I was working on a book, which was asking me to prove that some product is equal to nn. I had reduced the problem to proving a trigonometric identity, but I couldn't prove it although I spent much time on it. Then, I checked the solution, and it solves the problem with a quite different approach that doesn't even enter to trigonometric expressions at all. But, still, this nice identity that I hadn't encountered before must be true then! I was wondering if people can think of a direct proof for that?

Show this following identity: $$\sum_{i=1}^{n-1}\left\{\left(\cos{\frac{2\pi}{n}}-\cos{\dfrac{2\pi i}{n}}\right)\left[\left(\sum_{j=1}^{n}x_{j}\cos{\dfrac{2ij\pi}{n}}\right)^2+\left(\sum_{j=1}^{n}x_{j}\sin{\dfrac{2ij\pi}{n}}\right)^2\right]\right\} $$$$=\left(1-\cos{\dfrac{2\pi}{n}}\right)\left(\sum_{i=1}^{n}x_{j}\right)^2+n\cos{\dfrac{2\pi}{n}}\sum_{i=1}^{n}x^2_{i}-n\sum_{i=1}^{n}x_{i}x_{i+1},$$ where $x_{i}\in \Bbb R$ and $x_{n+1}=x_{1}.$

  • $\begingroup$ With your new condition $x_{n+1}=x_1$, the identity holds for $n=1,2$. $\endgroup$ – John Bentin Sep 11 at 8:33
  • $\begingroup$ A hint, if my arithmetic's correct. If $j\ne k$, the $x_jx_k$ coefficient is $2\cos\frac{2\pi i(j-k)}{n}\left(\cos\frac{2\pi}{n}-\cos\frac{2i\pi}{n}\right)$ on the left-hand side, and $2\left(1-\cos\frac{2\pi}{n}\right)-n\delta_{j+1,\,k}$ on the right hand-side. The $x_j^2$ coefficient is $\cos\frac{2\pi}{n}-\cos\frac{2i\pi}{n}$ on the left-hand side, and $1+(n-1)\cos\frac{2\pi}{n}$ on the right-hand side. $\endgroup$ – J.G. Sep 11 at 9:02
  • $\begingroup$ "I checked the solution, and it solves the problem with a quite different approach that doesn't even enter to trigonometric expressions at all." ... You should post (or at least summarize) that solution here. It may provide answerers with insights that can lead to alternative solutions without wasting time duplicating that effort, and it can alert answerers to the kind of solution you do not want to see. $\endgroup$ – Blue Sep 11 at 17:13

This is an application of the Plancherel-Parseval identities of the isometry properties of the Fourier transform, here in finite dimension. Essentially, a properly scaled variant of the discrete Fourier transform is unitary.

One comes to the use of the Fourier transform as the second factor on the left side is the norm square of components of the discrete Fourier transform. Using the convention $$ \hat x_m=\sum_{k=1}^nx_k\exp\left(-i\frac{2km\pi}{n}\right) =\sum_{k=1}^nx_k\cos\left(\frac{2km\pi}{n}\right) -i\sum_{k=1}^nx_k\sin\left(\frac{2km\pi}{n}\right) \tag{DFT} $$ the mentioned isometry is $$ n\langle {\bf x},{\bf y}\rangle =\langle {\bf \hat x},{\bf \hat y}\rangle. \tag{ISO} $$

The defining term in the claimed identity is the last one, which can be interpreted as scalar product of ${\bf x}$ with a shifted copy. Let $S$ be the left-shift operator, $(Sx)_i=x_{i+1}$, $(Sx)_n=x_1$, then the Fourier coefficients of it are $$ \widehat{(Sx)}_m =\sum_{k=1}^nx_{k+1}\exp\left(-i\frac{2km\pi}{n}\right) =\sum_{k=1}^nx_{k}\exp\left(-i\frac{2(k-1)m\pi}{n}\right) =\exp\left(i\frac{2m\pi}{n}\right)\,\hat x_m \tag1 $$ Thus we get for this last term $$ n\sum_{k=1}^nx_kx_{k+1}=n\langle {\bf x},S{\bf x}\rangle =\langle {\bf \hat x},\widehat{S{\bf x}}\rangle =\sum_{m=1}^n\cos\left(\frac{2m\pi}{n}\right)\,\left|\hat x_m\right|^2 \tag2 $$ This proves the non-trivial part of the claim.

All that remains is some window-dressing, essentially removing the outer terms of the last sum by using the identities $$ n\sum_{k=1}^nx_k^2=n\langle {\bf x},{\bf x}\rangle =\langle {\bf \hat x},{\bf \hat x}\rangle =\sum_{m=1}^{n-1}\left|\hat x_m\right|^2+\left(\sum_{k=1}^nx_k\right)^2 \tag3 $$ and $\hat x_0=\hat x_n=\sum_{k=1}^n x_k$.

In total the claimed identity is \begin{align} &\sum_{m=1}^{n-1}\left(\cos\left(\frac{2\pi}{n}\right)-\cos\left(\frac{2m\pi}{n}\right)\right)\,\left|\hat x_m\right|^2 \\ &=\cos\left(\frac{2\pi}{n}\right)\sum_{m=1}^{n-1}\left|\hat x_m\right|^2 -\sum_{m=1}^{n-1}\cos\left(\frac{2m\pi}{n}\right)\,\left|\hat x_m\right|^2 \\ &=\cos\left(\frac{2\pi}{n}\right)\left(\|\hat{\bf x}\|^2-|\hat x_n|^2\right) -\left(\langle {\bf \hat x},\widehat{S{\bf x}}\rangle-\left|\hat x_n\right|^2\right) \\ &=n\cos\left(\frac{2\pi}{n}\right)\sum_{k=1}^nx_k^2 +\left(1-\cos\left(\frac{2\pi}{n}\right)\right)\left(\sum_{k=1}^nx_k\right)^2 -n\sum_{k=1}^nx_kx_{k+1} \tag4 \end{align}

  • $\begingroup$ can you have simple methods?Thanks $\endgroup$ – function sug Sep 11 at 16:25
  • $\begingroup$ This is as simple as it gets, the Fourier transform is involved due to the terms on the left side, where the second factor is $(Re(\hat x_m)^2+Im(\hat x_m)^2)=|\hat x_m|^2$. What is really essential is the property (ISO) and equation (2), everything else is decoration around it, using that $\hat x_0=\hat x_n=\sum_{k=1}^n x_k$. $\endgroup$ – LutzL Sep 11 at 16:31

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