4
$\begingroup$

The following question is from a previous post. The reason that I am posting this again is because it had two questions which were not really related and hence, one of the questions was not answered. The question is from a book, "Mathematical Analysis - 2nd Edition" by Apostol. The question has three parts and is as follows:


  1. By equating imaginary parts in DeMoivre's formula, prove that $$\sin{\left(n\theta\right)} = \sin^n\theta \left\lbrace \binom{n}{1}\cot^{n - 1}\theta - \binom{n}{3} \cot^{n - 3}\theta + \binom{n}{5} \cot^{n - 5}\theta - + \cdots \right\rbrace$$

  1. If $0 < \theta < \dfrac{\pi}{2}$, prove that $$\sin{\left(\left( 2m + 1 \right)\theta\right)} = \sin^{2m + 1}\theta P_m(\cot^2\theta)$$ where $P_m$ is the polynomial of degree $m$ given by $$P_m\left( x \right) = \binom{2m + 1}{1}x^m - \binom{2m + 1}{3}x^{m-1} + \binom{2m + 1}{5}x^{m-2} - + \cdots$$ Use this to show that $P_m$ has zeros at $m$ distinct points $x_k = \cot^2\left( \dfrac{\pi k}{2m + 1} \right)$ for $k = 1, 2, \dots, m$.

  1. Show that the sum of zeros of $P_m$ is given by $$\sum\limits_{k=1}^{m} \cot^2\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$ and that the sum of theie squares is given by $$\sum\limits_{k=1}^{m} \cot^4\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right) \left( 4m^2 + 10m - 9 \right)}{45}$$

As far as the solution is concerned, I have been able to prove the first part and upto proving the existence of the polynomial in the second part. For later parts, I do not have any insights in proceeding towards the solution.

Help will be appreciated!

$\endgroup$
2
$\begingroup$

Hints


For part 2, since you already showed the existence of the polynomial s.t. $$\sin \left( 2m + 1 \right)\theta = \left(\sin^{2m + 1}\theta \right)P_m(\cot^2\theta)$$ replace $\theta$ with $\frac{k\pi}{2m+1}$ and $$\left(\sin^{2m + 1}\frac{k\pi}{2m+1} \right)\color{red}{P_m\left(\cot^2\frac{k\pi}{2m+1}\right)}=\sin \left(\left( 2m + 1 \right)\frac{k\pi}{2m+1}\right)=\sin{k\pi}=\color{red}{0}$$


For part 3, have you tried using Vieta's formulas? It is a simple application of $$\color{red}{\sum\limits_{k=1}^{m}x_k}=\sum\limits_{k=1}^{m}\cot^2\frac{\pi k}{2m+1}=\color{red}{-\frac{a_{m-1}}{a_m}}=-\frac{-\binom{2m+1}{3}}{\binom{2m+1}{1}}=\\ \frac{(2m+1)2m(2m-1)}{1\cdot2\cdot3}\cdot \frac{1}{2m+1}=\frac{m(2m-1)}{3}$$ For the 2nd part of $3$ use the fact that $$\sum\limits_{k=1}^{m}x_k^2=\left(\sum\limits_{k=1}^{m}x_k\right)^2-2\sum\limits_{k<t}x_kx_t=\left(-\frac{a_{m-1}}{a_m}\right)^2-2\left(\frac{a_{m-2}}{a_m}\right)$$

$\endgroup$
  • $\begingroup$ What about the second part? How will I show that the roots are what we have to show? $\endgroup$ – Aniruddha Deshmukh Feb 18 '18 at 10:52
  • $\begingroup$ @AniruddhaDeshmukh is it better now? $\endgroup$ – rtybase Feb 18 '18 at 11:18
  • 1
    $\begingroup$ Yes! It was pretty simple. I could not see it, though! $\endgroup$ – Aniruddha Deshmukh Feb 18 '18 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.