Disproof of Riemann Hypothesis

[dimension10]

I think I have managed to disprove the Riemann Hypothesis. Hope I have not made a calculation error.

 

[micromass]

So, if the Riemann hypothesis isn't true, can you give me a zero not on the critical line?

Anyway, you say in your proof:

Solving the first equation with respect to \Gamma(\frac{1}{2}+it) gives:

I don't quite see how you obtained the equation that comes after it. Your numerator and denominator are the same thing, thus that would imply \Gamma(\frac{1}{2}+it)=1 which isn't true for all t.

And I don't quite see why

\sum_{n=1}^{+\infty}{\frac{t}{2n}-atan(\frac{2t}{4n+1})}=+\infty

Fine, your calculator says it is, but you need to prove that this is the case. (and by the way, I think it converges).

 

[Dschumanji]

The biggest problem I see with this proof is that you are treating infinities as numbers instead of examining the limits of the expressions. I also agree with Micromass that some of the expressions you introduce are non sequitur.

 

[Ashwin_Kumar]

Simply in the equations, you have not proved the hypothesis wrong, and you have made a couple of errors.
When you write( i have put three dots instead because i don't wish to copy down the whole thing)

\Gamma(\frac{1}{2}+it)=...

You are in essence not correctly solving for the gamma function. Your derivation of one for the gamma function is quite impossible indeed. If you look at the next equation involving the cosine, your gamma function does not make sense at all. Your equations are contradicting themselves. The previous solution for the gamma function(the equation before you 'solve' the gamma function) contradicts your solution of one.
Furthermore, the summation is definitely not divergent, you can test that out yourselves.
Finally, give me an actual example. At this point, pi has nothing to do with it.
The riemann hypothesis maybe unproven as of yet, but nor has it been disproved. The paper simply does not disprove the hypothesis.

 

[dimension10]

Simply in the equations, you have not proved the hypothesis wrong, and you have made a couple of errors.
When you write( i have put three dots instead because i don't wish to copy down the whole thing)

\Gamma(\frac{1}{2}+it)=...

You are in essence not correctly solving for the gamma function. Your derivation of one for the gamma function is quite impossible indeed. If you look at the next equation involving the cosine, your gamma function does not make sense at all.

cotangent=tangent^-1
tangent=sin/cos

Your equations are contradicting themselves. The previous solution for the gamma function(the equation before you 'solve' the gamma function) contradicts your solution of one.

Where is the contradiction?

Furthermore, the summation is definitely not divergent, you can test that out yourselves.

A calculator showed it correct. And, a function can converge to infinity. This series does become infinity.

Finally, give me an actual example. At this point, pi has nothing to do with it.
The riemann hypothesis maybe unproven as of yet, but nor has it been disproved. The paper simply does not disprove the hypothesis.

Riemann Hypothesis has to do with the Gamma function, Z-function, Zeta function, Theta function... And these have to do with pi.

 

[dimension10]

The biggest problem I see with this proof is that you are treating infinities as numbers instead of examining the limits of the expressions. I also agree with Micromass that some of the expressions you introduce are non sequitur.

Even taking it as a limit, you get infinite.

 

[dimension10]

I don't quite see how you obtained the equation that comes after it. Your numerator and denominator are the same thing, thus that would imply \Gamma(\frac{1}{2}+it)=1 which isn't true for all t.

That is about the Theta function. Instead of the theta function itself, I used its definition. That part is a direct implication of the second equation there.

 

[dimension10]

Fine, your calculator says it is, but you need to prove that this is the case. (and by the way, I think it converges).

Ok. I understood. I'll try to do that. Thanks.

 

[micromass]

That is about the Theta function. Instead of the theta function itself, I used its definition. That part is a direct implication of the second equation there.

Well, I don't see how it is a direct implication. Can you elaborate? Do you agree that your implication implies that

\Gamma(\frac{1}{2}+it)=1

Also, I have a qual about i\infty that you seem to use. That notation doesn't really make sense in the complex numbers. The only way you can talk about \infty in the complex numbers is by looking at the Riemann-sphere. But e^x is not a meromorphic function, so you can't just say that e^{i\infty}=0.

What you could perhaps say is that

e^{i\infty}=\lim_{x\rightarrow}{e^{ix}}

but I'm afraid that this limit will not exist...

 

[micromass]

Ok. I understood. I'll try to do that. Thanks.

Don't bother, the series converges.

 

[Dschumanji]

Even taking it as a limit, you get infinite.

I don't understand how this is a justification for what you are doing. By letting infinity act as a number you end up with nonsensical conclusions such as the exponential function equaling zero. The exponential function is never equal to zero. The exponential function only approaches zero as its argument decreases without bound. This is only properly stated as a limit.

