Notices
Results 1 to 79 of 79

Thread: Riemann Hypothesis Proof !

  1. #1 Riemann Hypothesis Proof ! 
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Dear friends includiding This is not the end and Serpico Jr.,

    Thanks for hellping me out last time with the NP-P problem, now recently I visited Bangalore a city north of my State. At a Book store I caught hold of a Book by Prof. Stephen Hawkins, called the 'God created the Integers'.

    The reason I am posting this is , in the Book I found another Clay problem yet to be proved for the last 280 years called the 'Riemann Hypothesis. I came back spent around 3 hours or 4 in the night and found a proof. I came back to my home town Kochin and confirmed my proof. It is on the web http://www.riemann.co.in . In case you need to look at it please feel free to do so.

    Link: http://www.riemann.co.in.

    Comments are welcome, caveat: you should use your conjuring abilities before discussion., not like last tiem where we have to touch up everything.

    Thanks guys no hardfeelings.

    Mathew Cherian

    Mathew Cherian


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Masters Degree thyristor's Avatar
    Join Date
    Feb 2008
    Location
    Sweden
    Posts
    542
    What is the Rieman Hypothesis?


    373 13231-mbm-13231 373
    Reply With Quote  
     

  4. #3  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    The Riemann Hypothesis

    Define the function



    If you know calculus, you should see that this makes sense for real numbers s > 1. (This is a "p-series" in calculus terms.) If you know complex analysis, you should see that this makes sense for complex numbers s with Re(s) > 1. A wonderful fact is that this function has an analytic continuation to the entire complex plane, and it has exactly one simple pole at s = 1.

    There is a "functional equation" which relates ζ(s) to ζ(1-s). Namely, if we define



    then we have ξ(s) = ξ(1-s).

    Now it's not too hard to show that ζ(s) ≠ 0 for Re(s) > 1, and it follows from the functional equation that the only values of s with Re(s) < 0 for which ζ(s) has zeros are the negative even integers. So we have a pretty good understanding of what the function looks like outside of the vertical strip 0 ≤ Re(s) ≤ 1.

    The Riemann Hypothesis states that all of the zeros of ζ(s) in the strip 0 ≤ Re(s) ≤ 1 lie on the vertical line Re(s) = 1/2.
    Reply With Quote  
     

  5. #4  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Matthew:

    1. Your definition of the function ξ is incorrect. Riemann's ξ is a little different from my ξ, but they're very similar, and neither of them looks anything like Li. Li is related to π (the function, not the constant), which itself is related to certain integrals involving ξ.

    2. Your calculation that ξ(x) = x<sup>2</sup>/2a is absurd. You define a to be a function of x and then integrate it as if it were a constant. This is an error only second semester calculus students make.
    Reply With Quote  
     

  6. #5  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Thanks, your reply is much appreiciated, it definitely is analytic in the complex plane only that it reaches a limit of 6/pi^2. Also, what you have said has nothing to do with the Riemann's proof, which stipulates 1/2 as the real part of the solution.

    What you said are just excursions of the thoughts on Riemann's Hypothesis.

    The zeta function look like the reciprocal of the tan fuction. Moreover carefull here, we cannot have a definite zero on first differentiation may be close becacuse of the fact that it is not something like a hyperbola but a tan curve inverted reaching definite limits of 6/pi^2 as Euler directed. Zero is reched in my calculation on first derivative proving that Riemann was right in his conjuring of 1/2 as the root.

    ________________-----
    Mathew Cherian
    Reply With Quote  
     

  7. #6  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Dear Serpico Jr,

    Even when Riemann directed us that the solution lie around 1/2 nobody is sure whether he had a difinite proof for it. He just conjured like every experienced Mathematicians do. When we use the first principles to analyse his thoughts we see that the first derivative is close to zero when x = 1/2. Even though it doesn't make any difference whether I use it as a constant or a variable. It definitely is a constant for each number x we choose. You didn't see my caveat, we have to use our conjuring abilities not just the mathematical abilities to solve this behamouth which remain unresolved for 280 years.

    It is the block what we choose and what we don't. If we choose it as a variable we won't even be able to start in the first place we reach a degenerate state. Mathematics as we learn and as we practice make a whole lot of differnce. Our experience should tell as what we do with it, like we cannot invert non-linear funcitons. I feel you got the point clear.
    ________________________
    Mathew Cherian
    Reply With Quote  
     

  8. #7  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    it reaches a limit of 6/pi^2.
    You're confused here. You're thinking of the fact that ζ(2) = π<sup>2</sup>/6.

    The zeta function look like the reciprocal of the tan fuction.
    No, the reciprocal of tangent is cotangent, which looks a lot like tangent but nothing like zeta.


    Zero is reched in my calculation on first derivative proving that Riemann was right in his conjuring of 1/2 as the root.
    Your calculations are incorrect. You're dealing with the wrong functions.

    Matthew, you're going to have a real hard time convincing me you're right this time. Last time, you had the benefit of the fact that complexity theory is not my forte. This time, you're in trouble, as number theory is my forte.
    Reply With Quote  
     

  9. #8  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    Even though it doesn't make any difference whether I use it as a constant or a variable. It definitely is a constant for each number x we choose.
    Wow. You don't know calculus at all. The variable of integration cannot be treated as a constant. Sure, if you hold x constant, then a function of x is also constant. But if x is your variable of integration, it necessarily varies, and so you can't hold it constant.
    Reply With Quote  
     

  10. #9  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    You are right that reciprocal of tangent is cotangent.. For an exponential function it is like a gradient and a flatening. So the cotangent should be the downward slope followed by a slope which is the zeta function. My intention was just to show the shape of the zeta function.

    My proof depends on the fact that near 1/2 the the zeta function curves at right angle and slowly dips towards a limit which the second differential > 1/2 should give.

    When you dictate the gamma function you seeem to say that it is difficult to see the function below 1. In fact the log function is clearly visible below 2. Only that the gradients whose mode is positive might look negative.

    ______________________
    Mathew Cherian
    Reply With Quote  
     

  11. #10  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    I feel stupid--I've been calling you Matthew, not Mathew. Sorry.

