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

Give rough sketches of the function ${tan}^{k}\mathrm{\pi }$ for $0\beta €\mathrm{\pi }\beta €\frac{1}{4}\mathrm{Ο}$ in the two cases $k=1$ and $k\beta «1$.

(i)
Show that for any positive integer $n$
 ${\beta «}_{0}^{\frac{1}{4}\mathrm{Ο}}{tan}^{2n+1}\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi }={\left(\beta 1\right)}^{n}\left(\frac{1}{2}ln2+\underset{m=1}{\overset{n}{\beta }}\frac{{\left(\beta 1\right)}^{m}}{2m}\right),$ (β )

and deduce that

 $ln2=\beta \underset{m=1}{\overset{\mathrm{\beta }}{\beta }}\frac{{\left(\beta 1\right)}^{m}}{m}\phantom{\rule{0.3em}{0ex}}.$ (β )
(ii)
Show similarly that $\frac{\mathrm{Ο}}{4}=\beta \underset{m=1}{\overset{\mathrm{\beta }}{\beta }}\frac{{\left(\beta 1\right)}^{m}}{2m\beta 1}\phantom{\rule{0.3em}{0ex}}.$

The symbol $\beta «$ in the ο¬rst paragraph means βmuch 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 ${tan}^{k}\mathrm{\pi }$ 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 ${I}_{2n\beta 1}$ (in the obvious notation).

The series derived in part (ii) for $\frac{1}{4}\mathrm{Ο}$ is usually called Leibniz’ formula, although the general series for ${tan}^{\beta 1}x$ was written down by Gregory in 1671, two years before Leibniz. It was one of the ο¬rst explicit formulae for $\mathrm{Ο}$, 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 $\mathrm{Ο}$ 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 $\mathrm{Ο}\beta 2\mathrm{\beta }2$.

Solution to problem 22

For $0\beta €tan\mathrm{\pi }<\frac{1}{4}\mathrm{Ο}\phantom{\rule{0.3em}{0ex}}$, we have $tan\mathrm{\pi }<1$, so that the curve $y={tan}^{k}\mathrm{\pi }$ 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 $\left(\frac{1}{4}\mathrm{Ο},1\right)$.

Β Β Β Β Β

(i) To evaluate the integral, let

 ${I}_{2n+1}={\beta «}_{0}^{\frac{1}{4}\mathrm{Ο}}{tan}^{2n+1}\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}.$ ($\beta \beta $)

We shall express ${I}_{2n+1}$ in terms of ${I}_{2n\beta 1}$ using the relation ${tan}^{2}\mathrm{\pi }={sec}^{2}\mathrm{\pi }\beta 1$:

${I}_{2n+1}={\beta «}_{0}^{\frac{1}{4}\mathrm{Ο}}{tan}^{2n\beta 1}\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}\left(\beta 1+{sec}^{2}\mathrm{\pi }\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi }=\beta {I}_{2n\beta 1}+{\beta «}_{0}^{1}{u}^{2n\beta 1}\phantom{\rule{0.3em}{0ex}}\mathrm{d}u=\beta {I}_{2n\beta 1}+\frac{1}{2n}\phantom{\rule{0.3em}{0ex}}.$

To evaluate the second integral, set $u=tan\mathrm{\pi }$ so that $\mathrm{d}u={sec}^{2}\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}$.

Repeating the process gives

${I}_{2n+1}=\beta {I}_{2n\beta 1}+\frac{1}{2n}={I}_{2n\beta 3}\beta \frac{1}{2\left(n\beta 1\right)}+\frac{1}{2n}=\beta ―={\left(\beta 1\right)}^{n}{I}_{1}+\frac{1}{2n}\beta \frac{1}{2\left(n\beta 1\right)}+\beta ―+{\left(\beta 1\right)}^{n+1}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}.$

The above sum (starting with $1\beta \left(2n\right)$) is the same as that in overleaf, so it only remains to evaluate ${I}_{1}$, corresponding to $n=0$ in $\left(\beta \beta \right)$:

${I}_{1}={\beta «}_{0}^{\frac{1}{4}\mathrm{Ο}}tan\mathrm{\pi }\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi }=\beta ln\left(cos\mathrm{\pi }\right)|{}_{0}^{\frac{1}{4}\mathrm{Ο}}=\beta ln\left(1\beta \sqrt{2}\right)=\frac{1}{2}ln2\phantom{\rule{0.3em}{0ex}},$

as required.

We deduce the expression $\left(\beta ‘\right)$ 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 $n\beta \mathrm{\beta }$, we set ${I}_{2n+1}=0$ in which leads immediately to $\left(\beta ‘\right)$.

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

${I}_{0}={\beta «}_{0}^{\frac{1}{4}\mathrm{Ο}}1\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}\mathrm{\pi }=\frac{\mathrm{Ο}}{4}\phantom{\rule{0.3em}{0ex}}.$

Post-mortem

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

 $\frac{\mathrm{d}{tan}^{\beta 1}x}{\mathrm{d}x}=\frac{1}{1+{x}^{2}}=1\beta {x}^{2}+{x}^{4}\beta \beta ―$ ($\beta $)

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 ($\beta $) and setting $x=1$ is a bit delicate, since the series only converges for ${x}^{2}<1$.