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Problem 22: Β Gregory’s series (βœ“ βœ“) 1991 Paper II

Give rough sketches of the function tankπœƒ for 0 ≀ πœƒ ≀ 1 4Ο€ in the two cases k = 1 and k ≫ 1.

Show that for any positive integer n
∫ 01 4 Ο€ tan2n+1πœƒdπœƒ = (βˆ’1)n 1 2 ln2 + βˆ‘ m=1n(βˆ’1)m 2m , (†)

and deduce that

ln2 = βˆ’βˆ‘ m=1∞(βˆ’1)m m . (†)
Show similarly that Ο€ 4 = βˆ’βˆ‘ m=1∞ (βˆ’1)m 2m βˆ’ 1.


The symbol ≫ in the first 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 first paragraph, you have to see how the function tankπœƒ 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 I2n+1 and I2nβˆ’1 (in the obvious notation).

The series derived in part (ii) for 1 4Ο€ is usually called Leibniz’ formula, although the general series for tanβˆ’1x was written down by Gregory in 1671, two years before Leibniz. It was one of the first explicit formulae for Ο€, though Wallis had obtained a product formula in 1655 using a method similar to the method of this question, using sinkx in the integral. Previously, the value of 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 Ο€ β‰ˆ 2√2.

Solution to problem 22

For 0 ≀ tanπœƒ < 1 4Ο€, we have tanπœƒ < 1, so that the curve y = tankπœƒ 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 (1 4Ο€,1).

Β Β Β Β Β 

(i) To evaluate the integral, let

I2n+1 = ∫ 01 4 Ο€ tan2n+1πœƒdπœƒ. ( βˆ—βˆ—)

We shall express I2n+1 in terms of I2nβˆ’1 using the relation tan2πœƒ = sec2πœƒ βˆ’ 1:

I2n+1 = ∫ 01 4 Ο€ tan2nβˆ’1πœƒ(βˆ’1 + sec2πœƒ)dπœƒ = βˆ’I2nβˆ’1 + ∫ 01u2nβˆ’1du = βˆ’I 2nβˆ’1 + 1 2n.

To evaluate the second integral, set u = tanπœƒ so that du = sec2πœƒdπœƒ.

Repeating the process gives

I2n+1 = βˆ’I2nβˆ’1 + 1 2n = I2nβˆ’3 βˆ’ 1 2(n βˆ’ 1) + 1 2n = β‹― = (βˆ’1)nI 1 + 1 2n βˆ’ 1 2(n βˆ’ 1) + β‹― + (βˆ’1)n+11 2.

The above sum (starting with 1βˆ•(2n)) is the same as that in (†) overleaf, so it only remains to evaluate I1, corresponding to n = 0 in (βˆ—βˆ—):

I1 = ∫ 01 4 Ο€ tanπœƒdπœƒ = βˆ’ln(cosπœƒ)|01 4 Ο€ = βˆ’ln(1βˆ•2) = 1 2 ln2,

as required.

We deduce the expression (‑) overleaf for ln2 using the first line of the question. When n is very large, I2n+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 β†’βˆž, we set I2n+1 = 0 in (†) which leads immediately to (‑).

(ii) To obtain the formula for 1 4Ο€, we follow the above method using I2n instead of I2n+1. This time we have to calculate I0:

I0 = ∫ 01 4 Ο€1dπœƒ = Ο€ 4 .


You might think that the method of obtaining the formula for 1 4Ο€ in this question is rather indirect; one could instead just integrate the formula

dtanβˆ’1x dx = 1 1 + x2 = 1 βˆ’ x2 + x4 βˆ’β‹― (βˆ—)

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 (βˆ—) and setting x = 1 is a bit delicate, since the series only converges for x2 < 1.