Problems

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)

Solution: \begin{align*} \cosh x &= \frac12 (e^x + e^{-x}) \\ \sinh x &= \frac12 (e^x - e^{-x}) \\ \end{align*} \begin{align*} \cosh^4x -\sinh^4 x &= (\cosh^2x -\sinh^2 x)(\cosh^2x +\sinh^2 x) \\ &= \left ( \frac14 \left (e^{2x}+2+e^{-2x} \right)- \frac14 \left (e^{2x}-2+e^{-2x} \right) \right)(\cosh^2x +\sinh^2 x) \\ &= (\cosh^2x +\sinh^2 x) \\ &= \left ( \frac14 \left (e^{2x}+2+e^{-2x} \right)+ \frac14 \left (e^{2x}-2+e^{-2x} \right) \right) \\ &= \frac{1}{4} \left (2e^{2x}+2e^{-2x} \right) \\ &= \frac12 \left ( e^{2x}+e^{-2x} \right) \\ &= \cosh 2x \\ \\ \cosh^4x +\sinh^4 x &= \frac1{2^4}\left (e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x} \right)+\frac1{2^4}\left (e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x} \right) \\ &= \frac18 (e^{4x}+e^{-4x}) + \frac{3}{4} \\ &= \frac14 \cosh 4x + \frac34 \end{align*} \begin{align*} \cosh^n x &=\frac{1}{2^n} \left ( e^{x}+e^{-x} \right)^n \\ &= \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} e^{kx}e^{-(n-k)x} \\ &= \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} e^{2kx-nx} \\ &= \frac{1}{2^n} \left ( \binom{n}{n} \left(e^{nx}+e^{-nx} \right) + \binom{n}{n-1}\left(e^{(n-2)x}+e^{-(n-2)x} \right) + \cdots + \binom{n}{n-k} \left( e^{(n-2k)x}+e^{-(n-2k)x} \right) + \cdots \right) \\ &= \frac{1}{2^{n-1}} \cosh nx + \frac{1}{2^{n-1}} \binom{n}{n-1} \cosh (n-2)x + \cdots + \frac{1}{2^{n-1}} \binom{n}{n-k} \cosh (n-2k)x + \cdots \end{align*} ie \begin{align*} \cosh^{2m} x &= \frac{1}{2^{2m-1}} \cosh 2m x + \frac{2m}{2^{2m-1}} \cosh(2(m-1)x) + \cdots + \frac{1}{2^{2m-1}}\binom{2m}{k} \cosh (2(m-k)x) +\cdots+ \frac{1}{2^{2m-1}} \binom{2m}{m} \\ \sinh^{2m} x &= \frac{1}{2^{2m-1}} \cosh 2m x - \frac{2m}{2^{2m-1}} \cosh(2(m-1)x) + \cdots + (-1)^{k}\frac{1}{2^{2m-1}}\binom{2m}{k} \cosh (2(m-k)x) +\cdots+ (-1)^m\frac{1}{2^{2m-1}} \binom{2m}{m} \\ \cosh^{2m} x -\sinh^{2m} x &= \frac{m}{2^{2m-3}} \cosh (2(m-1)x) + \cdots + \frac{1}{2^{2m-2}} \binom{2m}{2k+1}\cosh(2(m-2k-1)x) + \cdots\\ \cosh^{2m} x +\sinh^{2m} x &= \frac{1}{2^{2m-2}} \cosh (2mx) + \cdots + \frac{1}{2^{2m-2}} \binom{2m}{2k}\cosh(2(m-2k)x) + \cdots \end{align*}

