I am reading an interesting paper One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational by Zudilin. We fix odd numbers $q$ and $r$, $q\geq r+4$ and a tuple $\eta_0,\eta_1,...,\eta_q$ of positive integer parameters satisfying the conditions $\eta_1\leq \eta_2\leq...\leq \eta_q<\eta_0/2$ and $$ \eta_1+\eta_2+...+\eta_q\leq \eta_0\left(\frac{q-r}{2}\right)\tag{1}$$ Define $$F_n:=\frac{1}{(r-1)!}\sum_{t=0}^\infty R_n^{(r-1)}(t)\tag{2}$$ and note that $R_n(t)=O(t^{-2})$. We put $m_j=\max\{\eta_r,\eta_0-2\eta_{r+1},\eta_0-\eta_1-\eta_{r+j}\}$ for $j=1,2,...,q-r$ and define the integer $$\Phi_n:=\prod_{\sqrt{\eta_0 n}<p\leq m_{q-r}n}p^{\varphi(n/p)}$$ where only primes enter the product and $$\varphi(x)=\min_{0\leq y<1}\left(\sum_{j=1}^{r}([y]+[\eta_0x-y]-[y-\eta_j x]-[(\eta_0-\eta_j)x-y]-2[\eta_j x])+\sum_{j=r+1}^{q}([(\eta_0-2\eta_j)x]-[y-\eta_j x]-[(\eta_0-\eta_j)x-y])\right)$$ where [.] denotes the ceiling function. Let $D_N$ denote the lcm of $1,2,...,N$.
Lemma $1$: ($2$) defines a linear form of $\zeta(r+2),\zeta(r+4),...,\zeta(q-2)$ with rational coefficients; moreover, $$ D_{m_1n}^r D_{m_2n... D_{m_{q-r}n}}.\Phi_n^{-1}.F_n\in\mathbb{Z}+\mathbb{Z}\zeta(r+2)+\mathbb{Z}\zeta(r+4)+...+\mathbb{Z}\zeta(q-2) \tag{3}$$ By Prime Number Theorem, $$\lim_{n\to\infty}\frac{\log D_{m_j n}}{n}=m_j,\ \ \ j=1,...,q-r.$$ We introduce the auxiliary function $$f_0(\tau)=r\eta_0\log(\eta_0-\tau)+\sum_{j=1}^{q} (\eta_j\log(\tau-\eta_j)-(\eta_0-\eta_j)\log(\tau-\eta_0+\eta_j)) -2\sum_{j=1}^r \eta_j\log \eta_j+\sum_{j=r+1}^q (\eta_0-2\eta_j)\log(\eta_0-2\eta_j)$$ defined in the $\tau$-plane with the cuts $(-\infty,\eta_0-\eta_1]$ and $[\eta_0,+\infty)$
Lemma $2$: Let $r=3$ and $\tau_0$ be a zero of the polynomial $$(\tau-\eta_0)^r(\tau-\eta_1)...(\tau-\eta_q)-\tau^r(\tau-\eta_0+\eta_1)...(\tau-\eta_0+\eta_q) $$ with Im $\tau_0>0$ and the maximum possible value of Re $\tau_0$. Assume Re $\tau_0<\eta_0$ and Im $f_0(\tau_0)\notin \pi \mathbb{Z}.$ Then $$\overline{\lim}_{n \to \infty} \frac{\log |F_n|}{n}=Re f_0(\tau_0)$$ If the sequence of linear forms on the left side of ($3$) assumes non-zero arbitrarily small values as $n$ increases, then in the case $r=3$ there are irrational numbers among $$ \zeta(5),\zeta(7),...,\zeta(q-4),\zeta(q-2) \tag{4}$$ Therefore the following holds:
Lemma $3$: Suppose that $r=3$ and in the above notation $C_0=-Re f_0(\tau_0)$, $$C_1=rm_1+m_2+...+m_{q-r}-\left(\int_0^1\varphi(x)\frac{\Gamma'(x)}{\Gamma(x)}dx-\int_0^{1/m_{q-r}}\varphi(x)\frac{dx}{x^2}\right)$$ If $C_0>C_1$, then at least of the numbers ($4$) is irrational. The line below Lemma $3$ reads: we put, $r=3,q=13$, $$\eta_0=91,\eta_1=\eta_2=\eta_3=27,\eta_4=29,\eta_5=30,\eta_6=31,...,\eta_{12}=37,\eta_{13}=38.$$ Then $$C_0=227.58019641...,C_1=226.24944266...$$
Question: Is the above result true only for $r=3$ or for other odd values of $r>3$ also? For example can we take $r=5,q=13$ and if we show that $C_0>C_1$ and with suitable choice of $\eta's$, can we say at least one of the four numbers $$ \zeta(7),\zeta(9),\zeta(11),\zeta(13)$$ is irrational? Rephrasing my question, Is Proposition $5$ in Arithmetic of linear forms involving odd zeta values true for $r=5$ and $q=13$?
