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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?

xotix
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2 Answers2

1

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

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

Dylan
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  • Yeah, that was a stupid mistake. Thanks – xotix Jan 31 '19 at 10:15
  • There is a small mistake. $c_0$ is not $0$. – Kavi Rama Murthy Jan 31 '19 at 10:18
  • @xotix Don't forget $c_0$! – Dylan Jan 31 '19 at 10:19
  • @KaviRamaMurthy I didn't say $c_0$ was $0$. Thanks for reminding us though! – Dylan Jan 31 '19 at 10:19
  • @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 $\sum_{k=-\infty}^\infty$? – xotix Jan 31 '19 at 12:15
  • Fourier series generally have $k$ go to $\infty$. If it's a real-valued series (with $\sin$ and $\cos$ instead) then you'd have $k \ge 0$. For both versions you have to compute $c_0$ – Dylan Jan 31 '19 at 12:20
  • I have $\sum_{k=-N}^N c_k e^{ikt}$ and $\frac{a_0}{2}+\sum_{k=1}^N a_k\cos(kt)+b_k\sin(kt)$ no? For $N \to \infty$ I get the fourier series, no? So I'd actually have to split the first sum to $c_0 + \sum_{k=1}^\infty c_k e^{ikt} + \sum_{k=-\infty}^{-1} c_k e^{ikt}$? – xotix Jan 31 '19 at 12:28
  • Yes, the constant term is "hidden" in there. But $c_k$ should be the same expression for all $k\ne 0$ – Dylan Jan 31 '19 at 12:45
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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 $

programmer
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  • Yeah I didn't really like that they continues with $|t|$ but I though Im not going to change it. Thanks – xotix Jan 31 '19 at 12:17