A curiosity about the Riemann Zeta Function....

In summary, a conversation on the Math Help Boards forum discussed the Riemann Zeta Function and its relationship to the Dirichlet Eta Function. The conversation included various integral and series representations of the functions, as well as methods for computing their values at certain points, such as $\zeta(0)$ and $\eta(0)$. Ultimately, the discussion focused on finding a way to demonstrate that $\lim_{s \to 0^{+}} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = \frac{1}{2}$ without relying on the fact that $\zeta(0) = -\frac{1}{2
  • #1
chisigma
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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 $\chi$ $\sigma$
 
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  • #2
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 \(\displaystyle z = 1\).
 
  • #3
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} $
 
  • #4
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 $\chi$ $\sigma$
 
  • #5
@ 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.
 
  • #6
Random Variable said:
@ 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

$\chi$ $\sigma$
 
  • #7
@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
.
 
  • #8
mathbalarka said:
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 \(\displaystyle \lim_{s\to 0^+} \eta(s) = \frac{1}{2}\)
 
  • #9
ZaidAlyafey said:
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 $\displaystyle \eta(0) = \frac{1}{2}$.
 
  • #10
mathbalarka said:
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 $\chi$ $\sigma$
 
  • #11
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
 

FAQ: A curiosity about the Riemann Zeta Function....

1. What is the Riemann Zeta Function?

The Riemann Zeta Function is a mathematical function that was first introduced by German mathematician Bernhard Riemann in 1859. It is defined as ζ(s) = 1 + 1/2^s + 1/3^s + 1/4^s + ... for all complex numbers s with a real part greater than 1.

2. Why is the Riemann Zeta Function important?

The Riemann Zeta Function is important because it has connections to many different areas of mathematics, including number theory, complex analysis, and even physics. It has been studied extensively in attempts to understand the distribution of prime numbers and has applications in cryptography and other fields.

3. What is the Riemann Hypothesis?

The Riemann Hypothesis is one of the most famous unsolved problems in mathematics. It states that all non-trivial zeros of the Riemann Zeta Function lie on the critical line with real part 1/2. This has significant implications for the distribution of prime numbers, but has yet to be proven.

4. How does the Riemann Zeta Function relate to the Prime Number Theorem?

The Prime Number Theorem is a well-known result in number theory that gives an approximation for the number of prime numbers less than a given number. The Riemann Zeta Function is closely related to this theorem through the Euler product formula, which links the values of the Zeta Function at certain points to the distribution of primes.

5. Are there any practical applications of the Riemann Zeta Function?

While the Riemann Zeta Function may seem abstract and theoretical, it does have several practical applications. It is used in cryptography to generate secure random numbers and has also been used in the study of quantum chaos. Additionally, a proof of the Riemann Hypothesis could have significant implications for number theory and cryptography.

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