Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Mathematica knows that:

$$\log (n)=\lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s)$$

Kind of tautological starting with logarithms, but I would like to know better why this limit works:

$${\Large \log (n)=\lim_{s\to 1+\frac{2 i \pi k}{\log (n)}} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s-i \Im(s))}$$

for $k$ an integer.

Solving it symbolically for some integer $k$ while leaving $n$ as a variable Mathematica says it is equal to zero. But setting $n$ to any value I get $\log (n)$.

Mathematica:

Table[Limit[Zeta[s - I*Im[s]]* Total[{1 - 1/(2)^(s - 1)}], 
  s -> (1 + (2*I \[Pi]*k/Log[2]))], {k, 1, 12}]
N[%, 12]

and:

Table[Table[
  Limit[Zeta[s - I*Im[s]]*Total[{1 - 1/(n)^(s - 1)}], 
   s -> (1 + (2*I \[Pi]*k/Log[n]))], {k, 1, 6}], {n, 1, 12}]
 N[%, 12]
share|improve this question
    
Could you post your code for the second problem ? Whatever I do, I always get $0$. –  Claude Leibovici Mar 1 at 13:00
    
I will add some Mathematica code. –  Mats Granvik Mar 1 at 13:03
    
This is really strange, indeed ! –  Claude Leibovici Mar 1 at 13:09

1 Answer 1

up vote 1 down vote accepted

I have no idea what Mathematica is up to, but if we write $s = \sigma + \frac{2i\pi k}{\log n} + it$, we have

$$\begin{align} \frac{1}{n^{s-1}} &= \exp \left((1-s)\log n\right)\\ &= \exp \left(\left(1-\sigma - \frac{2i\pi k}{\log n} - it\right)\log n\right)\\ &= \exp \left((1-\sigma-it)\log n\right)\cdot e^{-2i\pi k}\\ &= e^{(1-\sigma-it)\log n}\\ &= 1 - (\sigma-1+it)\log n + O\left((\sigma-1+it)^2\right). \end{align}$$

With

$$\zeta(s-i\Im s) = \zeta(\sigma) = \frac{1}{\sigma-1} + O(1),$$

we therefore have

$$\begin{align} \left(1 - \frac{1}{n^{s-1}}\right)\zeta(s-i\Im s) &= \left((\sigma-1+it)\log n + O\left((\sigma-1+it)^2\right)\right)\left(\frac{1}{\sigma-1} + O(1)\right)\\ &= \log n + \frac{it\log n}{\sigma-1} + O\left(\frac{t^2}{\sigma-1}\right) + o(1), \end{align}$$

and the limit does not generally exist. The limit exists and equals $\log n$ if $s$ approaches $1+ \frac{2i\pi k}{\log n}$ in such a way that

$$\lim \frac{t}{\sigma-1} = 0,$$

but along the ray $t = c(\sigma-1)$, the limit is $(1+ic)\log n$, and if $t$ approaches $0$ slower than $\sigma-1$, the expression is unbounded.

Apparently, for the nonexisiting limit, Mathematica just returned $0$ in the general case, but what made it produce the limit $\log n$ when given a specific $n$, I cannot guess.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.