John Wallis
(1616 - 1703)
An other infinite product concerning Pi
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 *I*_{0} and *I*_{1} for even
then odd *n*.
We have immediatly *I*_{0}=/2 and *I*_{1}=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 *cos*^{n}(x)<cos^{n-1}(x), so *n
-> cos*^{n}(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* |
*3,14**08 (2)* |
*n=10 000* |
*3,1415**1402 (4)* |
*n=100 000* |
*3,1415**847 (4)* |
Roughly *Log(n)*, not great...
Optismic Note...
Nothing stop us to add a little something in the sequence!
The *3/4* of the *3*rd formula at the top of the page comes
from an empirical observation. We already have :
*n=10* |
*3,14**07 (2)* |
*n=100* |
*3,1415**829 (4)* |
*n=1 000* |
*3,141592**555 (6)* |
*n=3 000* |
*3,1415926**42 (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* |
*3,1**008 (1)* |
*n=100* |
*3,1**376(1)* |
*n=1 000* |
*3,141**19 (3)* |
and even better if we add the famous *3/4* :
*n=10* |
*3,141**25 (2)* |
*n=100* |
*3,1415**89 (4)* |
*n=1 000* |
*3,1415926**2084 (7)* |
In the end, there are some ways to make this sequence useful!
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