    Anyway, Mathew, you don't have a proof. Nowhere in your paper do you talk about anything related to the zeta function. You believe that ξ and Li are equal, but they're not. And your integral is completely wrong. The whole reason people write Li instead of some explicit expression is because you cannot express Li in terms of elementary functions. But you manage to express it as x<sup>2</sup>/2a = x/2ln(x), which is in terms of elementary functions. So you're wrong. In fact, look--if I take the derivative of x/2ln(x), I get:

    (2ln(x)-2)/4(ln(x))<sup>2</sup> = ln(x)/2-1/2ln(x)

    This is not 1/ln(x), so your integration is wrong.
    Reply With Quote  
     

  12. #11  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    I feel stupid--I've been calling you Matthew, not Mathew. Sorry.

    Quote Originally Posted by serpicojr
    Anyway, Mathew, you don't have a proof. Nowhere in your paper do you talk about anything related to the zeta function. You believe that ξ and Li are equal, but they're not. And your integral is completely wrong. The whole reason people write Li instead of some explicit expression is because you cannot express Li in terms of elementary functions. But you manage to express it as x<sup>2</sup>/2a = x/2ln(x), which is in terms of elementary functions. So you're wrong. In fact, look--if I take the derivative of x/2ln(x), I get:

    (2ln(x)-2)/4(ln(x))<sup>2</sup> = ln(x)/2-1/2ln(x)

    This is not 1/ln(x), so your integration is wrong.
    Your expression is wrong, let me tell you x<sup>2</sup>/2a= x<sup>2</sup>/ln x<sup>x</sup>. You should note that your denominator is wrong.
    The numerator and substitution with rounding off errors is 0.693 - 0.707 which is close to 0.014 which is nothing other than zero. Then donominator of course is 4a<sup>2</sup>. It is not exactly zero because of the fact that it is not a curve that slopes downwards and then turns upwards rather it slopes downwards drasticaly then slants parallel to the x-axis converging on to 6/pi<sup>2</sup>.
    Your question why the first term of the numerator is +ve is because it is the modulus rather than the sign. A curve that grows upwards never can have a -ve growth.

    _______________________
    Mathew Cherian
    Reply With Quote  
     

  13. #12  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Li(x) is the convergent value of zeta function. They are equal at higher values of x. In fact zeta(x)/Li(x) = 1 as x tends higher.
    Proof is when ca function inflects at inflection point the f"(x)=0 which is the proof.

    Thanks.
    ___________________
    Mathew Cherian
    Reply With Quote  
     

  14. #13  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Also, you show throught the Gamma function that zeta(x) = zeta(x-1) which is true at higher values of x.

    Euler has shown it seems that the zeta function can be reduced to a time series = 1+ 1/2<sup>2</sup> + 1/3<sup>2</sup>+1/4<sup>2</sup>......+1/x<sup>2</sup>+.. for s=2.

    So my transformed function x<sup>2</sup>/2a is only the closed form state of the Euler Time series.

    This closed form state was what I initialy concieved and which I thought the only way we can prove Riemann's hypothesis, since then we can get the first derivative and substitute 1/2 and see whether it goes to zero, if it does then we are through, which it did.

    I have quoted Riemann in my paper where he stipulates that a root is what when we substitue for x in f ' (x) takes it to zero or the simply the inflection point which is 1/2 as I have shown. I have to say this is how we prove the root in calculus.

    In fact where ever f(x) is zero x is the root in calculus. In zeto function f(x) is also zero when x = 1/2 then we have x/a integrated between 2 and 1/2 a discontinuous integral.

    Thank you.

    _______________________
    Mathew Cherian
    Reply With Quote  
     

  15. #14  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    Your expression is wrong, let me tell you x<sup>2</sup>/2a= x<sup>2</sup>/ln x<sup>x</sup>. You should note that your denominator is wrong.
    Look:

    x<sup>2</sup>/2a = x<sup>2</sup>/2ln(x<sup>x</sup>) = x<sup>2</sup>/2xln(x) = x/2ln(x)

    ...+ve...-ve...
    Please don't use these abbreviations.

    In fact zeta(x)/Li(x) = 1 as x tends higher.
    No, π(x)/Li(x) tends to 1 as x tends to infinity.

    Also, you show throught the Gamma function that zeta(x) = zeta(x-1) which is true at higher values of x.
    No, I used gamma to relate zeta(s) to zeta(1-s) for any s.

    Euler has shown it seems that the zeta function can be reduced to a time series = 1+ 1/2<sup>2</sup> + 1/3<sup>2</sup>+1/4<sup>2</sup>......+1/x<sup>2</sup>+.. for s=2.
    This is true, but... this is by definition. Replace the power 2 with any s for Re(s) > 1, and you have zeta(s). See my post above.

    So my transformed function x<sup>2</sup>/2a is only the closed form state of the Euler Time series.
    I have no idea how you figure this.
    Reply With Quote  
     

  16. #15  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Quote Originally Posted by serpicojr
    Look:

    x<sup>2</sup>/2a = x<sup>2</sup>/2ln(x<sup>x</sup>) = x<sup>2</sup>/2xln(x) = x/2ln(x)
    I don't know why you do the above you can leave the denominator as the original, for two teasons. One is on differentiation of your simplified term the numerator though goes closer to zero, it is not as close as when we take the original term. Two, log function is not a linear function for one to multiply by x the log of x for logx<sup>x</sup>. I think you are taking the log function as linear which might not be all the more right may be because of that the numerator is though close to zero when x=1/2 but not closer as when the term is left alone by itself.

    All others seems to be OK, I have no objections.

    ___________________
    Mathew Cherian
    Reply With Quote  
     

  17. #16  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Quote Originally Posted by serpicojr
    This is true, but... this is by definition. Replace the power 2 with any s for Re(s) > 1, and you have zeta(s). See my post above.
    You are confusing s with x which is not true. s is a complex number depicting the elevation of the graph on the S-plane. It has got to do with root not the number. According to ones choice of s as 2,3, 4 ,..... the graph of the log function takes different position higher and higher.

