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

Give rough sketches of the function ${tan}^k𝜃$ 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 $n$

$${\int \; <!--nolimits-->}_0^{\frac{1}{4}\pi }{tan}^{2n+1}𝜃\mathrm{d}𝜃={\left(-1\right)}^n\left(\frac{1}{2}ln2+\sum _{m=1}^n\frac{{\left(-1\right)}^m}{2m}\right),$$

(†)

and deduce that

$$ln2=-\sum _{m=1}^{\infty }\frac{{\left(-1\right)}^m}{m}.$$

(†)

(ii)

Show similarly that $$\frac{\pi }{4}=-\sum _{m=1}^{\infty }\frac{{\left(-1\right)}^m}{2m-1}.$$

Comments

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 ${tan}^k𝜃$ 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

     

(i) To evaluate the integral, let

$$I_{2n+1}={\int \; <!--nolimits-->}_0^{\frac{1}{4}\pi }{tan}^{2n+1}𝜃\mathrm{d}𝜃.$$

($\ast \ast$)

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

$$I_{2n+1}={\int \; <!--nolimits-->}_0^{\frac{1}{4}\pi }{tan}^{2n-1}𝜃\left(-1+{sec}^2𝜃\right)\mathrm{d}𝜃=-I_{2n-1}+{\int \; <!--nolimits-->}_0^1{u}^{2n-1}\mathrm{d}u=-I_{2n-1}+\frac{1}{2n}.$$

To evaluate the second integral, set $u=tan𝜃$ so that $\mathrm{d}u={sec}^2𝜃\mathrm{d}𝜃$.

Repeating the process gives

$$I_{2n+1}=-I_{2n-1}+\frac{1}{2n}=I_{2n-3}-\frac{1}{2\left(n-1\right)}+\frac{1}{2n}=\cdots ={\left(-1\right)}^n{I}_1+\frac{1}{2n}-\frac{1}{2\left(n-1\right)}+\cdots +{\left(-1\right)}^{n+1}\frac{1}{2}.$$

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 $\left(\ast \ast \right)$:

$$I_1={\int \; <!--nolimits-->}_0^{\frac{1}{4}\pi }tan𝜃\mathrm{d}𝜃=-ln\left(cos𝜃\right)|{}_0^{\frac{1}{4}\pi }=-ln\left(1∕\sqrt{2}\right)=\frac{1}{2}ln2,$$

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 $I_0$:

$$I_0={\int \; <!--nolimits-->}_0^{\frac{1}{4}\pi }1\mathrm{d}𝜃=\frac{\pi }{4}.$$

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 $x^2<1$.