Alexander Aitken
(1895 - 1967)

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 t_n from x_n , 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 x_n converging towards L with an error e_n=x_n-L verifying

(1)

We are going to build a series t_n from x_n 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 e_n+1=Ae_n (2)
*Now, we have e_n=x_n-L by definition so we immediately obtain:

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

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

=x_n- (5)

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

L=x_n-

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

t_n=x_n- (7)

with t_n 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 e_n+1=(A+ß_n)e_n where ß_n tends to 0

*A small computation gives:

e_n=x_n+1-L-x_n+L=x_n
²e_n=e_n+2-2e_n+1+e_n=x_n+2-2x_n+1+x_n=²x_n

Let us express e_n and ²e_n , which will give x_n and ²x_n :

*On one hand, e_n+2=(A+ß_n+1)e_n+1=(A+ß_n+1)(A+ß_n)e_n and so

²e_n=e_n+2-2e_n+1+e_n=(A+ß_n+1)(A+ß_n)e_n-2(A+ß_n)e_n+e_n=(A-1)²e_n+ß_n'e_n

where ß_n'=Aß_n+Aß_n+1+ß_n+1ß_n-2ß_n tends to 0 at infinity

*On the other hand,

e_n=e_n+1-e_n=(A+ß_n)e_n-e_n=(A-1+ß_n)e_n

We replace now in (5) and we obtain :

t_n=x_n- and by dividing by x_n-L=e_n#0 :

And theorem is eventually proved, there is acceleration of the convergence.
It only remains to express x_n and ²x_n in t_n 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 !