31
$\begingroup$

Rather than relying on the consequences of Schanuel's conjecture, I set about using the same ideas Apery had used to construct integer arguments converging fast enough to show $\zeta(3)$ is irrational in a form Beukers had introduced. I'm sure someone out there can crack what I have so far.

I will be using the following facts:

Theorem 1: Suppose the complex-valued function $$\begin{align} f{(z)} = \begin{cases} -\left(\frac{1}{e}(1-z)^{1-\frac{1}{z}} \right)^q, & z\neq 0 \\ -1, & z = 0 \end{cases} \end{align}$$ has a power series with positive radius of convergence of the form $$f(z) = \sum_{n=0}^\infty b_n(q) z^n$$ Then $$b_n(q) = -\frac{q}{n} \sum_{k=1}^n \frac{b_{n-k}(q)}{k+1}, \quad b_0(q) = -1$$

Note that $b_n(q)$ is a polynomial of degree $n$.

Theorem 2: Let $m,q \in \mathbb{Z}$ and $m+q+1 \geq 0$; then $$\int_0^1 x^m \sin(\pi q x) \left(x^x (1-x)^{1-x}\right)^q \ dx = (-1)^{q+1} \pi e^q b_{m+q+1}(q)$$

The above can be shown by applying contour integration and residue theorem to the above function.

Theorem 3: For $n \in \mathbb{N} \cup \{0\}$, $\mathbb{P}$ be the set of primes, and let $$(n+1)!_\mathbb{P} = \prod_{p \in \mathbb{P}} p ^{\sum_{k\geq 0} \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor}$$ Then, for integer $q$, $(n+1)!_\mathbb{P} \cdot b_n(q)$ is an integer for $n \geq 0$.

This factorial like function is borrowed from Manjul Bhargava's work on the general factorial function.

Theorem 4 Let $n \in \mathbb{N} \cup \{0\}$; then $$(n+1)!_\mathbb{P}\sim e^{n(C-\gamma+o(1))}n^n$$ where $C = \sum_{p \in \mathbb{P}} \frac{\ln p }{(p-1)^2}$ and $\gamma$ is the Euler-Mascheroni constant.

If we let $P_n(x)$ be a polynomial of degree $n$ with integer coefficients and let $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) \left( x^x (1-x)^{1-x}\right)^q -1\right) \ dx$$

We have the following inequality, in the form of Dirichlet's irrationality criterion,

$$0 < \left|C_n \pi e^q - D_n \right| = \left|(n+q+2)!_\mathbb{P} (n+1) I_n \right|$$

where $C_n, D_n \in \mathbb{Z}$. Of course, we can apply Theorem 4, and have something more familiar to work with.

Question: Can we construct a polynomial $P_n(x)$ such that, for large $n$, $$\left|(n+q+2)!_\mathbb{P} (n+1) I_n \right| \to 0 \text{?}$$

If there does exist one, then, for $q \geq -2, q \neq 0$, the number $\pi e^q$ is irrational. Letting $q = 1, -1$, and the result follows.

I've been at this problem for some time, with no further progress. Frankly, I don't know what to do at all. If it helps, I've considered the shifted Legendre polynomials, as Beukers had done, though to no avail.

Most of what I've seen regarding the nature of constructing a polynomial is that it belongs to the family of orthogonal polynomials.

God bless.

$\endgroup$
3
$\begingroup$

This isn't really an answer as much as it is an "expanded" comment.

Consider, for integer $a$, $$P_n(x) = \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n = \sum_{m=0}^n \binom{n}{m} \binom{n+m}{m} (-ax)^m$$

Given $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) (x^x(1-x)^{1-x})^q - 1\right) \ dx$$ We have $$I_n \leq \int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx$$ where $a_q = \max_{x\in (0,1)}\{(x^x(1-x)^{1-x})^q\}$. Furthermore, we have $$\left|\int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx \right|= \left|\int_0^1 \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n\left( \sin(\pi q x) a_q - 1\right) \ dx\right|$$

$$= \left|\frac{1}{a^{n+1}n!} \int_{(0,1)\cup(1,a)} \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$ $$\leq \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{1}{n!a^{n+1}}\int_1^a \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$

Let $$S_n = \int_1^a \left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right)\frac{d^n}{d x^n} x^n (1-x)^n \ dx$$

So that we have $$ = \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right|$$

Now, observing the bound in question, applying Theorem 4, and letting $A = C - \gamma + o(1)$, we have $$\left| (n+q+2)!_\mathbb{P} (n+1) I_n \right|<\left|e^{An} e^{q+1} n^n \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}n^{q+1}(n+1)\left(\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right) \right| $$

If we ignore the $S_n$ term, we have that

$$\left| e^{q+1} \frac{n^n}{e^n n!} \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}\frac{n^{q+1}(n+1)}{b^n}\left(\frac{e^{A+1}\pi bq}{4a^2}\right)^n \frac{a_q}{a} \right|$$

where $b > 1 $. If we consider $a$ such that $ a^2 > \frac{e^{A+1}\pi bq}{4}$, and applying Stirling's approximation to the left-most term (-ish), for large $n$, then the whole expression above tends to $0$. Now, it is left to consider the $S_n$ term, though I have a bad feeling about it. :/

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.