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modif. 13/04/2013



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Alexander Aitken
(1895 - 1967)



A remarkable formula

Slices of life

Here is a very peculiar and too much underestimated man... I hope that this site will make its tremendous formula of acceleration of the convergence, Delta2, more famous !
But let's rather dive in the story of this uncommon mathematician:
Alexander Aitken was born in New Zealand in 1895. Studying languages and mathematics from 1913, he is wounded at the battle of the Somme during first world war. Schocked by this event, he sets in Edinburgh in Scotland after 3 months of hospital. Endowed with an extraordinary memory (he knows 2000 digits of Pi and is able to give any digit at the n-th place!!), he develops the famous
Delta2 accelerating series in an optimal way. But this first-rate memory too often calls him back the battle of the Somme and traumatizes him. He writes then his memories but they also aggravate his relative mental madness and he finally died in 1967.

About

We said Delta2 accelerates certain series in an optimal way . It is conceived to better work with geometrical sequences. With those series, it directly gives limit at the end of 3 iterations, otherwise it tries to guess the limit of this series. If it does not find it, it will increase at least convergence.
Delta2's properties are rather extraordinary, let us take for example series for

ln(1+x)=

valid on ]-1,1] as each knows. If one makes x=2, series is going to diverge, naturally. Well this dear Delta2 is going to make it converge during a little while !!! Look rather:
ln(3)=1,0986...
With the series at rank 8, we obtain -19,31 (!)
With Delta2 iterated twice, at rank 8, we obtain 1,0979
Later we go away from the real value. But the completely fake values in the series give a good value with the Delta2, it is somewhat fabulous, isn't it?
Generally,Delta2 accelerates all the series whose ratio of two consecutive differences converges to a limit included between 1 and -1, that is naturally the case of geometrical sequences.
Aitken's Delta2 is numerically very unstable because numerator and the denominator are close to 0 and it is so necessary to calculate with a great amount of digits ! The second member of the formula will be rather used, being more stable...

Proof

Formally, if we build tn from xn , this last one converging towards L, we say that there is acceleration of the convergence if we have:

That is easy to understand, intuitively...
We are going to build Delta2 and show that it verifies this property...

Let's take a series xn converging towards L with an error en=xn-L verifying

(1)

We are going to build a series tn from xn which will converge faster. That is not very long, and very interesting, you will see!

Construction :

Let us suppose that eq (1) is exact for every n (without going to infinity) , that is en+1=Aen (2)
*Now, we have en=xn-L by definition so we immediately obtain:

xn+2-xn+1=A(xn+1-xn) (3)

*Let xn=xn+1-xn. (4)
According to (2) and the definition of en , it comes xn+1-L=A(xn-L) and so L(1-A)=xn+1-Axn and finally :

=xn- (5)

*Let's compute 2xn=(xn)=xn+1-xn=xn+2-2xn+1+xn. (6)
Now, we have 1-A=1- so, we deduct from it:

L=xn-

But because the starting hypothesis (1) is only true at the limit, a new series is introduced:

tn=xn- (7)

with tn tending to L in the infinity

Proof of the theorem:

Being good architects, having built this series, let us now verify that this series indeed converges faster !
Because we are not anymore at the limit, let us write en+1=(A+ßn)en where ßn tends to 0

*A small computation gives:

en=xn+1-L-xn+L=xn
2en=en+2-2en+1+en=xn+2-2xn+1+xn=2xn


Let us express en and 2en , which will give xn and 2xn :

*On one hand, en+2=(A+ßn+1)en+1=(A+ßn+1)(A+ßn)en and so

2en=en+2-2en+1+en=(A+ßn+1)(A+ßn)en-2(A+ßn)en+en=(A-1)2enn'en

where ßn'=Aßn+Aßn+1n+1ßn-2ßn tends to 0 at infinity

*On the other hand,

en=en+1-en=(A+ßn)en-en=(A-1+ßn)en


We replace now in (5) and we obtain :

tn=xn- and by dividing by xn-L=en#0 :

And theorem is eventually proved, there is acceleration of the convergence.
It only remains to express xn and 2xn in tn like in (4) and (6) to find the formula of Aitken's Delta2. After this proof, I hope that everyone will have realized where the name Delta2 comes from !


Thanks to David Jelgersma for original translation
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