    Transformation for getting the closed form equation for the time series is given initialy in my paper. x<sup>x = e<sup>x ln x</sup> so on and so forth.


    ___________________
    Mathew Cherian
    Reply With Quote  
     

  18. #17  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    My statement that s has something to do with the root may not be right, also it depicts the graph moving upwards should be read as the graph moving sideways as s takes values 2,3,4,......

    _________________
    Mathew Cherian
    Reply With Quote  
     

  19. #18  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Quote Originally Posted by Mathew Cherian
    Quote Originally Posted by serpicojr
    Look:

    x<sup>2</sup>/2a = x<sup>2</sup>/2ln(x<sup>x</sup>) = x<sup>2</sup>/2xln(x) = x/2ln(x)
    I don't know why you do the above you can leave the denominator as the original, for two teasons. One is on differentiation of your simplified term the numerator though goes closer to zero, it is not as close as when we take the original term. Two, log function is not a linear function for one to multiply by x the log of x for logx<sup>x</sup>. I think you are taking the log function as linear which might not be all the more right may be because of that the numerator is though close to zero when x=1/2 but not closer as when the term is left alone by itself.
    Serpicojr is not treating the log function as linear, only using the identity log(a<sup>b</sup>) = b·log(a), which is true for all positive real numbers a and all real numbers b.

    Also, what is wrong with simplifying as Serpicojr did? Look:

    This is the graph of y = x^2*∕*(2xln(x)):
    http://www.mathsisfun.com/graph/function-grapher.php?func1=x^2/(2*x*ln(x))&xmin=-6&xmax=16&ymin=-6&ymax=6

    And this is the graph of y = x*∕*(2ln(x)):
    http://www.mathsisfun.com/graph/function-grapher.php?func1=x/(2*ln(x))&xmin=-6&xmax=16&ymin=-6&ymax=6

    The graphs are identical. Whatever you can do with one, you can with the other as well. Or (what’s more relevant here) whatever you cannot do with one, you cannot do with the other either!
    Reply With Quote  
     

  20. #19  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Mathew, you have to understand that your "proof" is deeply flawed. Let me reiterate:

    1. You do not prove anything about the zeta function or related functions in your paper. You claim to prove some properties of xi, but you claim that xi is the same as Li. Proving things about where Li or its derivatives are 0 does not tell you anything about where zeta is 0.

    2. Your calculations are full of errors.
    Reply With Quote  
     

  21. #20  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    To Jane Bennet and Serpico jr.,

    I didn't say that Serpico jr was wrong. The reason why I told you not to simplify my function is because on dfferentiation of the (u/v) function and substitution the numerator is closer to zero my way than Serpico jr's. I am not saying that it should be exactly zero for the fact that it is not a diminishing function and raising function where at the inflection point the first derivative goes to zero. Here it is a diminishing function followed by a curve and a slant which requires only that the inflection point is very close to zero (and diminishing again in this case).

    ___________________
    Mathew Cherian
    Reply With Quote  
     

  22. #21  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    You're ignoring the greater issue: your paper has nothing to do with the zeta function and is full of computational errors.
    Reply With Quote  
     

  23. #22  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    To Serpico Jr.:

    One thing I can admit is that s is the same as x only reason why s is used is because it is considered to be a complex number whose real part 2,3,4 etc; we are concerned with. The imaginary part we are not concerned in Rieman Hypothesis whic once again I am quoting here states the "zeta function whose s=2 has a root at 1/2 which is the real part of the root" which is what I am trying to prove.

    Take a Sine function. If Pi/2 is a root how do we get Pi/2. We differentiate Sine Pi/2 which is Cos Pi/2 where Cos Pi/2 =0 the first derivative and Pi/2 is considered as the root. Generaly speaking where ever on the x axis the first derivative equates to zero is the root of the equation which is a basic method which I am also eliciting here which Rieman also directed us but left undone.

    _________________
    Mathew Cherian
    Reply With Quote  
     

  24. #23  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    To Serpico jr.,

    S as I said is equal to x may not be correct. S is the base of the logarithm. So the natural log suffice.

    S is in the complex s plane and anything that grows is considered in the complex plane.

    _______________

    Mathew Cherian
    Reply With Quote  
     

  25. #24  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    The imaginary part we are not concerned in Rieman Hypothesis whic once again I am quoting here states the "zeta function whose s=2 has a root at 1/2 which is the real part of the root" which is what I am trying to prove.
    This is not the statement of the Riemann Hypothesis. This doesn't even make sense. The first post I make in this thread contains the correct statement of the Riemann hypothesis.

    Take a Sine function. If Pi/2 is a root how do we get Pi/2. We differentiate Sine Pi/2 which is Cos Pi/2 where Cos Pi/2 =0 the first derivative and Pi/2 is considered as the root.
    π/2 is not a root of sine. sin(π/2) = 1. π/2 is not a root of sine.

    Generaly speaking where ever on the x axis the first derivative equates to zero is the root of the equation which is a basic method which I am also eliciting here
    No. If f(x) is a complex-valued function, then b is a root of f if f(b) = 0. So π/2 is a root of cosine. 0 and π are roots of sine. π/2 is not a root of sine. This is not a "method". This is a definition.

    Mathew, really, your paper is wrong, wrong, wrong.
    Reply With Quote  
     

  26. #25  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Quote Originally Posted by serpicojr
    [No. If f(x) is a complex-valued function, then b is a root of f if f(b) = 0. So π/2 is a root of cosine. 0 and π are roots of sine. π/2 is not a root of sine. This is not a "method". This is a definition.]
    Let me answer this first since others need to be contested and takes more depth.

    In my Proof, 1/2 is the real part of the root and 0 is the imaginary part which is none other than the rate of change is zero which is none other than the inflection point.

    Now you check up the earlier post in which Jane Bennet graphed the function x<sup>2</sup>/2a. You move the cursor to 1/2 on the x axis and you will see in the bottom window the zero appearing.

    Also, 2 is also a root which shows the curvature.

    Thank you.

    __________________
    Mathew Cherian
    Reply With Quote  
     

  27. #26  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Mathew, you don't understand what I'm saying. What you're describing are critical points--zeros of the first derivative. The Riemann Hypothesis is a statement about roots, which are zeros of the function itself.

    Also, it's clear from the graphs that Jane posted that x = 1/2 is neither a root nor a critical point of that function. But it doesn't matter anyway, because the function in the graph is a function you obtained by faulty integration, and the function you were integrating doesn't have anything directly to do with the zeta function.

    It would really benefit you if you were to read and understand my criticisms. You're making very basic mistakes--you don't understand definitions and your computations are wrong. We'd be more than happy to help you sort things out, but first you have to accept the possibility that your proof is wrong, then you have to spend some time understanding my criticisms, and then you have to accept the fact that your proof is wrong.
    Reply With Quote  
     

  28. #27  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Dear Serpico Jr.,
    How can I accept that my proof is wrong, when I don't gamble with public display of my work. That demeans my professionalism right.

    First of all I am afraid to say that you don't consider the basic fact that root of an equation is the abasissa number that drives the Polyinomial to passivity. If yo have something beyond this I am prepared to listen which no Mathematician has ever published yet to my knowledge else I would have known it after an Engineering back ground and some work in solving problems in it.

    Of course sometimes the function that traces the basic function is also called the root. You won't see such roots in practice often but rarely some people call it root.

    Rieman Hypothesis is one where Rieman himself has directed posterity to use the first method to get the roots which is the general route.

    If you have something beyond this I will be very obliged to listen provided you are willing to spell it out first rather than make me say something that is right as wrong.

    I would appreciate it much if you graph it on Internet resources and see where the 1/2 goes to. If it is zero we are following the first principles of finding roots, which is what Rieman expected us to do.

    ____________________
    Mathew Cherian
    Reply With Quote  
     

  29. #28  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Read the following page for the definition of a root:

    http://en.wikipedia.org/wiki/Root_(mathematics)

    Please understand this simple concept and how it differs from your notion of root before continuing this discussion.
    Reply With Quote  
     

  30. #29  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Let me quote Rieman as translated and published by Pro. Stephen Hawkins in the Book 'God created integers', pp-823, para-4, which is the way I have also learned it during the time I was learing it during my Engineering.

    "A value x is a root of a function f(x) if f(x)=0".
    -Rieman.

    Now you go to "mathsisfun" website pointed out by Jane. You graph my function x<sup>2</sup>/2.a. Move the cursor to 0.5 and check up the value of x and y in the bottom box you will see that assuming computational errors it is zero for 1/2 on x which means 1/2 is a root and the rate shift is zero here because the tangent moves from one direction to the other.

    Rieman's definition of the root is widely followed world over and is standard. If f(x) has to be zero then it is ascertained by by finding f ' (x) which goes to zero which is none other than the inflection point. The meaning o inflection is that at inflection dy/dx=0.

    If you interpret it only as f(x)=0 then if a sine function is zero at origin does that mean it is a root of the sine function?. Sine function can come in two forms either as a 3 order or 2 nd order. It it is 3rd order then the bends at the begining and end along with the ben in the middle are roots. If it is second order the bend in the middle is the root for informations sake.

    I hope I didn't confuse you, but this is the reality.

    In Electirical Engineering this is clearly visible in passive circuit theory. An RLC circuit goes passive at the root where it behave just like a purely resistive circuit. If you look at the Frequency response we will see at the at the roots the function goes to zero which is just at freqency 1/2pi. sq.root rlc. At this point nothing goes from input to out put, and as the frequency shifts the transmission starts again.

    This is just an example why I say dy/dx should be equal to zero to figure out where f(x) goes to zero.

    In the Rieman zeta function integral 1/lnx is infefinite integral. If you integrate it as definite integral between 2 and 1/2 we stell get zero which is none other than saying f(x) is zero at 1/2. Mine is the standard method.

    Also if you take a quadratic function or second order function ax<sup>2</sup>+bx+c, it's roots are at -b+/-sp.root b<sup>2</sup>-4ac the whole divided by 2a. Here the roots will be at inflection points. This is just another example to show you that it is inflection points that depict the root for reasons of any need of meaning for it in use.

    _______________________
    Mathew Cherian
    Reply With Quote  
     

  31. #30  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    If f(x) has to be zero then it is ascertained by by finding f ' (x) which goes to zero which is none other than the inflection point. The meaning o inflection is that at inflection dy/dx=0.
    Simply not true. For example, let f(x) = x<sup>2</sup>+1. f(x) has two complex roots, ±i. f'(x) = 2x has one root, 0. 0 is not a root of f(x).

    I hope I didn't confuse you, but this is the reality.
    You are the confused one.
    Reply With Quote  
     

  32. #31  
    Forum Sophomore
    Join Date
    Jul 2007
    Location
    South Africa
    Posts
    196


    You can't use a = Li x, because this is not true for all values of x.
    It also matters what isn't there - Tao Te Ching interpreted.
    Reply With Quote  
     

  33. #32  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    To Serpico jr.,
    These sort of equations cannot occur. These are degenerae equations.

    I shall give you my Proof in a different way, now that you are not interested in going and graphing my f'(x) function.

    The second order equation for the log is given as follow,

    2f"(x)/2! + f'(x)+ (x-1)/(x-2)

    Now finding the roots of this equation we have,

    -1+/-Sq.root(b<sup>2</sup> - 4ac/2

    which is nothing other than 1/2 + 0 for 1/ln x.

    Now this is what I call the zeta function proof.

    _____________________
    Mathew Cherian
    Reply With Quote  
     

  34. #33  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    To Serpico Jr.,

    The roots should be read as (1/2, +/-0)

    _________________
    Mathew Cherian
    Reply With Quote  
     

  35. #34  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Mathew, I'd like to help you develop your math skills. We need to start at the very beginning, though. You display a poor grasp of mathematical proof. You do not seem to know how to interpret mathematical language. You have yet to present a coherent mathematical argument. It's impossible to do math without these skills, and it's a waste of our time to carry on this discussion if you aren't willing to learn them.
    Reply With Quote  
     

  36. #35  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Instead of concentrating on the Mthematics may be some at some involved level you are just criticising my Mathematics. Point 1) If you take a function if the graph say of f ' (x) will be the same as that of f(x) because it is the rate change the f ' (x) graphs.

    2) You are not willing to accept direct proof or the one through the dyanmical equations or even first principles of mathematics.

    3) You are not willing to graph the intergral and see for yourselves the results. Of course I haven't added the error estimate because our area of concern in the graph pretty much gives the result we are looking for.

    4) I also beleive that you haven't worked much with dynamical equations where by you even come with degenerate equations wiht roots on the ordinate which shows the equation is degenerate or the solution is on the origin which is only trivial.

    Once you take care of these factors my proof will be more visible to you. I am sorry I have to criticize a person who tried to work with my proof though I regret that you didn't co-operate fully to understand such a special function, properly.

    Regards,

    _________________
    Mathew Cherian
    Reply With Quote  
     

  37. #36  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Mathew Cherian, I’m keen to know what sort of math teachers you have/had. Did you really learn your problem-solving “skills” from them? :x
    Reply With Quote  
     

  38. #37  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    1) If you take a function if the graph say of f ' (x) will be the same as that of f(x) because it is the rate change the f ' (x) graphs.
    No. What does it mean if f'(x) = f(x), Mathew? How many functions satisfy that?

    2) You are not willing to accept direct proof or the one through the dyanmical equations or even first principles of mathematics.
    I am willing to accept correct proofs and only correct proofs. I have never seen you present anything close to a correct proof.

    3) You are not willing to graph the intergral and see for yourselves the results. Of course I haven't added the error estimate because our area of concern in the graph pretty much gives the result we are looking for.
    You can't leave things like error estimates out of your argument, especially if you make no mention of them in your argument, because then clearly your argument involves more than what you wrote down on paper. This is your responsibility.

    4) I also beleive that you haven't worked much with dynamical equations where by you even come with degenerate equations wiht roots on the ordinate which shows the equation is degenerate or the solution is on the origin which is only trivial.
    Your proof has nothing to do with dynamical systems or differential equations (not sure which you're trying to talk about).

    Once you take care of these factors my proof will be more visible to you.
    No, your proof is wrong. I've repeated my criticisms many times, you have not addressed them.

    I am sorry I have to criticize a person who tried to work with my proof though I regret that you didn't co-operate fully to understand such a special function, properly.
    This is an insult. I didn't cooperate fully to understand your proof? I did! It's you who failed to provide a proof!
    Reply With Quote  
     

  39. #38  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Now I am sorry if my words sounded as insults, I didn't mean it. My intentions were to give you more information on my work.

    Now please click the link below which gives you the graphs of 1/ln (x) and x/Ln(x<sup>2</sup>. There is a red graph underneath the blue one.

    http://www.mathsisfun.com/graph/function-grapher.php?func1=x/(ln(x^x))&func2=1/ln(x)&xmin=-12&xmax=12&ymin=-8&ymax=8

    You can see that the roots are falling on 1/2 where the function is 0.

    __________________
    Mathew Cherian
    Reply With Quote  
     

  40. #39  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Indeed, those are the same functions, but 1/ln(1/2) is not equal to 0. This would only be possible if ln(x) approached infinity as x approaches 1/2. This is not the case. In fact, ln is defined at 1/2, and we know that:

    ln(1/2) = -1+1/2-1/3+1/4-1/5+1/6...

    ln(1/2) = -1/2-1/2*2<sup>2</sup>-1/3*2<sup>3</sup>-1/4*2<sup>4</sup>-1/5*2<sup>5</sup>-1/6*2<sup>6</sup>...

    It's clear from either of these expressions that ln(1/2) is a real number, a negative one at that. And that means 1/ln(1/2) is not equal to 0.
    Reply With Quote  
     

  41. #40  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Dear Serpico Jr.,

    It is not 1/ln(1/2) it is integral 1/Ln(x) from 2 to 1/2 which is zero.
    _____________
    Mathew Cherian
    Reply With Quote  
     

  42. #41  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    It is not 1/ln(1/2) it is integral 1/Ln(x) from 2 to 1/2 which is zero.
    First of all, that's not what you said. You said that 1/ln(x) has a root at 1/2. But whatever.

    Second, what do you even mean by the integral from 1/2 to 2 of 1/ln(x)? This is an indefinite integral due to the singularity at 1, and the integral from 1/2 to 1 diverges to negative infinity while the integral from 1 to 2 divers to infinity. The standard way to do this is to interpret the integral as a Cauchy principal value. In any case, according to this definition, the integral from 1/2 to 2 is not 0--it is li(2)-li(1/2), and li(2) > 0 while li(1/2) < 2, whence li(2)-li(1/2) > 0.

    Third, this has nothing to do with the zeta function.

    Don't give up, Mathew. Devote all of your effort to understanding this. You need to understand that your proof makes no sense.
    Reply With Quote  
     

  43. #42  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    You don't have to do all what you have done to find the value of the zeta function from 2 to 1/2, even 1/2 to 2 will suffice but it is not ethical. It goes to zero.

    I have given a point form equation x<sup>2</sup>/2a. If you take the limits of this function from 2 to 1/2 you will get zero. It is like an S function half lieing negative and 1/2 positive. The sum going to zero.

    No arguments will suffice when you have the function in front of you giving 1/2 as the root.

    __________________
    Mathew Cherian
    Reply With Quote  
     

  44. #43  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    No arguments will suffice when you have the function in front of you giving 1/2 as the root.
    Get this though that skull of yours: you have not presented any functions which have 1/2 as root. It doesn't even matter, though, because none of the functions you have spoken about will tell you where the roots of zeta lie.

    (Also, the rest of this last post you made makes no sense.)

    What do you do for a living? I hope you don't have to do a lot of math.
    Reply With Quote  
     

  45. #44  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    The funection I had given clearly shows a root at 1/2. If you move the cursor to 1/2 it will show that it is converging on to zero.

    Also your need to use Cauchy's Principals to integrate this function is not really required when you can directly substitute into the equation got for the zeta function is the integral, one need to take only the limits.

