I'm currently struggling at calculation the Fourier series of the given function

$$\sin(x)^2 \cos(x)^3$$

Given Euler's identity, I thought that using the exponential approach would be the easiest way to do it.

What I found was: $$\frac{-1}{32}((\exp(2ix)-2\exp(2ix)+\exp(-2ix))(\exp(3ix)+3\exp(ix)+3\exp(-ix)+\exp(-3ix)))$$

Transforming it back, the result is:

$$ -\frac{1}{18}(\cos(5x)+\cos(3x)+2\cos(x))$$

(I've checked my calculations multiple times, I'm pretty sure it's correct.)

Considering the point $x = 0$ however, one can see that the series I found doesn't match the original function.

Could someone help me find my mistake?

  • $\begingroup$ Nice initials.... $\endgroup$ – Ron Gordon May 18 '13 at 17:54
  • $\begingroup$ Cheers, nice name :P $\endgroup$ – R.G. May 18 '13 at 18:00

1) Trigonometric identities: $$ \sin^2 x\cos^3x=(\sin x\cos x)^2\cos x=\left(\frac{\sin 2x}{2}\right)^2\cos x=\frac{1}{4}\sin^22x\cos x $$ $$ =\frac{1}{4}\left(\frac{1-\cos 4x}{2}\right)\cos x=\frac{\cos x}{8}-\frac{\cos 4x\cos x}{8} $$ $$ =\frac{\cos x}{8}-\frac{\cos 5x+\cos 3x}{16} $$ $$ =\frac{\cos x}{8}-\frac{\cos 3x}{16}-\frac{\cos 5x}{16} $$ 2) Complex exponential: $$ \sin^2x\cos^3x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2\left(\frac{e^{ix}+e^{-ix}}{2}\right)^3 $$ $$ =-\frac{1}{32}(e^{2ix}-2+e^{-2ix})(e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}) $$ $$ =-\frac{1}{32}(e^{5ix}+e^{3ix}-2e^{ix}-2e^{-ix}+e^{-3ix}+e^{-5ix}) $$ $$ =-\frac{1}{32}(2\cos 5x+2\cos 3x-4\cos x) $$ $$ =\frac{1}{16}(2\cos x-\cos 3x-\cos 5x) $$

Note: you made a mistake when you expanded $(e^{ix}-e^{-ix})^2$. I have no idea how you ended up with this $18$. You probably meant $16$.

  • $\begingroup$ Mh are you planning to do one with the cauchy produkt of the series ? :D $\endgroup$ – Dominic Michaelis May 18 '13 at 17:45
  • $\begingroup$ @DominicMichaelis He, he. Actually, I added the second one to show the OP where he/she went wrong. $\endgroup$ – Julien May 18 '13 at 17:46
  • $\begingroup$ Awesome, that's exactly where I went wrong :) Thanks so much for taking the time and providing such an elaborate answer! $\endgroup$ – R.G. May 18 '13 at 17:55
  • $\begingroup$ @R.G. You're welcome. Glad it helped. $\endgroup$ – Julien May 18 '13 at 17:58

I used trig idendities and came to \[ \sin(x)^2 \cdot \cos(x)^3 = \frac{1}{16} \cdot \left( 2\cos(x) - \cos(3x)-\cos(5x)\right)\] This surely works for $x=0$, and Mathematica gives the same.

  • $\begingroup$ Thanks a lot for your answer! $\endgroup$ – R.G. May 18 '13 at 17:56

You don't show enough for me to diagnose anything. I can tell you by using the double-angle and add-multiply formulae, I get

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

The way I did this is to rewrite the LHS as

$$\begin{align} \cos^3{x}-\cos^5{x} &= \cos{x} (\cos^2{x}-\cos^4{x})\\&= \frac12 \cos{x} (1+\cos{2 x}) - \frac14 \cos{x} (1+\cos{2 x})^2 \\ &= \frac12 \cos{x}(1+\cos{2 x}) - \frac14 \cos{x}(1+2 \cos{2 x} + \cos^2{2 x}) \\ &= \frac14 \cos{x} - \frac18 \cos{x} (1+\cos{4 x})\\ &= \frac18 \cos{x} - \frac{1}{16} \cos{3 x} - \frac{1}{16} \cos{5 x}\end{align}$$

The result follows.

  • $\begingroup$ Thanks a lot for your answer! $\endgroup$ – R.G. May 18 '13 at 17:57

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