Is it possible that $\sin^5 (x)$ doesn't have a Fourier series?

Question

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?

Answer 1

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\\ $

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

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

Answer 2

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.

Answer 3

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.