Doing excersies from Richard Hammings book about digital filters I've got two questions about them:

1) Fourier expansion of $g(x) = \sin^5(x)$. Provided answer is: $\sin^5(x)=5\sin(x)-20 \sin(3 x)+16\sin(5 x)$. Which is obviously contradicts well-known solution. Is it mistake in a book or author meant something another?

2)The second excersice which confusing me is expansion in compex notation of $g(x) = \exp^ {\alpha x}$. I got the following form of coefficients: $c_{k} = {\frac {1} {(2\pi)}} (\exp^{\alpha\pi} - \exp^{-\alpha\pi}){\frac {(-1)^k} {(\alpha-ik)}}$. These coefficients looks a little bit frightening to me and I don't know how to check them... Any ideas how to validate solution?

Thanks in advance!

  • $\begingroup$ look up "sinc function" in your textbooks or in Wikipedia or in Wolfram or somewhere. $\endgroup$ – robert bristow-johnson Dec 29 '13 at 14:23
  • $\begingroup$ Thank you for reply, Robert! But I can't understand your comment... This excersices were given without references to sinc function. Can you elaborate a little bit more?.. $\endgroup$ – Sharov Dec 29 '13 at 15:57
  • $\begingroup$ @robertbristow-johnson What on earth does the sinc function have to do with this? $\endgroup$ – Dilip Sarwate Dec 29 '13 at 21:44
  • $\begingroup$ i think the sinc function is related to $$\frac1{2\pi}\left[\frac{e^{(α-ik)x}}{α-ik}\right]_{x=-\pi}^{x=\pi}$$ $\endgroup$ – robert bristow-johnson Dec 30 '13 at 1:53

First, or second, with complex notation and the binomial theorem,

$$\sin^5 x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^5 = \frac{e^{i5x}-5e^{i3x}+10e^{ix}-10e^{-ix}+5e^{-3ix}-e^{-i5x}}{32i} = \frac{\sin(5x)-5\sin(3x)+10\sin x}{16}$$

On the other hand,

\begin{align} \sin(5x) &= Im(e^{i5x}) = Im([\cos x+i\sin x]^5) \\ &= 5\cos^4x\sin x - 10\cos^2x\sin^3x + \sin^5x\\ &= 5(1-2\sin^2 x+\sin^4 x)\sin x-10(1-\sin^2x)\sin^3x+ \sin^5x\\ &= 5\sin x - 20\sin^3x +16\sin ^5x \end{align}

so you got the coefficients right, but the notation reversed.

In the second task, is the star connected to convolution, cepstrum transforms or similar?

No, it is just $g(x)=\exp(α x)=e^{α x}$ defined on the interval $[-\pi, \pi]$. Then

\begin{align} c_k &=\frac1{2\pi}\int_{-\pi}^\pi g(x)e^{-ikx}dx =\frac1{2\pi}\int_{-\pi}^\pi e^{(α-ik)x}dx \\ &=\frac1{2\pi}\left[\frac{e^{(α-ik)x}}{α-ik}\right]_{x=-\pi}^{x=\pi}\\ \end{align}

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  • $\begingroup$ hey, Lutz. good to see you here. i remember you from Wikipedia. $\endgroup$ – robert bristow-johnson Dec 30 '13 at 1:52

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