10
$\begingroup$

Unfortunately, numerical data sugggest that is not possible to show that $\pi$ is irrational with the polynomials below. I've to search for another polynomial... But I've little faith that such polynomial exists. In the end there is a "explanation" why one of the polynomials defined below don't work.

We gonna do something very similar to what Beukers did to prove that $\zeta (2)$ is irrational. Sorry if is there any mistakes below.

Here, $\pi = 4-\frac{4}{3}+\frac{4}{5}+\cdots$

We have the series: $\frac{1}{1+x^2} = 1-x^2+x^4-\cdots$ \begin{align} I_n = 4\int_0^1\frac{x^{2n}}{1+x^2}\:dx &= 4\int_0^1 x^{2n} \sum_{k=0}^{\infty}(-1)^{k}x^{2k} \: dx\\ &=4\int_0^1 \sum_{k=0}^{\infty}(-1)^{k}x^{2k+2n} \: dx\\ &=4\sum_{k=0}^{\infty}(-1)^{k} \frac{1}{2k+2n+1} \end{align}

This shows that that for $n=0$ we have $I_0 = \pi$. For $n>0$ we have a piece of $\pi$ (hehe).

So we choose a polynomial with integers coefficients with all even powers: \begin{align} I_n = \frac{a_n\pi +b_n}{d_{2n}} \end{align} where $a_n,b_n$ are integers and $d_n$ is the less commom multiple of all consecutives natural numbers up to $2n$.

If it's possible to show that $d_{2n}I_n = a_n+b_n\pi \to 0$ as $n$ grows larger, that proves that $\pi$ is irrational.

The shifted legendre polynomial is defined as below:

$P_n(x) = \frac{1}{n!} \frac{d^n}{dx^n}(x^2-x)^n$.

The first $5$ are:

They are nice because it's easy to integrate $n$ times from $0$ to $1$, it makes the integral go to $0$ very fast and it's easy to find a upper bound for it. I first learned about them in irrationality proofs made by Beukers, where he uses these polynomials to show that $\zeta(2)$ and even $\zeta(3)$ are irrational. He also proves that $\pi$ is irrational but uses a different approach to what we are trying here.

The problem here is that these polynomial have odd power too, and that is not allowed here. So I wanted something like this:

$2x^2-1$

$ 6x^4-6x^2+1$

$ 20x^6-30x^4+12x^2-1$

I also created a polynomial of the form: $\frac{1}{(2n)!}\frac{d^{2n}}{dx^{2n}}(x^4-x^2)^n$, that satifies the property of having only even powers and integer coefficients, but this is terrible because it make the integral goes to infinity rather than $0$. Probably because the coefficients are very large since we must differentiate some many times.

I don't think we necessarily need to use a Legendre type polynomial. Maybe a polynomial with these properties mentioned to show that $\pi$ is irrational doesn't even exist. Any help will be appreciated.

Below are defined polynomials, $Q_n(x)$ and $P_n(x)$ with degrees $(2n)$ and $(4n)$ respectively.

$$Q_n(x) = \frac{d^{n}}{dx^{n}} \frac{(x-x^3)^{n}}{n!}$$

$Q_1(x) = 1-3x^2$

$Q_2(x) = 1 - 12 x^2 + 15 x^4$

$Q_3(x) = 1 - 30 x^2 + 105 x^4 - 84 x^6$.

Degree of $Q_n(x) = 2n$. Using the identity,

$$\int_a^b f^{(n)}(x)g(x) \:dx = (-1)^n\int_a^b f(x)g(x)^{(n)} \:dx $$ if $\left. f(x)g(x) \right|_a^b=0$:

$$\int_0^1 Q_n(x)/(1+x^2) = \frac{a_n \pi+b_n}{d_{2n}}.$$

$$ \int_0^1 Q_n(x)/(1+x^2) = (-1)^n \int_0^1 \frac{(x-x^3)^{n}}{n!} \frac{d^{n}}{dx^{n}} \frac{1}{1+x^2} $$

Another polynomial:

$$P_n(x) = \frac{d^{2n-1}}{dx^{2n-1}} \frac{(x-x^3)^{2n+1}}{(2n+1)(2n-1)!} $$

$P_1(x) = x^2 - 5 x^4 + 7 x^6 - 3 x^8$

$P_2(x) = 2 x^2 - 35 x^4 + 168 x^6 - 330 x^8 + 286 x^{10} - 91 x^{12}$

$P_3(x) = 3 x^2 - 126 x^4 + 1386 x^6 - 6435 x^8 + 15015 x^{10} - 18564 x^{12} + 11628 x^{14} - 2907 x^{16} $

$$\int_0^1 P_1(x)/(1+x^2) = 4\pi - \frac{88}{7} \approx 0.005057$$

$$\int_0^1 P_2(x)/(1+x^2) = 228\pi -\frac{70912}{99} \approx -0.000296$$

$$\int_0^1 P_3(x)/(1+x^2) = 3(4672\pi -\frac{6678272}{455} ) \approx 0.0000046480$$

