# Root objects and the simplest possible analytic continuation of the Riemann zeta function.

The equation I am trying to solve is:

$$\lim\limits_{k \rightarrow 3} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right)=0 \tag{1}$$

The simplest possible analytic continuation of the Riemann zeta function is: $$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right) \tag{2}$$ $$\mbox{ which appears to be true for }\Re(s)>0$$

So therefore I substituted all $$s$$ with $$x$$ except one of them like this: $$\lim\limits_{k \rightarrow 3} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^x}+ \frac{1}{k^{x - 1} \cdot (s - 1)}\right)=0 \tag{3}$$

Very crude rational approximations of logarithms are:

$$\log(1) = 0$$
$$\log(2) \approx 7/10$$
$$\log(3) \approx 11/10$$
(That is the level of precision I could afford computationally in this case.)

Notice that: $$\frac{1}{n^x}=\frac{1}{e^{x\log(n)}} \tag{4}$$
and substitute $$\log(n)$$ with the rational approximations for logarithms above.

Solving $$(3)$$ in Mathematica we can then write:

Clear[x, s];
Reduce[1/(E^Round[N[Log[1]], 10^-1])^x +
1/(E^Round[N[Log[2]], 10^-1])^x +
1/(E^Round[N[Log[3]], 10^-1])^x +
1/(E^Round[N[Log[3]], 10^-1])^(x - 1)/(s - 1) == 0, x]


This gives 11 Root objects subject to conditions. Picking the first Root object that Mathematica gives, we have:

x == 10 (2 I \[Pi] C[1] +
Log[Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &,
1]])


Latexifying it does not help much, but the changes I would do are to replace $$x$$ with $$s$$ and skip the term $$2 i \pi c_1$$ since I have understood that $$c_1$$ is an integer that can be zero.

So the equation that needs to be solved is:

s == 10 (Log[
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]])


Dividing by 10:

s/10 == Log[
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]]


Applying the exponential function we would have:

Exp[s/10] ==
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]


Now this is not solvable in Mathematica so we need a truncated series expansion of $$\exp(s)$$

$$\exp(s/10) \approx 1+s/10+\frac{(s/10)^2}{2}+\frac{(s/10)^3}{6}$$

$$1+\frac{s}{10}+\frac{1}{2} \left(\frac{s}{10}\right)^2+\frac{1}{6} \left(\frac{s}{10}\right)^3=\text{Root}\left[\text{\#1}^{11} (s-1)+\text{\#1}^4 (s-1)+s+e^{11/10}-1\&,1\right] \tag{5}$$

and this Mathematica can solve:

Reduce[(1 + s/10 + (s/10)^2/2 + (s/10)^3/6) ==
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1], s]


giving the first Root object starting as:

(-1 + Root[-1088391168000000000000000000000000000000000 +
362797056000000000000000000000000000000000 E^(11/10) +
544195584000000000000000000000000000000000 #1 +
295679600640000000000000000000000000000000 #1^2 + ...


Is this at all true or is it just overly complicated?

I mean we could use the same minimal analytic continuation of the zeta function and the crude rational approximations of the logarithms and only do the series expansion:

Clear[x, s];
Series[1/(E^Round[N[Log[1]], 10^-1])^s +
1/(E^Round[N[Log[2]], 10^-1])^s + 1/(E^Round[N[Log[3]], 10^-1])^s +
1/(E^Round[N[Log[3]], 10^-1])^(s - 1)/(s - 1), {s, 0, 5}]


This then would give:

$$\left(3-e^{11/10}\right)+\frac{1}{10} \left(-18+e^{11/10}\right) s+\left(\frac{17}{20}-\frac{101 e^{11/10}}{200}\right) s^2+\left(-\frac{279}{1000}-\frac{1699 e^{11/10}}{6000}\right) s^3+\left(\frac{8521}{120000}-\frac{82601 e^{11/10}}{240000}\right) s^4+\left(-\frac{29643}{2000000}-\frac{3968999 e^{11/10}}{12000000}\right) s^5+O[s]^6$$ and so on...

• Your first line doesn't make sense. Did you mean finding the zeros of $F(s)=\sum_{n=1}^3 n^{-s}+ 3^{1-s}/(s-1)$ ? If so the first step is to find $a,b$ such that for $\Re(s) \not \in [-a,b]$ then $F(s) \ne 0$ – reuns Mar 24 at 13:30

For $$k=3$$:

$$\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$$

For $$s=0$$:

$$1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}=0$$

Note for $$s=0$$, $$\sum\limits_{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=0$$ for all values of $$k$$.

For $$F(s)=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$$, the contour plots of $$\Re(F(s))=0$$ (blue curve) and $$\Im(F(s))=0$$ (orange curve) in Figure (1) below suggest the location of the complex roots of $$F(s)$$ are in a vertical strip centered near the origin. In Figure (1) below, the horizontal axis is the real axis and the vertical axis is the imaginary axis.

Figure (1): Contour Plots of $$\Re(F(s))=0$$ and $$\Im(F(s))=0$$ in blue and orange respectively

For $$G(s)=\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}$$, the contour plots of $$\Re(G(s))=0$$ (blue curve) and $$\Im(G(s))=0$$ (orange curve) in Figures (2) and (3) below illustrate how four of the zeros of $$G(s)$$ move closer to the first four non-trivial zeta zeros in the upper-half plane (red discrete portion of the plots) as the evaluation limit is increased from $$k=3$$ in Figure (2) to $$k=100$$ in Figure (3).

Figure (2): Contour Plots of $$\Re(G(s))=0$$ and $$\Im(G(s))=0$$ in blue and orange respectively with $$k=3$$

Figure (3): Contour Plots of $$\Re(G(s))=0$$ and $$\Im(G(s))=0$$ in blue and orange respectively with $$k=100$$

The following four figures illustrate how the four zeros of $$G(s)$$ associated with the first four non-trivial zeta zeros in the upper-half plane ($$\rho_1$$ to $$\rho_4$$) migrate from their initial values at $$k=3$$ to their values at $$k=100$$ in steps of $$\Delta_k=1$$.

Figure (4): Evaluation of $$G(s)$$ zero associated with $$\rho_4$$ from $$k=3$$ to $$k=100$$

Figure (5): Evaluation of $$G(s)$$ zero associated with $$\rho_3$$ from $$k=3$$ to $$k=100$$

Figure (6): Evaluation of $$G(s)$$ zero associated with $$\rho_2$$ from $$k=3$$ to $$k=100$$

Figure (7): Evaluation of $$G(s)$$ zero associated with $$\rho_1$$ from $$k=3$$ to $$k=100$$

Another fairly simple formula I've found useful is the following (see Riemann zeta function - Dirichlet series and Chapter 2 of Part II of "Theory of Functions" by Konrad Knopp):

$$\zeta(s)=\frac{1}{s-1}\sum_{k=1}^\infty\left(\frac{k}{(k+1)^s}-\frac{k-s}{k^s}\right),\quad \Re(s)>0$$

• The proof of $\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^k \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right)$ takes 1 line, the proof of your's takes 4 – reuns Jul 2 at 21:28