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I want to show that $\sum_{|j| < n} (n-|j|) \exp(ij\lambda)= \dfrac{\sin^2(\frac 1 2 n\lambda)}{\sin^2(\frac 1 2 \lambda)}$.

I know from Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$ \begin{equation} (1)\sum_{j=1}^{n-1} cos(j\lambda) = -\frac 1 2 + \frac{\sin(\frac{2n-1} 2 \lambda)}{2\sin(\frac \lambda 2)}. \end{equation}

and from the Hint in the first answer in How to show that $\frac{1}{\tan(x/2)}=2 \sum_{j=1}^{\infty}\sin(jx)$ in Cesàro way/sense? $$ (2)\frac d {dx} \sum_{j=1}^{n-1} sin(j\lambda) =\frac d {d\lambda} \frac{\cos({\frac \lambda 2})-\cos(\frac{n+1}2 \lambda)}{2\sin(\frac\lambda 2)} = \frac{n\sin(\frac{n+1} 2 \lambda)\sin(\frac \lambda 2)+\cos(\frac{n\lambda}2) - 1}{4\sin^2(\frac \lambda 2)}. $$

This is what i did so far: \begin{align} \sum_{|j| < n} (n-|j|) \exp(ij\lambda) &= n + 2\sum_{j=1}^{n-1}(n-j) cos(j\lambda) \\ &=n + 2n \sum_{j=1}^{n-1} cos(j\lambda) - \frac d {d\lambda}\sum_{j=1}^{n-1} sin(j\lambda)\\ &= \frac{n\sin(\frac{2n-1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 + \frac 1 2 n \sin(\frac {n+1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 \cos( \frac{n\lambda}2)}{\sin^2(\frac \lambda 2)}. \end{align}

If this and the proposition is true, it should be true that $$ n\sin(\frac{2n-1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 + \frac 1 2 n \sin(\frac {n+1} 2 \lambda)\sin(\frac \lambda 2) + \frac 1 2 \cos( \frac{n\lambda}2) - \sin^2(\frac {n\lambda} 2) = 0. $$

But for $n=2$, i get $$ 3\sin(\frac {3\lambda} 2) \sin(\frac \lambda 2) + \frac 1 2 + \frac 1 2 \cos(\lambda) - \sin^2 \lambda \ne 0. $$ (checked with Wolframalpha)

So there is probably something wrong with my calculations. But i checked everything twice and don't see my mistake. Do you see it?

EDIT

Trying to prove it as proposed in the second answer.

Let $z:=\exp(i\lambda)$ and $p(z):=\sum_{j=0}^{n-1} z^j$. First i want to show that $p(z)p(z^{-1})=\sum_{|j|<n} (n-|j|) z^j$.

It is \begin{align} \sum_{j=1}^{n-1} z^{j}&= p(z) - 1 = \frac{1-z^n}{1-z} - 1.\\ \sum_{j=1}^{n-1} z^{-j}&= p(z^{-1})-1 = \frac{1-z^{-n}}{1-z^{-1}} - 1.\\ \sum_{j=1}^{n-1} jz^{-j} &= i \frac d {d\lambda} p(z^{-1}) = \frac{-nz^{-n}+(n-1)z^{-n-1}+z^{-1}}{(1-z^{-1})^2}.\\ \sum_{j=1}^{n-1} jz^{j} &= -i \frac d {d\lambda} p(z) = \frac{-nz^{n}+(n-1)z^{n+1}+z}{(1-z)^2}.\\ \end{align} So i have \begin{align} \sum_{|j|<n} (n-|j|) z^j = n + n(\sum_{j=1}^{n-1} z^{j} + \sum_{j=1}^{n-1} z^{-j}) - \sum_{j=1}^{n-1} jz^{j} - \sum_{j=1}^{n-1} jz^{-j}. \end{align} If i now put in the values from above and substract $p(z)p(z^{-1})$ i should get $0$. But Wolfram Alpha doesn't agree, so i am again on the wrong track. What did i wrong?

The last step $p(z)p(z^{-1})= \frac{(z^{n/2}-z^{-n/2})^2}{(z^{1/2}-z^{-1/2})^2}$ was easy to show.

EDIT2

I did it now. It was a lot easier to show directly $p(z)p(z^{-1})=\sum_{|j|<n} (n-|j|) z^j$ without the geometric series partial sums values.

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It's often easier to solve these sorts of problems by setting $z=e^{i\lambda}$ and then simplifying a resulting polynomial expression. –  Thomas Andrews Nov 30 '12 at 14:32
    
You didn't distribute the $2$ when you expanded $2\sum (n-j)\cos j\lambda$ –  Thomas Andrews Nov 30 '12 at 15:38

2 Answers 2

up vote 2 down vote accepted

Letting $\alpha=\frac{\lambda}{2}$ to make the expressions cleaner, we get: $$\begin{align} \sum_{|j| < n} (n-|j|) \exp(ij\lambda) &= n + 2\sum_{j=1}^{n-1}(n-j) \cos j\lambda \\ &=n + 2n \sum_{j=1}^{n-1} \cos j\lambda - 2\frac d {d\lambda}\sum_{j=1}^{n-1} \sin j\lambda\\ &= \frac{n\sin(2n-1)\alpha\sin\alpha + \frac 1 2 - \frac 1 2 n \sin (n+1)\alpha \sin \alpha - \frac 1 2 \cos n\alpha}{\sin^2\alpha}. \end{align}$$ Note the minus signs in the numerator. I'm not sure if that solves the problem - there might be an error elsewhere. Wolfram Alpha says when $n=2$ the numerator is $\sin^2 2\alpha$, but $n=1$ doesn't agree.

The easier way to do this sort of problem is via the comment I gave above, using $z=e^{i\lambda}$.

Let $p(z)=\sum_{j=0}^{n-1} z^j$. Show that:$$p(z)p(z^{-1})=\sum_{|j|<n} (n-|j|) z^j$$

Now, $p(z)=\frac{z^n-1}{z-1}$, and $p(z^{-1})=z^{-(n-1)}p(z)$. Now show: $$p(z)p(z^{-1})= \frac{(z^{n/2}-z^{-n/2})^2}{(z^{1/2}-z^{-1/2})^2}$$

Then let $z=e^{i\lambda}$.

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I tried to proof it your way, but failed again... –  Runje Dec 2 '12 at 14:50
    
I edited my originial post where i tried to proof it. Or is there an easier way to prove it than i tried it? –  Runje Dec 2 '12 at 16:16
    
Thx Thomas, i did it now your way. –  Runje Dec 3 '12 at 14:46

The only thing that I see at first sight is that you didn't distribute properly the minus sign in front of the derivative.

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