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By the maximum modulus principle, since $\zeta(z)=\sum_{n=1}^{\infty}\frac{1}{n^z}$, with $\Re z>1$, is analytic inside the square Q of vertices $2,3,3+i$ and $2+i$, and continuous on Q, $$\max_{z\in Q}|\zeta(z)|=\max_{z\in \partial Q}|\zeta(z)|.$$

Question. How can we compute the maximum of $|\zeta(z)|$ on the square with vertices $2,3,3+i$ and $2+i$? Can you get this maximum following my outline or with a different way?

Taking Q as $\gamma_1 \cup \gamma_2 \cup\gamma_3 \cup\gamma_4$ with the counterwise orientation, my attempt is compute, for example on $\gamma_1$: $2\leq x\leq 3$ and $y=0$ then $|\zeta(z)|=|\zeta(x+i\cdot 0)|=\left|\sum_{n=1}^\infty \frac{1}{n^x}\right|=\frac{\pi^2}{6}$.

But on different arcs I don't know how compute the maximum, only bounds.

For example on $\gamma_2$: $0\leq y\leq 1$ and $x=3$ then $$|\zeta(z)|=|\zeta(3+iy)|\leq\sum_{n=1}^\infty \frac{\left|e^{-iy\log n}\right|}{n^3}\leq \zeta(3)$$, and using the reverse triangle inequality

$$|\zeta(3+iy)|\geq \left|\left|\sum_{n=1}^\infty \frac{\cos(y\log n)}{n^3}\right|-\left|\sum_{n=1}^\infty \frac{\sin(y\log n)}{n^3}\right|\right|.$$ Also I know that $\zeta(z)$ has a Euler's product $$\prod_{\text{p,prime}}\frac{1}{1-p^{-z}}.$$

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  • $\begingroup$ At least I want to see how to get an approximation of such maximum in a square. Thanks n advance. $\endgroup$ – user243301 Mar 15 '16 at 12:00
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In order to compute the maximum, you only need the bounds you mentioned. For all $z$ with $Re(z)\ge 2$ you have $$ \left| \frac{1}{n^z}\right|=\left| \frac{1}{n^{x+iy}}\right|= \left| \frac{1}{n^x}\right|\left| \frac{1}{n^{iy}}\right|=\left| \frac{1}{n^x}\right|=\frac{1}{n^x}, $$ where $x=Re(z)$, since $|n^{iy}|=|exp(i\ y\log(n))|=1$. Hence, for $z$ with $x=Re(z)\ge 2$, you have $$ |\zeta(z)|=\left|\sum_{n=1}^{\infty}\frac{1}{n^z}\right|\le \sum_{n=1}^{\infty}\left|\frac{1}{n^z}\right| =\sum_{n=1}^{\infty}\frac{1}{n^x}\le \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}=\zeta(2). $$ In particular, this holds for all $z$ in the square or on the boundary, so the maximum value $\frac{\pi^2}{6}$ is attained at $z=2$.

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  • $\begingroup$ Then I was confused, since I thought it was a problem hardest very thanks much for your answer.Your explanation is vey clear, very thanks much. $\endgroup$ – user243301 Mar 16 '16 at 8:40

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