# B Behavior of the zeta function

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1. Dec 7, 2016

### mustang19

Consider

Z(s)=Sum(1/N^s)

For n=1 to infinity.

Let s=(xi+1/2).

The divisor is then:

N^(xi+1/2)

This is equivalent to

(N^xi)(N^(1/2))

As n increases, the n^1/2 term will more greatly slow the increase of the divisor and accelerate z(s) away from zero. This means that zeroes will occur less often for larger n. The 1/2 term is necessary so that the zeta function z(s) will encounter fewer primes as it iterates.

Is this a valid understanding of z(s)' behavior?

2. Dec 7, 2016

### Simon Bridge

Well, your notation is inconsistent. That is usually a sign of an incomplete understanding.
Zeta Function: $\zeta (s) = \sum_{n=1}^\infty \frac{1}{n^s} : s\in \mathbb{C}$
... you are considering the case where $0<\Re(s)<1$.

There is a lot written on the zeta function, have you tried refining your question from these resources?
What do you mean by "valid" and "behaviour"?

3. Dec 7, 2016

### Staff: Mentor

There is no "zeta function for different n". Z(s) depends on s only.
That doesn't make sense.

4. Dec 7, 2016

### mustang19

Without the 1/2 term z(s) would be relatively smaller for larger n, thus finding more zeroes, which is not how primes behave.

I did not say there was a zeta function for different n. I am talking about n in the summation term.

Valid means true, behavior means the distribution of results.

Last edited: Dec 7, 2016
5. Dec 7, 2016

### Staff: Mentor

Without the 1/2 the zeta function doesn't have any zeros (apart from the trivial ones) - probably. You don't find "more zeros" if there is no zero at all.

6. Dec 7, 2016

### Simon Bridge

... my point was that your description did not make sense, and I needed you to clarify what you meant. So lets recap: by -
"Is this a valid understanding of z(s)' behavior?"
... you are asking if "is [statement] a true understanding of z(s) distribution of results?"
The "statement" being this:
... well, that's still pretty vague.
Don't know what you mean by "more greatly slow the increase in the divisor" - more greatly that what?
I think you need to go through the whole "statement" line by line and try to use less ambiguous terms so we don't have to guess what you are talking about.

As pointed out, z(s) does not depend on n but on s. Therefore, as written, "statement" is not true. You responded with:
OK I get that... when you said z(s) you actually meant "the terms in the sum that leads to z(s)". You didn't actually say that though - hence the misunderstanding.
This goes to the comment on notation in post #2. Try:
$$\zeta(s)=\sum_{n=1}^\infty \zeta_n(s) : \zeta_n(s)=\frac{1}{n^s}$$
... so, back to translating your question, you are actually wanting to ask: "is [statement] a true understanding of the distribution of $\zeta_n(s)$ with $n$?"
... ie. are you asking about how the individual $\zeta_n$ vary in the sum, treating each $\zeta_n$ as a continuation of an ordered series of complex functions?

The bit about primes still doesn't make sense, and you have not told us what you mean by "distribution of results" (Things get distributed with respect to something... what?) or what you mean by "results". How would you represent a "result"?
I mean, for fixed s, the output of $\zeta_n$ is distributed in the complex plane as $1/n^s$ ... this can be mapped out in the complex plane by allowing x and y to vary by $s=x+iy$.

Off the preamble in post #1: picking x=1/2, the zeta function terms become:

$$\zeta_n(y) = \frac{1}{\sqrt{n}}\frac{1}{n^{iy}} = \frac{1}{\sqrt{n}}e^{-iy\ln(n)}$$ ... errrr something like that right?
The effect of increasing $n$ is to decrease the modulus and rotate the phasor.
... But all this assumes I have understood you correctly, and I am not confident of that.

The process of trying to communicate an idea clearly will help you understand. I think you should try again: what is it that you are trying to understand?

Aside: it helps to write clear equations if you use LaTeX.

7. Dec 7, 2016

### mustang19

You're absolutely correct. Thank you.

Primes are less common among large numbers. The 1/2 term acts as a square root that causes z(s) to conform to this behavior.

Eliminating the 1/2 results in no zeroes being found, and adjusting the value of this term to something slightly greater will also cause it to deviate from the log2 behavior of the Chevyshev function. Remember that:

The concept of behavior is explained here for polynomials, although the concept could be applied to any function, and not just "end behavior".

http://www.varsitytutors.com/hotmath/hotmath_help/topics/end-behavior-of-a-function

8. Dec 8, 2016

### Simon Bridge

You're welcome. What about? I talked about more than one thing.... so once more, when you receive a direction about being vague, you respond with more vagueness.

... this does not make sense. I can guess what you are trying to say but why are you making me guess?!

... to what behaviour?

... that page describes the "end behaviour" for functions, using polynomials as an example.
That is what happens to $f(x):x\in\mathbb{R}$ as $x\to\pm\infty$ ... do you mean that you are interested in $\zeta_n(y)$ as $y\to\pm\infty$ (as described in post #6)?

Right now, every individual word in your original question has changed meaning from what you wrote.
I think you should start again - but it does strongly appear that you do not understand what you are trying to write about.

I am going to give up here.
I do not know what you are talking about and you don't seem to be able to tell me.
Good luck.

9. Dec 8, 2016

### mustang19

I understand Im not presenting a rigorous proof the riemann hypothesis. I am explaining why it cannot be resolved. Good day.

10. Dec 9, 2016

### mustang19

At least I want to know why the zeta function isn't presented in this simplified form:

Z(s) ~ 1/(n^(xi+1/2))

= 1/((n^xi)(sqrt(n)))
= (1/n^xi)(1/sqrt(n))
=(1/n^xi)(n^2)
=(n^2)/(n^xi)

From this simplification its clear that the "1/2" is really the square of n from which primes are factored out.

11. Dec 9, 2016

### jbriggs444

For one reason, because the right hand side does not depend on s and does depend on free variables n and xi, whatever those are supposed to be.

12. Dec 9, 2016

### Staff: Mentor

That doesn't make sense at all. You cannot ignore the summation over n, and the left hand side does not depend on n so the right hand side cannot depend on n either.

13. Dec 13, 2016

### mustang19

Its just an example.

Anyway mathematicians already understand the proof of RH they just don't have a rigorous way of expressing it. This is RH:

The logx term is simply from the prime number theorem. The square root term is x raised to the 1/2 power and just represents the fact that you only have to test numbers less than or equal to the square root of x for primes.

14. Dec 13, 2016

### PAllen

"Anyway mathematicians already understand the proof of RH they just don't have a rigorous way of expressing it."

Please provide a basis for your claim that ANYONE understands a proof of RH. A significant minority of mathematicians think it is likely untrue. Further, there is wide disagreement on what is the most likely broad direction a proof might take, with none coming close.

I would say your formula is a well known consequence of RH being true, rather than a statement of RH itself. Do you have a reference for the converse (that this error bound implies RH)?