Find the Fourier coefficient for the function $f(t)=t^3-2t$

Question

I'm trying to find the $a_5$ Fourier coefficient for

$f(t)=t^3-2t$ from $-π<t<π$

I think I'm supposed to get $a_5=0.32$. However, I don't know how to get there. If that is the right answer, how do I get to there? If that's not right, how do I do this problem to get to the right answer?

Thanks.

Do you know the formulas for the coefficients? – Pedro Oct 23 '13 at 01:26 — Pedro, Oct 23 '13 at 01:26
$$\frac{1}{\pi}\int_{-\pi}^\pi (t^3 - 2t)\cos (5t),dt$$ – Daniel Fischer Oct 23 '13 at 01:26 — Daniel Fischer, Oct 23 '13 at 01:26
@DanielFischer You got zero. – Felix Marin Oct 23 '13 at 01:29 — Felix Marin, Oct 23 '13 at 01:29
@FelixMarin Explain? I don't understand. (internet meme?) – Daniel Fischer Oct 23 '13 at 01:31 — Daniel Fischer, Oct 23 '13 at 01:31
@DanielFischer The integrand is odd. – Felix Marin Oct 23 '13 at 01:32 — Felix Marin, Oct 23 '13 at 01:32
@FelixMarin Oh, the value. Yes, all the $a_n$ are. – Daniel Fischer Oct 23 '13 at 01:33 — Daniel Fischer, Oct 23 '13 at 01:33

score 0 · Answer 1 · answered Oct 27 '13 at 00:48

You should have clarified what $a_5$ means. It could be the coefficient of $\cos 5t $ in one book, and the coefficient of $\sin 5t$ in another. If it's the former, the correct answer is $0$. Indeed, $t^3-2t$ is an odd function, and its Fourier series has sines only, no cosines. If it's the coefficient of $\sin 5t$, the value is $$ \frac{2}{\pi}\int_0^{\pi} (t^3-2t)\sin 5t\,dt =\frac{2}{125}(25\pi^2-56)\approx 3 $$

Find the Fourier coefficient for the function $f(t)=t^3-2t$

1 Answers1