0
$\begingroup$

Question: If if $n\in\mathbb{N}$ and $s\in \mathbb{C},$ say $s=\sigma+t\sqrt{-1},$ then Dirichlet Beta function is defined to be $$ \beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}; $$ which for Re(s)>1, has the Euler product representation over prime numbers $p:$ $$ L(s)=\prod_{p>2}{ 1\above 1.5pt 1-(-1)^{(p-1)/2}p^{-s}}; $$ equiv., $$ \prod_{p \equiv 1 \pmod 4}\frac{1}{1-p^{-s}} \prod_{p \equiv 3 \pmod 4}\frac{1}{1+p^{-s}}. $$ If $x$ is a positive number I write $L_x(s)$ where for the above product is restricted to those odd prime numbers that are less than or equal to $x.$ I set $$ m(s)=\operatorname*{min}_{x} L_x(s) $$ and $$ M(s)=\operatorname*{max}_{x} L_x(s) $$ Can we show that $m(s)$ and $M(s)$ exist and compute (estimate) there values ?


Since the product converges I am certain $L_x(s)$ is bounded and so has a maximum ? For example I think I have correctly that $$ \lim_{x \to \infty}L_x(1)=\frac{\pi}{4} $$ This is just the Euler product representation of $\frac{\pi}{4}$ and so $L_{x}(1)$ is the partial Euler product up to some magnitude $x.$ I would like to make the claim that over all $x$ that $\operatorname*{min}_{x} L_x(1)=\frac{3}{4} $ and that $\operatorname*{max}_{x} L_x(1)=\frac{15}{16}.$ That the minimum might be equal to $\frac{3}{4}$ might straightforward but even there I am not certain.


Similar but not quite the same question:

  1. Partial Euler Products
  2. how to prove that the partial euler product of primes less than or equal x is bounded below by log x
$\endgroup$
  • $\begingroup$ Do you mean to put $s=1$ in your definitions of $m$ and $M$? $\endgroup$ – Kimball Jul 10 at 3:00
  • $\begingroup$ @Kimball , no that was for a complex $s$ in general but I am most interested in the case $s=1.$ $\endgroup$ – Antonio Hernandez Maquivar Jul 10 at 3:03
  • $\begingroup$ In any case, because of the way the terms alternate, it's reasonable to expect the min and max occur when you have a minimal number of terms in the product. E.g., for $s=1$, the max is 1 (empty product) and we guess min is 1(1+1/3). One can probably prove this with a mix of computations and error bounds. $\endgroup$ – Kimball Jul 10 at 5:02
  • 1
    $\begingroup$ I think your question is understandable, though it will be a little clearer to write $m(s)$ and $M(s)$ as they depend on $s$. It might help you to understand what is going on to compute some partial Euler products and see how the partial products oscillate. The main issue to prove that the first 2 partial products are max and mins is that the primes don't perfectly alternate between 1 and 3 mod 4. But I checked for products with $p < 10000$ and it seems true. $\endgroup$ – Kimball Jul 11 at 2:06
  • 1
    $\begingroup$ The $p$s in the numerator of the double product are a typo. Also presumably one wants $s$ to be real for max and min to make sense (unless one invokes absolute values). $\endgroup$ – runway44 Jul 11 at 2:18
0
$\begingroup$

With $$\mathfrak{M}(s) = \sup_x\ \Re(\sum_{p \le x} (-1)^{(p-1)/2} p^{-s}), \qquad \Re(s) > 1$$

the $\Bbb{Q}$-linear independence of the $\log p$ implies $\lim_{k \to \infty} \mathfrak{M}(\sigma+it_k) = \sum_p p^{-\sigma}$ for a sequence $t_k\to \infty$ where each $t_k$ is the imaginary part of a zero of $\beta(s)-\zeta_2(\sigma)$. Moreover $\beta(s+it_k)$ converges uniformly to $\zeta(s)$ on $\Re(s) \ge 1+\delta$ and the same holds for the partial sums of their Euler product :

For $\epsilon> 0$ small enough, if $\mathfrak{M}(\sigma+it)$ is $\epsilon$-close to $ \sum_p p^{-\sigma}$ then $\beta(\sigma+it)$ is $\epsilon$-close to $\zeta_2(\sigma)=(1-2^{-\sigma})\zeta(\sigma)$ and $\beta'(\sigma+it)$ is $\epsilon$-close to $\zeta_2'(\sigma)$, and since $\beta''(s)$ is bounded it means there is a zero of $\beta(s)-\zeta_2(\sigma)$ at $s = \sigma+it+\frac{ \zeta_2(\sigma )- \beta(\sigma+it)}{\zeta_2'(\sigma)} + O(\epsilon^2) = \sigma+it + O(\epsilon)$.

In other words the rate of convergence of $\sup_{t \le T} \mathfrak{M}(\sigma+it) \to \sum_p p^{-\sigma}$ depends on the density of zeros of $\beta(s)-\zeta_2(\sigma)$ around $\Re(s)=\sigma$.

The latter is research level, there are some methods and results about it in the literature.

$\endgroup$

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.