\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2018.3.7 Question 7

1. We have \begin {align*} & \phantom {=} \frac {(\cot \theta + i)^{2n + 1} - (\cot \theta - i)^{2n + 1}}{2i} \\ & = \frac {\left (\cos \theta + i \sin \theta \right )^{2n + 1} - (\cos \theta - i \sin \theta )^{2n + 1}}{2i \sin ^{2n + 1} \theta } \\ & = \frac {\left (\cos (2n + 1) \theta + i \sin (2n + 1) \theta \right ) - (\cos (2n + 1) \theta - i \sin (2n + 1) \theta )}{2i \sin ^{2n + 1} \theta } \\ & = \frac {2i \sin (2n + 1) \theta }{2i \sin ^{2n + 1} \theta } \\ & = \frac {\sin (2n + 1) \theta }{\sin ^{2n + 1} \theta }, \end {align*}

   as desired.

   By applying the binomial expansion formula on the numerator, we have \begin {align*} & \phantom {=} (\cot \theta + i)^{2n + 1} - (\cot \theta - i)^{2n + 1} \\ & = \sum _{t = 0}^{2n + 1} \binom {2n + 1}{t} \cot ^t \theta \cdot i^{2n + 1 - t} - \sum _{t = 0}^{2n + 1} \binom {2n + 1}{t} \cot ^t \theta \cdot (-i)^{2n + 1 - t} \\ & = \sum _{t = 0}^{2n + 1} \binom {2n + 1}{t} \cot ^t \theta \cdot \left [i^{2n + 1 - t} - (-i)^{2n + 1 - t}\right ] \\ & = (-1)^n \cdot i \cdot \sum _{t = 0}^{2n + 1} \binom {2n + 1}{t} \cot ^t \theta \cdot i^{-t} \cdot \left [1 - (-1)^{1 - t}\right ]. \end {align*}

   Due to the existence of the final term, this means that only terms with even \(t\) will retain (give a 2), and odd \(t\)s will cancel. Hence, \begin {align*} & \phantom {=} (\cot \theta + i)^{2n + 1} - (\cot \theta - i)^{2n + 1} \\ & = (-1)^n \cdot i \cdot \sum _{t = 0}^{2n + 1} \binom {2n + 1}{t} \cot ^t \theta \cdot i^{-t} \cdot \left [1 - (-1)^{1 - t}\right ] \\ & = (-1)^n \cdot 2i \cdot \sum _{t = 0}^{n} \binom {2n + 1}{2t} \cot ^{2t} \theta \cdot i^{-2t} \\ & = 2i(-1)^n \cdot \sum _{t = 0}^{n} \binom {2n + 1}{2t} \cot ^{2t} \theta \cdot (-1)^t \\ & = 2i(-1)^n \cdot \sum _{t = 0}^{n} \binom {2n + 1}{2n - 2t + 1} \cot ^{2t} \theta \cdot (-1)^t \\ & = 2i(-1)^n \cdot \sum _{t = 0}^{n} \binom {2n + 1}{2t + 1} \cot ^{2 (n - t)} \theta \cdot (-1)^{n - t} \\ & = 2i \cdot \sum _{t = 0}^{n} \binom {2n + 1}{2t + 1} \cot ^{2 (n - t)} \theta \cdot (-1)^{t}. \end {align*}

   Hence, \begin {align*} & \phantom {=} \frac {\sin (2n + 1) \theta }{\sin ^{2n + 1} \theta } \\ & = \frac {2i \cdot \sum _{t = 0}^{n} \binom {2n + 1}{2t + 1} \cot ^{2 (n - t)} \theta \cdot (-1)^{t}}{2i} \\ & = \sum _{t = 0}^{n} \binom {2n + 1}{2t + 1} \cot ^{2 (n - t)} \theta \cdot (-1)^{t}. \end {align*}

   The left-hand side of the original equation is \begin {align*} \sum _{t = 0}^{n} \binom {2n + 1}{2t + 1} x^{n - t} \cdot (-1)^t. \end {align*}

   Let \(x = \cot ^2 \theta \), we have \[ \phantom {=} \frac {\sin (2n + 1) \theta }{\sin ^{2n + 1} \theta } = \sum _{t = 0}^{n} \binom {2n + 1}{2t + 1} x^{n - t} \cdot (-1)^t = 0. \]

   Therefore, we have \(\sin (2n + 1) \theta = 0\), and hence \((2n + 1) \theta = m \pi \) for \(m \in \ZZ \).

   To avoid duplicate solutions for \(x = \cot ^2 \theta \), we restrict \(\theta \in \left (0, \frac {\pi }{2}\right ]\), and hence \((2n + 1) \theta \in \left (0, \left (n + \frac {1}{2} \right )\pi \right ]\), and hence \(m = 1, 2, \ldots , n\).

   This solves to \(\theta = \frac {m\pi }{2n + 1}\) for \(m = 1, 2, \ldots , n\), and hence this gives exactly \[ x = \cot ^2 \left (\frac {m \pi }{2n + 1}\right ). \]

2. By Vieta’s Theorem, we will have \[ \sum _{m = 1}^{n} x_m = -\frac {-\binom {2n + 1}{3}}{\binom {2n + 1}{1}} = \frac {(2n + 1)(2n)(2n - 1)}{(2n + 1) \cdot 3 \cdot 2 \cdot 1} = \frac {n(2n - 1)}{3}, \] and since we have \[ x_m = \cot ^2 \left (\frac {m \pi }{2n + 1}\right ), \] we have \[ \sum _{m = 1}^{n} \cot ^2 \left (\frac {m \pi }{2n + 1}\right ) = \frac {n (2n - 1)}{3}. \]

3. For \(0 < \theta < \frac {1}{2}\pi \), we have \(0 < \sin \theta < \theta < \tan \theta \), and squaring this gives \[ 0 < \sin ^2 \theta < \theta ^2 < \tan ^2 \theta , \] and flipping to the reciprocal gives \[ 0 < \cot ^2 \theta < \frac {1}{\theta ^2} < \csc ^2 \theta = 1 + \cot ^2 \theta , \] which proves exactly what is desired.

   Therefore, we have \[ \sum _{m = 1}^{n} \cot ^2 \left (\frac {m\pi }{2n + 1}\right ) < \sum _{m = 1}^{n} \frac {1}{\left (\frac {m\pi }{2n + 1}\right )^2} < \sum _{m = 1}^{n} \left [1 + \cot ^2 \left (\frac {m\pi }{2n + 1}\right )\right ], \] and hence \[ \frac {n(2n - 1)}{3} < \sum _{m = 1}^{n} \frac {(2n + 1)^2}{m^2 \pi ^2} < \frac {2n(n + 1)}{3}, \] and hence \[ \frac {n(2n - 1) \pi ^2}{3 (2n + 1)^2} < \sum _{m = 1}^{n} \frac {1}{m^2} < \frac {2n(n + 1) \pi ^2}{3 (2n + 1)^2}. \]

   Take the limit as \(n \to \infty \), the strict inequalities become weak, and hence \[ \lim _{n \to \infty } \frac {n(2n - 1) \pi ^2}{3 (2n + 1)^2} \leq \sum _{m = 1}^{\infty } \frac {1}{m^2} \leq \lim _{n \to \infty } \frac {2n(n + 1) \pi ^2}{3 (2n + 1)^2}, \] and hence \[ \frac {2 \pi ^2}{3 \cdot 2^2} \leq \sum _{m = 1}^{\infty } \frac {1}{m^2} \leq \frac {2n \pi ^2}{3 \cdot 2^2}, \] and therefore \[ \frac {\pi ^2}{6} \leq \sum _{m = 1}^{\infty } \frac {1}{m^2} \leq \frac {\pi ^2}{6}, \] and hence \[ \sum _{m = 1}^{\infty } \frac {1}{m^2} = \frac {\pi ^2}{6}, \] as desired.