1. Just a thought that crossed my mind after reading about the Riemann hypothesis.

Riemann expanded the domain of the Zeta function towards all complex numbers except for the pole at 1. Then, after calculating the zero's for this function, he landed on the hypothesis about the non-trivial zero's all residing on the line with real part = 1/2.

These non-trivial zero's are all of the shape 1/2 +/- y i. Since they come in pairs, these imaginary points could be zero's of a real kwadratic function of the shape a^2 X + b X + c with (b^2 - 4ac) < 0. Injecting the non-trivial zero's of the shape 1/2 +/- y i would induce a set of functions of the shape:

X^2 - X + y^2 + 1/4

This of course doesn't reveal any better pattern than the complex non-trivial zeros, but I wondered whether this could imply that a real function that produces all the non-trivial zeros exists?  2.

3. What do you mean by "a function which produces zeros"?  4. Originally Posted by serpicojr
What do you mean by "a function which produces zeros"?
I think I didn't phrase it correctly. What I'm after is wether the complex values of the shape 1/2 +/- y i that generate a zero when plugged into the Zeta function, completely define the Zeta function (and the landscape it produces). Is it possible that another function than the Zeta function could produce the same complex values for their zero's?  5. Yes. Multiply the zeta function by any nonvanishing function, for example, the exponential function.  6. mmm, sounds like the weirerstrauss product theorem is needed here (do you know about it - its one of the cooler complex analysis results) from which you can derive the Hadamard product of the zeta function which is what you are asking about.  7. Originally Posted by river_rat
mmm, sounds like the weirerstrauss product theorem is needed here (do you know about it - its one of the cooler complex analysis results) from which you can derive the Hadamard product of the zeta function which is what you are asking about.
Thanks for your hint and apologies for a slow response. I wasn't familiar with the Weierstrass product theorem but looked it up on Wiki and it indeed is very much related to my question.

"The Hadamard product is a representation for the Riemann zeta function as a product over its nontrivial zeros. (...) Hadamard used the Weierstrass product theorem to derive this result."
In mathematics, the Weierstrass factorization theorem in complex analysis, named after Karl Weierstrass, asserts that entire functions can be represented by a product involving their zeroes. In addition, every sequence tending to infinity has an associated entire function with zeroes at precisely the points of that sequence.

A second form extended to meromorphic functions allows one to consider a given meromorphic function as a product of three factors: the function's poles, zeroes, and an associated non-zero holomorphic function.  8. Originally Posted by river_rat
mmm, sounds like the weirerstrauss product theorem is needed here (do you know about it - its one of the cooler complex analysis results) from which you can derive the Hadamard product of the zeta function which is what you are asking about.
Since we're on the subject and you refer to 'cool complex analysis results', I'd be grateful for your view on the following.

I wondered what would happen if we extend the zeta function and instead of letting the terms run from 1 to infinity now allowing it to run from -infinity to infinity. If you feed it with even natural numbers > 1 the outcome just doubles e.g. Zeta(2) becomes pi^2 / 3, etc. When you feed it with odd natural numbers > 1 it always gives you zero (we couldn't find any 'beauty' in these Zeta(odd) numbers anyway, best known result is that Zeta(3) has been proven irrational in the late 70's...). Natural numbers below 1 make the series diverge.

But we also know that Riemann managed to extend the Zeta domain towards the full complex plane by linking Zeta (-s) = 'a complicated looking formula' * Zeta (s+1) or what I also found: Zeta (1-s) = 'a still complicated looking formula' * Zeta (s).

Here's my question. Is this formula still valid for the extended 'zeta term reach' from -infinity to infinity? What happens to the location/pattern of the non-trival (and/or trivial) zero's?

P.S.1
I've tried to work this in Excel by just adding two Zeta Functions together; one is the original and the other one running from -1 to -n. Since Excel is not very friendly with complex faculties (needed for the 'complicated looking formula'), I've only managed to code the non-complex results from -100000 to +100000 (pretty straight forward).

