# Can we make do with a poor man's analytic continuation of the RZ function?

Just as an observation of what I saw in an informative video about the analytical continuation of the Riemann Zeta function, is it possible to define the continuation for complex numbers $s\in\mathbb{C}$ with $\Re(s)<1$ as simply $f(\zeta(f(s))$ where $f$ is a function that reflects its argument across the point $1+0i$ or across the line crossing 1 on the real axis?

There may be a way to actually formuate $f$. I would translate, then rotate $\pi$ radians, then translate back.

Ignoring reflections and the function $f$, is there a way to take $s\in\mathbb{C}$ with $\Re(s)<1$, transform it over to the other side where the traditional $\zeta$ function defined, evaluate it there, then transform it back?

There just seems to me to be some symmetry between where it's traditionally defined and where it's continued. I'm guessing the answer to my question is a big fat "No", because the formula for the analytic continuation looks rather complicated, but it does appear to be written in terms of the original $\zeta$ function.

• Riemann zeta function functional equation Commented Apr 12, 2017 at 3:05
• I'm not sure about the history of the phrase, but it is "make do" let me try to find a link. grammarist.com/usage/make-do-make-due Commented Apr 12, 2017 at 3:09
• After looking at the video again, it may be deceiving. It looks like the image of the continued zeta function is symmetric about the real part equal one half line, but this may not at all be the case. In fact, I should just try to prove this for myself with the functional equation when I get the chance. Commented Apr 12, 2017 at 6:09
• ...I'm not quite sure what you're asking, but I'm pretty sure the answer is "no." The zeta function's not radially symmetric. The alternating zeta function will give you a value for real part greater than zero, and the rest of the values can be obtained from the reflection formula. Commented Apr 12, 2017 at 6:15
• You should just look at the formula for the analytic continuation. The functional equation states that the xi function (which is related to, but different from the zeta function) has reflective symmetry about $\Im(z) = 1$. Commented Apr 12, 2017 at 7:43

You might want to look at the functional equation.

But before that, you should look at the simplest analytic continuation :

Define inductively $$a_0(n,s) =n^{-s}, \qquad a_{k+1}(n,s) = a_{k}(n,s)-a_k(n+1,s)$$ so that $$a_1(n,s) = n^{-s}-(n+1)^{-s}, \quad a_2(n,s) = n^{-s}-2(n+1)^{-s}+(n+2)^{-s}, \\ a_k(n,s) = \sum_{m=0}^{k} {k \choose m} (-1)^m (n+m)^{-s}$$

Then by summation by parts, using that $b_n = (-1)^{n+1} \implies b_n-b_{n+1} = 2b_n$ and $a_k(n,s) = \mathcal{O}(n^{-s-k})$ : $$\begin{array}{lll}\zeta(s) &=& \sum_{n=1}^\infty a_0(n,s)& \text{converges for } Re(s) > 1\\ \eta(s) &= & \sum_{n=1}^\infty (-1)^{n+1} a_1(n,s) & \text{converges for } Re(s) > 0\\ 2\eta(s) &= &\sum_{n=1}^\infty (-1)^{n+1} a_2(n,s)& \text{converges for } Re(s) > -1\\ & & &\vdots \\ 2^k\eta(s) &=& \sum_{n=1}^\infty (-1)^{n+1} a_{k+1}(n,s) & \text{converges for } Re(s) > -k\end{array}$$

• Thanks. Analytic continuation may be too difficult for me to understand. I know it has to do with defining a function where it is not yet defined, and that there is only one way it can be done so that a certain property of the function is preserved that has to do with the derivative. In any case, I have learned that the zeta function is symmetric about the critical line and outside the critical strip. Is it symmetric about the critical line and inside the critical strip? Commented Apr 13, 2017 at 19:57
• @SpencerParkin Let $E(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$ then $F(s) = E(1/2+s)$ is even ($F(s) = F(-s)$). We often look at the $\Xi$ function which is the same with the two poles removed, so it is analytic on the whole complex plane Commented Apr 13, 2017 at 20:08
• Is there an explicit rather than functional version of the continuation of the zeta function? Commented Apr 13, 2017 at 21:52
• So it looks like $F$ is radially symmetric about the origin. Commented Apr 13, 2017 at 22:00
• @SpencerParkin An analytic function is of the form $g(s) = \sum_{n=0}^\infty c_n s^n$, and it is even iff $\forall n, c_{2n+1} = 0$ so that $g(s) = \sum_{n=0}^\infty c_{2n} s^{2n}$ Commented Apr 13, 2017 at 22:32