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



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Carl-Louis-Ferdinand von Lindemann
(1852 - 1939)



Wow, now that's amazing...

is transcendental !!!

A brief look at his life

Carl Louis Ferdinand Lindemann was born in 12 avril 1852 in Hanovre. From 1870 to 1873, this great traveller did his studies at Göttingen, Erlangen, Munich, London and Paris. Then he teached at Wurzbourg (1877), Fribourg (1877-1883), Königsberg (1883-1893) and finnally Munich.
  In 1882, Lindemann publish Die Zahl Pi which brings an end to the problem of squaring a circle and 25 centuries of questionning !
But Lindemann also brought his first efforts towards geometry and after his great success on Pi,he will look at Fermat's big theorem for the rest of his life, without finding a solution...

Around - A few words on its transcendental

Remember that we say that a complex number (hence also reals) is said to be algebraic if it is a root of a non nul polynomial with integers coefficients and is said to be transcendental otherwise.
i is algebraic for example (well it will be useful later on...) since i is a root of the polynomial x4-1=0.
It was only in 1844 that the existance of transcendental terms was proved by Liouville.
  In1874, the great Georges Cantor proved thanks to his passion for set theory, that most of the reals are transcendental, or in fact that the set of algebraic real is countable (hence of size N !)
The transcendental of Pi is not as a meaningul result as we could think in the sense that it does not give us any practical interesting information on the decimals of Pi. Even more, as Cantor showed, the set of trancsendental is a lot more bigger than the one of algebraic, Pi, like any number taken at random, had it's chance to be with those! But since Lindemann's discover brought an end to one of the oldest problem in the world of mathematics, to know how to sqare a circle...
To draw a square (or a circle for the matter!) of the same area as a circle with a ruler and a compas lead to building a segment of length Pi with those same equipment. But I-can't-remember-who showed that only the additions, multiplication, roots and quotion could be build with the ruler and compas. Which is equivalent to the fact that Pi be the root of any polynomial with integers coefficients.
This dream (too beautiful!) died with Lindemann. His proved was heavely based on the method which Hermite used to prove the trancendance of e in 1873. He, himself predicted, after his exploit, that the method could be applied to Pi but even more complicate and did not have the courage to try...
Which will not be our case since the proof for the transcendance of e and then of Pi simplyfied by Weierstrass, Hilbert, Hurwitz and Gordan follow this paragraph. They are taken from Transcendental Number Theory by A. Baker and from le Fascinant Nombre Pi by J.P. Delahaye (The fascinating number Pi) (see Bibliography)
Note that if you can prove that e+ is rational or transcendental, you will have won the big prize since that problem is still not solved!

Proofs

e is transcendental:

  Let f(x) be a polynomial of degree m with reals coefficients. Take :

By doing integration by parts m times we then get :

(1)

Now, let f* be the polynomial f where the coefficients have been replaced by their absolute value.
By maximising the terms in the integral we get :

(2)

Now all the preliminary work has been done, lets start the battle and suppose that e is algebraic. In other terms, suppose that there exist an integer n>0 and non-zero q1,...,qn such that :

q0+q1e+...+qnen=0(3)

The reste consiste of making J=q0I(0)+q1I(1)+...+qnI(n)
I(t) does not change by definition and we choose f(x)=xp-1(x-1) p...(x-n) p with p a big prime integer.
By calculating J using the definition (1) and the hypothesis (3), we get :

where m is therefore the degree of the polynomial f i.e. (n+1)p-1=m.
The k variables between 1 and n being roots of order p each, we have f(j)(k)=0 if j<p and k>0 and the same for k=0 if j<p-1. Therefore for j#p-1 and k#0, either it's zero, or the derivative take out a  p! and so f (j)(k) is an integer divisible by p!. Further more, for the case j=p-1, by recurrence on n, we easily get :

f (p-1)(0)=(p-1)!(-1)np(n!) p

So, if we have p>n, that is p does not divide (n!), f(p-1)(0) is an integer divisible by (p-1)! but not by (p!).
And if we then take p>q0 , J is a non zero integer wich is divisible by (p-1)! hence, obviously, J(p-1)!
Here is the first inequality. Now all that remains is to find a contradiction.
For this, since k-n(2n) and m=(n+1)p-1, by maximising in f*, we get f*(k)(2n)m and so if we use (2) and the definition of J,

Jq1ef*(1)+...+qnnenf*(n)c p

since p>n, c being a constant independante of p. If we choose a large enough p the facctorial being more powerful than the power, the two inequality on J contradict each other.
And there we go, end of poof.
Obviously, we get a bit the feeling that it would work for any constant, but in truth, it's (1) and (2) that decides everything !

  is transcendental

Always by the absurd, suppose now that is algebraic and so =i is also.
Let d be the degree of the polynomial where is the solution. Since C is algebraic closed, this polynomial has d roots and denote 1=, 2,...,d all it's roots.
Then consider the minimal polynomial of , that is the smallest polynomial (non reductable) where is a root and whose coefficient are corprime, denote L it's dominant coefficientson, that is the term of the highest power.
Knowing, as our good old Euler proved, that exp(i)+1=0, we can write :

(1+exp(i1))(1+exp(i2))...(1+exp(id))=0

If we expand this last expression, we get the sum of 2d termes ex, where x is a set of values :

x=a11+a22+...+addai=0 ou 1

Suppose that n of those values x are null and denote them ß1,...,ßn.
hence the sum of the 2d termes is written :

q+exp(ß1)+...+exp(ßn)=0(4)

with q=2d-n.
As we did for e, the method is now to find two contradicting inequalities for J=I(ß1)+...+I(ßn), but this time with f(x)=Lnpxp-1(x-ß1)p...(x-ßn) p, p still representing a large prime number.
By using (1) and (4) in the definition of J, we get :

with as always m=(n+1)p-1.
If we take a closer look at the sum with k as an indices,we can see that, all the roots are playing the same role as in f(x), this sum is a symetric polynomial in 1,...,Lßn, that is it is invarient by permutation of those numbers.
But the theorem of algebra on symetric polynomials tell us that this kind of polynomial can be witten under the form of a pollynomial with coeeficien of the equation where 1,...,Lßn are roots. Hence this polynomial of coefficient is an integer, whose sum is on k as well.
Then the same reasoning as for e is applied. We have f (j)k)=0 when j<p, hence this sum of integers is divisible by p!. By calculating from the expression of f, we notice that it is also the case for the rational f (j)(0) if j#p-1. If j=p-1, we get :

f (p-1)(0)=(p-1)!(-L)np1...ßn) p

which is divisible by (p-1)! but not by p! for large enough p.
And so for p>q, we have J(p-1)!
But here again, the relation (2) give us the upper bound :

Jß1exp(ß1) f*(ß1)+...+ßnexp(ßn) f*(ßn)c p

with c being a constant independant of p.
But, Oh! the two inequality are incompatible for large enough p, hence i is transcendantal, so is transcendantal !

We don't really know what just happened, but the result is here!

The original proofs by Hermite and Lindemann are available in le Petit Archimède , but it's even longer and complicated, in fact, it's does not bare any resemblance of what we did just now...


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