Problems

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Solution:

1. \(A\), \(Q\), \(C\) lie on a straight line if \(q = \lambda a + (1-\lambda)c\) for some \(\lambda \in \mathbb{R}\), \begin{align*} && q &= \lambda a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)(a-c) \\ \Leftrightarrow && \frac{q - a}{c-a} &= 1-\lambda \\ \end{align*} therefore \(\frac{q-a}{c-a} \in \mathbb{R}\) \begin{align*} && \frac{q-a}{c-a} & \in \mathbb{R} \\ \Leftrightarrow && \left (\frac{q-a}{c-a} \right)^* &= \frac{q-a}{c-a} \\ \Leftrightarrow && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \end{align*} Given \(aa^* = cc^* = 1\), \begin{align*} && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \Leftrightarrow && q^*(c-a) - \frac{c}{a}+1 &= q \frac{a-c}{ca} - \frac{a}{c}+1 \\ \Leftrightarrow && (c-a)\l q^* +\frac{q}{ca}\r &= \frac{c}{a} - \frac{a}{c} \\ &&&= \frac{c^2-a^2}{ac} \\ \Leftrightarrow && q^* +\frac{q}{ca} &= \frac{c+a}{ac} \\ \Leftrightarrow && q^*ac +q &= a+c \end{align*}
2. Since \(Q\) lies on \(AC\) and \(BD\) we must have \begin{align*} &&& \begin{cases} q^*ac +q &= a+c \\ q^*bd +q &= b+d \\ \end{cases} \\ \Rightarrow && q^*(ac-bd) &= (a+c)-(b+d) \\ \Rightarrow && q(ac-bd) &= (b+d)ac-(a+c)bd \\ \Rightarrow && (q+q^*)(ac-bd) &= (a+c)(1-bd)+(b+d)(ac-1) \\ &&&=a-abd+c-bcd+abc-b+acd-d \\ &&&= a(1+cd)-b(1+cd)+c(1+ab)-d(1+ab) \\ &&&= (a-b)(1+cd)+(c-d)(1+ab) \end{align*}
3. If \(AB\) and \(CD\) meet at \(p\) we must have \(p^*ab + p = a+b\), ie \(p(1+ab) = a+b\) amd \(p(1+cd) = c+d\), so \begin{align*} && (q+q^*)(ac-bd) &= (a-b) \frac{c+d}{p} + (c-d) \frac{a+b}{p} \\ \Leftrightarrow && p(q+q^*)(ac-bd) &= (a-b)(c+d)+(c-d)(a+b) \\ &&&= ac+ad-bc-bd+ac+bc-ad-bd \\ &&&= 2(ac-bd) \\ \Leftrightarrow && p(q+q^*) &= 2 \end{align*}

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Solution:

1. \begin{align*} \frac{\left(\cot \theta + i\right)^{2n+1} -\left(\cot \theta - i\right)^{2n+1}}{2i} &= \frac{1}{\sin^{2n+1} \theta}\frac{\left(\cos \theta + i \sin \theta \right)^{2n+1} -\left(\cos\theta - i\sin \theta\right)^{2n+1}}{2i} \\ &= \frac{1}{\sin^{2n+1} \theta} \frac{e^{i(2n+1) \theta} - e^{-i(2n+1) \theta} }{2i} \\ &=\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} \end{align*} Notice that: \begin{align*} (\cot \theta + i)^{2n+1} - (\cot \theta - i)^{2n+1} &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}(i)^k \cdot \cot^{2n+1-k} \theta - \sum_{k=0}^{2n+1} \binom{2n+1}{k}(-i)^k \cdot \cot^{2n+1-k} \theta \\ &= \sum_{k=0}^{2n+1} \binom{2n+1}{k} \l i^k - (-i)^k \r \cdot \cot^{2n+1-k} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} \l i^{2l+1} - (-i)^{2l+1} \r \cdot \cot^{2n+1-(2l+1)} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} 2i \cdot \cot^{2(n-l)} \theta \\ &= 2i\sum_{l=0}^{n} \binom{2n+1}{2l+1} \cot^{2(n-l)} \theta \\ \end{align*} Therefore if \(\theta\) satisfies \(\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} = 0\) then \(z = \cot^2 \theta\) satisfies the equation. But \(\theta = \frac{m \pi}{2n+1}, m = 1, 2, \ldots, n\) are \(n\) distinct all the roots must be \(\cot^2 \frac{m \pi}{2n+1}\).
2. Notice that the sum of the roots will be \(\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{(2n+1)\cdot 2n \cdot (2n-1)}{3! \cdot (2n+1)} = \frac{n \cdot (2n-1)}{3}\) and so \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
3. For \(0 < \theta < \frac{1}{2}\pi\), \begin{align*} && 0 < \sin \theta < \theta < \tan \theta \\ \Leftrightarrow && 0 < \cot \theta < \frac{1}{\theta} < \frac{1}{\sin \theta} \\ \Leftrightarrow && 0 < \cot^2 \theta < \frac{1}{\theta^2} < \cosec^2 \theta = 1 + \cot^2 \theta\\ \end{align*} Therefore \begin{align*} && \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} < \sum_{n=1}^N \frac{(2N+1)^2}{n^2 \pi^2} < N + \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} \\ \Rightarrow && \frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \sum_{n=1}^N \frac{1}{n^2 \pi^2} < \frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \lim_{N \to \infty}\frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} < \lim_{N \to \infty}\frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \frac{1}{6} \leq \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} \leq \frac16 \\ \Rightarrow && \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6} \end{align*}

