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July 16th, 2013, 12:19 PM   #1
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Understanding the Riemann hypothesis

Quote:
 Originally Posted by CRGreathouse If I become a crank, I hope I can count on you to set me straight?
I hope I'll be able to but don't go nuts on the Riemann Hypothesis because I don't even get what they are asking!

July 16th, 2013, 12:45 PM   #2
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Re: Math Q&A Part 8

Quote:
Originally Posted by Hoempa
Quote:
 Originally Posted by CRGreathouse If I become a crank, I hope I can count on you to set me straight?
I hope I'll be able to but don't go nuts on the Riemann Hypothesis because I don't even get what they are asking!
Do you understand the prime-counting function $\pi(x)$ and the logarithmic integral $\operatorname{Li}(x)=\int_2^x\frac{dx}{\log x}$? The Riemann hypothesis is equivalent to

$|\pi(x)-\operatorname{Li}(x)|\le\frac{1}{8\pi}\sqrt x\log x$ for all x >= 2657.

Or to make it even easier, see
http://arxiv.org/abs/math/0008177

You don't have to understand the zeta function at all to understand RH. (Smart money says you'll have to understand it to follow the proof of RH... but we don't have one of those yet.)

July 16th, 2013, 02:01 PM   #3
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Re: Math Q&A Part 8

Quote:
 Originally Posted by CRGreathouse Do you understand the prime-counting function $\pi(x)$ and the logarithmic integral $\operatorname{Li}(x)=\int_2^x\frac{dx}{\log x}$? The Riemann hypothesis is equivalent to $|\pi(x)-\operatorname{Li}(x)|\le\frac{1}{8\pi}\sqrt x\log x$ for all x >= 2657. Or to make it even easier, see http://arxiv.org/abs/math/0008177 You don't have to understand the zeta function at all to understand RH. (Smart money says you'll have to understand it to follow the proof of RH... but we don't have one of those yet.)
Ah that makes more sense, but the article convinces me not to be a crank on a proof How would one find the results proposed? Inspection and than a proof?

 July 17th, 2013, 03:53 AM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Understanding the Riemann hypothesis I'd rather like to help one understand the Riemann Hypothesis than a consequence of it. It takes a lot to prove that the bound on Pi(x) - Li(x) is equivalent to Riemann Hypothesis. We have a function $\zeta(x)= \sum_{k = 1}^{\infty} \frac{1}{k^z}$ which is convergent only for Re(z) > 1. There are certain analytical methods which gives you some other function that agrees with the former one on Re(z) > 1 and also have it's own convergent values on some other domain. We can analytically continue zeta this way through the complex plane. RH says that all the zeros of zeta, except some which lies on the negative even numbers, are situated on Im(z) = 1/2. Balarka .
 July 17th, 2013, 04:25 AM #5 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Understanding the Riemann hypothesis mathbalarka meant Re(z)=1/2
July 17th, 2013, 05:23 AM   #6
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Re: Understanding the Riemann hypothesis

Quote:
 Originally Posted by mathbalarka I'd rather like to help one understand the Riemann Hypothesis than a consequence of it.
Ah, but that's merely a matter of perspective. From where I stand the RH *is* the result I gave above and the statement about the nontrivial zeros of the zeta function is merely a consequence.

'Of course' the two are logically equivalent.

Quote:
 Originally Posted by mathbalarka We have a function $\zeta(x)= \sum_{k = 1}^{\infty} \frac{1}{k^z}$ which is convergent only for Re(z) > 1. There are certain analytical methods which gives you some other function that agrees with the former one on Re(z) > 1 and also have it's own convergent values on some other domain. We can analytically continue zeta this way through the complex plane.
I don't think this is a useful explanation for Hoempa. If he doesn't understand analytic continuation, how is he going to understand that there are zeros on Re(x) = 1/2 at all, let alone have intuition about whether all would fall there or on Im(z) = 0?

July 17th, 2013, 06:18 AM   #7
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Re: Understanding the Riemann hypothesis

Quote:
 Originally Posted by CRGreathouse Ah, but that's merely a matter of perspective.
True. I'd like to work with the original RH rather than the PNT-form RH, anyways.

Quote:
 Originally Posted by CRGreathouse Of course' the two are logically equivalent.
Perhaps I don't see, but is it obvious?

Quote:
 Originally Posted by CRGreathouse If he doesn't understand analytic continuation, how is he going to understand that there are zeros on Re(x) = 1/2 at all, let alone have intuition about whether all would fall there or on Im(z) = 0?
Analytic continuation is a very easy concept, I don't think one should have problem understanding it. If F(x) has a complex domain D and G(x) has D', and F(x) = G(x) in D ? D', then F(x) is an extension to G(x) in D\{D'} and G(x) is an extension to F(x) in D'\{D}.

July 17th, 2013, 06:32 AM   #8
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Re: Understanding the Riemann hypothesis

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by CRGreathouse Of course' the two are logically equivalent.
Perhaps I don't see, but is it obvious?
Not at all obvious, but true. See Schoenfeld 1976.

Quote:
 Originally Posted by mathbalarka Analytic continuation is a very easy concept, I don't think one should have problem understanding it. If F(x) has a complex domain D and G(x) has D', and F(x) = G(x) in D ? D', then F(x) is an extension to G(x) in D\{D'} and G(x) is an extension to F(x) in D'\{D}.
We've both taken complex analysis, but I don't think Hoempa has. So I wouldn't assume that anything there is obvious.

For example, in the generality you present it, there's nothing special about a continuation -- there are beth_2 extensions of zeta to the values with Re(z) <= 1. It does make sense to talk about "the" analytic continuation of zeta, but only because of a manifestly nonobvious theorem.

July 17th, 2013, 06:40 AM   #9
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Re: Understanding the Riemann hypothesis

Quote:
 Originally Posted by CRGreathouse For example, in the generality you present it, there's nothing special about a continuation
Agree. Analyticity makes all the changes.

Quote:
 Originally Posted by CRGreathouse It does make sense to talk about "the" analytic continuation of zeta, but only because of a manifestly nonobvious theorem.
Well, the theorem there is not only obvious to me but somewhat trivial. However, everyone has a different choice of perspectives, I'd guess.

July 17th, 2013, 06:50 AM   #10
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Re: Understanding the Riemann hypothesis

Quote:
 Originally Posted by CRGreathouse We've both taken complex analysis, but I don't think Hoempa has.
Indeed, I haven't taken complex analysis.
Quote:
 Originally Posted by mathbalarka $\zeta(x)= \sum_{k = 1}^{\infty} \frac{1}{k^z}$
Is this to be $\zeta(z)$? Because you seem to be saying Re(z) > 1 so zeta should be at least a function of z I think.

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