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A curiosity about the Riemann Zeta Function...

chisigma

Well-known member
Feb 13, 2012
1,704
Recently some interesting material about the Riemann Zeta Function appeared on MHB and I also contributed in the post...


http://mathhelpboards.com/challenge-questions-puzzles-28/simplifying-quotient-7235.html#post33008


... where has been obtained the expression...

$\displaystyle \zeta (s) = \frac{1}{1-2^{1 - s}}\ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}\ (1)$

... that allows the effective computation of $\zeta(*)$ in the half plane where $\text{Re} (s) > 0$. It is well known that $\zeta (0) = - \frac{1}{2}$ so that is...

$\displaystyle \lim_{s \rightarrow 0 +} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = \frac{1}{2}\ (2)$

What I'm interested about is how to demonstrate (2) independently from (1), i.e. whitout the preliminary knowledge that $\zeta (0) = - \frac{1}{2}$. I spent many hours in attempts but without success (Emo)...


Kind regards


$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Actually, the fact that \(\displaystyle \lim_{s \to 0^+} \eta(s)\) is 1/2 has pretty much nothing to do with \(\displaystyle \zeta(0) = -1/2\). The former can be proved by using the fact that Grandi's series (i.e., \(\displaystyle \eta(0)\)) is basically the power series expansion of \(\displaystyle \frac1{1+z}\) at \(\displaystyle z = 1\).
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
There is the integral representation $\eta(s) = 2^{s-1} \displaystyle \int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt $ which is valid for all complex values of $s$.

See here.


Initially I had the sum of the residues expressed in terms of the Dirichlet eta function. Then I used that relationship to express it in terms of the Riemann zeta function.


So $ \displaystyle \eta(0) = \frac{1}{2} \int_{0}^{\infty} \frac{1}{\cosh \left(\frac{\pi t}{2} \right) }\ dt = \frac{1}{2}(1) = \frac{1}{2} $
 

chisigma

Well-known member
Feb 13, 2012
1,704
All right boys!... I realize that the series is known as 'Eta Function'...

$\displaystyle \eta(s) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{s}}\ (1)$

... and (1) is valid if $\text{Re} (s) > 0$. The series in (1) is 'alternate sign' and almost immediately after my post I remembered that some year ago I used the formula...

$\displaystyle \zeta (s) = \frac{\eta(s)}{1 - 2^{1-s}}\ (2)$

... for the computation of the function $\zeta(*)$ along the 'critical line' $\displaystyle s = \frac{1}{2} + i\ t$. A direct approach using (1) for $\text{Re} (s) = \frac{1}{2}$ has the drawback of very slow convergence, so that I used the so called 'Euler's Transformation' that consists in what follows: if we have an alternate signs series, then is...


$\displaystyle \sum_{n=0}^{\infty} (-1)^{n}\ a_{n} = \sum_{n=0}^{\infty} (-1)^{n}\ \frac{\Delta^{n} a_{0}}{2^{n+1}}\ (3)$

... where...

$\displaystyle \Delta^{n} a_{0} = \sum_{k=0}^{n} (-1)^{k}\ \binom{n}{k}\ a_{n-k}\ (4)$

Very well!... now we remember the binomial expansion...


$\displaystyle (1 - x)^{n} = \sum_{k=0}^{n} (-1)^{k}\ \binom{n}{k}\ x^{k}\ (5)$

... that for x=1 is 0 for any value on n... with the only exception of n=0 for which is [no matter if someone doesn't like it (Tongueout)...] $0^{0}=1$. The consequence is that for s=0 we have...


$\displaystyle \eta (0) = \sum_{n=0}^{\infty} \frac{(-1)^{n}\ 0^{n}} {2^{n+1}}= \frac{1}{2}\ (6)$

Is all that true?... probably yes... even if with the Riemann Zeta Function nothing is 100 % sure...


Kind regards


$\chi$ $\sigma$
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
@ chisigma

I'm curious what was unsatisfactory about my post.

I offered an integral representation of the Dirichlet eta function that defines the function everywhere (as opposed to a series representation that only defines the function for certain values), and a way to derive that representation that doesn't require any knowledge of the Riemann zeta function.

