Problems

Solution: \begin{align*} && y &= \cos(m \arcsin x) \\ && y' &= -m \sin (m \arcsin x) \cdot (1-x^2)^{-\frac12} \\ && y'' &= -m^2 \cos(m \arcsin x) \cdot (1-x^2)^{-1} -m \sin(m \arcsin x) \cdot (1-x^2)^{-\frac32} \cdot (-x) \\ &&&= -m^2 y (1-x^2)^{-1} + x(1-x^2)^{-1} y' \\ \Rightarrow && 0 &= (1-x^2)y'' - x y' + m^2y \\ \\ && 0 &= (1-x^2)y^{(3)} -2xy'' - xy''-y' + m^2y' \\ &&&= (1-x^2)y^{(3)} - 3xy'' + (m^2-1)y' \\ \\ && 0 &= (1-x^2)y^{(4)} - 2xy^{(3)} - 3xy^{(3)} - 3y^{(2)} + (m^2-1)y^{(2)} \\ &&&= (1-x^2)y^{(4)}- 5xy^{(3)} - (m^2-4)y^{(2)} \end{align*} Claim: \(0 = (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n)}\) Proof: (By induction) Clearly the first few base cases are true. Suppose it is true for some \(n\), then \begin{align*} && 0 &= (1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} + (m^2-n^2)y^{(n)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+3)} - 2xy^{(n+2)} - (2n+1)xy^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+3)} - (2n+3)xy^{(n+2)} + (m^2-n^2-2n-1)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+1+2)} - (2(n+1)+1)xy^{(n+1+1)} +(m^2-(n+1)^2)y^{(n)} \end{align*} And so we can conclude the result by induction. Notice that \begin{align*} && y(0) &= \cos(m 0) = 1 \\ && y'(0) &= -m\sin(m 0) = 0 \\ && y''(0) &= -m^2 y(0) = -m^2\\ \end{align*} Notice that \(y^{(n+2)}(0) + (m^2-n^2)y^{(n)} = 0\) so in particular all the odd terms will be \(0\) and the even terms will be \(1, -m^2, m^2(m^2-2^2), \cdots\), therefore \begin{align*} && \cos (m \arcsin x) &= 1 -\frac{m^2}{2!} x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 - \cdots \\ \Rightarrow && \cos(m \theta) &= 1 - \frac{m^2}{2!} \sin^2 \theta + \frac{m^2(m^2-2^2)}{4!} \sin^4 \theta \end{align*} Notice that if \(m\) is even, then at some point we will have \(m^2-m^2\) appearing in our expansion and all remaining terms will be zero. Therefore we will end up with a polynomial series, of degree \(m\) in \(\sin \theta\)

Solution: \begin{align*} && y &= \ln(x^2 -1) \\ && r &= \sqrt{x^2-1} \\ && \coth \theta &= x \\ && r &= \sqrt{\coth^2 \theta - 1} = \sqrt{\textrm{cosech}^2 \theta} = \textrm{cosech} \theta \\ && \frac{\d y}{\d x} &= \frac{2x}{x^2-1} \\ &&&= \frac{2 \coth \theta}{r^2} \\ &&&= \frac{2 \cosh \theta}{\sinh \theta \cdot r \cdot \textrm{cosech} \theta } \\ &&&= \frac{2 \cosh \theta}{r } \\ \\ && \frac{\d^2 y}{\d x^2} &= \frac{2(x^2-1)-4x^2}{(x^2-1)^2} \\ &&&= \frac{-2(1+x^2)}{r^2 \textrm{cosech}^2 r} \\ &&&= -\frac{2(1 + \coth^2 \theta) \sinh^2 \theta}{r^2} \\ &&&= -\frac{2(\sinh^2 \theta + \cosh^2 \theta)}{r^2} \\ &&&= -\frac{2 \cosh 2 \theta}{r^2} \\ \\ && \frac{\d^3 y}{\d x^3} &= \frac{-4x(x^2-1)^2-(-2x^2-2)\cdot2(x^2-1)\cdot 2x}{(x^2-1)^4} \\ &&&= \frac{-4x(x^2-1)+8x(x^2+1)}{(x^2-1)^3}\\ &&&= \frac{4x^3+12x}{(x^2-1)^3} \\ &&&=\frac{\sinh^3 \theta (4\coth^3 \theta + 12\coth \theta )}{r^3} \\ &&&=\frac{4\cosh^3 \theta + 12\cosh \theta \sinh^2 \theta}{r^3} \\ &&&= \frac{4 \cosh 3 \theta}{r^3} \\ \end{align*} Claim: \(\frac{\d^n y}{\d x^n} = (-1)^{n+1}\frac{2(n-1)!\cosh n \theta}{r^n}\) Proof: By induction. Base cases already proven \begin{align*} \frac{\d r}{\d x} &= \frac{x}{\sqrt{x^2-1}} = \frac{\coth \theta}{\textrm{cosech} \theta} = \cosh \theta \\ \frac{\d \theta}{\d x} &= - \sinh^2 \theta \\ \\ \frac{\d^{n+1} y}{\d x^{n+1}} &= (-1)^{n+1}(n-1)!\frac{\d}{\d x} \left ( \frac{2\cosh n \theta}{r^n}\right) \\ &= (-1)^{n+1}\frac{2 n \sinh n \theta \cdot r^n \cdot \frac{\d \theta}{\d x}- 2\cosh n \theta \cdot nr^{n-1} \frac{\d r}{\d x} }{r^{2n}} \\ &= (-1)^{n+2}\frac{2n( \cosh n \theta\cosh \theta + r\sinh n \theta \sinh^2 \theta) }{r^{n+1}} \\ &= (-1)^{n+2}n!\frac{2\cosh(n+1) \theta }{r^{n+1}} \\ \end{align*} We can think of this as \(\ln(x^2-1) = \ln(x+1)+\ln(x-1)\) and also note \(x \pm 1 = \coth \theta \pm 1 = \frac{\cosh \theta \pm \sinh \theta}{\sinh \theta} = \frac{e^{\pm \theta}}{\sinh \theta}\) \begin{align*} && \frac{\d^n}{\d x^n} \ln(x^2-1) &= (n-1)!(-1)^{n-1} \left ( \frac{1}{(x+1)^n} + \frac{1}{(x-1)^n} \right) \\ &&&= (-1)^{n-1}(n-1)! \left ( \frac{\sinh^n \theta}{e^{n\theta}} + \frac{\sinh^n \theta}{e^{-n\theta}} \right) \\ &&&= (-1)^{n-1} (n-1)!2\cosh n \theta \cdot \sinh^n \theta \\ &&&= (-1)^{n-1}(n-1)! \frac{2 \cosh n \theta }{r^n} \end{align*}

