John Wallis
(1616 - 1703)

Slice of Life

John Wallis was born in Ashford, he was the son of the town's chief educator officer. Quite precocious, he turns towards the mathematics at 15, but does not completly forget about physics where he gave laws on collision of hard bodies.... He is the first to use infinity properly (the famous symbol is one of his creation), nul, negative and fractional powers. Passioned by the infinte formulae and approximation that emerges from it( Arithmética infinitorum 1656), he forsaw the geometrical representation of complex numbers and the reprocity between the exponential and the natural logarithm. This fertile mind was also interested in astronomi, music and botanic and was one the founders of the Royal Society (1663). What a amazing and important man!

Around :

Of course, Wallis holds an important place in the legend of Pi since he was the first to discover the development of the famous constant in an infinite product of rationnal fractions. The convergence is poor, of course, but what a beautiful result!

Proof

A classical proof, but a simple model.
The original method of Wallis consisted of using the integrals which Wallis knew the result. He generalised at n=1/2 which gives the area of a quarter-circle of radius 1, hence /4.
In modern notation, let us introduce the equivalent integrals (change of variables x=sin(u) and x=cos(u)) given by Wallis :

Those integrals are in fact equal (change of variable x=/2-t) hence, we are only interested in I_n...
Let us do a little integration by part for n>1 :

but hence we get :
and finaly:

(1)

This recurrence relation alows to bring the calculation of I_n to that of I₀ and I₁ for even then odd n.
We have immediatly I₀=/2 and I₁=1 which gives us :

(2)

by immediate recurrence, then by a very ovious calculation! (use recurrence for example)
and similarly, we have :

(3)

But, on ]0,/2[, cos(x)<1 hence cosⁿ(x)<cos^n-1(x), so n -> cosⁿ(x) is decreasing and n->I_n is decreasing.
So we conclude that for n>1, we have :

I_2n+1<I_2n<I_2n-1

According to (2) we have :

so by the comparison test,

and finaly :

(4)

Great, no?

the first formula given at the top of the page is in fact the same, as we're going to see:


and, I hope that you will easily convince yourself, by an obvious reccurence (which I can't be bothered to write...).
So we finaly get:



according to (4), and there we go. The trick is done!

Note that we will need Wallis' formula for the proof of Stirling's formula, and that reciprocaly, it gives directly Wallis' formula by having:



Ah ! Wallis and Stirling, what a close and legendary liaison...

Trials

So, I've already said it, the convergence is poor, logarithmic...

n=10	3,0677 (0)
n=1000	3,1408 (2)
n=10 000	3,14151402 (4)
n=100 000	3,1415847 (4)

Roughly Log(n), not great...

Optismic Note...

Nothing stop us to add a little something in the sequence! The 3/4 of the 3rd formula at the top of the page comes from an empirical observation. We already have :

n=10	3,1407 (2)
n=100	3,1415829 (4)
n=1 000	3,141592555 (6)
n=3 000	3,141592642 (7)

Roughly 2Log(n) in convergence, nearly worthy seeing the simplicity of the sequence!

And the Delta2 of Aitken work well, we have in effect :

n=10	3,1008 (1)
n=100	3,1376(1)
n=1 000	3,14119 (3)

and even better if we add the famous 3/4 :

n=10	3,14125 (2)
n=100	3,141589 (4)
n=1 000	3,14159262084 (7)

In the end, there are some ways to make this sequence useful!