0
$\begingroup$

calculate the Fourier series of the $2\pi$-periodic continuation of

$$f(t):=|t|, \quad t\in[-\pi,\pi)\tag{1}$$

We know that

$$f(t)=\sum_{k=-N}^N c_k\cdot e^{ikt}\quad \&\quad c_k=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-ikt}dt\tag{2}$$

So let's calcualte the $c_k$.

$$c_k=\frac{1}{2\pi}\int_{-pi}^\pi|t|\cdot e^{-ikt}dt=\frac{2}{2\pi}\int_0^\pi t\cdot \cos(tk) dt=\frac{1}{\pi}[t\cdot\sin(tk)]_0^\pi+\frac{1}{\pi}[\cos(tk)]_0^\pi=0\frac{1}{\pi}(-1-1)=\frac{-2}{\pi}\tag{3}$$

Whereas we used the identity $\int_{-a}^a|t|e^{-ikt}dt=2\int_0^a t\cos(kt)dt$ Explanation (wheres the sub $t\to -t$ was used in the 2nd step)

So we get

$$f(t)=\sum_{k=-\infty}^\infty \frac{-2}{\pi}e^{ikt}\tag{4}$$

Sadly I don't have any solutions. Is that correct?

$\endgroup$
3
  • $\begingroup$ You can never have a nonzero value independent of $k$ for $c_k$, by Riemann Lebesgue Lemma. $\endgroup$ Jan 31, 2019 at 10:13
  • $\begingroup$ Yeah of course, that's a stupid mistake. But do you see any other issue here? $\endgroup$
    – xotix
    Jan 31, 2019 at 10:15
  • $\begingroup$ You will have to calculate $c_0$ seperately. $\endgroup$ Jan 31, 2019 at 10:17

2 Answers 2

1
$\begingroup$

Your integration is incorrect. Note that the anti-derivative of $\cos(kt)$ is $\frac{\sin(kt)}{k}$, so we have

\begin{align} \int t\cos(kt) dt &= t \color{red}{\frac{\sin(kt)}{k}} - \int\frac{\sin(kt)}{k}dt \\ &= t\color{red}{\frac{\sin(kt)}{k}} + \color{red}{\frac{\cos(kt)}{k^2}} \end{align}

Inserting the limits gives

$$ c_k = \frac{1}{\pi}\frac{\cos(k\pi)-1}{k^2} = \begin{cases} 0, && k \text{ even} \\ \dfrac{-2}{\pi k^2}, && k \text{ odd} \end{cases} $$

$\endgroup$
8
  • $\begingroup$ Yeah, that was a stupid mistake. Thanks $\endgroup$
    – xotix
    Jan 31, 2019 at 10:15
  • $\begingroup$ There is a small mistake. $c_0$ is not $0$. $\endgroup$ Jan 31, 2019 at 10:18
  • $\begingroup$ @xotix Don't forget $c_0$! $\endgroup$
    – Dylan
    Jan 31, 2019 at 10:19
  • $\begingroup$ @KaviRamaMurthy I didn't say $c_0$ was $0$. Thanks for reminding us though! $\endgroup$
    – Dylan
    Jan 31, 2019 at 10:19
  • $\begingroup$ @Dylan I am a bit confused to the borders of the sum. if I have $\sum_{k=1}^\infty$ it's clear that I have to calculate $c_0$ but when do I have $\sum_{k=-N}^N$ and when do I have $\sum_{k=1}^\infty$ and when do I have $\sum_{k=-\infty}^\infty$? $\endgroup$
    – xotix
    Jan 31, 2019 at 12:15
1
$\begingroup$

The right side of 2nd step of your substitution might be better expanded, assuming $a \ge 0$, as

$ \int_{-a}^0 |t| (\cos(st) + i \sin(st)) \ dt \\ = \int_{-a}^0 (-t) (\cos(st) + i \sin(st)) \ dt \ [ \ \because |t| = -t, t \le 0 \ ] \\ = \int_0^a t (\cos(-st) + i \sin(-st)) \ dt \ [ \ \text{substituting} \ t \to -t \ ] \\ = \int_0^a t (\cos(st) - i \sin(st)) \ dt $

$\endgroup$
1
  • $\begingroup$ Yeah I didn't really like that they continues with $|t|$ but I though Im not going to change it. Thanks $\endgroup$
    – xotix
    Jan 31, 2019 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.