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.
About
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...
Proof
It would be better to say construction, because
it is not a real mathematical proof...
A suite of regular polygons is considered P_{0 },P_{1}, P_{2 }... P_{n} of respectively 2^{2 }, 2^{3},...,
2^{n+2} sides, having
 it is important  the same perimeter L
The following figure, with A_{n }B_{n} =C_{n} and OH_{n} =r_{n} .
Let us look for a relation of recurrence between r_{n+1} and r_{n} knowing that r_{n} tends towards the radius of the circle.
One knows that 2^{n+2}C_{n} =L because L is the perimeter of the polygone P_{n} of side C_{n}.
This relation being valid for every nN,we have also 2^{2}C_{0}=L.
For P_{0} , we got a square OH_{0 }=C_{0}/2=r_{0}, so r_{0}=C_{0}/2=L/(2.2^{2})=L/8
Let E the middle of the small arc A_{n}B_{n} . the segment which joins the circles, the middles A_{n+1} and B_{n+1} of [EA_{n}] and [EB_{n}] is the side of P_{n+1}. All the history is going to be geometrical, then let us
concentrate!
We have A_{n+1 }B_{n+1}=C_{n+1}=L
/ 2^{n+3} : Indeed,
by the theorem of this dear Thales,
In the rightangled triangle OEA_{n+1} (Because OA_{n}E is isosceles), we have A_{n+1}H_{n+1}^{2}=EH_{n+1 }*H_{n+1}O
But let us show it if it is not evident!
We have on the first hand EO^{2}=(EH_{n+1}+H_{n+1}O)^{2}=EH_{n+1}^{2}+2EH_{n+1}*H_{n+1}O+H_{n+1}O^{2}
So .
On the other hand, A_{n+1}H_{n+1}^{2}=A_{n+1}E^{2 } EH_{n+1}^{2} by Pythagore and A_{n+1}H_{n+1}^{2}=OA_{n+1}^{2 } OH_{n+1}^{2}, so
we have :
Now always by pythagore EO^{2 }=A_{n+1}E^{2 }+ OA_{n+1}^{2 }
So A_{n+1}H_{n+1}^{2}=EH_{n+1}*H_{n+1}O (openly saddened for heaviness of notations!!)
Now
and EH_{n+1}=H_{n+1}H_{n} (obvious by Thales!)
=OH_{n+1}OH_{n} =r_{n+1}r_{n} and again, H_{n+1}O=r_{n+1}
So :
Well, Here we are, our relation of recurrence!
It is moreover a polynomial in r_{n+1}, that is obviously positive.
The only positive root of the polynomial is extracted and we obtain:
When n increases, the polygone P_{n} tends
to become merged with the circle of perimeter L=8r_{0}=2r_{n} (2*radius...)
so :
Interesting, no?
And not so bad in term of efficiency!
Trials
Let us look at it closer...
Expression is reminiscent of the area of a rectangle of r_{n+1} and r_{n+1}r_{n} sides. The geometrical series of the areas of these rectangles
would be of reason 1/4. A priori, the relation between r_{n+1} and r_{n} should also have to behave as a geometrical series, and
convergence should be linear (Log (r_{n})
=a*n+b)... Let us verify by taking L=8, and so r_{0}=1 (the choice of L does not influence the result because the relation between r_{n+1} and r_{n} 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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