Problems

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*}

Solution: The initial volume of water in \(A\) is: \begin{align*} \pi \int_0^4 x^2 \, \d y &= \pi \int_0^4 y \d y \\ &= \pi [ \frac{y^2}{2}]_0^4 \\ &= 8\pi \end{align*} We assume that no water is in the tube as it is `thin'. Therefore we must have: \begin{align*} && 8\pi &= \pi \int_0^{h-1} x^2 \d y +\pi \int_0^{h} x^2 \d y \\ &&&= \pi \int_0^{h-1} y \d y +\pi \int_0^{h} \l \sinh \frac{x}{2}\r^2 \d y \\ &&&= \pi \left [\frac{y^2}{2} \right]_0^{h-1} + \pi \int_0^h \frac{-1+\cosh y}{2}\d y \\ &&&= \pi \frac{(h-1)^2}{2} + \pi \left [ -\frac{y}{2} +\frac{\sinh y}{2}\right]_0^h \\ &&&= \pi \frac{(h-1)^2}{2} -\pi \frac{h}{2} + \pi \frac{\sinh h}{2} \\ \Rightarrow && 0 &= h^2-2h+1-h+\sinh h -16 \\ &&&= h^2 -3h+\sinh h - 15 \\ \Rightarrow && 15 &= h^2 -3h+\sinh h \end{align*}