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Boris Gourévitch
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modif. 13/04/2013

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René Descartes
(1596 - 1650)

The original method of isoperimeters

Slices of life

René Descartes was born at Hague in Touraine exactly on March 31, 1596. He was certainly the one of the most famous French philosopher! But he was not only that...
After a licence of right in 1616, he chooses the military profession in Holland then in Duke de Bavaria service until 1620. Back in in France in 1625, he draws up there his philosophical works - famous, but it is not the object of this site! - and publish scientific works about optics, astronomy, biology and especially geometry. In 1631, appears Geometry in that it defines the Cartesian coordinates of a point. Let us note by the way that it is for Descartes that we have to the habit to represent quantities known by the first letters of the alphabet a, b, c, d and the unknowns by x, y, z.... He dies in 1650.


After his death, the method of isoperimeters will be found in his papers. It consists in making the opposite of Archimede's method it should to tell to determine the radius of a circle of which the perimeter is fixed in advance. It is a wholly geometrical construction...


It would be better to say construction, because it is not a real mathematical proof...
A suite of regular polygons is considered P0 ,P1, P2 ... Pn of respectively 22 , 23,..., 2n+2 sides, having - it is important - the same perimeter L

The following figure, with An Bn =Cn and OHn =rn .
Let us look for a relation of recurrence between rn+1 and rn knowing that rn tends towards the radius of the circle.
One knows that 2n+2Cn =L because L is the perimeter of the polygone Pn of side Cn. This relation being valid for every nN,we have also 22C0=L.
For P0 , we got a square OH0 =C0/2=r0, so r0=C0/2=L/(2.22)=L/8
Let E the middle of the small arc AnBn . the segment which joins the circles, the middles An+1 and Bn+1 of [EAn] and [EBn] is the side of Pn+1. All the history is going to be geometrical, then let us concentrate!

We have An+1 Bn+1=Cn+1=L / 2n+3 : Indeed, by the theorem of this dear Thales,

In the right-angled triangle OEAn+1 (Because OAnE is isosceles), we have An+1Hn+12=EHn+1 *Hn+1O
But let us show it if it is not evident!
We have on the first hand EO2=(EHn+1+Hn+1O)2=EHn+12+2EHn+1*Hn+1O+Hn+1O2
So .
On the other hand, An+1Hn+12=An+1E2 - EHn+12 by Pythagore and An+1Hn+12=OAn+12 - OHn+12, so we have :

Now always by pythagore EO2 =An+1E2 + OAn+12
So An+1Hn+12=EHn+1*Hn+1O (openly saddened for heaviness of notations!!)


and EHn+1=Hn+1Hn (obvious by Thales!)
=OHn+1-OHn =rn+1-rn and again, Hn+1O=rn+1
So :

Well, Here we are, our relation of recurrence!
It is moreover a polynomial in rn+1, that is obviously positive.
The only positive root of the polynomial is extracted and we obtain:

When n increases, the polygone Pn tends to become merged with the circle of perimeter L=8r0=2rn (2*radius...)
so :

Interesting, no?
And not so bad in term of efficiency!


Let us look at it closer...
Expression is reminiscent of the area of a rectangle of rn+1 and rn+1-rn sides. The geometrical series of the areas of these rectangles would be of reason 1/4. A priori, the relation between rn+1 and rn should also have to behave as a geometrical series, and convergence should be linear (-Log (rn) =a*n+b)... Let us verify by taking L=8, and so r0=1 (the choice of L does not influence the result because the relation between rn+1 and rn is homogeneous in L) there:

n=5 3,1422 (2)
n=10 3,1415932 (5)
n=50 28 decimals exact
n=100 60 decimals exact

Totaly, a good small convergence in 3n / 5, here is that is very honorable!

Acceleration of the convergence

What there is well with Aitken's Delta2, it is that there is always an acceleration, so small is it. But then when it is gigantic, How euphoric! Let us look at attempts:

  Without Aitken With Aitken With Aitken iterated
n=5 3,1422 (2) 3,14159508 (5) 3,1415926559 (8)
n=10 3,1415932 (5) 3,141592653592 (10) 16 decimals exact
n=20 3,1415926535903 (10) 23 decimals exact 35 decimals exact
n=50 28 decimals exact 59 decimals exact 90 decimals exact

It is incredible! Aitken multiplies by more 2 the performance of the series which reaches a convergence of 1.2n !
I feel that it is the best result obtained with Aitken for series converging toward Pi.
And look at the results with Aitken iterated (Delta2 is applied twice) because of the limit preciseness of my computer (100 decimals) and the sensibility of Delta2, it is even possible to obtain a better result.
We reache with Aitken iterated an upper preciseness in 1.6n and that goes better!

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