    ______________
    Mathew Cherian
    Reply With Quote  
     

  46. #45  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Your integration is incorrect. It doesn't have anything to do with zeta, in any case.

    If you put the cursor on the x-axis at x = 1/2, then yes, it says y = 0. But that's because that's where the cursor is. If you put it on the graph where x = 1/2, you get something else. Furthermore, the graph clearly does not intersect the x-axis at x = 1/2.
    Reply With Quote  
     

  47. #46  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Dear Jane Benett,
    If you think I had Professors like Weber or Gauass or Lagrange to tackle a problem like this you mistake me. I had normal modern day Profs. who taught me and get interested in these sort of things.

    We are mortal Humans unlike Riemann or Euiler or Gauss, we endowed with normal intlligence but endowed with the panorama of their work.

    Please refer to my new posting on my website http//www.riemann.co.in for an elaborate exposition of Riemann Hypothesis proof.

    Thanks for the grahphing web.

    ________________
    Mathew Cherian
    Reply With Quote  
     

  48. #47  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Quote Originally Posted by Mathew Cherian
    If you think I had Professors like Weber or Gauass or Lagrange to tackle a problem like this you mistake me. I had normal modern day Profs. who taught me and get interested in these sort of things.
    I never thought anything of that sort. In fact, I was – and still am – of the belief that the teachers who taught you maths belong to the same category of teachers as those who taught William McCormick physics.

    I think it’s time this thread was moved to Pseudoscience. We don’t have a Pseudomathematics area but the Pseudoscience will just do for this thread.
    Reply With Quote  
     

  49. #48  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    The zeta function is not equal to, as you state in your paper:

    x<sup>2</sup>/2a+π/ln(x)

    I'm not sure what you mean by π. If by π you mean π(x), the prime counting function, then this is incorrect, as π(x) is discontinuous at many values of x, and hence x<sup>2</sup>/2a+π(x)/ln(x) must be, too. On the other hand, the zeta function has a single discontinuity.

    If you just mean the constant π, then this is again wrong. The zeta function is meromorphic in the whole complex plane, with only a simple pole at 1. If x<sup>2</sup>/2a+π/ln(x) = (x/2+π)/ln(x) were meromorphic in the plane, then so would be 1/ln(x), and hence so would be ln(x). But this is famously not the case.

    So your representation for the zeta function cannot be right for very fundamental reasons--the functions you propose fail to satisfy some of the most basic analytic properties that we know the zeta function satisfies.
    Reply With Quote  
     

  50. #49  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    I don't think you got the zeta function right till now. It is the integral of 1/ln(x) and the function I de.rived just mimics tha

    It is differentiable in the whole function so is analytic.

    Mathew Cherian
    Reply With Quote  
     

  51. #50  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    Dear Jane Benet,

    Thank You.

    Mathew Cherian
    Reply With Quote  
     

  52. #51  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    I don't think you got the zeta function right till now. It is the integral of 1/ln(x)
    Zeta is not the integral of 1/ln(x).
    Reply With Quote  
     

  53. #52  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    It is differentiable in the whole function so is analytic.
    Real differentiability and complex differentiability are different things. Real differentiability doesn't imply analyticity, for example. Besides, I was talking about a function being analytic in the entire plane, which is saying more than just saying a function is analytic. Your integral of 1/ln(x) is indeed analytic, but it cannot be extended to be analytic in the whole plane, as otherwise 1/ln(x) would be analytic in the whole plane, and then ln(x) would be. But ln(x) is not.
    Reply With Quote  
     

  54. #53  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    The definition of zeta function is integral.1/ln(x) which is also called as Li(x).

    Yesterday I remember you telling me zeta function is meramorphic and has a root at 1.

    Now you think of a frequency which is 1/T. The smallest fraction of a frequency is P1/2Pi which is 1/2 which is the root. If you can reach there I think you will get a better understanding of what I am trying to say.

    Thank You.

    Mathew Cherian[/img]
    Reply With Quote  
     

  55. #54  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Mathew Cherian
    The definition of zeta function is integral.1/ln(x) which is also called as Li(x).
    No. This is not true. The zeta function is not the integral of 1/ln(x). I've given you a simple reason why it's true--zeta is meromorphic in the whole plane, so its derivative is meromorphic in the whole plane; if its derivative were 1/ln(x), then 1/ln(x) is meromorphic in the whole plane; the reciprocal of a function meromorphic in the whole plane is again meromorphic in the whole plane, and so ln(x) must be meromorphic in the whole plane, which it is not.

    However, you correctly state that the integral of 1/ln(x) is Li(x). This implies zeta and Li(x) are two different things.

    Yesterday I remember you telling me zeta function is meramorphic and has a root at 1.
    You know, one of the nice things about forums is you can check what people said in the past. If you decided to check what I said about the behavior of zeta at 1, you'd note that I said zeta has a pole at 1. This is similar to but not the same as 1 being a root of a function--since 1 is a pole of zeta, 1 is a root of 1/zeta.

    Please, if you're going to quote me, get it right.

    Now you think of a frequency which is 1/T. The smallest fraction of a frequency is P1/2Pi which is 1/2 which is the root. If you can reach there I think you will get a better understanding of what I am trying to say.
    This makes no sense at all. Please explain yourself more carefully.

    On second thought, don't bother.
    Reply With Quote  
     

  56. #55  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    I will advice you to please refer the page I mentioned in my paper on my website in the Reference section the Book I mentioned is by Stephen Hawkins which will be easily available even now in UK.

    ______________________
    Mathew Cherian
    Reply With Quote  
     

  57. #56  
    Forum Freshman
    Join Date
    Mar 2008
    Location
    India
    Posts
    53
    [

    You know, one of the nice things about forums is you can check what people said in the past. If you decided to check what I said about the behavior of zeta at 1, you'd note that I said zeta has a pole at 1. This is similar to but not the same as 1 being a root of a function--since 1 is a pole of zeta, 1 is a root of 1/zeta..
    I liked it. Once I tried to tell you about the f'(x) being a root which is nothing but the pole. The zeros are the roots of the numerator and poles the roots of the denominator function.

    Also the zeta function is reduced to a second order differential equation which means it has only two roots.

    Now you think of a frequency which is 1/T. The smallest fraction of a frequency is P1/2Pi which is 1/2 which is the root. If you can reach there I think you will get a better understanding of what I am trying to say.
    In the same vain to my explanation above, zeta function is downward sloping function, with each of the root pairs depicting a polynomial. For different frequencies applied we have different sine functions and each cycle of the frequency has a root just at 1/2.

    _____________________
    Mathew Cherian
    Reply With Quote  
     

  58. #57  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Here is hopefully a real proof of the Riemann hypothesis: http://arxiv.org/abs/0807.0090v2

    PDF file: http://arxiv.org/PS_cache/arxiv/pdf/...807.0090v2.pdf
    Reply With Quote  
     

  59. #58  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Interesting... This guy is a serious mathematician--the names he drops at the beginning of his paper are some heavyweight number theorists. I'd read the paper, but it quickly becomes too analytic for my tastes. Did you see this announced somewhere, Jane?
    Reply With Quote  
     

  60. #59  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Yeah, it was posted at the Math Help Forum: http://www.mathhelpforum.com/math-he...ypothesis.html (scroll down to the last post in the thread)
    Reply With Quote  
     

  61. #60  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Oof, I feel like it's my duty to join that forum and fix the misconceptions that TPH is spreading. RH for number fields is not solved--RH for number fields implies RH for the zeta function! And RH is indeed number theory. It's largely of analytic interest, but even the algebraists will find uses for it.
    Reply With Quote  
     

  62. #61  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Quote Originally Posted by serpicojr
    Oof, I feel like it's my duty to join that forum and fix the misconceptions that TPH is spreading.
    Yes, please do! TPH is a pompous old twit on that forum – it’s time someone put him to rights.
    Reply With Quote  
     

  63. #62  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    Sorry I was away for sometime. I saw your postings though since I was somewhere else and forgot my password, I couldn't make my comments on your posts.

    I saw your post on XianLi(mispelt I think) and I saw you were all euphoric about it. Please check the following website for the state of that proof.
    http://science.slashdot.org/article....52248&from=rss

    I followed Janets direction and went to check the proof. It is full of Analysis. The first four lines I saw there were some errors like he taking the limits between 1 and n and I stoped.

    My question is if one proves it through Analysis, how are they ever going to prove the practical direction of this conjencture unless one has some graphical proof for it.

    I have also to clarify that for Riemann Hypothesis s=2 means he directed us only to the 2nd order part of the whole Zeta function. After that it can be even a 3rd order in which case the s=3.
    So if you refer to the graph in my website riemann.co.in you will see that I have the second order part clearly graphed.

    I also saw another Purdue Mathematician has a 124 page proof. I haven't seen any comments on it.

    __________________
    Mathew Cherian
    ____________________--
    Mathew Cherian
    Reply With Quote  
     

  64. #63  
    Forum Freshman
    Join Date
    Jan 2007
    Posts
    46
    Guys, I think one has to first look at the zeta function as the complex-contour integral function that it really is:



    There's no ambiguity there. That is exactly what it is. The lollipop symbol at the bottom of the integral is called a "Hankel contour". It's tough to understand at first: It's a branch-cut contour over the principal branch of the integrand. Note the complicated expression reduces to the Euler sum when Re(s)>1. Now, that integral expression is the "analytic continuation" of the Euler sum to the half-plane Re(s)<= 1.

    Keep in mind when someone talks of the zeta function, they mean that integral expression and not just the Euler sum which it reduces to when Re(s)>1.

    It's complex-analysis mumbo-jumbo I know. But that is the only way to fully understand what the zeta function is and what the Riemann Hypothesis means.
    Reply With Quote  
     

  65. #64  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    The equation Zetaman has given looks spooky. According to Prof. Hopkins it is for when s=2 that Riemann predicted the roots to be 1/2 for the real part. Then why all the mumo-jumbo about gamma discounting and all when he says he has a negative numerator when s is subtracted from 1.

    _________________-
    Mathew Cherian
    Reply With Quote  
     

  66. #65  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    I don't think I have a problem in understanding the Riemann Zeta function. Comprehensive and easy explanation of it is given by Prof. Hopkins in his book "God created integers". I don't think there can be a simpler explanation other than this.

    So many might have tried to explain or state the problem in many different ways, but the exact statement of the problem I believe is given by Prof. Hopkins. Have a look at it and don't be diluded by other writings which can take one to the wrong directions.

    _______________
    Mathew Cherian
    Reply With Quote  
     

  67. #66  
    Forum Freshman
    Join Date
    Jan 2007
    Posts
    46
    The equation I gave above is the equation Riemann gave in his 1859 paper although he used an even more spooky-looking integral:



    Same dif except I'm reversing the direction of the path. I'm curious about the book; I'll try and find a copy.
    Reply With Quote  
     

  68. #67  
    Forum Freshman
    Join Date
    Mar 2007
    Posts
    32
    Quote Originally Posted by imag94
    I don't think I have a problem in understanding the Riemann Zeta function. Comprehensive and easy explanation of it is given by Prof. Hopkins in his book "God created integers". I don't think there can be a simpler explanation other than this.

    So many might have tried to explain or state the problem in many different ways, but the exact statement of the problem I believe is given by Prof. Hopkins. Have a look at it and don't be diluded by other writings which can take one to the wrong directions.

    _______________
    Mathew Cherian
    You mean Prof. Hawking?
    where are we? what are we?
    Reply With Quote  
     

  69. #68  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    Yes indeed, Prof. Stepen Hawkings and the book is "God created Integers", start at pp 882, I think if my memory is correct.

    Thanks.

    ____________
    Mathew Cherian
    Reply With Quote  
     

  70. #69  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    Can someone teach me how to enter all the symbols in formulas in this Forum reply boxes. Is it like entering the tags and placing the verbal name of the symbol. Like integral as
    Code:
    integral
    or or [img]integral[/img]

    Thank you.
    _____________
    Mathew Cherian[/tex]
    Reply With Quote  
     

  71. #70  
    Forum Ph.D.
    Join Date
    Apr 2008
    Posts
    956
    Type [tex] \int [/tex] – that will give you (the integral sign). Or [tex] \displaystyle \int [/tex] for (a bigger one).

    This is called TeX (a variant of LaTeX). There is a sticky at the top of the index page for this Mathematics subforum, which you might want to check out. Please do. It’s always a pleasure to me to see my hard work being appreciated.
    Reply With Quote  
     

  72. #71  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    Thanks Jane, but is there some way we can generate them without learing the codes? It would have been easier if you provide these symbol tables from which we can select the symbols as and when we require it.

    The way you explain the coding part nullifies the effort required to create the formulas in the first place. It is easier to get to solutions than to generate these formulas in codes for it's public consumption.

    Sorry for the indulgence.

    Nice work and I appreciate it.


    __________________
    Mathew Cherian
    Reply With Quote  
     

  73. #72  
    Suspended
    Join Date
    Apr 2008
    Posts
    2,176
    Quote Originally Posted by JaneBennet
    Quote Originally Posted by Mathew Cherian
    If you think I had Professors like Weber or Gauass or Lagrange to tackle a problem like this you mistake me. I had normal modern day Profs. who taught me and get interested in these sort of things.
    I never thought anything of that sort. In fact, I was – and still am – of the belief that the teachers who taught you maths belong to the same category of teachers as those who taught William McCormick physics.

    I think it’s time this thread was moved to Pseudoscience. We don’t have a Pseudomathematics area but the Pseudoscience will just do for this thread.

    Jane until you can explain the false theory of attraction as a force. I would use another analogy.

    My good teachers were good. My bad teachers were bad.





    Sincerely,


    William McCormick
    Reply With Quote  
     

  74. #73  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by JaneBennet
    Here is hopefully a real proof of the Riemann hypothesis: http://arxiv.org/abs/0807.0090v2

    PDF file: http://arxiv.org/PS_cache/arxiv/pdf/...807.0090v2.pdf
    I haven't seen this noted elsewhere in this thread so I guess I should interject that this paper has been withdrawn. There was, not surprisingly given the difficulty of the problem, a serious mistake in the original version.

    The Riemann Hypothesis remains a very open problem.

    For those with interest here is a statement of the problem by Enrico Bombieri that is the official form of the problem for purposes of the Millenium Problems of the Clay Institute.

    http://www.claymath.org/millennium/R...is/riemann.pdf
    Reply With Quote  
     

  75. #74  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    To Dr. Rocket,

    The Riemann Hypothesis is well explained by Prof. Stephen Hawkings in his book 'God created Integers' pp 822 which is fluid, clear and presented like an expert and I don't see any reason to look at other presentations.

    I pointed out what you found out in an earlier thread.

    Thanks.

    _________________
    Mathew Cherian
    Reply With Quote  
     

  76. #75  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by imag94
    To Dr. Rocket,

    The Riemann Hypothesis is well explained by Prof. Stephen Hawkings in his book 'God created Integers' pp 822 which is fluid, clear and presented like an expert and I don't see any reason to look at other presentations.

    I pointed out what you found out in an earlier thread.

    Thanks.

    _________________
    Mathew Cherian
    One reason to look at Bombieri's piece is because of the perspective that you might gain from it. Bombieri is probably as knowledgeable about the Riemann Hypothesis as anyone in the world, and one can make a good case that he is in fact the world's expert on the subject.

    You might also profitably look at Edward's book Riemann's Zeta Function and at Riemann's original paper on the subject. That paper is available, translated from the original German at the Clay Mathematics Institute web site, which is where Bombieri's description is also posted. The Riemann Hypothesis is one of the Milenium Problems, the solution to which carries a $1 million prize. It is the only one of the Milenium problems that was also a Hilbert Problem.
    Reply With Quote  
     

  77. #76  
    Forum Sophomore
    Join Date
    Jul 2007
    Location
    South Africa
    Posts
    196
    The complexity of the formulas in Xian-Jin Li 's paper are staggering. The graph theoretic approach to the equivalent of the RH just requires "order of" computations of sums as its' most complex part.

    The graph theoretic way you have sums in which the primes contribute zero odd cycles and one even cycle, the Farrey sequence equivalent has a sum where the primes contribute the largest number, so the idea is to use two equivalences so that the one way's deficiencies is easily compensated for by the other one.
    It also matters what isn't there - Tao Te Ching interpreted.
    Reply With Quote  
     

  78. #77  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    With due respect to all of you who goes in search of what Riemann Hypothesis is, what I have to say is once a good grasp of the problem is attained and a solution is proposed then it no longer is a problem that need further study for the Solver.

    I appreciate your caring nature for the fellow human being who is interested in this problem or similar problems where one has to get a clear grasp of the problem before solving it.

    But for me personaly it is no longer a problem without a solution rather I believe from the solution one has to row further ahead to find out what it actually does in each field may be in EE, Physics and even in Social Science. I mean the regularity present in nature in it's near and deeper indirections.

    Thanks for being considerate but I feel you should spend more time on my proof if you haven't comprehended it yet. At least you should check the first graph I have presented and integrate it between 2 and 1/2 and see that it is zero what Riemann himself predicted. Also you can look in the same graph and see that at 3 the integral just falls at 3 meaning there are 3 primes before 3 including 3 and excluding 0 which also is what Riemann and others like Euler predicted. See if this makes sense now when you read my paper.

    Thanks guys.

    _______________________
    Mathew Cherian
    Reply With Quote  
     

  79. #78  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    To give you some more excitement, I feel you should probe further into see the association of RH proof in Astronomy.

    _________________
    Mathew Cherian
    Reply With Quote  
     

  80. #79  
    Forum Freshman
    Join Date
    Jul 2008
    Posts
    10
    One more thing one should keep in mind when one wanders away from the original problem and it's explanation by others other than the problem formulators is that in later times even Mathematics has taken very deviant indirections through modern Analysis which obscures the real phenomenon on which the scales of Mathematics is fitted.

    __________________
    Mathew Cherian
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •