Year: 2012
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
In this question, $\vert x \vert <1$ and you may ignore issues of convergence.
\begin{questionparts}
\item
Simplify
\[
(1-x)(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n})\,,
\]
where $n$ is a positive integer,
and deduce that
\[
\frac1{1-x}
= (1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) + \frac {x^{2^{n+1}}}{1-x}\,.
\]
Deduce further that
\[
\ln(1-x) = - \sum_{r=0}^\infty \ln \left (1+ x ^{2^r}\right)
\,,
\]
and hence that
\[
\frac1 {1-x} = \frac 1 {1+x} + \frac {2x}{1+x^2} + \frac {4x^3}{1+x^4}
+\cdots\,.
\]
\item
Show that
\[
\frac{1+2x}{1+x+x^2} = \frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4}
+ \frac {4x^3-8x^7}{1-x^4+x^8} + \cdots\,.
\]
\end{questionparts}\begin{questionparts}
\item
\begin{align*}
(1-x)&(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\
&= (1-x^2)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) \\
&= (1-x^4)(1+x^4) \cdots (1+x^{2^n}) \\
&= 1-x^{2^{n+1}} \\
\end{align*}
Therefore,
\begin{align*}
&& \frac{1}{1-x} - \frac{x^{2^{n+1}}}{1-x} &= (1+x)(1+x^2)\cdots(1+x^{2^n}) \\
\Rightarrow && \frac{1}{1-x} &=(1+x)(1+x^2)\cdots(1+x^{2^n})+ \frac{x^{2^{n+1}}}{1-x} \\
\Rightarrow && -\ln (1-x) &= \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\
\Rightarrow && \ln(1-x) &= - \sum_{r=0}^{\infty} \ln (1+x^{2^r}) \\
\underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{1-x} &= \sum_{r=0}^{\infty} \frac{2^r x^{2^r-1}}{1+x^{2^r}} \\
&&&= \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots
\end{align*}
\item Consider
\begin{align*}(1+x+x^2)&(1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) \\
&= (1+x^2 + x^4)(1-x^2+x^4) \cdots (1-x^{2^n}+x^{2^{n+1}}) \\
&= (1-x^{2^{n+1}}+x^{2^{n+2}}) \\
\end{align*}
Therefore,
\begin{align*}
&& \frac{1}{1+x+x^2} &= (1-x+x^2)(1-x^2+x^4)\cdots(1-x^{2^n}+x^{2^{n+1}}) + \frac{x^{2^{n+1}}}{1+x+x^2} -\frac{x^{2^{n+2}}}{1+x+x^2} \\
\Rightarrow && -\ln(1+x+x^2) &= \sum_{r=0}^\infty \ln (1 - x^{2^r}+x^{2^{r+1}}) \\
\underbrace{\Rightarrow}_{\frac{\d}{\d x}} && -\frac{1+2x}{1+x+x^2} &= \sum_{r=0}^{\infty} \frac{ -2^r x^{2^r-1}+2^{r+1}x^{2^{r+1}-1}}{1 - x^{2^r}+x^{2^{r+1}}} \\
&&&= \frac{-1+2x}{1-x+x^2}+\frac{-2x+4x^3}{1-x^2+x^4} + \frac{-4x^3+8x^7}{1-x^4+x^8} + \cdots
\end{align*}
Which is the desired result when we multiply both sides by $-1$
\end{questionparts}
Three quarters of the candidates attempted this question, making it the second most popular, and one of the two most successful. Generally, part (i) was successfully attempted, though at times marks were dropped through insufficient explanation and quite a few struggled to deal with the "remainder term". Some candidates expanding the brackets worked with the second and third, the fourth brackets etc., only including the first bracket last. About half the candidates considered the product of all of the denominators in part (ii) and replicated the method for the first part, whilst others used the results from part (i), replacing by and employing the factorisations of sums and differences of cubes. Full marks were not uncommon on this question, nor were half marks.