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Boris Gourévitch
The world of Pi - V2.57
modif. 13/04/2013



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Takebe Katahiro
(1664 - 1739)



A beautiful algorithm

Bit's of his life

This name probably does not ring a bell for most people because Takebe Katahiro was a japanese mathematician, which was not common, especially at that time. Born in 1664 from a samourai familly, the mutation of a still feudal Japan could only encourage Takebe to complete a samourai formation that was very incomplete..... very quickly he follows the teaching of Seki Takakazu (? - 1708), the most brilliant mathematician of Japan!
The mathematicians of this country did not know of the occidental science. Using littles sticks instead of numbers and without trigonometry (!), Seki and his disciple will build a new science in Japan, developping algebra and analysis. Takebe dies in 1739.

Around :

Takebe is interested in Pi, and even calculate 41 decimals using the Archimede method and a polygon with 1024 sides, which is very a,azing considering the numerical system used !
This does not stop Takebe to offer us a very interesting formula, since our japanese friend is the first to have expressed the square of a circle arc in the from of an inifite sum in his Classique de Tetsujutsu (1722) .
The proofs were not rigourous back then (unfortunatly!), we only have the numerical method that he used to get his formula...
We easily get out of it an algorithm with modern notation then a sum with very interesting performance. Since I could not find anywhere a proof, I looked myself and found one that is not hard, but I'm still looking for a more elegant method!

Proof

Let f(u)= and Un=. We get :
hence from Alembert's critiria, the serie converges for u<1. (In passing, with the formula Moivre/Stirling, we can easily show that the serie does not converge for u=1)
The function being even, it is therefore defined on (-1,1).
We are going to find the exacte expression of f by a classical procedure but that can be classsified as "brute force, typical of first years"!
Let's look for the differential equation that satisfies f !
We have :


so calculate for u[0,1) :


  hence y=f(u) satisfy :


Integrate between 0 and x(0,1) :



but, y'(0)=0 (u=0 in the power serie of the limit f'(u)) so we integrate once more :


Finnaly for r>1 : take U0=4. We have :


which brings an end to the proof! It goes without further saying that the two forms of Un are easily deduced one and the other from imediate recurrence(so immediate that I'm not even going to write it down!)

Trials

Not only the formula is quite beutiful but it's performance is far from rubish! It is quite evident from the ratio form that the convergence is linear.
= in our case here. This means with large enough n, the sequence Un nearly behave like a geometric serie of ratio . Moivre/Stirling equivalence gives us furthermore :


whoes convergence is a little bit faster than the linear convergence...
Theoretically, we can even construct thanks to the different values of r convergant sequences nearly linear as fast as we can!
Let's check all of this with small trials (for r=2) :

n=1 3,055050
n=5 3,1399 (1)
n=10 3,141568 (4)
n=20 3,141592643 (7)
n=50 17 correct decimals
n=100 33 correct decimals
n=200 64 correct decimals

which gives us a convergence of around n / 3, not too bas...

Now for r=3 :
n=5 3,14157 (4)
n=10 3,141592644 (7)
n=50 33 correct decimals
n=100 63 correct decimals

Convergence of roughly 2n/3
 
And then for r=6 :
n=5 3,141592646 (7)
n=10 13 correct decimals
n=50 61 correct decimals

Convergence 1.2n ?

Finnaly for a bit of fun, r=12
n=5 11 correct decimals
n=10 20 correct decimals
n=50 92 correct decimals

We start flirting with a convergence of 2n, but it's surprising that the convergence seems to slow down when she should be a but better than the linear convergence....
You should notice that for r=2 or r=3, we get a series of rational that only needs an extraction of a root at the end, interesting!

Acceleration of convergence

As always with linear convergence (that is sequences that resemble more and more to a geometric sequence), the Delta2 of Aitken shoulf be terribly efficiant. But it's a bit deceiving....

For r=2 :

  Without Aitken With Aitken
n=2 3,112 (1) 3,132 (1)
n=5 3,1399 (1) 3,141428 (3)
n=10 3,141568 (4) 3,14159179 (5)
n=20 3,141592643 (7) 3,141592653479 (9)
n=50 17 correct decimals
19 correct decimals
n=100 33 correct decimals
35 correct decimals
n=200 64 correct decimals
66 correct decimals

no need to carry on further with the demonstration, the Delta2 only adds 2 more decimals to the result, which is not very conclusive!

For r=6
n=5 3,141592646 (7) 3,14159265325 (9)
n=10 13 correct decimals 15 correct decimals
n=50 61 correct decimals 65 correct decimals

Just a bit better, even if there 4 more decimals for n=50, this start to be become a bit more interesting. I can unfortunatly not push the calculation firther for the moment...

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