Degree of $ P_n(x) = 4n+4$:

$$\int_0^1 P_n(x)/(1+x^2) = \frac{a_n\pi+b_n}{d_{4n+4}} $$

$$ \int_0^1 P_n(x)/(1+x^2) = (-1)^n \int_0^1 \frac{(x^3-x)^{2n+1}}{(2n+1)(2n-1)!} \frac{d^{2n-1}}{dx^{2n-1}}\frac{1}{1+x^2}. $$

Finding the $n$ th derivative of $1/(1+x^2)$

\begin{align} &\frac{1}{1+x^2} = \frac{i}{2(x+i)} - \frac{i}{2(x-i)}\\ &\frac{d^n}{dx^n} \frac{1}{1+x^2} = (-1)^n\frac{n!\:i}{2(x+i)^{n+1}} - (-1)^n \frac{n!\:i}{2(x-i)^{n+1}}\\ \end{align}

Chosing $Q_n(x)$, which has a degree $2n$:

\begin{align} I_n &=\int_0^1 Q_n(x)/(1+x^2)\\ &= (-1)^n \int_0^1 \frac{(x-x^3)^{n}}{n!} \frac{d^{n}}{dx^{n}} \frac{1}{1+x^2}\\ &= (-1)^n \int_0^1 \frac{(x-x^3)^{n}}{n!}\left( (-1)^n\frac{n!\:i}{2(x+i)^{n+1}}- (-1)^n \frac{n!\:i}{2(x-i)^{n+1}} \right) \end{align}

Multiplying by $i$ and taking the absolute value:

\begin{align} |iI_n| = |I_n| &=\left| (-1)^n \int_0^1 \frac{(x-x^3)^{n}}{n!}\left( (-1)^{n+1}\frac{n!}{2(x+i)^{n+1}} - (-1)^{n+1} \frac{n!}{2(x-i)^{n+1}} \right) \right|\\ &=\int_0^1 \frac{(x-x^3)^{n}}{n!}\left( \frac{n!}{2(x+i)^{n+1}} - \frac{n!}{2(x-i)^{n+1}} \right) \\ &=\int_0^1 {(x-x^3)^{n}}\left( \frac{1}{2(x+i)^{n+1}} - \frac{1}{2(x-i)^{n+1}} \right). \\ \end{align}

Removing the $i$ above: \begin{align} \frac{1}{2(x+i)^{n+1}} - \frac{1}{2(x-i)^{n+1}} =\frac{(x-i)^{n+1} -(x+i)^{n+1}}{2(1+x^2)} \end{align} writing in polar form $x+i = |z|e^{i\theta}:$ \begin{align} \frac{(x-i)^{n+1} -(x+i)^{n+1}}{2(1+x^2)} &= \frac{(e^{-i\theta})^{n+1} - (e^{i\theta})^{n+1}} {|z|^{n+1}}\\ &=\frac{\cosh[ (n+1) \theta ] } {|z|^{n+1}}\\ &=\frac{\cosh[ (n+1) \arctan 1/x ] } {(\sqrt{1+x^2})^{n+1}}\: \text{because $x>0$}. \end{align}

\begin{align} |I_n| &= \int_0^1 {(x-x^3)^{n}} \frac{\cosh[ (n+1) \arctan 1/x ] } {(\sqrt{1+x^2})^{n+1}}\\ &= \int_0^1 \frac{(x-x^3)^{n}} {(\sqrt{1+x^2})^{n+1}} \frac{1}{2} \left( \frac{1}{e^{(n+1)\arctan 1/x}} + e^{(n+1)\arctan 1/x} \right)\\ &= \int_0^1 \left( \frac{(x-x^3)}{\sqrt{1+x^2}} \right)^n \frac{1}{\sqrt{1+x^2}} \left( \frac{1} { {(e^{\arctan 1/x})}^n } \frac{1}{e^{\arctan 1/x}} + {(e^{\arctan 1/x})}^n e^{\arctan 1/x} \right) \\ &= \int_0^1 \left( \frac{x-x^3}{e^{\arctan 1/x}\sqrt{1+x^2}} \right)^n \frac{1}{e^{\arctan 1/x}\sqrt{1+x^2}} + \left( \frac{(x-x^3)(e^{\arctan 1/x})}{\sqrt{1+x^2}} \right)^n \frac{e^{\arctan 1/x}}{\sqrt{1+x^2}} \end{align}

The maximum of

$$\left( \frac{(x-x^3)(e^{\arctan 1/x})}{\sqrt{1+x^2}} \right)^n$$

is $\approx 1.03^n$ (thanks wolfram) which implies that $Q_n(x)$ can't be used for proving the irrationality of $\pi$.

