Why π is irrational (Part 2 of 5)

In this post I introduce the function f(x) and look at some of its properties.

Table of Contents
1. Outline
2. A polynomial and its derivatives
3. A new function
4. The integral is an integer (pending)
5. The size of the integral (pending)

I’ll go ahead and be a little blunt. Let’s define f(x) as

Alright, so what is n? That turns out not to matter, as long as it’s a postive integer. We will prove several things about f(x) that don’t depend on n; later on, we’ll see that the contradiction we are looking for appears only when n is “large enough,” but this won’t be a problem.

Note that f(x) is symmetric about the line :

This symmetry will come into play later on.

Some derivatives
We now show that f(i)(0) and f(i)(π) are always integers. We first note that the numerator of f(x) can be written as

a polynomial of degree n through 2n whose coefficients are all integers.

f(i)(0)
Note that the first n-1 derivatives of f(x) have no constant term, so those derivatives are all 0, which is an integer. The same goes for the 2n+1th and all further derivatives–once we take the 2n+1th derivative of a plynomial of degree 2n (such as f(x)), we have 0. So all we need to think about now are the n+1th through 2nth derivatives.

As we noted above, f(x) is a sum of terms of the form

where and ci is an integer. Taking the mth derivative reduces this to

This term only becomes significant when we’ve taken the n+ith derivative. The term then becomes

which is clearly an integer.

f(i)(π)
Note that

since f(x) is symmetric about the line , so

We can repeat the process as many times as we want. Thus, the even derivatives of f(x) are symmetrix about the line . Since, as we have shown, f(i)(0) is always an integer, f(i)(π) is always an integer as well.

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