Problem 22: Β Gregory’s series ($\mathrm{\beta \x9c\x93}$ $\mathrm{\beta \x9c\x93}$) 1991 Paper II

Give rough sketches of the function ${tan}^{k}\mathrm{\pi \x9d\x9c\x83}$ for $0\beta \x89\u20ac\mathrm{\pi \x9d\x9c\x83}\beta \x89\u20ac\frac{1}{4}\mathrm{{\rm O}\x80}$ in the two cases $k=1$ and $k\beta \x89\xab1$.

- (i)
- Show that for any positive integer $n$
and deduce that

$$ln2=\beta \x88\x92\underset{m=1}{\overset{\mathrm{\beta \x88\x9e}}{\beta \x88\x91}}\frac{{\left(\beta \x88\x921\right)}^{m}}{m}\phantom{\rule{0.3em}{0ex}}.$$ (β ) - (ii)
- Show similarly that
$$\frac{\mathrm{{\rm O}\x80}}{4}=\beta \x88\x92\underset{m=1}{\overset{\mathrm{\beta \x88\x9e}}{\beta \x88\x91}}\frac{{\left(\beta \x88\x921\right)}^{m}}{2m\beta \x88\x921}\phantom{\rule{0.3em}{0ex}}.$$

Comments

The symbol $\beta \x89\xab$ in the ο¬rst paragraph means βmuch greater than’, so for the second sketch $k$ is a large number.

This is a good question. In part (i) you are told what to do (in not-very-easy stages) and part (ii) tests your understanding of what you have done and why you have done it by asking you to apply the method to a different but essentially similar problem.

In the ο¬rst paragraph, you have to see how the function ${tan}^{k}\mathrm{\pi \x9d\x9c\x83}$ changes when $k$ increases. You need only a rough sketch to show that you have understood the important point. This should be done by thought, not by means of a calculator.

If you are stuck with the integral of the second paragraph, you might like to think in terms of a recurrence formula, i.e. a formula relating ${I}_{2n+1}$ and ${I}_{2n\beta \x88\x921}$ (in the obvious notation).

The series derived in part (ii) for $\frac{1}{4}\mathrm{{\rm O}\x80}$ is usually called Leibniz’ formula, although the general series for ${tan}^{\beta \x88\x921}x$ was written down by Gregory in 1671, two years before Leibniz. It was one of the ο¬rst explicit formulae for $\mathrm{{\rm O}\x80}$, though Wallis had obtained a product formula in 1655 using a method similar to the method of this question, using ${sin}^{k}x$ in the integral. Previously, the value of $\mathrm{{\rm O}\x80}$ could only be estimated geometrically, by (for example) approximating the circumference of a circle by the edges of an inscribed regular polygon. Using a square gives $\mathrm{{\rm O}\x80}\beta \x89\x882\mathrm{\beta \x88\x9a}2$.

Solution to problem 22

For $0\beta \x89\u20actan\mathrm{\pi \x9d\x9c\x83}<\frac{1}{4}\mathrm{{\rm O}\x80}\phantom{\rule{0.3em}{0ex}}$, we have $tan\mathrm{\pi \x9d\x9c\x83}<1$, so that the curve $y={tan}^{k}\mathrm{\pi \x9d\x9c\x83}$ is close to zero (i.e. much smaller than 1) when $k$ is large. This is illustrated in the ο¬gure which shows three cases: for $k=1$ the graph is mildly curved; for larger $k$ the graph hugs the $x$-axis before taking off. The graphs all pass through the point $\left(\frac{1}{4}\mathrm{{\rm O}\x80},1\right)$.

(i) To evaluate the integral, let

We shall express ${I}_{2n+1}$ in terms of ${I}_{2n\beta \x88\x921}$ using the relation ${tan}^{2}\mathrm{\pi \x9d\x9c\x83}={sec}^{2}\mathrm{\pi \x9d\x9c\x83}\beta \x88\x921$:

