3 problems found
Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\), \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\), \(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.
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 \(\sin \theta\)
Prove that \(\cos3\theta=4\cos^{3}\theta-3\cos\theta\). Show how the cubic equation \[ 24x^{3}-72x^{2}+66x-19=0\tag{*} \] can be reduced to the form \[ 4z^{3}-3z=k \] by means of the substitution \(y=x+a\) and \(z=by\), for suitable values of the constants \(a\) and \(b\). Hence find the three roots of the equation \((*)\), to three significant figures. Show, by means of a counterexample, or otherwise, that not all cubic equations of the form \[ x^{3}+\alpha x^{2}+\beta x+\gamma=0 \] can be solved by this method.
Solution: \begin{align*} \cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta \sin \theta \\ &= (2\cos^2\theta-1)\cos \theta - 2\cos \theta \sin^2 \theta \\ &= 2\cos^3\theta-\cos \theta - 2\cos \theta(1- \cos^2 \theta) \\ &= 4\cos^3 \theta - 3\cos \theta \end{align*} \begin{align*} 0 &= 24x^{3}-72x^{2}+66x-19 \\ &= 24(y+1)^3-72(y+1)^2+66(y+1)-19 \\ &= 24(y^3+3y^2+3y+1)-72(y^2+2y+1)+66(y+1)-19\\ &= 24y^3+(72-144+66)y+(24-72+66-19) \\ &= 24y^3-6y-1 \\ &= 24b'^3z^3 - 6b'z - 1 \\ &= \frac{2}{\sqrt{3}}(4 z^3 -3z) - 1 \\ \end{align*} Therefore if \(b = \sqrt{3}, a = 1\), we have: \(4z^3 - 3z = \frac{\sqrt{3}}{2}\) So if \(z = \cos \theta \Rightarrow \cos 3\theta = \frac{\sqrt{3}}2 \Rightarrow 3 \theta = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6} \Rightarrow \theta = \frac{\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}\). Since \(\frac{\cos x}{\sqrt{3}} < 1\) we only need to approximate the first part to 2 significant figures. Therefore: \begin{align*} \sqrt{3} &\approx 1 + \frac{1}{1 + \frac{1}{2+\frac11}} = \frac{7}{4} = 1.75\\ \cos \tfrac{\pi}{18} &\approx \cos \frac{1}{6} \approx 1 - \frac{1}{2} \frac{1}{6^2} = \frac{71}{72} \approx 1 - \frac{1}{70} = 1 - 0.014 = 0.986 \\ \frac{\cos \tfrac{\pi}{18}}{\sqrt{3}} & \approx \frac{.986}{1.75} = 0.57 \\ \end{align*} Final answers: \(1.57, 0.803, 0.629\). We wouldn't be able to solve \(x^3 + 1= 0\) using this method, as we would have 2 non-real roots
The function \(\mathrm{f}\) is defined by \[ \mathrm{f}(x)=ax^{2}+bx+c. \] Show that \[ \mathrm{f}'(x)=\mathrm{f}(1)\left(x+\tfrac{1}{2}\right)+\mathrm{f}(-1)\left(x-\tfrac{1}{2}\right)-2\mathrm{f}(0)x. \] If \(a,b\) and \(c\) are real and such that \(\left|\mathrm{f}(x)\right|\leqslant1\) for \(\left|x\right|\leqslant1\), show that \(\left|\mathrm{f}'(x)\right|\leqslant4\) for \(\left|x\right|\leqslant1\). Find particular values of \(a,b\) and \(c\) such that, for the corresponding function \(\mathrm{f}\) of the above form \(\left|\mathrm{f}(x)\right|\leqslant1\) for all \(x\) with \(\left|x\right|\leqslant1\) and \(\mathrm{f}'(x)=4\) for some \(x\) satisfying \(\left|x\right|\leqslant1\).
Solution: Let \(f(x) = ax^2 + bx + c\) then \begin{align*} f'(x) &= 2ax + b \\ f(0) &= c \\ f(1) &= a+b+c \\ f(-1) &= a-b+c \\ f(1)+f(-1) &= 2(a+c) \\ f(1)-f(-1) &= 2b \\ f'(x) &= x(f(1)+f(-1)) + \frac12 (f(1) - f(-1)) - 2f(0)x \end{align*} as required. Since \(f'(x)\) is a straight line, the maximum value is either at \(1, -1\) or it's constant and either end suffices. \begin{align*} |f'(1)| & \leq |f(1)|\frac{3}{2} + |f(-1)| \frac12 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\ &= 4 \\ \\ |f'(-1)| & \leq |f(1)|\frac{1}{2} + |f(-1)| \frac32 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\ &= 4 \\ \end{align*} Therefore \(|f'(x)| \leq 4\). Suppose \(|f'(x)| = 4\) for some value in \(x \in [-1,1]\), then it must be either \(-1\) or \(1\). If \(f'(1) = 4\) then \(f(1) = 1, f(-1) = 1, f(0) = -1\) so \(f(x) = 1+ k(x^2-1) \Rightarrow f(x) = 1+2(x^2-1) = 2x^2 -1\). If \(f'(-1) = 4\) then \(f(1) = -1, f(-1) = -1, f(0) = 1 \Rightarrow f(x) = -2x^2 + 1\)