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Solution:

1. \begin{align*} && I &= \int_0^1 \frac{f(x^{-1})}{1+x} \d x \\ u = x^{-1}, \d u = -x^{-2} \d x: &&&= \int_{\infty}^1 \frac{f(u)}{1+\frac{1}{u}} \frac{-1}{u^2} \d u \\ &&&= \int_1^{\infty} \frac{f(u)}{u(1+u)} \d u \\ &&&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(u)}{u(u+1)} \d u \\ \\ \text{if} f(x) = f(x+1)\, \forall x && &=\sum_{n=1}^{\infty} \int_{0}^1 \frac{f(x+n)}{(x+n)(x+n+1)} \d x \\ &&&= \sum_{n=1}^\infty \int_0^1 \frac{f(x)}{(x+n)(x+n+1)} \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n+1)}\r \d x \\ &&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \l \frac{1}{x+n} - \frac{1}{x+n+1} \r\r \d x \\ &&&= \int_0^1 f(x) \l \frac{1}{x+1} \r \d x \\ &&&= \int_0^1\frac{f(x)}{x+1} \d x \\ \end{align*}
2. Since the fractional part is periodic with period \(1\), we can say \begin{align*} && \int_0^1 \frac{\{ x^{-1} \} }{1+x} \d x &= \int_0^1 \frac{\{ x\}}{x+1} \d x \\ &&&= \int_0^1 \frac{x}{x+1} \d x \\ &&&= \int_0^1 1-\frac{1}{x+1} \d x \\ &&&= [x - \ln (1+x) ]_0^1 \\ &&&= 1 - \ln 2 \end{align*} \begin{align*} && \int_0^1 \frac{\{ 2x^{-1}\}}{1+x} \d x &= \int_0^1 \frac{\{2x\}}{x+1} \d x \\ &&&= \int_0^{1/2} \frac{2x}{x+1} \d x +\int_{1/2}^{1} \frac{2x-1}{x+1} \d x \\ &&&= 2\int_0^1 \frac{x}{x+1} \d x + \int_{1/2}^1 \frac{-1}{x+1} \d x \\ &&&= 2 - 2\ln 2 - \l \ln 2 - \ln \tfrac32 \r \\ &&&= 2 - 4 \ln 2 + \ln 3 \\ &&&= 2 + \ln \tfrac {3}{16} \end{align*}
3. \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x\r \end{align*} Consider for \(f\) periodic with period \(1\) \begin{align*} \int_0^1 \frac{f(x^{-1})}{1-x} \d x &= \int_1^{\infty} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{f(u)}{u(u-1)} \d u \\ &= \sum_{n=1}^{\infty}\int_{0}^{1} \frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} \sum_{n=1}^{\infty}\frac{f(u)}{(u+n)(u+n-1)} \d u \\ &= \int_{0}^{1} f(u) \sum_{n=1}^{\infty}\l\frac{1}{u+n-1} - \frac{1}{u+n} \r\d u \\ &= \int_0^1 \frac{f(u)}{u} \d u \end{align*} So we have \begin{align*} && \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x \r \\ &&&= \frac12 \int_0^1 \frac{\{ x \}}{x} \d x - \frac12 (1 - \ln 2) \\ &&&= \frac12 - \frac12 + \frac12 \ln 2 \\ &&&= \frac12 \ln 2 \end{align*}

