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As a math project I have to find the Fourier series of this function over $[-\pi,\pi]$ and I have tried integration for the coefficients and also complex numbers with the binomial theorem and what both of these methods had in common is that the series equaled to 0. Does this mean that it has no Fourier series?

mathreadler
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Maaa09
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    No, it certainly does have a (nonzero) Fourier series. – Qiaochu Yuan Apr 22 '19 at 06:49
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    You will have to show some working so that we can see why you think the coefficients are all zero. – ancient mathematician Apr 22 '19 at 06:49
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    Remember that $\sin x=-\frac i2(e^{ix}-e^{-ix})$. – Angina Seng Apr 22 '19 at 06:54
  • Given what you have told us, both approaches should work (and give a non-zero series). Could you show us some of your working, so we can help you to identify the mistake? – preferred_anon Apr 22 '19 at 07:05
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    Since compound angle formulae obtain $\cos mx,,\sin mx$ as polynomial functions of $\cos x,,\sin x$ (which can be proved with or without complex numbers), you can write $\sin^n x$ as a linear combination of such compound angle formulae, without using Fourier methods. Therefore, your Fourier methods should reproduce this result. If you're not getting $\sin^5 x=\frac{\sin 5x+10\sin x-5\sin 3x}{16}$, please edit in your working. – J.G. Apr 22 '19 at 07:07
  • Something went wrong in your integration. Surely for example $$\int_{-\pi}^{\pi}\sin^5x\sin x,dx>0$$ for the integrand is non-negative everywhere and vanishes at isolated points only. Implying that $b_1$, the coefficient of $\sin x$ in the Fourier series, must also be positive. – Jyrki Lahtonen Apr 22 '19 at 07:36

3 Answers3

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If it is a project, I suspect we should not give you too much help on it.

However you can use the hint of Lord Shark the Unknown:

$$\sin(\alpha) = -\frac i 2 \left(e^{i\alpha} - e^{-i\alpha}\right)$$

It is a very good start.

mathreadler
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So here are the two things I tried and tell me where I made the mistake:

First the integration one. Since it is an odd function, the series should look like this: $$f(x) = \sum_{n=1}^\infty b_nsin(nx)$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi sin^5(x)sin(nx)dx$$ And according to WolframAplha this will be zero for every $n$ because of the $sin(n\pi)$: $$\int_{-\pi}^\pi\sin^5(x)\sin(nx)\mathrm dx=-\frac{240\sin(n\pi)}{n^6-35n^4+259n^2-225}$$ The other method looked like this: $$sin^5(x) = \Biggr(\frac{e^{ix}-e^{-ix}}{2i}\Biggr)^5=\frac{(e^{ix}-e^{-ix})^5}{-32} = \sum_{n=0}^5 \binom{5}{n}(e^{ix(5-n)})(-(e^{-ix}))^n$$ And for this I tried a for loop in MATLAB, and in the end the sum was zero.

Maaa09
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  • For $n \in {1,3,5 }$ , $$\frac{240\sin(n\pi)}{n^6-35n^4+259n^2-225}$$ is a $0 \over 0$ form. Evaluate the limit and you'll find your Fourier coefficients. – N74 Apr 22 '19 at 10:37
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The easy way

$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\ \sin^5 x = \frac {(e^{ix} - e^{-ix})^5}{32i}\\ \frac {e^{5ix} -5 e^{3ix} + 10 e^{ix} - 10 e^{-ix} + 5e^{-3ix} + e^{-5ix}}{32i}\\ \frac {\sin 5x -5 \sin 3x + 10 \sin x}{16}$

The harder way:

$\sin 5x = 5\cos^4 x\sin x - 10\cos^2 x\sin^3 x + \sin^5 x \\ \sin 5x = 5(1-2\sin^2 x + \sin^4 x)\sin x - 10(1-\sin^2 x)\sin^3 x + \sin^5 x \\ \sin 5x = 5\sin x - 20 \sin^3 x + 16\sin^5 x\\ \sin^5 x = \frac 1{16}\sin 5x -\frac {5}{16} \sin x + \frac {20}{16} \sin^3 x\\ $

$\sin 3x = 3\sin x - 4\sin^3 x\\ \sin^3 x = \frac {3\sin x - \sin 3x}{4}$

$\sin^5 x = -\frac {5}{16} \sin x + \frac {20}{16} (\frac 34\sin x - \frac {1}{4}\sin 3x)\\ \sin^5 x = \frac 1{16}\sin 5x -\frac {5}{16} \sin 3x + \frac {10}{16} \sin x$

user317176
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  • I just noticed that instead of $32i$ I used $-32$. MATLAB now returned the right value, thank you :D – Maaa09 Apr 22 '19 at 11:05