Solution: \begin{align*} && f(x) &= \sin 2x \cos x \\ &&&= \frac12 \l \sin 3x + \sin x \r \\ \Rightarrow && f^{(1988)}(x) &= \frac12 \l 3^{1988} (-1)^{994} \sin 3x+ (-1)^{994} \sin x \r \\ &&&= \boxed{\frac12 \left (3^{1998} \sin 3x + \sin x \right)} \\ \\ f^{(1988)}(x) = 0: && 0 &= 3^{1988} \sin 3x + \sin x \\ \Rightarrow && 0 &= 3^{1988} ( 3\sin x-4\sin^3 x) + \sin x \\ \Rightarrow && 0 &= \sin x \left (1+3^{1989}-4\cdot 3^{1988}\sin^{2} x \right) \end{align*} Since \(\sin x\) will first contribute a zero when \(x = \frac{\pi}{2}\) we focus on the second bracket, in particular, we need: \begin{align*} && \sin^2 x &= \frac{3}{4} \left ( 1 + \frac{1}{3^{1988}} \right) \\ \Rightarrow && \sin x &= \frac{\sqrt{3}}2 \left (1 + \frac{1}{2 \cdot 3^{1988}} + \cdots \right ) \end{align*} Since near \(\frac{\pi}{3}\), \begin{align*} \sin (\frac{\pi}{3} + \epsilon) &= \sin \frac{\pi}{3} \cos \epsilon + \cos \frac{\pi}{3} \sin \epsilon \\ &\approx \frac{\sqrt{3}}{2} (1-\epsilon^2 + \cdots ) + \frac{1}{2}(\epsilon + \cdots) \\ &= \frac{\sqrt{3}}2 + \frac12 \epsilon + \cdots \end{align*} Therefore by comparison we can see that \(x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} 3^{-1988}\) will be a very good approximation for the root.

Solution:

1. If \(f(x) = \alpha x\) then \(f^{-1}(x) = \frac{1}{\alpha}x\). \begin{align*} && \frac{\d^{n-1}}{\d y^{n-1}} [y - \alpha y]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [(1 - \alpha)^n y^n] \\ && &= (1-\alpha)^n n! y \\ \Rightarrow && y + \sum_{n=1}^{\infty} \frac{1}{n!}\frac{\d^{n-1}}{\d y^{n-1}} [y - \alpha y]^n &= y +\sum_{n=1}^{\infty} (1-\alpha)^ny \\ &&&= y + y\l \frac{1}{1-(1-\alpha)}-1 \r \\ &&&= \frac{1}{\alpha}y \end{align*} Where we can sum the geometric progression if \(|1-\alpha| < 1 \Leftrightarrow 0 < \alpha < 2\)
2. Suppose that \(f(x) = x-\frac14x^3\). We would like to find \(f^{-1}(\frac12)\). \begin{align*} && \frac{\d^{n-1}}{\d y^{n-1}} [y - (y+\frac14 y^3)]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [\frac1{4^n} y^{3n}] \\ && &= \frac{1}{4^n} \frac{(3n)!}{(2n+1)!} y^{2n+1} \\ \Rightarrow && f^{-1}(\frac12) &= \frac12 + \sum_{n=1}^{\infty} \frac{1}{4^n} \frac{(3n)!}{n!(2n+1)!} \frac{1}{2^{2n+1}} \\ &&&= \frac12 + \sum_{n=1}^{\infty} \frac{(3n)!}{n!(2n+1)!} \frac{1}{2^{4n+1}} \\ \end{align*} Since when \(n = 0\) \(\frac{0!}{0!1!} \frac{1}{2^{0+1}} = \frac12\) we can include the wayward \(\frac12\) in our infinite sum and so we have the required result.
3. Consider \(f(x) = x - e^{\lambda x}\) we are interested in \(f^{-1}(0)\). \begin{align*} && \frac{\d^{n-1}}{\d y^{n-1}} [y - (y-e^{\lambda y})]^n &= \frac{\d^{n-1}}{\d y^{n-1}} [e^{n\lambda y}] \\ &&&= n^{n-1} \lambda^{n-1}e^{n \lambda y} \\ \Rightarrow && f^{-1}(0) &= \sum_{n=1}^\infty \frac{1}{n!} n^{n-1} \lambda^{n-1} \end{align*} We don't care about convergence, but it's worth noting this has a radius of convergence of \(\frac{1}{e}\) (ie this series is valid if \(|\lambda| < \frac1e\)).