Pi Squared is Irrational/Proof 2

Theorem

Pi squared ($\pi^2$) is irrational.

Proof

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Aiming for a contradiction, suppose $\pi^2$ is rational.

Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.

Let us define a polynomial:

$\ds \map f x = \frac {\paren {1 - x^2}^n} {n!} = \sum_{m \mathop = n}^{2 n} \frac {c_m} {n!} x^m$

for $c_m \in \Z$.

$\map f x = \map f {-x}$ and so we get:

\(\ds \map {f^{\paren k} } x = \paren {-1}^k \map {f^{\paren k} } {-x}\)

\(=\)

\(\ds \begin {cases} \ds \sum_{m \mathop = n}^{2 n} \frac{k!} {n!} \binom m k c_m x^{m - k} & : 0 \le k \le n - 1 \\ \ds \sum_{m \mathop = k}^{2 n} \frac {k!} {n!} \binom m k c_m x^{m - k} & :n \le k \le 2 n \end {cases}\)

\(\ds \map {f^{\paren k} } 1 = \paren {-1}^k \map {f^{\paren k} } {-1}\)

\(=\)

\(\ds \begin {cases} 0 & : 0 \le k \le n - 1 \\ \dfrac {k!} {n!} c_k & : n \le k \le 2 n \end {cases}\)

hence they are all integers.

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Now let:

$\ds A_n = \int_{-1}^1 \map f x \map \cos {\dfrac {\pi x} 2} \rd x = 2 \int_0^1 \map f x \map \cos {\dfrac {\pi x} 2} \rd x$

Repeated integration by Parts gives:

\(\ds A_n\)

\(=\)

\(\ds \frac {2^2} \pi \intlimits {\map {f^{\paren 0} } x \map \sin {\dfrac {\pi x} 2} } 0 1 + \frac {2^3} {\pi^2} \intlimits {\map {f^{\paren 1} } x \map \cos {\dfrac {\pi x} 2} } 0 1 - \cdots \pm \frac {2^{2 n + 2} } {\pi^{2 n + 1} } \intlimits {\map {f^{\paren {2 n} } } x \map \sin {\dfrac {\pi x} 2} } 0 1 \pm \frac {2^{2 n + 2} } {\pi^{2 n + 1} } \int_0^1 \map {f^{\paren {2 n + 1} } } x \map \sin {\dfrac {\pi x} 2} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac {2^2} \pi \intlimits {\map {f^{\paren 0} } x \map \sin {\dfrac {\pi x} 2} } 0 1 - \frac {2^4} {\pi^3} \intlimits {\map {f^{\paren 2} } x \map \sin {\dfrac {\pi x} 2} } 0 1 + \frac {2^6} {\pi^5} \intlimits {\map {f^{\paren 4} } x \map \sin {\dfrac {\pi x} 2} } 0 1 - \cdots + \paren {-1}^n \frac {2^{2 n + 2} } {\pi^{2 n + 1} } \intlimits {\map {f^{\paren {2 n} } } x \map \sin {\dfrac {\pi x} 2} } 0 1\)

\(\ds \)

\(=\)

\(\ds \frac {4} \pi \map {f^{\paren 0} } 1 - \frac {4^2} {\pi^3} \map {f^{\paren 2} } 1 + \frac {4^3} {\pi^5} \map {f^{\paren 4} } 1 - \cdots + \paren {-1}^n \frac {4^{n + 1} } {\pi^{2 n + 1} } \map {f^{\paren {2 n} } } 1\)

\(\ds \leadsto \ \ \)

\(\ds \frac \pi 4 A_n\)

\(=\)

\(\ds \sum_{k \mathop = 0}^n \paren {-\frac 4 {\pi^2} }^k \map {f^{\paren {2 k} } } 1\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^n \paren {-\frac {4 q} p}^k \map {f^{\paren {2 k} } } 1\)

Now let:

$\ds N_n = \frac \pi 4 p^n A_n = \sum_{k \mathop = 0}^n p^{n - k} \paren {-4 q}^k \map {f^{\paren {2 k} } } 1$

Since $\map f x >0$ and $\map \cos {\dfrac {\pi x} 2} > 0$ on the open interval $\openint {-1} 1$:

$A_n > 0$

Hence $N_n$ is a positive integer.

Nevertheless, for $x \in \closedint {-1} 1$:

$0 \le \map f x \le \dfrac 1 {n!}$

and:

$0 \le \map \cos {\dfrac {\pi x} 2} \le 1$

Hence:

$0 < A_n < \dfrac 2 {n!}$

Therefore:

$0 < N_n < \dfrac \pi 2 \dfrac {p^n} {n!}$

From Power over Factorial:

$\ds \lim_{n \mathop \to \infty} \dfrac {p^n} {n!} = 0$

It follows from the Squeeze Theorem that:

$\ds \lim_{n \mathop \to \infty} N_n = 0$

Hence for sufficiently large $n$, $N_n$ is strictly between $0$ and $1$.

This contradicts the supposition that $N_n$ is an integer.

It follows that $\pi^2$ is irrational.

$\blacksquare$