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Solution:

1. Just before collision \(n-1\): Velocity of \(P\) is \(u_{n-2}\) Velocity of \(Q\) is \(-v_{n-2}\) \begin{align*} COM: && mu_{n-2}+km(-v_{n-2}) &= mu_{n-1}+kmv_{n-1} \\ \Rightarrow && u_{n-2}-kv_{n-2} &= u_{n-1}+kv_{n-1} \\ NEL: && v_{n-1}-u_{n-1} &= -e((-v_{n-2})-u_{n-2}) \\ \Rightarrow && v_{n-1}-u_{n-1} &= e(v_{n-2}+u_{n-2}) \end{align*} \begin{align*} &&kv_{n-1} &= u_{n-2} - kv_{n-2}-u_{n-1} \\ &&kv_{n-1}&= ku_{n-1}+kev_{n-2}+keu_{n-2} \\ \Rightarrow && kv_{n-2}(1+e) &= u_{n-2}(1-ke)-u_{n-1}(1+k) \\ \Rightarrow && kv_{n-1}(1+e) &= u_{n-1}(1-ke)-u_{n}(1+k) \\ && k(1+e)v_{n-1}-k(1+e)u_{n-1} &= k(1+e)e(v_{n-2}+u_{n-2}) \\ \Rightarrow && u_{n-1}(1-ke)-u_{n}(1+k)-k(1+e)u_{n-1} &= e(u_{n-2}(1-ke)-u_{n-1}(1+k))+k(1+e)eu_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n + ((ke-1)+k(1+e)-e(1+k))u_{n-1} \\ &&& \quad \quad + (e(1-ke)+k(1+e)e)u_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n- (1-k)(1+e)u_{n-1} +e(1+k)u_{n-2} \end{align*}
2. \(u_0 = A + B\) \begin{align*} &&& \begin{cases}u_0 - kv_0 &= kv_1 + u_1 \\ \frac12 (u_0+v_0) &= v_1 - u_1 \\ \end{cases} \\ \Rightarrow && (1+k)u_1 &= u_0 - kv_0 - \frac{k}{2}(u_0 + v_0) \\ \Rightarrow && u_1 &= \frac{1}{k+1} \l u_0 (1-\frac{k}{2}) - \frac32 k v_0 \r \\ &&&= \frac{67}{70} u_0 - \frac{3}{70} v_0 \end{align*} Therefore \(A+B = u_0, \frac{49A+50B}{70} = \frac{67}{70} u_0 - \frac{3}{70} v_0\) \begin{align*} && A+B &= u_0 \\ && 49A+50B &= 67u_0 - 3v_0 \\ \Rightarrow && 50u_0 - A &= 67u_0 - 3v_0 \\ \Rightarrow && A &= -17u_0 + 3v_0 \\ && B &= 18u_0 - 3v_0 \end{align*} If \(0 < 6u_0 < v_0\), then \(B < 0\) and as \(n \to \infty\) we will find that \(\l \frac57 \r^n\) dominates \(\l \frac7{10} \r^n\) and so our velocity will be negative and the particle will change direction

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Solution:

