Showing 1-3 of 3 problems
Show that for positive integer \(n\), \(x^n - y^n = (x-y)\displaystyle\sum_{r=1}^{n} x^{n-r} y^{r-1}\).
- Let \(\mathrm{F}\) be defined by
\[ \mathrm{F}(x) = \frac{1}{x^n(x-k)} \quad \text{for } x \neq 0,\, k \]
where \(n\) is a positive integer and \(k \neq 0\).
- Given that
\[ \mathrm{F}(x) = \frac{A}{x-k} + \frac{\mathrm{f}(x)}{x^n}, \]
where \(A\) is a constant and \(\mathrm{f}(x)\) is a polynomial, show that
\[ \mathrm{f}(x) = \frac{1}{x-k}\left(1 - \left(\frac{x}{k}\right)^n\right). \]
Deduce that
\[ \mathrm{F}(x) = \frac{1}{k^n(x-k)} - \frac{1}{k}\sum_{r=1}^{n} \frac{1}{k^{n-r}x^r}. \]
- By differentiating \(x^n \mathrm{F}(x)\), prove that
\[ \frac{1}{x^n(x-k)^2} = \frac{1}{k^n(x-k)^2} - \frac{n}{xk^n(x-k)} + \sum_{r=1}^{n} \frac{n-r}{k^{n+1-r}x^{r+1}}. \]
- Hence evaluate the limit of
\[ \int_2^N \frac{1}{x^3(x-1)^2} \; \mathrm{d}x \]
as \(N \to \infty\), justifying your answer.
Let \(a_n\) be the coefficient of \(x^n\) in the series expansion,
in ascending powers of \(x\), of
\[\displaystyle
\frac{1+x}{(1-x)^2(1+x^2)}
\,,
\]
where \(\vert x \vert <1\,\).
Show, using partial fractions,
that either \(a_n =n+1\) or \(a_n = n+2\) according to the value of \(n\).
Hence find a decimal approximation, to nine significant figures,
for the fraction \( \displaystyle \frac{11\,000}{8181}\).
\newline
[You are not required to justify the accuracy of your approximation.]
Show Solution
\begin{align*}
&& \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\
\Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\
\Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\
\Rightarrow && 1 &= B \\
\Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\
\Rightarrow && A &= -D \\
\Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\
\Rightarrow && C &= \frac12\\
\Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\
\Rightarrow && 3 &= -6A+6 \\
\Rightarrow && A,D &=-\frac12,\frac12 \\
\Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\
&&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k}
\end{align*}
Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively.
\begin{align*}
\frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\
&= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\
&= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\
& \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\
&= 1.34457890 + \frac{12}{10^{10}} + \cdots
\end{align*}
\begin{align*}
&& \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\
&& &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\
\Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\
&&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right )
\end{align*}
Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf
The function \(\mathrm{f}\) is defined, for \(x\neq1\) and \(x\neq2\) by
\[
\mathrm{f}(x)=\frac{1}{\left(x-1\right)\left(x-2\right)}
\]
Show that for \(\left|x\right|<1\)
\[
\mathrm{f}(x)=\sum_{n=0}^{\infty}x^{n}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n}
\]
and that for \(1<\left|x\right|<2\)
\[
\mathrm{f}(x)=-\sum_{n=1}^{\infty}x^{-n}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n}
\]
Find an expression for \(\mbox{f}(x)\) which is valid for \(\left|x\right|>2\).
Show Solution
\begin{align*}
&& \f(x) &= \frac1{(x-1)(x-2)} \\
&&&=\frac{1}{x-2} -\frac{1}{x-1} \\
\end{align*}
Therefore, for \(|x| < 1\)
\begin{align*}
&& \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\
&&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac{1}{1-x} \\
&&&= \sum_{n=0}^{\infty} x^n - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2
\end{align*}
where both geometric series converge since \(|x| < 1\) and \(|\frac{x}{2}| < 1\)
When \(1 < |x|< 2 \Rightarrow |\frac{1}{x}| < 1\), we must have:
\begin{align*}
&& \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\
&&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac1{x}\frac{1}{1-\frac{1}{x}} \\
&&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \frac{1}{x} \sum_{n=0}^{\infty} x^{-n} \\
&&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \sum_{n=1}^{\infty} x^{-n} \\
\end{align*}
Finally, when \(|x| > 2\), ie \(|\frac{2}{x}| < 1\) we have
\begin{align*}
&& \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\
&&& =\frac1{x} \frac{1}{1-\frac{2}{x}} - \frac{1}{x}\frac{1}{1-\frac{1}{x}} \\
&&&= \frac1{x} \sum_{n=0}^{\infty} \l \frac{2}{x} \r^n - \sum_{n=1}^{\infty}x^{-n} \\
&&&= \sum_{n=1}^{\infty} 2^{n-1} x^{-n} - \sum_{n=1}^{\infty}x^{-n} \\
\end{align*}