# Math Help - asymptotic relation to a function

1. ## asymptotic relation to a function

How do you find an asymptotic relation to a given type of function. I was reading a paper by P. Sebah and X. Gourdon , and in that paper they gave the asymptotic relation

$\ln (\zeta (sn))\sim \frac{1}{2^{sn}}$

also it is mentioned that to get 20 digits, only

$\frac{20}{\log_{10}(4)}\approx 33$

are required. I didn't know how get these results. Can someone help me?

2. ## Re: asymptotic relation to a function

Originally Posted by mathematicaphoenix
How do you find an asymptotic relation to a given type of function. I was reading a paper by P. Sebah and X. Gourdon , and in that paper they gave the asymptotic relation

$\ln (\zeta (s_n))\sim \frac{1}{2^{s_n}}$

also it is mentioned that to get 20 digits, only

$\frac{20}{\log_{10}(4)}\approx 33$

are required. I didn't know how get these results. Can someone help me?
What is $sn$ or $s_n$ ?

CB

3. ## Re: asymptotic relation to a function

Originally Posted by mathematicaphoenix
How do you find an asymptotic relation to a given type of function. I was reading a paper by P. Sebah and X. Gourdon , and in that paper they gave the asymptotic relation

$\ln (\zeta (s_n))\sim \frac{1}{2^{s_n}}$

also it is mentioned that to get 20 digits, only

$\frac{20}{\log_{10}(4)}\approx 33$

are required. I didn't know how get these results. Can someone help me?
Remembering the 'Euler product' ...

$\frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}})$ (1)
... You derive immediately ...

$\ln \zeta(s) = - \sum_{p} \ln (1-\frac{1}{p^{s}}) =$

$= \sum_{p} (\frac{1}{p^{s}} + \frac{1}{2\ p^{2s}} + \frac{1}{3\ p^{3s}} +...)$ (2)

Now observing (2) it is not too difficult to see that…

$\lim_{s \rightarrow + \infty} 2^{s}\ \ln \zeta(s) = 1$ (3)

… and (3) is equivalent to write…

$\ln \zeta(s) \sim \frac{1}{2^{s}}$ (4)

Kind regards

$\chi$ $\sigma$

4. ## Re: asymptotic relation to a function

Originally Posted by CaptainBlack
What is $sn$ or $s_n$ ?

CB
It's $sn$

5. ## Re: asymptotic relation to a function

Originally Posted by chisigma
Remembering the 'Euler product' ...

$\frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}})$ (1)
... You derive immediately ...

$\ln \zeta(s) = - \sum_{p} \ln (1-\frac{1}{p^{s}}) =$

$= \sum_{p} (\frac{1}{p^{s}} + \frac{1}{2\ p^{2s}} + \frac{1}{3\ p^{3s}} +...)$ (2)

Now observing (2) it is not too difficult to see that…

$\lim_{s \rightarrow + \infty} 2^{s}\ \ln \zeta(s) = 1$ (3)

… and (3) is equivalent to write…

$\ln \zeta(s) \sim \frac{1}{2^{s}}$ (4)

Kind regards

$\chi$ $\sigma$
Thanks chisigma. But I am still fuzzy about how you got
$\lim_{s \rightarrow + \infty} 2^{s}\ \ln \zeta(s) = 1$
NOTE: There is a typing mistake in the original question. Correct form is
$\ln \zeta (sn)\sim \frac{1}{2^{sn}}$

Also can you help me on how to get
$\frac{20}{\log_{10}4}$
this is the approximation to ge 20 digits in base 10!!

Thanks

6. ## Re: asymptotic relation to a function

Writing 'term by term' we have...

$2^{s}\ \ln \zeta(s)= - 2^{s}\ \sum_{p} \ln (1-\frac{1}{p^{s}}) =$

$= \frac{2^{s}}{2^{s}} + \frac{2^{s}}{2\ 2^{2 s}} + \frac{2^{s}}{3\ 2^{3 s}} + ...$

$+ \frac{2^{s}}{2^{s}} + \frac{2^{s}}{2\ 3^{2 s}} + \frac{2^{s}}{3\ 3^{3 s}} + ...$

$+...$

$+ \frac{2^{s}}{p^{s}} + \frac{2^{s}}{2\ p^{2 s}} + \frac{2^{s}}{3\ p^{3 s}} + ...$ (1)

... and observing (1) You can see that if n tends to infinity the first term remains 1 and all other terms tend to 0...

How good is the approximation for s 'large enough' given by the expression $\ln \zeta(s) \sim \frac{1}{2^{s}}$ must be verified, if possible, by direct computation...

Kind regards

$\chi$ $\sigma$

7. ## Re: asymptotic relation to a function

Thank you very much. I got it now.