Solution: Let \(t = \sinh^2 y\), then \begin{align*} && \sinh(2x) &= \cosh y \tag{1}\\ && \sinh(2y) &= 2 \cosh x \tag{2} \\ \\ && \cosh(2x) &= 2 \cosh^2 x -1 \\ (2): &&&= \frac12 \sinh^2(2y) -1 \\ && 1 &= \left (\frac12 \sinh^2(2y) -1 \right)^2 - \cosh^2 y \\ &&&= \frac14 \sinh^4(2y)-\sinh^2(2y)+1-\cosh^2 y \\ \Rightarrow && 0 &= \frac14 (\cosh^2 (2y)-1)^2- (\cosh^2 (2y)-1) - \cosh^2 y \\ &&&= \frac14 \left ( \left (1+2\sinh^2 y \right)^2-1 \right)^2 -\left ( \left (1+2\sinh^2 y \right)^2 -1\right) - (1 + \sinh^2 y ) \\ &&&= \frac14 \left ( 1 + 4t+4t^2 -1\right)^2 - \left ( 1+4t+4t^2-1\right) - (1 + t) \\ &&&= \frac14 (4t + 4t^2)^2 - (4t+4t^2)-1-t \\ &&&= 4(t+t^2)^2 - 4t^2-5t-1 \\ &&&= 4t^4+8t^3+4t^2-4t^2-5t-1 \\ &&& = 4t^4+8t^3-5t-1 \\ &&&= (t+1)(4t^3+4t^2-4t-1) \end{align*} Since \(\sinh^2 y\) is positive, we must be a root of the second cubic. Let \(f(t) = 4t^3+4t^2-4t-1\), then \(f(0) = -1\) and \(f'(t) = 12t^2+8t-4 = 4(t+1)(3t-1)\), so we have turning points at \(-1\) and \(\frac13\). Since \(f(-1) = 3 > 0\) and \(f(0) < 0\) we must have exactly one root larger than zero. Therefore there is a unique root. \(f(0.7) = -0.468 < 0\) \(f(0.8) = 0.408 > 0\) since \(f\) is continuous and changes sign, the root must fall in the interval \((0.7, 0.8)\). Let \(t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}\), and \(t_0 = 0.75\), then \begin{align*} t_0 &= 0.75 \\ t_1 &= 0.7571428571 \\ t_2 &= 0.7570684728 \\ t_3 &= 0.7570684647 \end{align*} So \(t \approx 0.757068\), \(\sinh y \approx 0.870097\), \(y \approx 0.786474\), \(x \approx 0.546965\)

Solution: \begin{align*} && y & =x\frac{\mathrm{d}y}{\mathrm{d}x}-\cosh\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d x} + x\frac{\d ^2 y}{\d x^2} - \sinh \left ( \frac{\d y}{\d x} \right) \frac{\d^2 y}{\d x^2} \\ \Rightarrow && 0 &= \frac{\d^2 y}{\d x^2} \left ( x - \sinh \left ( \frac{\d y}{\d x}\right)\right) \end{align*} Therefore \(\frac{\d^2y}{\d x^2} = 0\) or \( x - \sinh \left ( \frac{\d y}{\d x}\right) = 0\) as required. \begin{align*} && \frac{\d ^2 y}{\d x^2} &= 0 \\ \Rightarrow && y &= ax + b \\ \\ && 0 &= x - \sinh \left ( \frac{\d y}{\d x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \sinh^{-1} (x) \\ \Rightarrow && y &= x \sinh^{-1} x - \sqrt{x^2+1} + C \end{align*} Since it is necessary the solution satisfies one of those equations, we just need to check if either of these types of solutions work for our differential equation, ie \begin{align*} && ax + b &\stackrel{?}{=} ax - \cosh(a) \\ \Rightarrow && b &= -\cosh(a) \\ \Rightarrow && y &= ax -\cosh(a) \\ \\ && x \sinh^{-1} x - \sqrt{x^2+1} + C &\stackrel{?}{=} x\sinh^{-1} x - \cosh ( \sinh^{-1} x) \\ &&&= \sinh^{-1} x -\sqrt{x^2+1} \\ \Rightarrow && C &= 0 \end{align*} Therefore the general solutions are, \(y = ax - \cosh(a)\) and \(y = x \sinh^{-1} x - \sqrt{x^2+1}\)

Solution:

