Problems

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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)