Finding $\eta(0)$ then became equivalent to evaluating a simple definite integral.
 

chisigma

Well-known member
Feb 13, 2012
1,704
@ chisigma

I'm curious what was unsatisfactory about my post.

I offered an integral representation of the Dirichlet eta function that defines the function everywhere (as opposed to a series representation that only defines the function for certain values), and a way to derive that representation that doesn't require any knowledge of the Riemann zeta function.

Finding $\eta(0)$ then became equivalent to evaluating a simple definite integral.
If the scope is to compute the $\eta(*)$ evaluating a 'simple definite integral' the integral You propose [extended from 0 to $\infty$...] isn't quite 'simple'... much more 'pratical' may be is the following integral...

$\displaystyle \eta(s) = \frac{1}{\Gamma(s)}\ \int_{0}^{1} \int_{0}^{1} \frac{\{- \ln (x\ y)\}^{s-2}}{1 + x\ y}\ d x\ dy\ (1)$

The scope of my post was only to answer to my curiosity...

Kind regards

$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
@chisigma, you seem to be mixing up summability theorems and ideas together.

The way you showed the Grandi's series converges up to 1/2 through Euler acceleration stands, but not in the Cauchy sense. And it has absolutely nothing to do with the fact that zeta at 0 is -1/2 or eta at 0 is 1/2. The formula RV showed is indeed correct and a doable analytic continuation in the complex plane.

You must see the difference between ACs and Summability methods.

Balarka
.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Actually, the fact that \(\displaystyle \lim_{s \to 0^+} \eta(s)\) is 1/2 has pretty much nothing to do with \(\displaystyle \zeta(0) = -1/2\) ...
If \(\displaystyle \zeta(s) (1-2^{1-s})=\eta(s)\)

Does not that already prove \(\displaystyle \lim_{s\to 0^+} \eta(s) = \frac{1}{2}\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
If \(\displaystyle \zeta(s) (1-2^{1-s})=\eta(s)\)

Does not that already prove \(\displaystyle \lim_{s\to 0^+} \eta(s) = \frac{1}{2}\)
I think he meant that you can't infer from that relationship that $ \displaystyle \lim_{s \to 0^{+}} \sum_{s=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = \frac{1}{2}$.

What you can infer is that $\displaystyle \eta(0) = \frac{1}{2}$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Actually, the fact that \(\displaystyle \lim_{s \to 0^+} \eta(s)\) is 1/2 has pretty much nothing to do with \(\displaystyle \zeta(0) = -1/2\). The former can be proved by using the fact that Grandi's series (i.e., \(\displaystyle \eta(0)\)) is basically the power series expansion of \(\displaystyle \frac1{1+z}\) at \(\displaystyle z = 1\).
Adding curiosity to curiosity I can say that the Italian mathematician and philosoph Guido Grandi was born in the year 1671 in Cremona, the same town in North Italy where I was born 280 years later!... the most famous of his 'inventions' is probably the following series...

$\displaystyle S = \sum_{n=0}^{\infty} (-1)^{n} = 1 - 1 + 1 - 1 + ...\ (1)$

Grandi himself wrote that his scope was to do an example of alternating signs series that doesn't converge. Successive [very poor...] 'mathematicians' however tried to 'demonstrate' the convergence of (1) in very fun way like these...


a) $\displaystyle S = (1 - 1) + (1 - 1) + ... = 0 + 0 + ... = 0$

b) $\displaystyle S = 1 - (1 - 1) - (1 - 1) - ... = 1 - 0 - 0 -... = 1$

c) $\displaystyle S = 1 - (1 - 1 + 1 -...) = 1 - S \implies S = \frac{1}{2}$

No comments!:cool:...


Kind regards


$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
I know the derivation of Grandi. Although the result stands (not in Cauchy's world) in the Abelian sense, his derivation doesn't. The best way to converge the divergent Grandi's series is to use Abelian theorems.

Try this, it will help : Divergent series - Wikipedia, the free encyclopedia