Year: 2022
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Hyperbolic functions
No solution available for this problem.
One question was attempted by well over 90% of the candidates two others by about 90%, and a fourth by over 80%. Two questions were attempted by about half the candidates and a further three questions by about a third of the candidates. Even the other three received attempts from a sixth of the candidates or more, meaning that even the least popular questions were markedly more popular than their counterparts in previous years. Nearly 90% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
You may assume that all infinite sums and products in this question converge.
\begin{questionparts}
\item Prove by induction that for all positive integers $n$,
\[ \sinh x = 2^n \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \sinh\!\left(\frac{x}{2^n}\right) \]
and deduce that, for $x \neq 0$,
\[ \frac{\sinh x}{x} \cdot \frac{\dfrac{x}{2^n}}{\sinh\!\left(\dfrac{x}{2^n}\right)} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right)\,. \]
\item You are given that the Maclaurin series for $\sinh x$ is
\[ \sinh x = \sum_{r=0}^{\infty} \frac{x^{2r+1}}{(2r+1)!}\,. \]
Use this result to show that, as $y$ tends to $0$, $\dfrac{y}{\sinh y}$ tends to $1$.
Deduce that, for $x \neq 0$,
\[ \frac{\sinh x}{x} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdots\,. \]
\item Let $x = \ln 2$. Evaluate $\cosh\!\left(\dfrac{x}{2}\right)$ and show that
\[ \cosh\!\left(\frac{x}{4}\right) = \frac{1 + 2^{\frac{1}{2}}}{2 \times 2^{\frac{1}{4}}}\,. \]
Use part (ii) to show that
\[ \frac{1}{\ln 2} = \frac{1 + 2^{\frac{1}{2}}}{2} \times \frac{1 + 2^{\frac{1}{4}}}{2} \times \frac{1 + 2^{\frac{1}{8}}}{2} \cdots\,. \]
\item Show that
\[ \frac{2}{\pi} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2+\sqrt{2}}}{2} \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \cdots\,. \]
\end{questionparts}
This was the fourth most popular question being attempted by more than four fifths of the candidates, with a moderate degree of success scoring a mean of 9/20. Part (i) suffered from incorrect flows of logic in the inductive and base cases, as well as failure to mention anything about not dividing by zero. In part (ii) many ignored the instruction to use the Maclaurin series, and used de L'Hopital's Rule to their cost, and some ignored the higher order terms. Part (iii) was generally well done, though the most common error was not justifying the evaluation of the product using a geometric series in the exponent. For part (iv), the best attempted route was to use an imaginary substitution which led to mostly successful solutions. Some candidates attempted to prove an analogous trigonometric identity using similar arguments to the previous parts, however losing marks for not sufficiently fleshing out the details, and some attempted to use Osborn's Rule, often with insufficient justification or stating that it was being used. Once the identity was achieved, the calculation was generally done well if the candidate progressed this far.