I was solving the exercise questions of the book "Mathematical Analysis - 2nd Edition" by Tom Apostol and I came across the questions mentioned below. While I was able to solve a few questions, the others I did not even get any hint of!

1. (a) By equating imaginary parts in DeMoivre's Formula prove that $$\sin {n\theta} = \sin^n\theta \left\lbrace \binom{n}{1} \cot^{n - 1}\theta - \binom{n}{3} \cot^{n - 3}\theta + \binom{n}{5} \cot^{n - 5}\theta - + \cdots \right\rbrace$$ (b) If $0 < \theta < \dfrac{\pi}{2}$, prove that $$\sin{\left( 2m + 1 \right)\theta} = \sin^{2m+1}\theta . P_m\left( \cot^2 \theta \right)$$ where $P_m$ is the polynomial of degree $m$ given by $$P_m(x) = \binom{2m + 1}{1} x^m - \binom{2m + 1}{3} x^{m - 1} + \binom{2m + 1}{5} x^{m - 2} - + \cdots$$ Use this to show that $P_m$ has zeros at $m$ distinct points $x_k = \cot^2 \left( \dfrac{k\pi}{2m + 1} \right)$ for $k = 1, 2, \dots, m$. (c) Show that the sum of zeros of $P_m$ is given by $$\sum\limits_{k = 1}^{m} \cot^2 \dfrac{k\pi}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$ and that the sum of there squares is given by $$\sum\limits_{k = 1}^{m} \cot^4 \dfrac{k\pi}{2m + 1} = \dfrac{m\left( 2m - 1 \right) \left( 4m^2 + 10m - 9 \right)}{45}$$

  1. Prove that $z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right)$ for all complex $z$. Use this to derive the formula $$\prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$

As far as the solutions are concerned, I am through with the 1st part of 1st question and even half of the second part. But, in the second question, proving the zeros and their sum (and the sum of their squares) is getting really difficult. I am not getting any sort of hint as to how to prove it further.

And for the second question, I could do the first half part since it was essentially finding the $n$ roots of unity. But for the second part, I have nearly proved everything but what was asked. Many times I came to the conclusion that $$\prod\limits_{k = 1}^{n} \sin \dfrac{k\pi}{n} = 0$$ which is obvious because at $k = n$, we have a term of $\sin \pi$ which is equal to $0$. I am not getting how to remove that last term from the product using the result we just proved above!

Help will be appreciated!

  • $\begingroup$ For the future, try to avoid asking multiple questions in one post. $\endgroup$ – rtybase Feb 13 '18 at 22:46
  • $\begingroup$ @rtybase Surely, I will take of this from next time! $\endgroup$ – Aniruddha Deshmukh Feb 14 '18 at 6:09

With the 2nd question, 2nd part, you are asked to

$$z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right) \Rightarrow \prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$

Note that when $k=n$ $$e^{\frac{2ki\pi}{n}}=e^{2i\pi}=1$$ also $$z^n-1=(z-1)(z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1)$$ altogether $$\color{red}{(z-1)}(z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1)=\color{red}{(z-1)}\prod\limits_{k = 1}^{n-1} \left( z - e^{\frac{2ki\pi}{n}} \right)$$ which is $$z^{n-1}+z^{n-2}+z^{n-3}+...+z^2+z+1=\prod\limits_{k = 1}^{n-1} \left( z - e^{\frac{2ki\pi}{n}} \right)$$ and substituting $z=1$ $$n=\prod\limits_{k = 1}^{n-1} \left( 1 - e^{\frac{2ki\pi}{n}} \right)= \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \left( e^{-\frac{ki\pi}{n}} - e^{\frac{ki\pi}{n}} \right)=\\ (2i)^{n-1} (-1)^{n-1} \cdot \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \left(\frac{ e^{\frac{ki\pi}{n}} - e^{-\frac{ki\pi}{n}}}{2i} \right)=\\ 2^{n-1} (-i)^{n-1} \cdot \prod\limits_{k = 1}^{n-1} e^{\frac{ki\pi}{n}} \sin{\left(\frac{k\pi}{n}\right)}=\\ 2^{n-1} (-i)^{n-1} e^{\sum\limits_{k=1}^{n-1}\frac{ki\pi}{n}} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=\\ 2^{n-1} (-i)^{n-1} e^{\frac{i\pi}{n}\frac{n(n-1)}{2}} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=\\ 2^{n-1} (-i)^{n-1} i^{n-1} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}=2^{n-1} \cdot \prod\limits_{k = 1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)}$$

  • $\begingroup$ Any hints for 1st question? $\endgroup$ – Aniruddha Deshmukh Feb 14 '18 at 6:10

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