$${I}_{2n+1}={\beta \x88\xab}_{0}^{\frac{1}{4}\mathrm{{\rm O}\x80}}{tan}^{2n\beta \x88\x921}\mathrm{\pi \x9d\x9c\x83}\phantom{\rule{0.3em}{0ex}}\left(\beta \x88\x921+{sec}^{2}\mathrm{\pi \x9d\x9c\x83}\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi \x9d\x9c\x83}=\beta \x88\x92{I}_{2n\beta \x88\x921}+{\beta \x88\xab}_{0}^{1}{u}^{2n\beta \x88\x921}\phantom{\rule{0.3em}{0ex}}\mathrm{d}u=\beta \x88\x92{I}_{2n\beta \x88\x921}+\frac{1}{2n}\phantom{\rule{0.3em}{0ex}}.$$To evaluate the second integral, set $u=tan\mathrm{\pi \x9d\x9c\x83}$ so that $\mathrm{d}u={sec}^{2}\mathrm{\pi \x9d\x9c\x83}\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi \x9d\x9c\x83}\phantom{\rule{0.3em}{0ex}}$.

Repeating the process gives

$${I}_{2n+1}=\beta \x88\x92{I}_{2n\beta \x88\x921}+\frac{1}{2n}={I}_{2n\beta \x88\x923}\beta \x88\x92\frac{1}{2\left(n\beta \x88\x921\right)}+\frac{1}{2n}=\beta \x8b\u2015={\left(\beta \x88\x921\right)}^{n}{I}_{1}+\frac{1}{2n}\beta \x88\x92\frac{1}{2\left(n\beta \x88\x921\right)}+\beta \x8b\u2015+{\left(\beta \x88\x921\right)}^{n+1}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}.$$The above sum (starting with $1\beta \x88\x95\left(2n\right)$) is the same as that in $\left(\beta \x80\right)$ overleaf, so it only remains to evaluate ${I}_{1}$, corresponding to $n=0$ in $\left(\beta \x88\x97\beta \x88\x97\right)$:

$${I}_{1}={\beta \x88\xab}_{0}^{\frac{1}{4}\mathrm{{\rm O}\x80}}tan\mathrm{\pi \x9d\x9c\x83}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi \x9d\x9c\x83}=\beta \x88\x92ln\left(cos\mathrm{\pi \x9d\x9c\x83}\right)|{}_{0}^{\frac{1}{4}\mathrm{{\rm O}\x80}}=\beta \x88\x92ln\left(1\beta \x88\x95\sqrt{2}\right)=\frac{1}{2}ln2\phantom{\rule{0.3em}{0ex}},$$as required.

We deduce the expression $\left(\beta \x80\u2018\right)$ overleaf for $ln2$ using the ο¬rst line of the question. When $n$ is very large, ${I}_{2n+1}$ is very small, being the area under a graph which is almost zero for almost all of the range of integration. In the limit $n\beta \x86\x92\mathrm{\beta \x88\x9e}$, we set ${I}_{2n+1}=0$ in $\left(\beta \x80\right)$ which leads immediately to $\left(\beta \x80\u2018\right)$.

(ii) To obtain the formula for $\frac{1}{4}\mathrm{{\rm O}\x80}$, we follow the above method using ${I}_{2n}$ instead of ${I}_{2n+1}$. This time we have to calculate ${I}_{0}$:

$${I}_{0}={\beta \x88\xab}_{0}^{\frac{1}{4}\mathrm{{\rm O}\x80}}1\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi \x9d\x9c\x83}=\frac{\mathrm{{\rm O}\x80}}{4}\phantom{\rule{0.3em}{0ex}}.$$

Post-mortem

You might think that the method of obtaining the formula for $\frac{1}{4}\mathrm{{\rm O}\x80}$ in this question is rather indirect; one could instead just integrate the formula

$$\frac{\mathrm{d}{tan}^{\beta \x88\x921}x}{\mathrm{d}x}=\frac{1}{1+{x}^{2}}=1\beta \x88\x92{x}^{2}+{x}^{4}\beta \x88\x92\beta \x8b\u2015$$ | ($\beta \x88\x97$) |

term by term and get the result immediately by setting $x=1$. The virtue of the method used in the question is that it gives an explicit form (an integral) of the remainder after $n$ terms of the series. We were able to show, by means of a sketch, that the remainder tends to zero as $n$ tends to inο¬nity; in other words, we showed that the series converges. Although the sketch method of proof is a bit crude, it can easily be made more rigorous once the concept of integration is more carefully deο¬ned. On the other hand, integrating ($\beta \x88\x97$) and setting $x=1$ is a bit delicate, since the series only converges for ${x}^{2}<1$.