Year: 2010
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
About 80% of candidates attempted at least five questions, and well less than 20% made genuine attempts at more than six. Those attempting more than six questions fell into three camps which were those weak candidates who made very little progress on any question, those with four or five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th or even 8th questions as an "insurance policy" against a solution that seemed strong but wasn't, or else for entertainment!
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
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.\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 $\sin \theta$
Just over 60% attempted this question, achieving moderate success. The opening result was well done, but the two similar equations foundered frequently on incorrect differentiation. If these two were correctly obtained, then the conjecture and induction were usually correct. Appreciating that the final expression was actually a polynomial, and what this entails, passed most by.