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Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation

$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots + \dfrac{(-1)^{n}}{\dbinom{2n+1}{1}} =0 $$

Hence prove that

$$\sum_{r=1}^{\infty} \dfrac{1}{r^2}=\dfrac{\pi^{2}}{6}$$

I was stumped on the first sight. Then I tried using complex numbers but it was in vein. I further tried simplifying the equation, but since it contains only half of the binomial coefficients, I wasn't able to get a simpler equation.

Any help will be appreciated.
Thanks.

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  • $\begingroup$ Two things about your equation: do we assume that the sum continues until the $x^1$ term, or is there also an $x^0=1$ at the end? Did you mean to write $-\ldots=0$, i.e. $\pm$ in alternation? $\endgroup$ – Demosthene May 16 '15 at 20:58
  • $\begingroup$ @Demosthene Yes, the signs are alternate and it continues till the constant term i.e., $(-1)^{n} \dfrac{1}{\dbinom{2n+1}{1}}$ $\endgroup$ – Henry May 16 '15 at 22:03
  • $\begingroup$ @Samurai I have suggested an edit for your title. In the future, please try to be more descriptive; lots of questions here involve "humongous equations", and that won't help anyone who might be interested in this question to find it. $\endgroup$ – Eric Tressler May 16 '15 at 22:10
  • $\begingroup$ @EricTressler k, thanks :) $\endgroup$ – Henry May 16 '15 at 22:12
  • $\begingroup$ Since this is a polynomial $p(x)$ of degree $n$, it has at most $n$ roots. You only need to verify that the $n$ roots given are distinct and that your polynomial is $0$ on them; i.e. you can take an arbitrary $x_k = \cot^2\left(\frac{k\pi}{2n+1}\right)$ for $1 \leq k \leq n$, and show that $p(x_k) = 0$. $\endgroup$ – Eric Tressler May 16 '15 at 22:30
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We have

$$\frac{e^{m i \theta}}{\sin^m \theta} = \frac{(\cos \theta + i \sin \theta)^m}{\sin^m \theta} = (\cot \theta + i)^m $$

for $m \in \mathbb{N}$ and $\theta \in \mathbb{R}$. Take $m=2n+1$ and $\theta = k\pi/(2n+1)$ to get

$$\frac{(-1)^k}{\sin^{2n+1}\theta} = \sum_{r=0}^{2n+1} \binom{2n+1}{r}i^r \cot^{2n+1-r} \theta . $$

The imaginary part comes from the summands for which $r$ is odd, hence

$$0 = \sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s (\cot^2\theta)^{n-s}. $$

Equivalently, the polynomial

$$\sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s x^{n-s} $$

has $\cot^2 \theta$ as a root. Divide by $2n+1$ to get the relation in the question. For the second part, take the coefficient of $x^{n-1}$ to get

$$\sum_{k=0}^n (\cot \frac{k\pi}{2n+1})^2 = \frac{1}{2n+1}\binom{2n+1}{3}. $$

Now divide by $2n(2n-1)$ and take a limit using $\cot \theta = 1/\theta + O(\theta)$.

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Like Sum of tangent functions where arguments are in specific arithmetic series,

$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$

Multiplying the denominator & the numerator by $\cot^{2n+1}x$ we get,

$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\cot^{2n}x-\binom{2n+1}3\cot^{2n-2}x+\cdots}{\cot^{2n+1}x-\binom{2n+1}2\cot^{2n-1}x+\cdots}$$

If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is any integer

$\implies x=\dfrac{r\pi}{2n+1}$ where $r=-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$

So, the finite roots of $$\binom{2n+1}1y^{2n}-\binom{2n+1}3y^{2(n-1)}+\cdots=0$$

are $y=\cot x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

But $\cot(-A)=-\cot A$

So, the finite roots of $$\binom{2n+1}1z^n-\binom{2n+1}3z^{(n-1)}+\cdots=0$$

are $z=\cot^2x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=1,2\cdots,n\pmod{2n+1}$

The rest has already been taken care of in the other answer

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