P.S.2
A quick check with the Google calculator shows a non-symmetrical pattern between f.i. Zeta(0,5 + i) for:

term n=2 => 1 / (2^(0,5 + i)) = 0,543934044 - 0,451813851 i
term n=-2 => 1 / ((-2)^(0,5 + i)) = -10,4552855 - 12,5870105 i   9. The function you propose would be equal to:

(1+e^isπ)zeta(s)

Actually, there's an ambiguity in the definition of (-n)^s for positive integers n, so I just went with (-1)^s = (e^πi)^s = e^isπ, and then I'm assuming this is extended by multiplicativity to all negative integers. So you really wouldn't get anything new.

Studying the Euler product:

zeta(s) = ∏(1-p^-s)^-1

(where this is a product over primes) it kind of looks like the 1+e^isπ term might belong, or perhaps even something like (1-e^isπ)^-1. And, if we didn't know better, we might propose that this somehow "completes" the zeta function (so that it knows about primes and -1, which could be a placeholder for the "prime at infinity").

But we do know better: the appropriate factor to multiply zeta by is π^(-s/2) Gamma(s/2). Calling the resulted completed zeta function xi(s), i.e.:

xi(s) = zeta(s) π^(-s/2) Gamma(s/2)

we get the functional equation in a very simple form:

xi(1-s) = xi(s)

This representation is very natural, as it comes from a Mellin transform of a (modified) theta function, and this is how we get a lot of L-functions in general.

Do you know the classical theory of the zeta function and Dirichlet L-functions? (I mean like what's contained in Davenport's Multiplicative Number Theory, or even what's in Tom Apostol's Intro to Analytic Number Theory.) If not, I'd suggest learning it before coming up with your own modifications of the zeta function.[/img]  10. Also, let me say that the Weierstrass product in and of itself doesn't tell us where the zeros are. Rather, it's a key ingredient in relating sums over primes to sums over zeros of the zeta function. Now the more we know about the locations of zeros (and the more this information looks like RH), the better the asymptotic formulas (e.g., for the prime number theorem) we can derive from this. And, in fact, the error estimates for said formulas are in a sense equivalent to knowledge about the location of the zeros--if we were able to come up with an estimate like:

psi(x) = x + O(x^(1/2)log x)

we'd have RH (where psi(x) = ∑ log p for prime powers p^n ≤ x). But this isn't the way to go about solving RH--it's more likely that we'll derive this estimate from RH than that we'll derive RH from this estimate.  11. Originally Posted by serpicojr
(...)
But we do know better: the appropriate factor to multiply zeta by is π^(-s/2) Gamma(s/2). Calling the resulted completed zeta function xi(s), i.e.:

xi(s) = zeta(s) π^(-s/2) Gamma(s/2)

we get the functional equation in a very simple form:

xi(1-s) = xi(s)

This representation is very natural, as it comes from a Mellin transform of a (modified) theta function, and this is how we get a lot of L-functions in general.

Do you know the classical theory of the zeta function and Dirichlet L-functions? (I mean like what's contained in Davenport's Multiplicative Number Theory, or even what's in Tom Apostol's Intro to Analytic Number Theory.) If not, I'd suggest learning it before coming up with your own modifications of the zeta function.[/img]
Many thanks for taking the trouble to explain, Serpicojr. I'm sincerely impressed by your knowledge of the subject and will first study more before making any new post on this.   12. The idea keeps circling around in my head. One more try... Zeta(s) = sum(1/(n^s), n = 1...infinity)

Suppose we make a similar series for only the negative integers:

Zota(s) = sum(1/(n^s), n = -infinity...-1)

Then I postulate that Zota(s) + Zeta(s) = xi(s) (as already derived by Serpicojr in the previous post for the 'completed' Zeta function with terms running from -infinity...infinity)

Now, to extend the xi(s) function into the entire complex plane (except for pole at 1), the appropriate factor to multiply Zeta by is:

π^(-s/2) Gamma(s/2)

So we get:

xi(s) = zeta(s) π^(-s/2) Gamma(s/2)

and we get the functional equation:

xi(1-s) = xi(s)

Given this natural result for xi(s) and it being the sum of Zota(s) + Zeta(s), I've two questions:

1. Can we derive the appropriate factor for Zota(s) by smartly manipulating (subtracting?) the xi(s) and zeta(s) series?

2. Will the non-trivial zero's of Zeta(s) be perfectly complemented by equivalent 'peaks' in Zota(s) to sustain xi(1-s) = xi(s) ?

