So, a few days ago I found a proof for the irrationality of π2 and I thought I would share it with you all. The proof was originally done by Adrien-Marie Legendre in the 7th edition of his book elements de geometrie (1808). The version of this proof that I’ll be posting is more modern, and was completed in 1946 by Carl Siegel and published in his book: Transcendental Numbers. In case you don’t want to read it or the math gets too harry to read (I’ve been there with this proof), the proof is one by contradiction, where if we assume that π2 is a rational number and sub it in rational form into an equation with integer coefficients, we would expect an integer value to come out, but instead get number between 0 and 1 which is obviously not an integer.This proof is pretty long, so here we go!
Where A(x) and B(x) are polynomials of degree n, such that for x≈0 the ratio becomes close to exact.
Now, assume we can also write this equation
Since both A(x) and B(x) are of degree n, we can write:
By the direct multiplication of B(x) with the power series of ex we get
Suppose we want R(x) to start with the 2n+1 term, that is let us assume the first 2n+1 coefficients of R(x) vanish. This will allow R(x) to vanish as fast as x2n+1, which is faster than x itself.
We now have the three equations
Notice that we have 2n+1 equations, but 2n+2 coefficients so despite the restriction of R(x) we can still have nontrivial solutions. Now, we begin our exploration of differential operators. We let
Observe that the condition for R(x) to approach zero when x approaches zero is vital in that
So taking R(x) to zero for R(0) will allow the operators to commute. We now have to find a way to extend the definition for the inverse operator for all n as opposed to the singular definition we have now. Just to get a feel for the inverse operator for other values of n, let’s see how the n=2 case looks.
This integral has a nice geometric interpenetration. With t on the y-axis and s on the x-axis, the region we are integrating is under the line s=t, meaning we can make the substitution:
Which changes the integral to:
Which is again one dimensional. You can do the induction proof on your own, but this is already going to be too long, so I’ll just give the generalization for all n.
We can use this to investigate integrals we’ll come across later. Now, we must investigate functions in the form:
We can see that taking the first derivative gives
Taking the second derivative begins to show a pattern for nth derivatives, as:
Again, I’ll skip the induction and just say that
Back to our formula for R(x),
We can say this because A(x) was assumed to be degree n, and therefore disappears when differentiated n+1 times. From before, we can get a formula for the term containing B(x) by using the formula we just derived and setting λ=1.
Which means we can immediately write
Since R(x) starts with the 2n+1 term, we can say that:
Plus all of the other terms of degree n+1 and higher, where r1 is the first coefficient of R(x). Because B(x) was assumed to be a polynomial of degree n,
Will have to be degree n as well, as one term will contain 1B(x), and the others will all contain some constant times some derivative of B(x) which will be a lower degree polynomial. If we Say that
Then we may only keep the leading term of R(x) in our expansion. If we let e^x take on its power expansion again, we see the only term that need be considered again is .So,
We can use a similar method to derive A(x) (multiply through the original expression for R(x) by e^(-x) and then do the work with derivatives we did here) which works out to:
Now, recall that
Using our solution for B(x) that we just derived, we can see that
Letting u=(s/x), du=1/x, we see that
If we let t=1-u and dt=-du then,
If we change the dummy variable of integration in the line
To t, add it to the previous expression and then divide by 2, we see that:
Changing the form a bit, we get
We can now sub in a value for x to make further simplifications on this integral, let x=iπ.
Some quick simplification yields:
Now, we’re not evaluating this. I just want you to note that depending on the parity of n, the expression for , we can get either a positive or negative number, but not zero. Therefore, . Now, we have to do a bit of manipulation with the equation that we had for R(x) in the beginning.
Immediately it is clear that
Remember, our original first alternate expression for R(x),
Altering it to fit the previous expression gives
For t=-s and ds=-dt,
again for u=x-t, du=-dt
Which makes it clear that
Now we can get two sets of equations for A(x) and B(x)
Subtracting the second from the first give
This can be true for all x if and only if the brackets vanish identically, that is:
These are equivalent statements, and they can be brought together by again setting .
Since A(x) is a polynomial in x, we can expand it from here and see where this fact leaves us.
Where the final term in A(-x) depends on the parity of n. Adding these expressions together as before gives:
where the final term is either of degree n or n-1. If we let [m] be the integer part of a number m, and change the variable from x^2 to u, we can see:
Now, here comes the end, suppose that π2 is rational, then with x=iπ and u=-π, it follows that
Since all of the coefficients a_i were assumed to be integers, then
where j is some integer, positive or negative. Since we know that
Finally, from before,
Taking limits, we can see that both the coefficient and the integral approach zero as n approaches infinity, so we can always chose sufficient n such that
Which of course implies
This means that there should be an integer between zero and one, which is an obvious contradiction. This is sufficient to show that π2 is irrational, and it follows that π is irrational
Have a happy pi month :)