# Minimum and maximum of a partial Euler product?

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:

• Do you mean to put $s=1$ in your definitions of $m$ and $M$? – Kimball Jul 10 at 3:00
• @Kimball , no that was for a complex $s$ in general but I am most interested in the case $s=1.$ – Antonio Hernandez Maquivar Jul 10 at 3:03
• 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. – Kimball Jul 10 at 5:02
• 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. – Kimball Jul 11 at 2:06
• 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). – runway44 Jul 11 at 2:18

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 $$\Re(s)=\sigma$$.