Please be gentle...   13. I think you may be confused by must post about "completing" the zeta function. We do not have that:

zeta(s)+zota(s) = xi(s)

One way to see this is that, if the above were true, then since:

zeta(s)+zota(s) = (1+e<sup>isπ</sup>)zeta(s)

xi(s) = zeta(s)π<sup>-s/2</sup>Gamma(s/2)

We would have, dividing through by zeta(s):

1+e<sup>isπ</sup> = π<sup>-s/2</sup>Gamma(s/2)

This is impossible, as the function on the left has zeros, while the function on the right has none (similarly, the function on the right has poles, while the function on the left doesn't).

I think my original intent was to suggest to you that zeta(s)+zota(s), unfortunately, doesn't tell us much about zeta(s).[/img]  14. I think my original intent was to suggest to you that zeta(s)+zota(s), unfortunately, doesn't tell us much about zeta(s)
And that was exactly what I was after... I had hoped that by extending the terms of the Zeta function towards the negative integers, we might find something 'natural'. Then Zeta + Zota would behave like: 'ugly1' + 'ugly2' = 'beautiful'. For the non-complex domain Zota(odd) + Zeta(odd) = 0 and Zota(even) + Zeta(even) = 2 * Zeta (or Zota). I hoped to find some similar beauty in the complex domain.

Since we're still struggling with the proof that all non-trivial zero's of the Zeta have a real part 1/2, I hoped we might be able to shift focus to studying the Zota complement instead and suddenly find a disguised pattern or proof...

Another broken dream.

Thanks for your answer anyway, Serpicojr   15. Originally Posted by accountabled
For the non-complex domain Zota(odd) + Zeta(odd) = 0 and Zota(even) + Zeta(even) = 2 * Zeta (or Zota). I hoped to find some similar beauty in the complex domain.
I mean, there is an expression relating zeta to zeta+zota:

zeta(s)+zota(s)=(1+e<sup>isπ</sup>)zeta(s)

And this implies the statement about odd and even values of s. This is pretty!  16. Originally Posted by serpicojr Originally Posted by accountabled
For the non-complex domain Zota(odd) + Zeta(odd) = 0 and Zota(even) + Zeta(even) = 2 * Zeta (or Zota). I hoped to find some similar beauty in the complex domain.
I mean, there is an expression relating zeta to zeta+zota:

zeta(s)+zota(s)=(1+e<sup>isπ</sup>)zeta(s)

And this implies the statement about odd and even values of s. This is pretty!
That's pretty indeed! Would this also be true for f.i. s= 1/2 (since you'd get complex numbers in Zota)?

If I do the maths correctly:

zota(s)/zeta(s) = e<sup>isπ</sup>

It should be possible to derive a similar prime product formula for Zota(s) as we have for Zeta(s) and see how that relates to e<sup>isπ</sup> i.e.:

"zota(s) as prime product" / ∏(1-p^-s)^-1 = e<sup>isπ</sup>  17. Yes, this is valid for any complex value of s. However, note that you wouldn't be able to use these expressions to calculate zeta(s) for s < 1, at least not without estimation and some serious justification...

You got the correct value for the ratio of zeta and zota. Since zota is just a function times zeta, it kind of has a product over primes... It's just the product for zeta times this new function (e<sup>isπ</sup>).  18. Still couldn't rest it, so one more try...

