Year: 2018
Paper: 3
Question Number: 2
Course: LFM Pure
Section: Differentiation
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
The sequence of functions
$y_0$, $y_1$, $y_2$, $\ldots\,$ is defined by $y_0=1$ and, for $n\ge1\,$,
\[
y_n = (-1)^n \frac {1}{z} \, \frac{\d^{n} z}{\d x^n}
\,,
\]
where $z= \e^{-x^2}\!$.
\begin{questionparts}
\item Show that
$\dfrac{\d y_n}{\hspace{-4.7pt}\d x} = 2x y_n -y_{n+1}\,$ for $n\ge1\,$.
\item Prove by induction that, for $n\ge1\,$,
\[
y_{n+1} = 2x y_n -2ny_{n-1}
\,.
\]
Deduce that, for $n\ge1\,$,
\[
y_{n+1}^2 - {y}_n {y}_{n+2}
= 2n
(y_n^2 - y_{n-1}y_{n+1})
+ 2 y_n^2
\,.
\]
\item Hence show that $y_{n}^2
- y^2_{n-1} y^2_{n+1} > 0$ for $n \ge 1$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
\frac{\d y_n}{\d x} &= \frac{\d}{\d x} \l (-1)^n e^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r \r \\
&= (-1)^n 2xe^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r + (-1)^n e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\
&= 2xy_n - (-1)^{n+1} e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\
&= 2xy_n - y_{n+1}
\end{align*}
\item $y_0 = 1$, $y_1 = (-1) e^{x^2} \cdot (-2x) \cdot e^{-x^2} = 2x$, $y_2 = e^{x^2} \frac{\d^2}{\d x^2} \l e^{-x^2}\r = e^{x^2} \frac{\d }{\d x}\l -2xe^{-x^2} \r = e^{x^2} \l -2e^{-x^2}+4x^2e^{-x^2}\r = 4x^2-2$. Therefore $2xy_1 - 2y_0 = 2x \cdot 2x - 2\cdot1 = 4x^2-2 = y_2$ so our statement is true for $n=1$. Assume the statement is true for $n=k$, then
\begin{align*}
&& y_{k+1} &= 2xy_k - 2ky_{k-1} \\
\frac{\d }{\d x}: && \frac{\d y_{k+1}}{\d x} &= 2\frac{\d}{\d x}\l xy_k \r - 2k\frac{\d y_{k-1}}{\d x} \\
\Rightarrow && 2xy_{k+1}-y_{k+2} &= 2y_k+2x \l 2xy_k-y_{k+1}\r - 2k \l 2xy_{k-1}-y_k \r \\
\Rightarrow && y_{k+2} &=2y_k+ 4x \cdot y_{k+1}-(4x^2+2k)y_k+2x \cdot 2k y_{k-1} \\
&&&= 4x \cdot y_{k+1}-(4x^2+2(k+1))y_k+2x \l2xy_k - y_{k+1} \r \\
&&&= 2x \cdot y_{k+1} -2(k+1)y_k
\end{align*}
Therefore since our statement is true for $n=1$ and if it is true for $n=k$ it is true for $n=k=1$, therefore by the principle of mathematical induction it is true for all $n \geq 1$.
Since $2x = \frac{y_{n+1}+2ny_{n-1}}{y_n}$ for all $n$, we must have
\begin{align*}
&& \frac{y_{n+1}+2ny_{n-1}}{y_n} &= \frac{y_{n+2}+2(n+1)y_{n}}{y_{n+1}} \\
\Leftrightarrow && y_{n+1}^2+2ny_{n-1}y_{n+1} &= y_ny_{n+2}+2ny_n^2+2y_n^2 \\
\Leftrightarrow && y_{n+1}^2-y_ny_{n+2} &= 2n(y_n^2-y_{n-1}y_{n+1})+2y_n^2
\end{align*}
\item Consider the functions $f_n(x) = y_{n}^2-y_{n-1}y_{n+1}$ then clearly $f_{n+1} = 2nf_{n} + 2y_n^2 \geq f_{n}$ so to prove $f_n(x) > 0$ for $n \geq 1$ it suffices to prove it for $n = 1$. But $f_1 = y_1^2 - y_0y_{2} = (2x)^2-(4x^2-2) = 2 > 0$ so we are done.
\end{questionparts}
With about 89% attempting this, it was the second most popular question, and with a mean score of nearly 11/20, the second most successful. Part (i) was generally well-handled, and although most scored some marks on the proof by induction in part (ii), candidates often struggled to complete it, many of them because they attempted to use the original definition of the functions, which rarely led to success, rather than employ the result of part (i). Some candidates noticed that the proof by induction in part (ii) was equivalent to proving 2 by induction. This gave them a simpler base case but did not significantly simplify the inductive step. For the deduction in part (ii), it was very common for the first result of part (ii) to be squared, leading to pages of algebra, although they were often then successful. Part (iii) was surprisingly poorly attempted as few candidates realised that a proof by induction (or equivalent) was required.