$P_n(x)$ goes to zero and we can do all this work again, but what will happen is that maximum of $P_n(x)$ multiplied by the $LCM$ of ${1,2,...,4n}$ will be greater than $1$, because of the degree $4n$. I'll try to show that, since we have all the tools. Also, I believe that polynomials of the form

$$\frac{d^{an+c_1}}{dx^{an+c_1}} \frac{P(x)^{bn+c_2}}{(an+c_1)!}$$

where both $a,b$ and $c's$ are constants, is that $c's$ are useless, it's only matters the degree of the polynomial and addding or subtracting constants doesn't change that.

Choosing $P_n(x)$, which has degree $4n+4$:

\begin{align} J_n &= \int_0^1 P_n(x)/(1+x^2) \\ &= (-1)^n \int_0^1 \frac{(x-x^3)^{2n+1}}{(2n+1)(2n-1)!} \frac{d^{2n-1}}{dx^{2n-1}}\frac{1}{1+x^2} \\ &= (-1)^n \int_0^1 \frac{(x-x^3)^{2n+1}}{(2n+1)(2n-1)!} \left( (-1)^{2n-1}\frac{(2n-1)!\:i}{2(x+i)^{2n}} - (-1)^{2n-1} \frac{(2n-1)!\:i}{2(x-i)^{2n}} \right)\\ &= (-1)^n \int_0^1 \frac{(x-x^3)^{2n+1}}{(2n+1)(2n-1)!} \left( -\frac{(2n-1)!\:i}{2(x+i)^{2n}} + \frac{(2n-1)!\:i}{2(x-i)^{2n}} \right) \end{align}

Doing the same as before: \begin{align} |J_n| &= \int_0^1 \frac{(x-x^3)^{2n+1}}{(2n+1)(2n-1)!} \left( \frac{(2n-1)!}{2(x+i)^{2n}} - \frac{(2n-1)!}{2(x-i)^{2n}} \right)\\ &=\int_0^1 \frac{ (x-x^3)^{2n+1} }{ 2n+1 } \left( \frac{1}{2(x+i)^{2n}} - \frac{1}{2(x-i)^{2n}} \right)\\ &=\int_0^1 \frac{ (x-x^3)^{2n+1} }{ 2n+1 } \frac{\cosh(2n\arctan 1/x)}{(\sqrt{1+x^2})^{2n}}\\ &=\int_0^1 \frac{ (x-x^3)^{2n+1} }{ (2n+1)(\sqrt{1+x^2})^{2n} } \frac{1}{2} \left( \frac{1}{e^{(2n)\arctan 1/x}} + e^{(2n)\arctan 1/x} \right)\\ &=\int_0^1 \frac{ (x-x^3)^{2n+1}\sqrt{1+x^2} } { (2n+1)(\sqrt{1+x^2})^{2n+1} } \frac{1}{2} \left( \frac{ e^{\arctan 1/x} }{ e^{(2n+1)\arctan 1/x} } + \frac{ e^{(2n+1)\arctan 1/x} }{ e^{\arctan 1/x} } \right)\\ &= 1/2 \int_0^1 \frac{ e^{\arctan 1/x} \sqrt{1+x^2} }{2n+1} \left( \frac{x-x^3} { e^{\arctan 1/x} \sqrt{1+x^2} } \right)^{2n+1} + \frac{ \sqrt{1+x^2} } {(2n+1)e^{\arctan 1/x}} \left( \frac{ (x-x^3)e^{\arctan 1/x} } {\sqrt{1+x^2}} \right)^{2n+1} \end{align}

I was wrong about why this polynomial would fail, i thought it was bacause of the $LCM(1,2,...,4n)$ would be greater than $1$ when multiplied with the maximum of the integral, but what happened is that we have the exactly same function that appeared using $Q_n(x)$. I think there is no polynomial with such properties that can be used to prove that $\pi \notin \mathbb{Q}$ using this approach.

If anyone have any polynomial in mind...

Well, I think that's all folks.

$\endgroup$
  • $\begingroup$ "We gonna do something very similar to what Beukers did". But you could just do the same. Then $\zeta(2)=\pi^2/6$ is irrational. Now assume hat $\pi=a/b$ is rational. $\endgroup$ – Dietrich Burde Apr 14 '19 at 18:11
  • 7
    $\begingroup$ I know, but I just want to find another proof of the irrationality of $\pi$. $\endgroup$ – Pinteco Apr 14 '19 at 22:05
  • 2
    $\begingroup$ To turn this into a proof of $\pi\not\in\mathbb{Q}$, you need bounds for $a_n\pi+b_n$ and $d_{4n+4}$. $\endgroup$ – Jack D'Aurizio May 6 '19 at 17:40
  • $\begingroup$ @JackD'Aurizio $d_n<e^{An}$ for some $A>1$, and I'm working on the estimation for the integral but the $i$ brings a little trouble. I'm more worried about the polynomial (that might not even exist), but we shall see. $\endgroup$ – Pinteco May 8 '19 at 4:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.