Pi is Irrational/Proof 1

Proof
Aiming for a contradiction, suppose that $\pi$ is rational.

Then from Existence of Canonical Form of Rational Number:
 * $\exists a \in \Z, b \in \Z_{>0}: \pi = \dfrac a b$

Let $n \in \Z_{>0}$.

We define the polynomial function:


 * $\forall x \in \R: f \left({x}\right) = \dfrac {x^n \left({a - b x}\right)^n} {n!}$

We differentiate this $2 n$ times, and then we build:


 * $\displaystyle F \left({x}\right) = \sum_{j \mathop = 0}^n \left({-1}\right)^j f^{\left({2 j}\right)} \left({x}\right) = f \left({x}\right) + \cdots + \left({-1}\right)^j f^{\left({2 j}\right)} \left({x}\right) + \cdots + \left({-1}\right)^n f^{(2 n)} \left({x}\right)$

That is, $F \left({x}\right)$ is the alternating sum of $f$ and its first $n$ even derivatives.

First we show that:
 * $(1): \quad F \left({0}\right) = F \left({\pi}\right)$

From the definition of $f \left({x}\right)$, and our supposition that $\pi = \dfrac a b$, we have that:


 * $\forall x \in \R: f \left({x}\right) = b^n \dfrac {x^n \left({\pi - x}\right)^n} {n!} = f \left({\pi - x}\right)$

Using the Chain Rule, we can apply the Principle of Mathematical Induction to show that, for all the above derivatives:


 * $\forall x \in \R: f^{\left({j}\right)} \left({x}\right) = \left({-1}\right)^j f^{\left({j}\right)} \left({\pi - x}\right)$

In particular, we have:


 * $\forall j \in \left\{{1, 2, \ldots, n}\right\}: f^{\left({2 j}\right)} \left({0}\right) = f^{\left({2 j}\right)} \left({\pi}\right)$

From the definition of $F$, it follows that:
 * $F \left({0}\right) = F \left({\pi}\right)$

Next we show that:
 * $(2): \quad F \left({0}\right)$ is an integer.

We use the Binomial Theorem to expand $\left({a - bx}\right)^n$:
 * $\displaystyle \left({a - bx}\right)^n = \sum_{k \mathop = 0}^n \binom n k a^{n - k} \left({-b}\right)^k x^k$

By substituting $j = k + n$, we obtain the following expression for $f$:
 * $\displaystyle f \left({x}\right) = \frac 1 {n!} \sum_{j \mathop = n}^{2 n} \binom n {j - n} a^{2 n - j} \left({-b}\right)^{j - n} x^j$

Note the following:
 * The coefficients of $x^0, x^1, \ldots, x^{n - 1}$ are all zero
 * The degree of the polynomial $f$ is at most $2 n$.

So we have:
 * $\forall j < n: f^{\left({j}\right)} \left({0}\right) = 0$
 * $\forall j > 2 n: f^{\left({j}\right)} \left({0}\right) = 0$

But for $n \le j \le 2 n$, we have:
 * $\displaystyle f^{\left({j}\right)} \left({0}\right) = \frac {j!} {n!} \binom n {j - n} a^{2 n - j} \left({-b}\right)^{j - n}$

Because $j \ge n$, it follows that $\dfrac {j!} {n!}$ is an integer.

So is the binomial coefficient $\dbinom n {j - n}$ by its very nature.

As $a$ and $b$ are both integers, then so are $a^{2 n - j}$ and $\left({-b}\right)^{j - n}$.

So $f^{\left({j}\right)} \left({0}\right)$ is an integer for all $j$, and hence so is $F \left({0}\right)$.

Next we show that:
 * $(3): \quad \displaystyle \dfrac 1 2 \int_0^\pi f \left({x}\right) \sin x \, \mathrm d x = F \left({0}\right)$

As $f \left({x}\right)$ is a polynomial function of degree $n$, it follows that $f^{\left({2n + 2}\right)}$ is the null polynomial.

This means:
 * $F'' + F = f$

Using the Product Rule and the derivatives of sine and cosine, we get:
 * $\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)' = f \left({x}\right) \sin x$

By the Fundamental Theorem of Calculus, this leads us to:
 * $\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \, \mathrm d x = \frac 1 2 \left[{\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)}\right]_{x \mathop = 0}^{x \mathop = \pi}$

From Sine and Cosine are Periodic on Reals, we have that $\sin 0 = \sin \pi = 0$ and $\cos 0 = - \cos \pi = 1$.

So, from $F \left({0}\right) = F \left({\pi}\right)$ (see $(1)$ above), we have:
 * $\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \ \mathrm d x = F \left({0}\right)$

The final step:

On the interval $\left({0 \,.\,.\, \pi}\right)$, we have from Sine and Cosine are Periodic on Reals that $\sin x > 0$.

So from $(2)$ and $(3)$ above, we have that $F \left({0}\right)$ is a positive integer.

Now, we have that:
 * $\left({x - \dfrac \pi 2}\right)^2 = x^2 - \pi x + \left({\dfrac \pi 2}\right)^2$

and so:
 * $x \left({\pi - x}\right) = \left({\dfrac \pi 2}\right)^2 - \left({x - \dfrac \pi 2}\right)^2$

Hence:
 * $\forall x \in \R: x \left({\pi - x}\right) \le \left({\dfrac \pi 2}\right)^2$

Also, from Boundedness of Sine and Cosine, $0 \le \sin x \le 1$ on the interval $\left({0 \,.\,.\, \pi}\right)$.

So, by the definition of $f$:
 * $\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \, \mathrm d x \le \frac {b^n} {n!} \left({\frac \pi 2}\right)^{2 n + 1}$

But this is smaller than $1$ for large $n$, from Power Series over Factorial.

Hence, for these large $n$, we have $F \left({0}\right) < 1$, by $(3)$.

This is impossible for the positive integer $F \left({0}\right)$.

So our assumption that $\pi$ is rational must have been false.