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Peter

#1 h May 25, 2007, 12:24 AM • 3 Y

Y by UpvoteFarm, Adventure10, Mango247

Show that $\pi$ is irrational.

This post has been edited 1 time. Last edited by Peter, Aug 30, 2007, 3:09 PM

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#2 h May 25, 2007, 12:24 AM • 3 Y

Y by UpvoteFarm, Adventure10, Mango247

Here's an outline of a proof, with three questions :

We assume by sake of contradiction that $\pi = \frac pq,$ with $p$ and $q$ positive coprime integers. We then consider $I_n = \int_0^{p/q} \frac {x^n(p-qx)^n}{n!} \sin x dx, n \in \mathbb{N}$ and show that $I_n$ satisfies
a) $I_n$ is an integer;
b) $I_n > 0;$
c) $\lim_{n \to +\infty} I_n = 0.$

Historical remark(s) : Lambert has showed in 1761 that $\pi$ is irrational, Legendre has proved in 1794 that $\pi^2$ is irrational, Lindemann has proved in 1882 that $\pi$ is transcendental.

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3615 posts

Peter

#3 h May 25, 2007, 12:24 AM • 3 Y

Y by UpvoteFarm, Adventure10, Mango247

Please can you prove your claim?

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#4 h May 25, 2007, 12:24 AM • 3 Y

Y by UpvoteFarm, Adventure10, Mango247

Okay, let's go.

(With the notations of my previous post,) at first, for $0 \leq x \leq \frac{p}{q},$ we have $0 \leq x(p-qx) = \frac p{2q}(p-\frac p{2q}q) = \frac{p^{2}}{4q},$ and so,
$0 \leq I_{n}\leq \frac 1{n!}\int_{0}^{p/q}\left(\frac{p^{2}}{4q}\right)^{n}dx = \frac{\pi}{n!}\left( \frac{p^{2}}{4q}\right)^{n}.$
Therefore, the sequence $(I_{n})$ converges and $\lim_{n \to+\infty}I_{n}= 0.$
Then, since for $x \in [0, \pi],$ we have $x^{n}(p-qx)^{n}\sin x \geq 0,$ for a given positive integer $n,$ we have
$I_{n}= \frac 1{n!}\int_{0}^{\pi}x^{n}(p-qx)^{x}\sin x dx \geq \frac 1{n!}\int_{\pi/4}^{3 \pi/4}x^{n}(p-qx)^{n}\sin x dx \geq \frac 1{n!}(3\pi/4-\pi/4)\left(\frac p{4q}(p-\frac p{4q}q) \right)^{n}\frac 1{\sqrt{2}}= \frac{\pi}{2n!}\left(\frac{3p^{2}}{16q}\right)^{n}> 0.$
Thus, $\forall n \in \mathbb{N}, I_{n}> 0.$
Let us finally check that, for any positive integer $n, I_{n}$ is an integer.
Let $P_{n}= x^{n}(p-qx)^{n}.$ $P_{n}$ is a polynomial with degree $2n$ and $0$ and $\frac{p}{q}$ are roots of order $n$ of $P_{n}$ and so, for $0 \leq k \leq n,$ roots of order $n-k$ of $P_{n}^{(k)}.$ Especially, $P_{n}^{(k)}(0)$ and $P_{n}^{(k)}(\frac{p}{q})$ are, for $0 \leq k \leq n,$ integers. Similarly, since $\text{deg}P_{n}= 2n,$ for $k \geq 2n+1, P_{n}^{(k)}\geq 0$ and especially, $P_{n}^{(k)}(0)$ and $P_{n}^{(k)}(\frac{p}{q})$ are, for $k \geq 2n+1,$ integers.
Let $k$ be an integer such that $n \leq k \leq 2n.$
$\frac 1{n!}x^{n}(p-qx)^{n}= \frac 1{n!}^{n}\sum_{i=0}^{n}\binom ni p^{n-i}(-1)^{i}q^{i}x^{i}= \sum_{i=0}^{n}\frac{\binom ni}{n!}p^{n-i}(-1)^{i}q^{i}x^{n+i}= \sum_{k=n}^{2n}\frac{\binom n{k-n}}{n!}p^{2n-k}(-1)^{k-n}q^{k-n}x^{k}.$
We then know that
$P_{n}^{(k)}(0) = k! \times (\text{coefficient of }x^{k}) = (-1)^{k-n}\frac{k!}{n!}\binom n{k-n}p^{2n-k}q^{k-n},$
which shows that $P_{n}^{(k)}(0)$ is an integer. Then, since $P_{n}(\frac{p}{q}-x) = P_{n}(x),$ we can deduce that $P_{n}^{(k)}(\frac{p}{q})$ is an integer.
We have thus showed that for any natural integer $k, P_{n}^{(k)}$ and $P_{n}^{(k)}(\frac{p}{q})$ are integers.
Let us then show that $I_{n}$ is an integer.
A first integration by parts yields : $I_{n}= [-P_{n}(x) \cos x]_{0}^{p/q}+\int_{0}^{p/q}P'_{n}(x) \cos x dx.$
$\cos$ takes integer values in $0$ and $\frac{p}{q}= \pi,$ like $P_{n}.$ Hence,
$I_{n}\in \mathbb{Z}\iff \int_{0}^{p/q}P'_{n}(x) \cos x dx \in \mathbb{Z}.$
A second integration by parts yields : $\int_{0}^{p/q}P'_{n}(x) \cos x dx = [P'_{n}(x) \sin x]_{0}^{p/q}-\int_{0}^{p/q}P''_{n}(x) \sin x dx.$
$\sin$ takes integer values in $0$ and $\frac{p}{q}= \pi,$ like $P'_{n}$ and
$I_{n}\in \mathbb{Z}\iff \int_{0}^{p/q}P''_{n}(x) \sin x dx \in \mathbb{Z}.$
Repeating the integration by parts and since $\sin$ and $\cos$ take integer values in $0$ and $\pi$ like the successive derivatives of $P_{n},$ we deduce that :
$I_{n}\in \mathbb{Z}\iff \int_{0}^{p/q}P_{n}^{(2n)}(x) \sin x dx \in \mathbb{Z}.$
But,
$\int_{0}^{p/q}P_{n}^{(2n)}(x) \sin x dx = \int_{0}^{p/q}\frac 1{n!}(-q)^{n}(2n)! \sin x dx = 2(-q)^{n}(2n)(2n-1)\cdots (n+1) \in \mathbb{Z}.$
So for any natural integer, $I_{n}$ is an integer, positive from what we've done above. We deduce that for any natural $n, I_{n}\geq 1.$ This last thing contradicts the fact that the sequence $(I_{n})$ converges towards $0.$
Therefore, $\pi$ is irrational.

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#5 h Aug 30, 2007, 2:35 PM • 2 Y

Y by Adventure10, UpvoteFarm

Peter wrote:

Show that $\pi$ is irrational.

The reference for this problem "CRUX, Problem A240, Mohammed Aassila" is not correct.

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