First, notice that the centre of mass will lie directly below \(A\) and will be \(\frac{m}{M+m}\) of the way between \(O\) and \(P\). Therefore \(\sin \beta = \frac{m}{M+m}\). The cosine rule states that: \begin{align*} && AP^2 &= a^2 + a^2 - 2a^2 \cos \angle AOP \\ \Rightarrow && \frac{AP^2}{a^2} &= 2 - 2 \sin \beta \\ &&&= \frac{2M+2m-2m}{M+m} \\ &&&= \frac{2M}{M+m} \\ \Rightarrow && \frac{AP}{a} &= \sqrt{\frac{2M}{M+m}} \end{align*} Considering conservation of energy, we have: Rotational kinetic energy for the disc: \(\frac12 I \dot{\theta}^2\) Kinetic energy for the particle: \(\frac12 m (\dot{\theta} \sqrt{2-2\sin \beta} a)^2 = (1- \sin \beta)ma^2 \dot{\theta}^2\) GPE: The important thing is the vertical location of \(G\). The triangle \(OAG\) will still have angle \(\beta\) at \(A\). The vertical height below is: \(\cos \theta \cdot AG = \cos \theta a \cos \beta\). The distance from when \(\theta = 0\) will be \(a \cos \beta (1- \cos \theta)\) and so the GPE will be \((M+m)ga \cos \beta ( 1- \cos \theta)\) we can therefore say by conservation of energy: \[ \frac12 I \dot{\theta}^2 + (1- \sin \beta)ma^2 \dot{\theta}^2+(M+m)ga \cos \beta ( 1- \cos \theta) \] is constant. Suppose \(m = \frac32 M\) and \(I = \frac32 Ma^2\) then differentiating the constant wrt to \(\theta\) gives \(\sin \beta = \frac{m}{M+m} = \frac{3}{5}, \cos \beta = \frac45\) \begin{align*} && 0 &= \frac12 \frac32 M a^2 \cdot 2 \dot{\theta}\ddot{\theta} + (1- \sin \beta)\frac32M a^2 2 \dot{\theta}\ddot{\theta} + (M+\frac32M) ga \cos \beta \sin \theta \cdot \dot{\theta} \\ \Rightarrow && 0 &= \frac32 \ddot{\theta} + 3(1-\sin \beta) \ddot{\theta} + \frac{5}{2}\frac{g}{a} \cos \beta \sin \theta \\ &&&= (\frac32 + \frac65) \ddot{\theta} + \frac{2g}{a} \sin \theta \\ &&&= \frac{27}{10} \ddot{\theta} + \frac{2g}{a} \sin \theta \end{align*} If \(\theta\) is small, we can approximate this by: \(\frac{27}{10} \ddot{\theta} + \frac{2g}{a} \theta = 0\) which will have period \(\displaystyle 2 \pi \sqrt{\frac{27a}{10\cdot2 g}} = 2 \pi \sqrt{\frac{3a}{5g}}\) as required.

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Solution:

\begin{align*} \text{N2}(\swarrow): &&T +mg \cos \alpha &= m \frac{V^2}{b} \\ \end{align*} So the string goes slack when \(bg\cos \alpha = V^2 \Rightarrow V = \sqrt{bg \cos \alpha}\). Once the string goes slack, the particle moves as a projectile. It's initial speed is \(V\binom{-\cos \alpha}{\sin \alpha}\) and it's position is \(\binom{b\sin \alpha}{b\cos \alpha}\): \begin{align*} && \mathbf{s} &= \binom{b\sin \alpha}{b\cos \alpha}+Vt \binom{-\cos \alpha}{\sin \alpha} + \frac12 gt^2 \binom{0}{-1} \\ &&&= \binom{b\sin \alpha - Vt \cos \alpha}{b\cos \alpha + Vt \sin \alpha - \frac12 gt^2} \\ |\mathbf{s}|^2 = b^2 \Rightarrow && b^2 &= \left ( \binom{b\sin \alpha}{b\cos \alpha}+Vt \binom{-\cos \alpha}{\sin \alpha} + \frac12 gt^2 \binom{0}{-1} \right)^2 \\ &&&= b^2 + V^2t^2+\frac14 g^2 t^4 -gb\cos \alpha t^2-V\sin \alpha gt^3 \\ \Rightarrow && 0 &= V^2t^2 + \frac14 g^2 t^4 - V^2 t^2- V \sin \alpha g t^3 \\ &&&= \frac14 g^2 t^4 - V \sin \alpha gt^3 \\ \Rightarrow && gT &= 4V \sin \alpha \end{align*} The particle will have velocity \(\displaystyle \binom{-V \cos \alpha}{V \sin \alpha - 4V \sin \alpha} = \binom{-V \cos \alpha}{-3V \sin \alpha}\) so the angle \(\beta\) will satisfy \(\tan \beta = 3 \tan \alpha\). The particle will come to an instantaneous rest if all the momentum is destroyed, ie if the particle is travelling parallel to the string. \begin{align*} && 3 \tan \alpha &= \frac{b\cos \alpha + Vt \sin \alpha - \frac12 gt^2}{b\sin \alpha - Vt \cos \alpha} \\ &&&= \frac{\frac{V^2}{g}+\frac{4V^2\sin^2\alpha}{g} - \frac{8V^2\sin^2 \alpha}{g}}{\frac{V^2\sin \alpha}{g \cos \alpha} - \frac{4V^2 \sin \alpha \cos \alpha}{g}} \\ &&&= \frac{1 -4\sin^2 \alpha}{\tan \alpha(1 - 4\cos^2 \alpha)} \\ \Leftrightarrow&& 3 \frac{\sin^2 \alpha}{1-\sin^2 \alpha} &= \frac{1- 4 \sin^2 \alpha}{-3+4\sin^2 \alpha} \\ \Leftrightarrow && -9 \sin^2 \alpha + 12 \sin^4 \alpha &= 1 - 5 \sin^2 \alpha + 4 \sin^4 \alpha \\ \Leftrightarrow && 0 &= 1+4 \sin^2 \alpha - 8\sin^4 \alpha \\ \Leftrightarrow && \sin^2 \alpha &= \frac{1 + \sqrt{3}}4 \end{align*} (taking the only positive root)