1. \begin{align*} && f(x) &= \ln (1+e^x) \\ && f'(x) &= \frac{1}{1+e^x} \cdot e^x \\ &&&= \frac{e^x}{1+e^x} \\ \Rightarrow && \ln [f'(x)] &= x - \ln (1+e^x) \\ &&&= x - f(x) \\ \\ \Rightarrow && \frac{f''(x)}{f'(x)} &= 1 - f'(x) \\ \Rightarrow && f''(x) &= f'(x) - [f'(x)]^2 \\ && f'''(x) &= f''(x) - 2f'(x) f''(x) \\ && f^{(4)}(x) &= f'''(x) - 2[f''(x)]^2-2f'(x)f'''(x) \end{align*} \begin{align*} f(0) &= \ln 2 \\ f'(0) &= \tfrac12 \\ f''(0) &= \tfrac12 -\tfrac14 \\ &= \tfrac 14 \\ f'''(0) &= \tfrac14 - 2 \tfrac12 \tfrac 14 \\ &= 0 \\ f^{(4)}(0) &= -2 \cdot \tfrac1{16} \\ &= -\frac18 \end{align*} Therefore \(f(x) = \ln 2 + \tfrac12 x + \tfrac18 x^2 - \frac1{8 \cdot 4!} x^4 + O(x^5)\)
2. As \(x \to 0\), \(g(x) \to \infty\) therefore there can be no power series about \(0\). But as \(x \to 0, x g(x) \not \to \infty\) as \(\frac{x}{\sinh x}\) is well behaved. We can also notice that \(x g(x)\) is an even function, since \(\cosh x\) is even and \(\frac{x}{\sinh x}\) is even, therefore the power series will consist of even powers of \(x\) \begin{align*} \lim_{x \to 0} \frac{x}{\sinh x \cosh 2 x} &= \lim_{x \to 0} \frac{x}{\sinh x} \cdot \lim_{x \to 0} \frac{1}{\cosh2 x} \\ &= 1 \end{align*} Notice that \begin{align*} \frac{x}{\sinh x \cosh 2 x} &= \frac{4x}{(e^x - e^{-x})(e^{2x}+e^{-2x})} \\ &= \frac{4x}{(2x + \frac{x^3}{3} + \cdots)(2 + 4x^2 + \frac43 x^4 + \cdots )} \\ &= \frac{1}{1+\frac{x^2}{6}+\frac{x^4}{5!} + \cdots } \frac{1}{1 + 2x^2 + \frac23 x^4 + \cdots } \\ &= \left (1-(\frac{x^2}{6} + \frac{x^4}{5!})+ (\frac{x^2}{6} )^2 + O(x^6)\right) \left (1-(2x^2+\frac23 x^4)+ (2x^2)^2 + O(x^6)\right) \\ &= \left (1 - \frac16 x^2 + \frac{7}{360} x^4 + O(x^6) \right) \left (1 - 2x^2+ \frac{10}3x^4 + O(x^6) \right) \\ &= 1 - \frac{13}{6} x^2 + \frac{1327}{360}x^4 + O(x^6) \end{align*}

Solution: \begin{align*} && 0 &= u^2 + 2u \sinh x -1 \\ &&&= u^2 + u(e^x-e^{-x})-e^{x}e^{-x} \\ &&&= (u-e^{-x})(u+e^x) \\ \Rightarrow && u &= e^{-x}, -e^x \end{align*} \begin{align*} && 0 &= \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1 \\ \Rightarrow && \frac{\d y}{\d x} &= e^{-x}, -e^x \\ \Rightarrow && y &= -e^{-x}+C, -e^x+C \\ y(0) = 0: && C &= 1\text{ both cases } \\ y'(0) > 0: && y &= 1-e^{-x} \end{align*} \begin{align*} && 0 &= \sinh x u^2 + 2u -\sinh x \\ \Rightarrow && u &= \frac{-2 \pm \sqrt{4+4\sinh^2 x}}{2\sinh x} \\ &&&= \frac{-1 \pm \cosh x}{\sinh x} = - \textrm{cosech }x \pm \textrm{coth}x \\ \\ && 0 &= \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x \\ \Rightarrow && \frac{\d y}{\d x} &= - \textrm{cosech }x \pm \textrm{coth}x \\ \Rightarrow && y &= -\ln \left ( \tanh \frac{x}{2} \right) \pm \ln \sinh x+C \end{align*} For \(x \to 0\) to be defined, we need \(+\), so \begin{align*} && y &= \ln \left (\frac{\sinh x}{\tanh \frac{x}{2}} \right) + C \\ && y &= \ln \left (\frac{2\sinh \frac{x}{2} \cosh \frac{x}{2}}{\tanh \frac{x}{2}} \right)+C \\ &&&= \ln \left (2 \cosh^2 x \right) + C \\ y(0) = 0: && 0 &= \ln 2+C \\ \Rightarrow && y &= \ln(2 \cosh^2 x) -\ln 2 \\ && y &= 2 \ln (\cosh x) \end{align*}