k\sin{kx})$

I'm learning about Fourier series and need help with this problem:

Find the Fourier series of $$f(x) = \frac{a_0}{2} + \sum_{k = 1}^{n}(a_k\cos{kx} + b_k\sin{kx})$$

My thoughts:

The above trigonometric polynomial is the nth partial sum of the Fourier series for $$f$$. Taking the sum to infinity we get the Fourier series of $$f$$. Is it that simple or am I missing something here?

• Why don't you try to find the Fourier transform the usual way? Use the orthogonality of trigonometric functions. Mar 18, 2016 at 13:16
• We have not seen orthogonality of trigonometric functions yet in class. This problem can certainly be solved with another method. Mar 18, 2016 at 13:23
• This is a classic. Similar to "Find the Laurent series for $z^{-1}$ around $z=0$." -- Note that in this task, $n$ is fixed. -- If you meant $a_k=0=b_k$ for $k>n$, then your statement is correct. Mar 18, 2016 at 13:33
• Ok, then you should convince yourself that this already is in the appropriate form, i.e. the function $f(x)$ is expressed as a Fourier series. It just has a finite number of coefficients. If you would formally calculate its Fourier series, you would get the identical form of $f(x)$. Mar 18, 2016 at 13:40
• @LutzL The analogy withe the Laurent series of $z^{-1}$ was indeed helpful, thank you. I'm still having difficulties to convince myself that the function $f(x)$ is already in the appropriate form. All of the examples that we have seen in class had an infinite sum, the fact that the sum is finite here is disturbing me. Mar 18, 2016 at 13:58

\begin{align} \int _{-\pi }^{\pi }\sin (nx)\sin (mx)\mathrm{d}x&=\pi \delta _{m,n}\\ \int _{-\pi }^{\pi }\cos (nx)\cos (mx)\mathrm{d}x&=\pi \delta _{m,n}\\ \int _{-\pi }^{\pi }\sin (nx)\cos (mx)\mathrm{d}x&=0 \end{align} which can be easily verified by using basic trigonometric identities, such as product-to-sum formulas. Let $$f(x)=\frac{c_0}{2}+\sum _{m=1}^{+\infty }(c_m\cos (mx)+d_m\sin (mx))$$ be the Fourier series of $f$. Then the coefficients for your $f(x)$ satisfy
\begin{align} c_m&=\frac{1}{\pi }\int _{-\pi }^{\pi }f(x)\cos (mx)\mathrm{d}x\\ &=\frac{1}{\pi}\sum _{k=1}^{n }\left(a_k\int _{-\pi }^{\pi }\cos (kx)\cos (mx)\mathrm{d}x+b_k\int _{-\pi }^{\pi }\sin (kx)\cos (mx)\mathrm{d}x\right)\\ &=\frac{1}{\pi}\sum _{k=1}^{n }\left(a_k \cdot \pi \delta _{k,m}+b_k\cdot 0\right)=a_m \text{ for m\leq n and 0 otherwise} \end{align}
The same goes for $c_0=a_0$ and $d_m=b_m$ and this proves explicitly that the Fourier series of $f(x)$ is $f(x)$. The $\delta$ symbol is the Kronecker delta.