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**Anthonny** In the next step it says 0 is equal to that equation and I understand why its equal to 0. However, I do not see why it ends with (-1)^m. I think I see why it would be a '1' since the 'i' was factored out, but I do not understand why it is negative and why it's raised to the mth power.

In the next equation, a function p(t) is defined. I was wondering why (cotx)² needs to be a one-to-one function in the interval in order for this to happen. I also do not see why the equation before it must equal 0 for this to happen.

This $\displaystyle (-1)^m$ comes from the fact that $\displaystyle i^{2m+1}=(-1)^m i $. Let me rewrite the expansion of $\displaystyle (\cot x + i)^{2m+1}$:

$\displaystyle (\cot x+i)^{2m+1}=\sum_{k=0}^{2m+1}{2m+1\choose k}(\cot x)^{2m+1-k}i^k$ $\displaystyle =\sum_{p=0}^m {2m+1\choose 2p}(\cot x)^{2m+1-2p}i^{2p}+\sum_{p=0}^m {2m+1\choose 2p+1}(\cot x)^{2m+1-(2p+1)}i^{2p+1}$

(separating terms with even and odd indices) and $\displaystyle i^{2p}=(-1)^p$, $\displaystyle i^{2p+1}=(-1)^p i$, hence

$\displaystyle (\cot x+i)^{2m+1}=\sum_{p=0}^m (-1)^p{2m+1\choose 2p}(\cot x)^{2m+1-2p}$ $\displaystyle +i\sum_{p=0}^m (-1)^p{2m+1\choose 2p+1}(\cot x)^{2(m-p)}$.

Equating imaginary parts, we get

$\displaystyle \frac{\sin((2m+1)x)}{(\sin x)^{2m+1}}=\sum_{p=0}^m (-1)^p{2m+1\choose 2p+1}(\cot x)^{2m+1-(2p+1)}=P\big((\cot x)^2\big)$,

where we define the polynomial $\displaystyle P(t)=\sum_{p=0}^m (-1)^p{2m+1\choose 2p+1}t^{m-p}$.

For $\displaystyle r=1,\ldots,m$, the real number $\displaystyle x_r=\frac{r\pi}{2m+1}$ satisfies $\displaystyle 0<x_r<\frac{\pi}{2}$, hence $\displaystyle \sin x_r\neq 0$, and $\displaystyle \sin((2m+1)x_r)=0$, so that the previous equation shows that $\displaystyle (\cot x_r)^2$ is a root of $\displaystyle P$.

The degree of $\displaystyle P$ is $\displaystyle m$, hence it has at most $\displaystyle m$ roots. Furthermore, since $\displaystyle 0<x_1<x_2<\cdots<x_m<\frac{\pi}{2}$, we have $\displaystyle \cot x_1>\cot x_2>\cdots>\cot x_m>0$, hence $\displaystyle (\cot x_1)^2>(\cot x_2)^2>\cdots>(\cot x_m)^2>0$, which shows that these $\displaystyle m$ roots are distinct. Therefore they are the $\displaystyle m$ roots of $\displaystyle P$.

For any polynomial $\displaystyle P(t)=a_0+a_1 t+\cdots +a_m t^m$ with $\displaystyle m$ roots $\displaystyle t_1,\ldots,t_m$, one can factorize $\displaystyle P(t)=\lambda(t-t_1)(t-t_2)\cdots(t-t_m)$. Expanding this expression (mentally) and gathering the terms of common degree, we see that the term of degree $\displaystyle m$ is just $\displaystyle a_m t^m=\lambda t^m$ and the term of degree $\displaystyle m-1$ is $\displaystyle a_{m-1}t^{m-1}=-\lambda(t_1+\cdots+t_m)t^{m-1}$. Thus we have$\displaystyle t_1+\cdots+t_m=-\frac{a_{m-1}}{a_m}$.

In the case of our polynomial, this gives $\displaystyle \sum_{r=1}^m \big(\cot\frac{r\pi}{2m+1}\big)^2 = \frac{{2m+1\choose 3}}{{2m+1\choose 1}}=\frac{2m(2m+1)}{6}$.

On the other hand, we have, for $\displaystyle 0<x<\frac{\pi}{2}$, $\displaystyle \cot^2x<\frac{1}{x^2}<1+\cot^2x$. Applying this to $\displaystyle x=x_r$ gives two bounds for $\displaystyle \sum_{r=1}^m \frac{1}{x_r^2}$, hence for $\displaystyle \sum_{r=1}^m \frac{1}{r^2}$ by factorizing a few terms out. Letting $\displaystyle m$ go to infinity, both bounds are easily seen to converge toward $\displaystyle \frac{\pi^2}{6}$, hence the conclusion.

Feel free to ask for more clarifications.