3
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ Why don't you try to find the Fourier transform the usual way? Use the orthogonality of trigonometric functions. $\endgroup$
    – eyedropper
    Mar 18, 2016 at 13:16
  • $\begingroup$ We have not seen orthogonality of trigonometric functions yet in class. This problem can certainly be solved with another method. $\endgroup$
    – glpsx
    Mar 18, 2016 at 13:23
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 18, 2016 at 13:33
  • $\begingroup$ 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 $f(x)$. $\endgroup$
    – eyedropper
    Mar 18, 2016 at 13:40
  • $\begingroup$ @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. $\endgroup$
    – glpsx
    Mar 18, 2016 at 13:58

1 Answer 1

2
$\begingroup$

By the orthogonality of trigonometric functions I meant the following identities

$$ \begin{align} \int _{-\pi }^{\pi }\sin (nx)\sin (mx)\mathrm{d}x&=\pi \delta _{m,n}\\ \int _{-\pi }^{\pi }\cos (nx)\cos (mx)\mathrm{d}x&=\pi \delta _{m,n}\\ \int _{-\pi }^{\pi }\sin (nx)\cos (mx)\mathrm{d}x&=0 \end{align} $$ 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.

$\endgroup$
1
  • $\begingroup$ Thank you for your detailed and clear explanation. It was extremely helpful and understandable. It also gave me another way of thinking about this problem. $\endgroup$
    – glpsx
    Mar 21, 2016 at 22:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .