PeterJ Posted January 23, 2012 Share Posted January 23, 2012 Not sure if this the right place for the question. Please move this post if not. Is there anyone here able to help me understand how R's Z-function is used to create his famous landscape? I will never understand the Z-function but am happy to treat it as a black box. What I'm trying to grasp, among other things, is what numbers have to be used as inputs in order to produce the necessary outputs. My thought is that each output value must have an input value associated with it. Yet I've never seen a discussion of the inputs, only the outputs. So maybe I'm misunderstanding something very basic. Just a discussion in natural language would be fine. I am not a mathematician. Thanks. Link to comment Share on other sites More sharing options...

Shadow Posted January 23, 2012 Share Posted January 23, 2012 (edited) I'm no expert on this, frankly I'd say we're at about the same level, but I'm next to certain you get the landscape just by plotting the zeta function. The Z function is a complex function, taking as input and returning as output a complex number. I would imagine taking the absolute value (which always returns a positive real number regardless of the input) of the outputs would give you the landscape you're talking about, ie. the landscape is made out of all the points with coordinates (Re(z), Im(z), Abs[Zeta(z)]), where z is any complex number. If you go through all the complex numbers, you get the whole landscape. But again, this is only a conclusion drawn from common sense and what little I know about the function. I would prefer someone more familiar with the Zeta function to confirm my reasoning. Also, I'm not sure what level of mathematics you're familiar with, so I hope the above is understandable enough. Edited January 23, 2012 by Shadow Link to comment Share on other sites More sharing options...

PeterJ Posted January 24, 2012 Author Share Posted January 24, 2012 Thanks. Not many people seem to know much about this. So do we enter each integers in turn, each with a range of imaginary components, and the landscape includes all possible complex values? (Or a sample of all values). I'm wondering if it is possible to list the input values that create non-trivial zeros. Link to comment Share on other sites More sharing options...

imatfaal Posted January 24, 2012 Share Posted January 24, 2012 Thanks. Not many people seem to know much about this. So do we enter each integers in turn, each with a range of imaginary components, and the landscape includes all possible complex values? (Or a sample of all values). I'm wondering if it is possible to list the input values that create non-trivial zeros. I think if you could list the non-trivial zeros you would be well on the way to a millennium million dollar prize - but I could be wrong, this is all a bit beyond me. If you can get past the horrific layout Prof M du Sautoy's website http://www.musicoftheprimes.com/01intro.htm is a good stage by stage build up to the question on the make up of the landscape the landscape is situated on the complex plane. each complex number has a real and an imaginary component - example 2 + 3i or 2 + 3(sqrt(-1). on the landscape each point is uniquely designated by how far you go in the real 'direction' East/West and how far in the imaginary 'direction' North/South. For 2+3i you would start at the origin and move 2 units East and 3 North, for -2-3i you would go 2 west and 3 south, etc. Each point on the landscape has a result that is also a complex number! this means you need two dimensions to show in the input and 2 more to show the output - 4 dimensions just do not work with the human brain so we can break it down in a few ways. You can produce two landscapes one of which shows the real output and the other shows the imaginary output. or you could produce a single graph which shows the absolute value of the complex number (the distance you end up from the origin when you move east/west then north/south). Link to comment Share on other sites More sharing options...

PeterJ Posted January 24, 2012 Author Share Posted January 24, 2012 I think if you could list the non-trivial zeros you would be well on the way to a millennium million dollar prize - but I could be wrong, this is all a bit beyond me. Not so easy I think. There will always be one more, and it might not be on the line. If you can get past the horrific layout Prof M du Sautoy's website http://www.musicofth...com/01intro.htm is a good stage by stage build up to the question on the make up of the landscape the landscape is situated on the complex plane. each complex number has a real and an imaginary component - example 2 + 3i or 2 + 3(sqrt(-1). on the landscape each point is uniquely designated by how far you go in the real 'direction' East/West and how far in the imaginary 'direction' North/South. For 2+3i you would start at the origin and move 2 units East and 3 North, for -2-3i you would go 2 west and 3 south, etc. Each point on the landscape has a result that is also a complex number! this means you need two dimensions to show in the input and 2 more to show the output - 4 dimensions just do not work with the human brain so we can break it down in a few ways. You can produce two landscapes one of which shows the real output and the other shows the imaginary output. or you could produce a single graph which shows the absolute value of the complex number (the distance you end up from the origin when you move east/west then north/south). Thanks. I've read du Sautoy's book a couple of time but it's not much help on this issue. I did not know that the inputs were being plotted with the outputs. I thought the landscape was created from only the outputs. Are you sure about this? Link to comment Share on other sites More sharing options...