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Solution:

1. \begin{align*} && \mathbb{P}(Y_k \leq y) &= \sum_{j=k}^n\mathbb{P}(\text{exactly }j \text{ values less than }y) \\ &&&= \sum_{j=k}^m \binom{m}{j} y^j(1-y)^{n-j} \end{align*}
2. This is the number of ways to choose a committee of \(m\) people with the chair from those \(m\) people. This can be done in two ways. First: choose the committee in \(\binom{n}{m}\) ways and choose the chair in \(m\) ways so \(m \binom{n}{m}\). Alternatively, choose the chain in \(n\) ways and choose the remaining \(m-1\) committee members in \(\binom{n-1}{m-1}\) ways. Therefore \(m \binom{n}{m} = n \binom{n-1}{m-1}\) \begin{align*} (n-m) \binom{n}{m} &= (n-m) \binom{n}{n-m} \\ &= n \binom{n-1}{n-m-1} \\ &= n \binom{n-1}{m} \end{align*} \begin{align*} f_{Y_k}(y) &= \frac{\d }{\d y} \l \sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m} \r \\ &= \sum^{n}_{m=k} \l \binom{n}{m}my^{m-1}\left(1-y\right)^{n-m} -\binom{n}{m}(n-m)y^{m}\left(1-y\right)^{n-m-1} \r \\ &= \sum^{n}_{m=k} \l n \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n \binom{n-1}{m} y^{m}\left(1-y\right)^{n-m-1} \r \\ &= n\sum^{n}_{m=k} \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n\sum^{n+1}_{m=k+1} \binom{n-1}{m-1} y^{m-1}\left(1-y\right)^{n-m} \\ &= n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \end{align*} \begin{align*} &&1 &= \int_0^1 f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1} \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \Rightarrow && \frac{1}{n \binom{n-1}{k-1}} &= \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \end{align*}
3. \begin{align*} && \mathbb{E}(Y_k) &= \int_0^1 y f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k}(1-y)^{n-k} \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k+1-1}(1-y)^{n+1-(k+1)} \d y \\ &&&= n \binom{n-1}{k-1} \frac{1}{(n+1) \binom{n}{k}}\\ &&&= \frac{n}{n+1} \cdot \frac{k}{n} \\ &&&= \frac{k}{n+1} \end{align*}

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Solution: \begin{align*} &&G_X(t) &= \mathbb{E}(t^N) \\ &&&= \sum_{k=0}^{\infty} \mathbb{P}(X = k) t^k \\ \Rightarrow && G_X(1) &= \sum_{k=0}^{\infty} \mathbb{P}(X = k) \\ \Rightarrow && G_X(-1) &= \sum_{k=0}^{\infty} (-1)^k\mathbb{P}(X = k) \\ \Rightarrow && \frac12 (G_X(1) + G_X(-1) &= \sum_{k=0}^{\infty} \frac12 (1 + (-1)^k) \mathbb{P}(X = k) \\ &&&= \sum_{k=0}^{\infty} \mathbb{P}(X =2k) \end{align*}

