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Problem 22:  Gregory’s series ($✓$ $✓$) 1991 Paper II

Give rough sketches of the function ${tan}^{k}𝜃$ for $0\le 𝜃\le \frac{1}{4}\pi$ in the two cases $k=1$ and $k\gg 1$.

(i)
Show that for any positive integer $n$
 ${\int }_{0}^{\frac{1}{4}\pi }{tan}^{2n+1}𝜃\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃={\left(-1\right)}^{n}\left(\frac{1}{2}ln2+\sum _{m=1}^{n}\frac{{\left(-1\right)}^{m}}{2m}\right),$ (†)

and deduce that

 $ln2=-\sum _{m=1}^{\infty }\frac{{\left(-1\right)}^{m}}{m}\phantom{\rule{0.3em}{0ex}}.$ (†)
(ii)
Show similarly that $\frac{\pi }{4}=-\sum _{m=1}^{\infty }\frac{{\left(-1\right)}^{m}}{2m-1}\phantom{\rule{0.3em}{0ex}}.$

The symbol $\gg$ in the ﬁrst paragraph means ‘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 ﬁrst paragraph, you have to see how the function ${tan}^{k}𝜃$ 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 ${I}_{2n-1}$ (in the obvious notation).

The series derived in part (ii) for $\frac{1}{4}\pi$ is usually called Leibniz’ formula, although the general series for ${tan}^{-1}x$ was written down by Gregory in 1671, two years before Leibniz. It was one of the ﬁrst explicit formulae for $\pi$, 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 $\pi$ 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 $\pi \approx 2\surd 2$.

Solution to problem 22

For $0\le tan𝜃<\frac{1}{4}\pi \phantom{\rule{0.3em}{0ex}}$, we have $tan𝜃<1$, so that the curve $y={tan}^{k}𝜃$ is close to zero (i.e. much smaller than 1) when $k$ is large. This is illustrated in the ﬁgure 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 $\left(\frac{1}{4}\pi ,1\right)$. (i) To evaluate the integral, let

 ${I}_{2n+1}={\int }_{0}^{\frac{1}{4}\pi }{tan}^{2n+1}𝜃\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃\phantom{\rule{0.3em}{0ex}}.$ ($\ast \ast$)

We shall express ${I}_{2n+1}$ in terms of ${I}_{2n-1}$ using the relation ${tan}^{2}𝜃={sec}^{2}𝜃-1$:

${I}_{2n+1}={\int }_{0}^{\frac{1}{4}\pi }{tan}^{2n-1}𝜃\phantom{\rule{0.3em}{0ex}}\left(-1+{sec}^{2}𝜃\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃=-{I}_{2n-1}+{\int }_{0}^{1}{u}^{2n-1}\phantom{\rule{0.3em}{0ex}}\mathrm{d}u=-{I}_{2n-1}+\frac{1}{2n}\phantom{\rule{0.3em}{0ex}}.$

To evaluate the second integral, set $u=tan𝜃$ so that $\mathrm{d}u={sec}^{2}𝜃\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃\phantom{\rule{0.3em}{0ex}}$.

Repeating the process gives

${I}_{2n+1}=-{I}_{2n-1}+\frac{1}{2n}={I}_{2n-3}-\frac{1}{2\left(n-1\right)}+\frac{1}{2n}=\cdots ={\left(-1\right)}^{n}{I}_{1}+\frac{1}{2n}-\frac{1}{2\left(n-1\right)}+\cdots +{\left(-1\right)}^{n+1}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}.$

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

${I}_{1}={\int }_{0}^{\frac{1}{4}\pi }tan𝜃\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃=-ln\left(cos𝜃\right)|{}_{0}^{\frac{1}{4}\pi }=-ln\left(1∕\sqrt{2}\right)=\frac{1}{2}ln2\phantom{\rule{0.3em}{0ex}},$

as required.

We deduce the expression $\left(‡\right)$ overleaf for $ln2$ using the ﬁrst 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 $n\to \infty$, we set ${I}_{2n+1}=0$ in $\left(†\right)$ which leads immediately to $\left(‡\right)$.

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

${I}_{0}={\int }_{0}^{\frac{1}{4}\pi }1\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}𝜃=\frac{\pi }{4}\phantom{\rule{0.3em}{0ex}}.$

Post-mortem

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

 $\frac{\mathrm{d}{tan}^{-1}x}{\mathrm{d}x}=\frac{1}{1+{x}^{2}}=1-{x}^{2}+{x}^{4}-\cdots$ ($\ast$)

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 inﬁnity; 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 deﬁned. On the other hand, integrating ($\ast$) and setting $x=1$ is a bit delicate, since the series only converges for ${x}^{2}<1$.