Any help would be highly appreciated. Thank you!
There isn't a lot of thought that goes into the verification. It's pure computation at this point.
I'll include code that computes $C_0$. Computing $C_1$ is similarly a direct computation (but it requires integrating the digamma function).
Now
term1andterm2are the two polynomials$$ (x - 38) \cdot (x - 37) \cdot (x - 36) \cdot (x - 35) \cdot (x - 34) \cdot (x - 33) \cdot (x - 32) \cdot (x - 31) \cdot (x - 30) \cdot (x - 29) \cdot (x - 91)^{3} \cdot (x - 27)^{3} $$ and $$ (x - 62) \cdot (x - 61) \cdot (x - 60) \cdot (x - 59) \cdot (x - 58) \cdot (x - 57) \cdot (x - 56) \cdot (x - 55) \cdot (x - 54) \cdot (x - 53) \cdot (x - 64)^{3} \cdot x^{3}. $$ (I just had sage output the tex for these polynomials for me).
We continue.
The roots are
Continuing,
The root $\tau_0$ is given by
$$ \tau_0 = 87.4790054197046 + 3.32820691055980i.$$
Finally, we directly compute $f_0$.
I observe that there are some differences in the less significant digits. If I use 200 bits of precision (instead of the default 53 bit double precision implicit above) ((second parenthetical: this can be done by using
f = ComplexField(200)[x](f)instead ofCC[x](f)above)), then I get$$ C_0 \approx 227.58019641270392421956170302923769732274712647908924133109. $$
This agrees with the claimed $C_0$. It is also possible to bound the numerical error due to precision, but that's a separate topic.
I looked at computing $C_1$ again today and give my incomplete code. When I wrote the first version of this answer, I tried to be clever about computing $\varphi(x)$ (which is merely a somewhat complicated step function, and which we can in principle identify what intervals it takes which values).
Today, I'm not clever and I'll give a brute force answer. I actually get something very close to what I had last week (perhaps I'm consistently wrong somewhere, in which case I hope this partial code leads someone to identify my mistake).
To compute $\varphi(x)$, I look on a mesh of $10000$ different $y$ values and identify the minimum.
This is just an approximation. (Actually, I populated a table with precomputed values for numerical integration. I used something equivalent to
brutephito populate the table).For the first integral, we note that $\psi$ is an increasing, differentiable function, and thus
$$ \int_0^1 \varphi(x) d \psi = \int_0^1 \varphi(x) \psi'(x) dx. $$
The correct way to do this would be to break $[0, 1]$ into the intervals where $\varphi(x)$ is constant and explicitly integrate $\psi'(x)$ on each of those intervals (which is just $\psi(x)$). But again, today we use brute force and I use numerical Riemann-Stieltjes integration. (Aside: I also implemented the trapezoidal rule in Riemann integration, and it gave me the same answer).
The second integral is similar. I use the trapezoidal rule for numerical integration, explicitly.
We put this all together.
This approximation says that $C_1 \approx 183$, which is rather far from the claimed value. I note that brute-forcing $\varphi$ in the way I've done it should overestimate the first integral, which should result in subtracting too large of a number, and so any errors from $\varphi$ computation all make the resulting $C_1$ estimate too small.
I would be surprised if the brute approximation to $\varphi$ was enough to cause such a big difference. Nonetheless, I hope that this incomplete solution is helpful.