Year: 2018
Paper: 2
Question Number: 5
Course: UFM Pure
Section: Taylor series
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1505.3
Banger Comparisons: 4
In this question, you should ignore issues of convergence.
\begin{questionparts}
\item Write down the binomial expansion, for $\vert x \vert<1\,$, of
$\;\dfrac{1}{1+x}\,$ and deduce that
%. By considering
%$
%\displaystyle \int \frac 1 {1+x} \, \d x
%\,,
%$
%show that
\[
\displaystyle
\ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n
\,
\]
for $\vert x \vert <1 \,$.
\item Write down the series expansion in powers of $x$ of
$\displaystyle \e^{-ax}\,$.
Use this expansion to show that
\[
\int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x
\,\d x = \ln(1+a)
\ \ \ \ \ \ \ (\vert a \vert <1)\,.
\]
\item
Deduce the value of
\[
\int_0^1 \frac{x^p - x^q}{\ln x} \, \d x
\ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1)
\,.
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\
\Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\
&&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\
\Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n}
\end{align*}
\item \begin{align*}
&& e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\
\Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\
&&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\
&&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\
&&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\
&&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\
&&&= \ln (1+a)
\end{align*}
\item \begin{align*}
&& \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\
e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\
&&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\
&&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\
v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\
&&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\
&&&= \ln \left ( \frac{1+p}{1+q} \right)
\end{align*}
\end{questionparts}
Of the pure questions this was the one that attracted the poorest responses in general, with a significant proportion of attempts scoring 0. In the first part of the question the majority of candidates did not include the constant of integration and so did not produce a fully justified solution. The expansion and substitution in the second part of the question was done well in general, although many candidates attempted to expand the as well as the (1 − ), after which they were unable to complete the integral. In many of these cases there were then unjustified jumps to the series expansion found in part (i). Answers to part (iii) were generally much better than part (ii), although some substitution errors were seen. Most of the marks lost in this part were because candidates failed to spot the connection with the previous part. It is worth noting that many candidates did not attempt part (iii), having failed to complete part (ii) successfully. It is likely that some additional marks could have been scored by these candidates had they attempted this final part.