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I am told to find the Fourier series of $f(t)=t^3$ on the interval $-1<t<1$ where $T=2$, thus $L=1$ and $\omega = \pi$.

Since $f(t)$ is odd, I need only find the Fourier Sine Series which is $f(t)=\sum^{\infty}_{n=1} b_n \sin(n\omega t)$ where $b_n =\frac{2}{L}\int^{L}_{0} f(t)\sin(n\omega t)$

Using integration by parts (confirmed using MATLAB and Wolfram) I found $b_n$ to be:

$$2\cos(n\pi)\bigg[\frac{-1}{n\pi}+\frac{6}{n^3\pi ^3}\bigg]$$ $$=\frac{2(-1)^{n+1}}{n\pi}+\frac{12(-1)^n}{n^3\pi^3}$$

Thus, the Fourier series of $f(t)$ is:

$$\sum^{\infty}_{n=1} \sin(\pi nt)\bigg[\frac{2(-1)^{n+1}}{n\pi}+\frac{12(-1)^n}{n^3\pi^3}\bigg]$$

However, when I use Wolfram to find the Fourier Sine Series to check, it shows a similar but different answer (https://www.wolframalpha.com/input/?i=FourierSineSeries%5Bt%5E3,t,5%5D)

For example, for $n=1$, my result would obtain $f(t)=\sin(t)[\frac{2}{\pi}-\frac{12}{\pi^3}]$ whereas Wolfram says it's $2(\pi^2-6)\sin(t)$.

Not sure where I have gone wrong.

Teddy38
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Patrick Robertson
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  • Up until line 5 you are using $\sin(n\pi t)$ as an orthogonal system, while afterwards (wolfram included) you switch to $\sin(nt)$. This, of course, changes the coefficients. –  Oct 16 '17 at 06:05
  • Okay I changed that, but still does not get the right answer. It seems that my answer is different by a factor of $\pi^3$. If I multiply my answer by that then it's correct. Might be a fault in my integration but I don't see how, considering I checked with MATLAB and Wolfram. – Patrick Robertson Oct 16 '17 at 06:15
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    As I said, you've asked Wolframalpha for the expansion in $\sin (nt)$ over $[0,\pi]$ (or, which is the same for $t^3$, over $[-\pi,\pi]$), not in $\sin(n\pi t)$ over $[0,1]$. Therefore it is calculating something different. –  Oct 16 '17 at 06:28
  • But even when plotting in MATLAB it isn't correct. – Patrick Robertson Oct 16 '17 at 06:32
  • Don't worry figured it out thanks – Patrick Robertson Oct 16 '17 at 08:10

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