Let us suppose that infinity and negative infinity are numbers and that the exponential function does equal zero at negative infinity. The only reason you say this in your proof is because the expression:

\sum_{n=1}^{\infty}{\frac{t}{2n}-atan(\frac{2t}{4n+1})}=\infty

is equal to infinity when t is equal to infinity. You then go on to conclude that the denominator in your equations must necessarily equal zero because zero times a number is zero, right? Not exactly. If you let t equal infinity in the other expressions of your denominator you will ultimately end up with the expression:

0\cdot\infty

This expression is undefined! Don't believe me? You implicitly assume the following identity in your proof:

\infty = \frac{c}{0}

I'm going to ignore the fact that this expression makes absolutely no sense and is undefined for the moment. You are implicitly allowing division by zero to be defined by the normal rules of division. This means I can multiply both sides of the equation by zero and cancel out the zeros on the right side of the equation. This means that zero times infinity can equal anything we desire since c is any number. I could also multiply both sides of the equation by zero and let the right side just equal zero since the normal rules of multiplication apply. This is a contradiction to the previous result. Ultimately the expression must be nonsense if we let the normal rules of multiplication and division apply.

I hope this helps you understand the problems of treating infinity as a number (and the problems of division by zero).

 

[micromass]

Things like

\frac{c}{0}=\infty

are actually ok is complex analysis if we are working in the Riemann sphere (and if c is not zero). What it actually is saying is that the function

f(x)=\frac{c}{x}

is meromorphic and takes on the value \infty in 0.

However, I don't think the OP really meant all of this...

 

[Dschumanji]

Things like

\frac{c}{0}=\infty

are actually ok is complex analysis if we are working in the Riemann sphere (and if c is not zero). What it actually is saying is that the function

f(x)=\frac{c}{x}

is meromorphic and takes on the value \infty in 0.

However, I don't think the OP really meant all of this...

I have heard of the Riemann sphere, but have no background in complex analysis. Would zero times infinity be defined on the Riemann sphere, if so what would it be? I would think it would still give answers which are undefined.

 

[micromass]

I have heard of the Riemann sphere, but have no background in complex analysis. Would zero times infinity be defined on the Riemann sphere, if so what would it be? I would think it would still give answers which are undefined.

Things like 0\cdot \infty, \infty+\infty, \infty\cdot \infty and 0/0 will remain undefined. See http://en.wikipedia.org/wiki/Riemann_sphere

 

[Dschumanji]

Things like 0\cdot \infty, \infty+\infty, \infty\cdot \infty and 0/0 will remain undefined. See http://en.wikipedia.org/wiki/Riemann_sphere

Ah, thank you, Micromass!

 

[dimension10]

Things like 0\cdot \infty, \infty+\infty, \infty\cdot \infty and 0/0 will remain undefined. See http://en.wikipedia.org/wiki/Riemann_sphere

Isn't \infty\times \infty=\infty and \infty+\infty=\infty

 

[disregardthat]

Isn't \infty\times \infty=\infty and \infty+\infty=\infty

Well, what are you referring to by "\infty"? In the extended reals, only the latter is true.

 

[dimension10]

Ok, I don't know what's wrong with my calculator but with excell sheets, I got a graph which strongly suggests convergence. I have attached that file here. The sum seems to be around t /times /sqrt{3}

 

[Nebuchadnezza]

I also made a discovery ! This graph here shows that clearly \sum_{k=1}^{\infty }\frac{1}{k} = 0

http://www.wolframalpha.com/input/?i=1/x++++from+0+to+110

...

You simply can't deduse facts from simple looking graphs, even if the graph looks like one thing, the truth could be something completely different. What you has to do, is prove it! Not just say that it is true, because of a graph that might be true.

 

[micromass]

Isn't \infty\times \infty=\infty and \infty+\infty=\infty

Not in the complex numbers.

 

[3.1415926535]

I also made a discovery ! This graph here shows that clearly \sum_{k=1}^{\infty }\frac{1}{k} = 0

http://www.wolframalpha.com/input/?i=1/x++++from+0+to+110

No it doesn't. It shows that \lim_{x\rightarrow \infty}\frac{1}{x}=0
Here is the graph for the sum

http://www.wolframalpha.com/input/?i=Sum+1/x

 

[micromass]

No it doesn't. It shows that \lim_{x\rightarrow \infty}\frac{1}{x}=0
Here is the graph for the sum

http://www.wolframalpha.com/input/?i=Sum+1/x

I think you didn't pick up the sarcasm

 

[disregardthat]

I think you didn't pick up the sarcasm

You'll get warned for less!

