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| Mathew Cherian |
 Posted: Fri May 23, 2008 10:48 am Post subject: Riemann Hypothesis Proof ! |
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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
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| thyristor |
| Posted: Fri May 23, 2008 12:05 pm Post subject: |
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What is the Rieman Hypothesis?
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| serpicojr |
| Posted: Fri May 23, 2008 2:00 pm Post subject: |
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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. |
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| serpicojr |
| Posted: Fri May 23, 2008 2:09 pm Post subject: |
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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) = x2/2a 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. |
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| Mathew Cherian |
| Posted: Fri May 23, 2008 2:20 pm Post subject: |
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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.
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| Mathew Cherian |
| Posted: Fri May 23, 2008 2:29 pm Post subject: |
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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.
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| serpicojr |
| Posted: Fri May 23, 2008 2:37 pm Post subject: |
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| Mathew Cherian wrote: |
| it reaches a limit of 6/pi^2. |
You're confused here. You're thinking of the fact that ζ(2) = π2/6.
| Quote: |
| 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.
| Quote: |
| 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. |
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| serpicojr |
| Posted: Fri May 23, 2008 2:40 pm Post subject: |
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| Mathew Cherian wrote: |
| 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. |
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| Mathew Cherian |
| Posted: Fri May 23, 2008 3:07 pm Post subject: |
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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.
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| serpicojr |
| Posted: Fri May 23, 2008 3:20 pm Post subject: |
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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 x2/2a = 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))2 = ln(x)/2-1/2ln(x)
This is not 1/ln(x), so your integration is wrong. |
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| Mathew Cherian |
| Posted: Fri May 23, 2008 4:29 pm Post subject: |
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I feel stupid--I've been calling you Matthew, not Mathew. Sorry.
| serpicojr wrote: |
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 x2/2a = 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))2 = 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 x2/2a= x2/ln xx. 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 4a2. 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/pi2.
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.
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| Mathew Cherian |
| Posted: Fri May 23, 2008 4:35 pm Post subject: |
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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.
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| Mathew Cherian |
| Posted: Fri May 23, 2008 8:09 pm Post subject: |
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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+ 1/22 + 1/32+1/42......+1/x2+.. for s=2.
So my transformed function x2/2a 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.
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| serpicojr |
| Posted: Fri May 23, 2008 9:21 pm Post subject: |
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| Mathew Cherian wrote: |
| Your expression is wrong, let me tell you x2/2a= x2/ln xx. You should note that your denominator is wrong. |
Look:
x2/2a = x2/2ln(xx) = x2/2xln(x) = x/2ln(x)
Please don't use these abbreviations.
| Quote: |
| In fact zeta(x)/Li(x) = 1 as x tends higher. |
No, π(x)/Li(x) tends to 1 as x tends to infinity.
| Quote: |
| 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.
| Quote: |
| Euler has shown it seems that the zeta function can be reduced to a time series = 1+ 1/22 + 1/32+1/42......+1/x2+.. 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.
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| So my transformed function x2/2a is only the closed form state of the Euler Time series. |
I have no idea how you figure this. |
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| Mathew Cherian |
| Posted: Sat May 24, 2008 12:00 am Post subject: |
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| serpicojr wrote: |
Look:
x2/2a = x2/2ln(xx) = x2/2xln(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 logxx. 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.
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