a>0$. (ii) If we set $a=-1$, $b=1$ in the formula of~(i), we obtain \[\int_{-1}^{1}\frac{dx}{1+x^{2}} \overset{?}{=} -\int_{-1}^{1}\frac{dt}{1+t^{2}}\] Explain this apparent failure of the method of integration by substitution. (iii) Write the result of (i) in terms of $\tan^{-1}$ and prove it using standard trigonometric identities. \end{question} \begin{question}\label{Question, define logarithm}\label{Q52} In this question we give an alternative treatment of the logarithm so no properties of the exponential or logarithmic function should be used. You should quote all the theorems that you use, paying particular attention to those on integration. We set \[l(x)=\int_{1}^{x}\frac{1}{t}\,dt.\] (i) Explain why $l:(0,\infty)\rightarrow{\mathbb R}$ is a well defined function. (ii) Use the change of variable theorem for integrals to show that \[\int_{x}^{xy}\frac{1}{t}\,dt=l(y)\] for all $x,\, y>0$. Deduce that $l(xy)=l(x)+l(y)$. (iii) Show that $l$ is everywhere differentiable with $l'(x)=1/x$. (iv) Show that $l$ is a strictly increasing function. (v) Show that $l(x)\rightarrow\infty$ as $x\rightarrow\infty$. \end{question} \begin{question}\label{Q53} In the lectures we deduced the properties of the logarithm from those of the exponential. Reverse this by making a list of the properties of the exponential, define $\exp$ as the inverse function of $\log$ (explaining carefully why you can do this) and using the properties of $\log$ found in the previous question (Question~\ref{Question, define logarithm}). \end{question} \begin{question}$\negthickspace^{\star}$\label{Q54} {\bf [The first mean value theorem for integrals]} Suppose $g:[a,b]\rightarrow{\mathbb R}$ is a continuous function such that $g(x)\geq 0$ for all $x\in[a,b]$. If $f:[a,b]\rightarrow{\mathbb R}$ is continuous explain why we can find $k_{1}$ and $k_{2}$ in $[a,b]$ such that \[f(k_{1})\leq f(x)\leq f(k_{2})\] for all $x\in[a,b]$. Deduce carefully that \[f(k_{1})\int_{a}^{b}g(x)\,dx\leq \int_{a}^{b}f(x)g(x)\,dx \leq f(k_{2})\int_{a}^{b}g(x)\,dx\] and show, stating carefully any theorem that you need, that there exists a $c\in[a,b]$ such that \[\int_{a}^{b}f(x)g(x)\,dx=f(c)\int_{a}^{b}g(x)\,dx.\] Does this result remain true if $g(x)\leq 0$ for all $x\in[a,b]$? Does this result remain true if we place no restrictions on $g$ (apart from continuity). In each case give a proof or a counter example. (The first mean value theorem for integrals is used in the numerical analysis course.) \end{question} \begin{question} Use the integral form of the remainder\label{Q55} in Taylor's theorem (i.e. Theorem~\ref{Integral Taylor}) to obtain the power series expansion for $\sin$. \end{question} \begin{question}{\bf [The binomial theorem]}\label{Q56} If $1>x\geq t\geq 0$ show that \[\frac{x-t}{1+t}\leq x.\] If $-10$ such that $\sum_{n=1}^{\infty}a_{n}$ diverges show that we can find $b_{n}>0$ such that $b_{n}/a_{n}\rightarrow 0$ but $\sum_{n=1}^{\infty}b_{n}$ diverges. Given $a_{n}>0$ such that $\sum_{n=1}^{\infty}a_{n}$ converges show that we can find $b_{n}>0$ such that $b_{n}/a_{n}\rightarrow \infty$ but $\sum_{n=1}^{\infty}b_{n}$ converges. [These two results show that it is futile to look for some sort of `supercharged ratio test' to decide the convergence of all possible series.] \end{question} \begin{question}\label{Q61} (i) By writing \[ r(r+1)\dots (r+m-1)=A_{m}\big(r(r+1)\dots (r+m)-(r-1)r\dots (r+m-1)\big),\] where $A_{m}$ is to found explicitly, compute $\sum_{r=1}^{N}r(r+1)\dots (r+m-1)$. Deduce that \[N^{-m-1}\sum_{r=1}^{N}r(r+1)\dots (r+m-1)\rightarrow\frac{1}{m+1}.\] (ii) Show that \[r(r+1)\dots (r+m-1)-r^{m}=P(r)\] where $P$ is a polynomial of degree less than $m$. Show using (i) and induction, or otherwise, that \[N^{-m-1}\sum_{r=1}^{N}r^{m}\rightarrow\frac{1}{m+1}.\] (iii) Use dissections of the form \[{\mathcal D}=\{0,\ a/n,\ 2a/n,\ \dots,\ a\}\] to compute \[\int_{0}^{a}x^{m}\,dx\] directly from the definition. (iv) Use dissections of the form \[{\mathcal D}=\{br^{n},\ br^{n-1},\ br^{n-2},\ \dots,\ b\}\] with $0 0$}\\ 0&\text{if $\lambda=0$}\\ -L&\text{if $\lambda<0$} \end{cases} \end{equation*} \noindent [Note that $G$ is not continuous at $0$. This is an indication of the unintuitive behaviour which infinite integrals can exhibit.] \end{question} \begin{question}$\negthickspace^{\star}$\label{exercise, bounded variation} (i) Suppose $f_{1},\, f_{2}:[a,b]\rightarrow{\mathbb R}$\label{Q64} are increasing and $g=f_{1}-f_{2}$. Show that there exists a $K$ such that, whenever \[a=x_{0}\leq x_{1}\leq x_{2}\leq\dots\leq x_{n}=b\] we have \[\sum_{j=1}^{n}|g(x_{j})-g(x_{j-1})|\leq K.\] (ii) Let $g:[-1,1]\rightarrow{\mathbb R}$ be given by $g(x)=x^{2}\sin x^{-4}$ for $x\neq 0$, $g(0)=0$. Show that $g$ is once differentiable everywhere but that $g$ is not the difference of two increasing functions. \end{question} \begin{center} {\bf Final Note To Supervisors} \end{center} Let me reiterate my request for corrections and improvements \emph{particularly to the exercises}. The easiest path for me is e-mail. My e-mail address is \verb+twk@dpmms+. \end{document}