Zeta(s) = ∏(1-p<sup>-s</sup>)<sup>-1</sup>

Zota(s) = ∏(-1-(-p)<sup>-s</sup>)<sup>-1</sup>

Since for Re > 1 we showed that zota(s)/zeta(s) = e<sup>isπ</sup> we can now write:

∏(-1-(-p)<sup>-s</sup>)<sup>-1</sup>
---------------- = e<sup>isπ</sup>
∏(1-p<sup>-s</sup>)<sup>-1</sup>

Unfortunately we're not allowed to enter the domain Re(s) < 1 (I was already dreaming of making s = 1/iπ which would produce a formula that yielded e from the primes, but fearing Serpicojr's honed mathematical sword, I decided to quickly drop that idea... 8) ).

There's also not so much to explore on the Re(s) > 1 side, since I found that if Re(s) > 1, "the sieved right-hand side approaches 1, which follows immediately from the convergence of the Dirichlet series for ζ(s)." I assume the same it true for Zota(s).

The division between Zota and Zeta results in 1 for any even integer s and -1 for any odd integer s. When I plug in s=1.5 or s=2.5 it becomes -i and i respectively.

Any other ideas?  19. The product for zota is not valid... A necessary condition for an infinite product to converge is for the terms to converge to 1. This will never be the case for your product for zota. Even formally, an infinite product of -1's is not defined, and so we wouldn't be able to express your product for zota as a Dirichlet series.

zota(s) does not approach a limit as Re(s) goes to infinity. Since zeta(s) approaches 1 and zota(s)=zeta(s)e<sup>isπ</sup>, we have that zota(s) is asymptotic to e<sup>isπ</sup> as s goes to infinity; for s real, this basically traces out the unit circle in the complex plane.

As for a pretty formula for e in terms of the primes, I wouldn't be surprised if something like that existed, but I don't know anything off the top of my head.

What book are you working out of? I'm assuming the quote is from a book.  20. Originally Posted by serpicojr
The product for zota is not valid... A necessary condition for an infinite product to converge is for the terms to converge to 1. This will never be the case for your product for zota. Even formally, an infinite product of -1's is not defined, and so we wouldn't be able to express your product for zota as a Dirichlet series.

zota(s) does not approach a limit as Re(s) goes to infinity. Since zeta(s) approaches 1 and zota(s)=zeta(s)e<sup>isπ</sup>, we have that zota(s) is asymptotic to e<sup>isπ</sup> as s goes to infinity; for s real, this basically traces out the unit circle in the complex plane.

What book are you working out of? I'm assuming the quote is from a book.
Serpicojr,

I wasn't actually working from a book, but found that sentence on this website when I tried to derive the Euler product for Zota.

http://en.wikipedia.org/wiki/Proof_o..._zeta_function

I do see now that Zota(s) is an amibiguous series. I could have know that a negative base when raised to a power can be a tricky animal. I always liked the following trick:

-1 = 3rd Root(-1) = (-1)<sup>1/3</sup> = (-1)<sup>2/6</sup> = 6th Root((-1)<sup>2</sup>) = 1

I now also have grasped your point about for s being real, it traces out the unit circle in the complex plane.

This means that if I want to discover something new, I need to feed the equation with complex numbers with Re(s) > 1. But even if I would manage to extend Zota or Zeta into the full complex domain (with Re(s) also < 1), we know that using (zota(s) * xxx) /(zeta(s) * yyy) would induce a division by zero for some s of the shape 1/2+yi... This brings us back to the "0*0 =?"-thread and maybe I should therefore try the Riemann spheres or one of the Wheels... :wink:

Anyway,

Many thanks for teaching me a lot of new insights!   21. My pleasure! Let me suggest looking at Tom Apostol's Introduction to Analytic Number Theory. It talks about the zeta function in addition to similar functions called Dirichlet L-functions. He proves Dirichlet's Theorem on Primes in Arithmetic Progressions and the Prime Number Theorem in there--these are classic applications of Dirichlet L-functions and the zeta function.  Bookmarks
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