 

[3.1415926535]

I think you didn't pick up the sarcasm

Pardon me, I didn't know Nebuchadnezza was being sarcastic

 

[micromass]

Pardon me, I didn't know Nebuchadnezza was being sarcastic

Me neither, but his statement made so little sense that he had to be

 

[Ashwin_Kumar]

anyway, i don't think that you C A N take 1/0 to be infinity because by using several different functions(not just f(x)=1/x), you get different results, and technically the value is totally undefined. So moving back to the equations stated in the paper, your result would simply be undefined(and no, you cannot use that to say that pi is undefined). In several functions, to derive at the value for 1/0 you use a limit, and they yield different answers. Thus we cannot take the general assumption that 1/0 is infinite, although that seems logically true.
Meanwhile, the summation does indeed seem to have a limit somewhere, and in the end if you do the whole calculation, then congratulations. You have invented a new way to find the value of pi!(sarcasm mark here)
I do not understand how you are defining your theta function.
There are several errors here, but i feel the main error would be your assumption that the series diverges and the result would be infinity(Even if it was, the function could diverge infinitely downwards, yielding a negative result. And if a series converged and the result was infinite... well that totally makes no sense at all. series can converge with the x-value at infinity but not the y-value at infinity. It simply means that the result is divergent, so your idea that it can converge to infinity is wrong).
And it D O E S make a difference, by the way, if you represent infinity as a limit rather than a number itself, and when you say "the result is still infinity", you are missing the point. You leave it as a limit, you make your infinite summation into a limit rather than replacing it with infinity)
Also, euler never wrote that that gamma function would result in 1! That is just plain mad.
(Also, don't forget that the zeta function is a complex plane. So you have a problem if you say infinity times infinity is infinity etc)

 

[dimension10]

anyway, i don't think that you C A N take 1/0 to be infinity because by using several different functions(not just f(x)=1/x), you get different results, and technically the value is totally undefined. So moving back to the equations stated in the paper, your result would simply be undefined(and no, you cannot use that to say that pi is undefined). In several functions, to derive at the value for 1/0 you use a limit, and they yield different answers. Thus we cannot take the general assumption that 1/0 is infinite, although that seems logically true.
Meanwhile, the summation does indeed seem to have a limit somewhere, and in the end if you do the whole calculation, then congratulations. You have invented a new way to find the value of pi!(sarcasm mark here)
I do not understand how you are defining your theta function.
There are several errors here, but i feel the main error would be your assumption that the series diverges and the result would be infinity(Even if it was, the function could diverge infinitely downwards, yielding a negative result. And if a series converged and the result was infinite... well that totally makes no sense at all. series can converge with the x-value at infinity but not the y-value at infinity. It simply means that the result is divergent, so your idea that it can converge to infinity is wrong).
And it D O E S make a difference, by the way, if you represent infinity as a limit rather than a number itself, and when you say "the result is still infinity", you are missing the point. You leave it as a limit, you make your infinite summation into a limit rather than replacing it with infinity)
Also, euler never wrote that that gamma function would result in 1! That is just plain mad.
(Also, don't forget that the zeta function is a complex plane. So you have a problem if you say infinity times infinity is infinity etc)

1/0=infintiy is a convention. Anyway, I found out there was a mistake. It does converge. Something's wrong with the calculator.

 

[dimension10]

Ok, so it converges. I have attached an "extension" to the Riemann Hypothesis here. It can also be found in the first attachment, but that one's last few equations are flawed since it does converge.

 

[3.1415926535]

Ok, so it converges. I have attached an "extension" to the Riemann Hypothesis here. It can also be found in the first attachment, but that one's last few equations are flawed since it does converge.

As I can see you have made a mistake in the 2nd line of the third page

sin(\frac{\pi }{4}+\frac{\pi it}{2})=sin(\pi(\frac{2it+1}{4}))\neq sin(\pi(\frac{4it+1}{4}))

 

[micromass]

anyway, i don't think that you C A N take 1/0 to be infinity because by using several different functions(not just f(x)=1/x), you get different results, and technically the value is totally undefined. So moving back to the equations stated in the paper, your result would simply be undefined(and no, you cannot use that to say that pi is undefined). In several functions, to derive at the value for 1/0 you use a limit, and they yield different answers. Thus we cannot take the general assumption that 1/0 is infinite, although that seems logically true.

We can take 1/0=\infty[/tex] and we do that in the Riemann sphere. Taking different functions, like 1/x, 1/x², etc. will all yield infinity in zero. This has been rigorously defined. See "meromorphic functions". It's a very useful concept!

 

[Ashwin_Kumar]

Ok, I think that would help micro mass. Don't know much about the riemann sphere myself. And dimension, that is an incredulously silly mistake you made there.