imatfaal Posted January 24, 2012 Share Posted January 24, 2012 Not so easy I think. There will always be one more, and it might not be on the line. Thanks. I've read du Sautoy's book a couple of time but it's not much help on this issue. I did not know that the inputs were being plotted with the outputs. I thought the landscape was created from only the outputs. Are you sure about this? The inputs (real component and imaginary component of the complex number s) are the plane / the flat surface / the x and y coordinates /the base of the landscape - the output is the height (the z coordinate) above the plane. As there is only one dimension (height) left after the East/West North/South are taken by the input we cannot display both real and imaginary components of the output on a graph. A way of visualizing this is to take a sheet of graph paper - draw axes with the original at the centre, the vertical (NS) y-axis is the imaginary component of the input, the horizontal x-axis is the real. At each point of the graph paper you place a column rising up from the desk this is the sole remaining z-axis and the one that varies interestingly Link to comment Share on other sites More sharing options...

PeterJ Posted January 25, 2012 Author Share Posted January 25, 2012 Right. This is what I've been misunderstanding. Now I need to go think some more. Many thanks. Very helpful. Link to comment Share on other sites More sharing options...

DrRocket Posted January 27, 2012 Share Posted January 27, 2012 Not sure if this the right place for the question. Please move this post if not. Is there anyone here able to help me understand how R's Z-function is used to create his famous landscape? I will never understand the Z-function but am happy to treat it as a black box. What I'm trying to grasp, among other things, is what numbers have to be used as inputs in order to produce the necessary outputs. My thought is that each output value must have an input value associated with it. Yet I've never seen a discussion of the inputs, only the outputs. So maybe I'm misunderstanding something very basic. Just a discussion in natural language would be fine. I am not a mathematician. Thanks. The Riemann zeta function is a function of a complex variable that starts with [math] \zeta (s) = \sum_{n=1}^{\infty} \frac {1}{n^s}[/math] It is easy to see that this converges for [math] re(s)>1[/math] From that point you need to understand more of complex analysis and the notion of analytic continuation. It is the analytic continuation of the zeta function that is needed for the Riemann hypothesis. But to your basic question the input to this function is a complex number [math]s[/math] and the output is the complex number that is the value of the zeta function at that point. It is not something defined by any closed-form expression and that is one reason that the Riemann hypothesis is such a difficult problem (it is NOT the primary reason why it is difficult since mathematicians works with other functions lacking a closed form expression all the time). To really get into the problem you would need to read something like Titschmarsh's book The Theory of the Riemann Zeta-Function. Link to comment Share on other sites More sharing options...

PeterJ Posted January 28, 2012 Author Share Posted January 28, 2012 (edited) Thank you doctor. I think I've sorted my problem now. I missed the obvious, as is so easy to do. For me anyway. One thing. How does one convert the complex number that is the output into a single value for the the vertical z-axis? Edited January 28, 2012 by PeterJ Link to comment Share on other sites More sharing options...

Shadow Posted January 28, 2012 Share Posted January 28, 2012 Any way that works and gives you some meaningful information, for example the absolute value ([math]|a+bi| = \sqrt{a^2 + b^2} \in \mathbb{R}[/math]) Link to comment Share on other sites More sharing options...

PeterJ Posted January 30, 2012 Author Share Posted January 30, 2012 Okay, I think I'm getting there. So, the nontrivial zeros are positions where the input is real part 1/2, with some value for i, where z is zero. Is that it? Link to comment Share on other sites More sharing options...

Shadow Posted January 30, 2012 Share Posted January 30, 2012 If by z you mean "output", then yes. Link to comment Share on other sites More sharing options...

PeterJ Posted January 30, 2012 Author Share Posted January 30, 2012 Hooray. Got it at last. Thanks. Link to comment Share on other sites More sharing options...

DrRocket Posted January 30, 2012 Share Posted January 30, 2012 Okay, I think I'm getting there. So, the nontrivial zeros are positions where the input is real part 1/2, with some value for i, where z is zero. Is that it? No. The "trivial zeros" of the Riemann zeta function are the negative real even integers -2, -4,-6, .... They are "trivial" only by comparison with other zeros of the zeta function, and the proof that these points are zeros relies on some knowledge of the zeta function, and in particular a functional equation that it satisgies. Any zero of the zeta function that is not one of trivial zeros is called "non-trivial". The Riemann hypothesis is the conjecture that the real part of any non-trivial zero is 1/2. There are suggestive results indicating that the conjecture may be true, but no one knows. So the hypothesis is that to input a value s to the zeta function and receive 0 as an output, either s must be a negative even integer or else s must have real part 1/2. The Riemann hypothesis has been a major, in some minds the major, open problem in mathematics since David Hilbert's address of 1900 -- the Riemann hypothesis is one of the original "Hilbert Problems", the solution of any one of which would get you great recognition in the mathematical community. It is the only one of the original Hilbert Problems that is also a Milenium Problem, the solution to which would carry a $1 million prize. If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann hypothesis been proven? -- David Hilbert Link to comment Share on other sites More sharing options...

PeterJ Posted January 30, 2012 Author Share Posted January 30, 2012 The Riemann hypothesis is the conjecture that the real part of any non-trivial zero is 1/2. There are suggestive results indicating that the conjecture may be true, but no one knows. So the hypothesis is that to input a value s to the zeta function and receive 0 as an output, either s must be a negative even integer or else s must have real part 1/2. Thanks. I thought that was what I said, but maybe not. Now I just need to understand how these outputs tell us anything about the prime numbers. I suspect that my maths is not up to that. Link to comment Share on other sites More sharing options...

DrRocket Posted January 31, 2012 Share Posted January 31, 2012 Thanks. I thought that was what I said, but maybe not. Now I just need to understand how these outputs tell us anything about the prime numbers. I suspect that my maths is not up to that. See the Euler product formula for the zeta function: http://en.wikipedia.org/wiki/Riemann_zeta_function Link to comment Share on other sites More sharing options...

imatfaal Posted January 31, 2012 Share Posted January 31, 2012 Peter - I think what DocRock was getting at was that you were stating the hypothesis as a given fact - whereas it is of course a hypothesis. If we knew and could prove that all non-trivial zeros were on the line with a real component of 1/2 then it would be the reimann theory - proving that the conjecture is true is the challenge. Link to comment Share on other sites More sharing options...

PeterJ Posted January 31, 2012 Author Share Posted January 31, 2012 Hmm. Thanks for the link but all this is over my head. I do not understand what the significance of the zeros is. Something to do with the prime distribution, but that's as vague as it is for me. I have managed a heuristic proof of the Twin Primes conjecture, yet still cannot understand Euler's mathematics. Annoying, since I'd like to pursue the topic further but cannot. Peter - I think what DocRock was getting at was that you were stating the hypothesis as a given fact - whereas it is of course a hypothesis. If we knew and could prove that all non-trivial zeros were on the line with a real component of 1/2 then it would be the reimann theory - proving that the conjecture is true is the challenge. No, it's okay. I know it's an hypothesis. Link to comment Share on other sites More sharing options...

DrRocket Posted January 31, 2012 Share Posted January 31, 2012 Hmm. Thanks for the link but all this is over my head. I do not understand what the significance of the zeros is. Something to do with the prime distribution, but that's as vague as it is for me. I have managed a heuristic proof of the Twin Primes conjecture, yet still cannot understand Euler's mathematics. Annoying, since I'd like to pursue the topic further but cannot. The twin prime conjecture and the Riemann Hypothesis are two long-standing open problems in mathematics. It should not surprise you that you cannot find proofs for either -- neither can anyone else. The Riemann Hypothesis is widely regarded as the most important and difficult problem in all of mathematics. If you want to pursue these topics further, the first baby step is get a PhD in mathematics -- they really are that difficult. Link to comment Share on other sites More sharing options...

PeterJ Posted January 31, 2012 Author Share Posted January 31, 2012 Well, I've managed a correct heuristic proof of the TPC according to the mathematician with whom I consulted, and I think I can make it rigorous when I get around to it. So I wonder whether a PhD would really be necessary. Probably, but not certainly. Sometimes it helps to be naive. But you're probably right in respect of the RH. Link to comment Share on other sites More sharing options...

imatfaal Posted February 1, 2012 Share Posted February 1, 2012 Well, I've managed a correct heuristic proof of the TPC according to the mathematician with whom I consulted, and I think I can make it rigorous when I get around to it. So I wonder whether a PhD would really be necessary. Probably, but not certainly. Sometimes it helps to be naive. But you're probably right in respect of the RH. Not wishing to sound like a misery-guts but the twin-prime conjecture is something that people like Erdos, Polignac, well practically every number theorist have tried their hand at - so if I had to bet on the chances of someone finding a proof without any real basic background in maths; well, I would bet dollars to donuts that you haven't. Link to comment Share on other sites More sharing options...

PeterJ Posted February 5, 2012 Author Share Posted February 5, 2012 You'd better argue this one out with Dr. Booker at the Uni of Bristol. His view is that that it is correct but not rigorous. It works, but there is an almost insignificant theoretical chance that it might breakdown eventually. It's not such a difficult problem if it's looked at mechanically. I've tried to take the same appraoch to RH but don't understand it well enough. Probably never will. Link to comment Share on other sites More sharing options...

DrRocket Posted February 5, 2012 Share Posted February 5, 2012 Well, I've managed a correct heuristic proof of the TPC according to the mathematician with whom I consulted, and I think I can make it rigorous when I get around to it. So I wonder whether a PhD would really be necessary. Probably, but not certainly. Sometimes it helps to be naive. But you're probably right in respect of the RH. There is no such thing as a "correct heuristic proof". You either have a proof or you don't. If a "mathematician" told you differently then he is not much of a mathematician. 1 Link to comment Share on other sites More sharing options...

ajb Posted February 5, 2012 Share Posted February 5, 2012 The Riemann Hypothesis is widely regarded as the most important and difficult problem in all of mathematics. At the risk of making myself look foolish, why is the Riemann Hypothesis important? It is a seemingly simple statement that has turned out to be very hard to prove, but that cannot be a reason why it is important. The hypothesis links complex functions with primes and tools from algebraic geometry have been developed. To my naive mind, the hypothesis seems important to number theorists; it deals with the distribution of primes. There are generalisations in analytic number theory, the so-called L-functions. So my question really is, should non-number theorists really care about this? Link to comment Share on other sites More sharing options...

DrRocket Posted February 6, 2012 Share Posted February 6, 2012 At the risk of making myself look foolish, why is the Riemann Hypothesis important? It is a seemingly simple statement that has turned out to be very hard to prove, but that cannot be a reason why it is important. The hypothesis links complex functions with primes and tools from algebraic geometry have been developed. To my naive mind, the hypothesis seems important to number theorists; it deals with the distribution of primes. There are generalisations in analytic number theory, the so-called L-functions. So my question really is, should non-number theorists really care about this? Like many problems in mathematics I think that the importance of the problem lies at least as much with the anticipated new ideas that will be required to solve the Riemann Hypothesis as with the result in and of itself. This was true of Wiles proof of the Fermat theorem by way of proving the Shimura-Taniyama-Weil conjecture. It was also true Pereleman's proof of the Poincare Conjecture and the more general geometrization conjecture of Thurston using Hamilton's Ricci Flow. As you note the Riemann hypothesis is connected to the theory of L-funtions, and while the implications are primarily stated in terms of number theory, it is, of course, analytic number theory and I personally view it as an extremely difficult problem in complex analysis. So I anticipate that any proof will involve fundamentally new ideas and techniques in analysis. There is some thought, going back to Hilbert and Polya, that a successful proof might be found through the spectrum of some suitably constructed operator on a Hilbert space. No matter how one views the problem, I think it is safe to say that after so many years of determined yet unsuccessful to solve it, any successful attack on the Riemann Hypothesis will involve fundmentally new ideas and techniques and open up new avenues of inquiry into mathematics of which we cannot at this time even conceive. If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann hypothesis been proven? -- David Hilbert http://www.claymath.org/millennium/Riemann_Hypothesis/riemann.pdf http://www.claymath.org/millennium/Riemann_Hypothesis/Sarnak_RH.pdf 1 Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now