1. \begin{align*} 1 &= \sum_r \mathbb{P}(Y = r) \\ &= \sum_{k=0}^\infty k \cdot \mathbb{P}(X = 2k) \\ &= k \cdot \frac12 \l e^{-\lambda(1-1) } + e^{-\lambda(1+1) }\r \\ &= \frac{k}{2}(1+e^{-2\lambda}) \end{align*} Therefore \(k = \frac{2}{1+e^{-2\lambda}} = e^{\lambda} \frac{1}{\cosh \lambda}\) \begin{align*} && G_X(t) + G_X(-t) &= \sum_{k=0}^\infty \mathbb{P}(X = k)t^k(1^k + (-1)^k) \\ &&&= \sum_{k=0}^\infty \mathbb{P}(X = k)t^k(1^k + (-1)^k) \\ &&&= 2\sum_{k=0}^\infty \mathbb{P}(X = 2k)t^{2k} \\ &&&= 2\sum_{k=0}^\infty \frac{1}{k}\mathbb{P}(Y = 2k)t^{2k} \\ &&&= \frac{2}{k}G_Y(t) \\ \Rightarrow && G_Y(t) &= k \cdot \frac{G_X(t) + G_X(-t)}{2} \\ &&&= k\frac{e^{-\lambda(1-t)} + e^{-\lambda(1+t)}}{2} \\ &&&= \frac{e^\lambda}{\cosh \lambda} \frac{e^{-\lambda} (e^{\lambda t}+e^{-\lambda t}) }{2} \\ &&&= \frac{\cosh \lambda t}{\cosh \lambda} \end{align*} Since \(\mathbb{E}(Y) = G_Y'(1)\) and \begin{align*} && G_Y'(t) &= \frac{\lambda \sinh \lambda t}{\cosh \lambda t} \\ \Rightarrow && G_Y'(1) &= \lambda \tanh \lambda \\ &&&< \lambda \end{align*} since \(\tanh x < 1\)
2. \begin{align*} && \frac14 \l G_X(t) + G_X(it) +G_X(-t) + G_X(-it) \r &= \sum_{k=0}^\infty \mathbb{P}(X=k)t^k (1 + i^k + (-1)^k + (-i)^k) \\ &&&= \sum_{k=0}^\infty \mathbb{P}(X = 4k)t^{4k} \\ &&&= \frac{G_Z(t)}{c} \end{align*} Since \(G_Z(1) = 1\) we must have \(c = \frac1{\frac14 \l G_X(1) + G_X(i) +G_X(-1) + G_X(-i) \r}\) \begin{align*} && c &= \frac{4e^{\lambda}}{e^{\lambda} + e^{-\lambda} + e^{i\lambda} + e^{-i\lambda}} \\ &&&= \frac{2e^{\lambda}}{\cosh \lambda + \cos \lambda} \\ && G_Z(t) &= c \cdot \frac14 \l e^{-\lambda(1-t)}+e^{-\lambda(1-it)}+e^{-\lambda(1+t)}+e^{-\lambda(1+it)} \r \\ &&&= \frac{ce^{-\lambda t}}{4} \l 2\cosh \lambda t + 2 \cos \lambda t\r \\ &&&= \frac{\cosh \lambda t + \cos \lambda t}{\cosh \lambda + \cos \lambda} \end{align*} We are interested in \(G_Z'(1)\) so: \begin{align*} && G_Z'(t) &= \frac{\lambda (\sinh \lambda t - \sin \lambda t)}{\cosh \lambda + \cos \lambda } \end{align*} Considering various values of \(\lambda\), it makes sense to look at \(\lambda = \pi\) (since \(\cos \lambda = -1\) and the denominator will be small). From this we can see: \begin{align*} G'_Z(1) &= \frac{\pi (\sinh \pi-0)}{\cosh \pi-1} \\ &= \frac{\pi}{\tanh \frac{\pi}{2}} > \pi \end{align*} So \(\mathbb{E}(Z)\) is larger than \(\lambda\) for \(\lambda = \pi\) (and probably many others)