Year: 2013
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Taylor series
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
In this question, you may ignore questions of convergence.
Let $y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,$. Show that
\[
(1-x^2)\frac {\d y}{\d x} -xy -1 =0
\]
and prove that, for any positive integer $n$,
\[
(1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}}
-(n+1)^2 \frac{\d^ny}{\d x^n}=0\,
.
\]
Hence obtain the Maclaurin series for $ \dfrac {\arcsin x}{\sqrt{1-x^2}}\,$, giving the general term for odd and for even powers of $x$.
Evaluate the infinite sum
\[
1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots +
\frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,.
\]\begin{align*}
&& y &= \frac{\arcsin x}{\sqrt{1-x^2}} \\
&& \frac{\d y}{\d x} &= \frac{(1-x^2)^{-1/2} \cdot (1-x^2)^{1/2}-\arcsin x \cdot (-x)(1-x^2)^{-1/2}}{1-x^2} \\
&&&= \frac{1+ xy}{1-x^2} \\
\Rightarrow && 0 &= (1-x^2) \frac{\d y}{\d x} -xy-1\\ \\
\frac{\d^n}{\d x^{n+1}}: && 0 &= \left ( (1-x^2) y' \right)^{(n+1)} - (xy)^{(n+1)} \\
\Rightarrow && 0 &= (1-x^2)y^{(n+2)} + \binom{n+1}{1}(1-x^2)^{(1)}y^{(n+1)}+\binom{n+1}{2} (1-x^2)^{(2)}y^{(n)} - (xy^{(n+1)} +\binom{n+1}{1} y^{(n)} ) \\
&&&= (1-x^2)y^{(n+2)}+\left ( (n+1)\cdot(-2x)-x \right)y^{(n+1)} + \left ( \frac{(n+1)n}{2} \cdot (-2)-(n+1) \right)y^{(n)} \\
&&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( (n+1)n+(n+1)\right)y^{(n)} \\
&&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( n+1\right)^2y^{(n)} \\
\end{align*}
Since $y(0) = 0, y'(0) = 1$ we can look at the recursion: $y^{(n+2)} - (n+1)^2y^{(n)}$ for larger terms, ie $y^{(2k)}(0) = 0$
$y^{(1)}(0) = 1, y^{(3)}(0) = (1+1)^2 \cdot 1 = 2^2, y^{(5)}(0) = (3+1)^2 y^{(3)} = 4^2 \cdot 2^2$ and $y^{(2k+1)}(0) = (2k)^2 \cdot (2k-2)^2 \cdots 2^2 \cdot 1^2 = 2^{2k} \cdot (k!)^2$. Therefore
\begin{align*}
&& \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} x^{2k+1} \\
\\
\Rightarrow && \frac{\arcsin \frac12}{\sqrt{1-\left (\frac12 \right)^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} 2^{-2k-1}\\
&&&= \frac12 \sum_{k=0}^{\infty} \frac{ (k!)^2}{(2k+1)!} \\
&&&= \frac12 \left ( 1 + \frac1{3!} + \frac{2^2}{5!} + \cdots+ \right) \\
\Rightarrow&& S &= 2 \frac{2\frac{\pi}{6}}{\sqrt{3}} = \frac{2\pi}{3\sqrt{6}}
\end{align*}
This was the second most popular question, attempted by six out of every seven candidates, with only marginally less success than its predecessor. The first differential equation was proved correctly and many successfully completed the general result by induction, although there were some problems with the initial case. Some had difficulty finding the correct coefficients for the odd powers of in the Maclaurin series but the last part produced a variety of errors and few correct answers. Such errors included sin, forgetting to divide by, and attempting to evaluate the series using.