 

[dimension10]

As I can see you have made a mistake in the 2nd line of the third page

sin(\frac{\pi }{4}+\frac{\pi it}{2})=sin(\pi(\frac{2it+1}{4}))\neq sin(\pi(\frac{4it+1}{4}))

I think you are correct. So I would need to cut that short again. I have attached that here.

 

[micromass]

You could delete your entire document altogether, because the very first step is not valid. Or you haven't given a reason why it is valid.

Also "Euler and others showed..." should have a reference to where you found the result.

 

[Hurkyl]

On the Riemann sphere (or equivalently, the projective complex numbers), \infty \cdot \infty = \infty. But \infty + \infty is undefined as mm said.



Sometimes, other compactifications of the complex numbers are used. One common one is to let both the real and imaginary parts be extended real numbers. We can extend functions by continuity to get identities such as (for complex x):

1. e^{x - \infty} =0 [/itex ]<br /> [*]\log (x + i \infty) = \infty + i \frac{\pi}{2}

<br /> And contour integrals are often defined to use such points as endpoints. e.g. to write the inverse Laplace transform as an integral.

 

[Ashwin_Kumar]

So technically this is no longer a disproof nor a proof.

 

[Ashwin_Kumar]

I think the paper should be scrapped because it makes no sense anymore. It no longer disproves the hypothesis. Your first steps are incorrect, once again I feel that the gamma function is wrong. And your definition of the theta function, once again, does not seem to make sense. And because the paper ends with no definite result, it is meaningless.

 

[dimension10]

You could delete your entire document altogether, because the very first step is not valid. Or you haven't given a reason why it is valid.

Also "Euler and others showed..." should have a reference to where you found the result.

I just substituted the theta function and the Euler-Mascheroni constant.

So technically this is no longer a disproof nor a proof.

Yes.

I think the paper should be scrapped because it makes no sense anymore. It no longer disproves the hypothesis. Your first steps are incorrect,

I just substituted the theta function and the Euler Mascheroni Constant.

once again I feel that the gamma function is wrong.

Where is it wrong?

And your definition of the theta function, once again, does not seem to make sense.

Go to http://en.wikipedia.org/wiki/Theta_function. That is the definition of theta function I used. The general one, not the special case.

And because the paper ends with no definite result, it is meaningless.

Well but it seems to be much easier do prove the end result rather than the original Riemann Hypoothesis.

 

[amitjohar]

The only way to disprove Riemann Hypothesis is to give a counter example where the non-trivial zero of the Riemman Zeta Function does not have half as its real part. Its that simple. The disproof should not be more than a line or 2 at most. It is the proof that is expected to be very lengthy. To date the first 10 trillion non-trivial zeros of the zeta function are confirmed to be on the critical line. What you have done is really not a disproof.

 

[dimension10]

The only way to disprove Riemann Hypothesis is to give a counter example where the non-trivial zero of the Riemman Zeta Function does not have half as its real part. Its that simple. The disproof should not be more than a line or 2 at most. It is the proof that is expected to be very lengthy. To date the first 10 trillion non-trivial zeros of the zeta function are confirmed to be on the critical line. What you have done is really not a disproof.

That's why its an extension.

 

[amitjohar]

I believe that the Riemann Hypothesis is true. There is a one million dollar prize for anyone who can prove the Riemann Hypothesis. But there is no prize for the disproof of Riemman Hypothesis. This shows that RH is likely true. Why else is no prize offered for disproof of RH?

 

[pmsrw3]

I believe that the Riemann Hypothesis is true. There is a one million dollar prize for anyone who can prove the Riemann Hypothesis. But there is no prize for the disproof of Riemman Hypothesis. This shows that RH is likely true. Why else is no prize offered for disproof of RH?

Ah, the classic reductio ad praemium method of proof...

 

[micromass]

I believe that the Riemann Hypothesis is true. There is a one million dollar prize for anyone who can prove the Riemann Hypothesis. But there is no prize for the disproof of Riemman Hypothesis. This shows that RH is likely true. Why else is no prize offered for disproof of RH?

A prize is offered for the solution of the Riemann Hypothesis. So even a counterexample could get you 1000000$.

 

[pmsrw3]

A prize is offered for the solution of the Riemann Hypothesis. So even a counterexample could get you 1000000$.

Well, now we have a problem. By the method of proof advanced by amitjohar, we now have shown that the Riemann hypothesis is both true and false.

 

[micromass]

Well, now we have a problem. By the method of proof advanced by amitjohar, we now have shown that the Riemann hypothesis is both true and false.

Aha! So we have shown mathematics to be inconsistent!

 

[pmsrw3]

Aha! So we have shown mathematics to be inconsistent!

Isn't there a prize for that?