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I'm really not sure about this Product-to-sum identity on wiki.

See:

I cannot find this anywhere on the web - does anybody know a reference? Certainly the one wiki gives does not cover it.

I'm guessing it is used for something like $\cos(x)\cos(5x)$ (easier done with the other identities mentioned) to convert into a sum but I'm not sure really. I don't understand the $S=\{1,-1\}^n$ bit.

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  • $\begingroup$ $S$ is the set of all $n$-tuple with coordinates $-1$ or $1$. $\endgroup$
    – Vincenzo Tibullo
    Commented Jul 20, 2014 at 12:36

1 Answer 1

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Note that the Wikipedia formula you are referring to contains a wrong extra factor of $\frac12$ (this can be seen already by setting $n=1$). More formal proof in the general case: \begin{align} 2^n\prod_{k=1}^{n}\cos\theta_k&=\prod_{k=1}^{n}\left(e^{i\theta_k}+e^{-i\theta_k}\right)=\prod_{k=1}^{n}\left(\sum_{\epsilon_k=\pm 1}e^{i\epsilon_k\theta_k}\right)=\\&=\sum_{\epsilon_1,\ldots,\epsilon_n=\pm 1}e^{i\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right)}=\\ &=\frac12\sum_{\epsilon_1,\ldots,\epsilon_n=\pm 1}\left(e^{i\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right)}+e^{-i\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right)}\right)=\\ &=\sum_{\epsilon_1,\ldots,\epsilon_n=\pm 1}\cos\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right). \end{align}

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  • $\begingroup$ @onepound: One thing that may add some intuition as to why O.L. suggests this is that the factor of $2^{n-1}$ is suggestive: by multiplying both sides by $2^n$, the only functions involved are all of the form $2 \cos \phi$. $\endgroup$
    – Semiclassical
    Commented Jul 20, 2014 at 12:52
  • $\begingroup$ so now if I substitute cos(x)cos(5x) into the corrected version would I get 1/4 e^(-4 i x)+1/4 e^(4 i x)+1/4 e^(-6 i x)+1/4 e^(6 i x) or 1/2 cos(4 x)+1/2 cos(6 x) ? The e1,...,en=+/-1 makes e1theta1+ e2theta2+ e3theta3...e(n)theta(n) is that correct? $\endgroup$
    – onepound
    Commented Jul 20, 2014 at 14:16
  • $\begingroup$ @onepound if you substitute that ($n=2$, $\theta_1=x$, $\theta_2=5x$) then in the corrected version on the right you will have $$\frac1{2^2} \left[\cos (+x+5x)+\cos (+x-5x)+\cos (-x+5x)+\cos (-x-5x)\right]= \frac{\cos6x+\cos4x}{2}$$ $\endgroup$
    – Start wearing purple
    Commented Jul 20, 2014 at 15:18
  • $\begingroup$ okay O.L. thanks for clearing that up and showing how to use the identity. Maybe I'll have to be a bit more careful with what's on wiki in future! $\endgroup$
    – onepound
    Commented Jul 20, 2014 at 19:10

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