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- Thread starter dimension10
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In summary, the conversation revolves around a proof attempting to disprove the Riemann Hypothesis by using the Gamma function and other mathematical concepts. The conversation includes discussions about the validity of certain equations and the treatment of infinities as numbers. Ultimately, the proof is deemed to be incorrect as it does not properly disprove the Riemann Hypothesis.

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Anyway, you say in your proof:

Solving the first equation with respect to [itex]\Gamma(\frac{1}{2}+it)[/itex] 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 [itex]\Gamma(\frac{1}{2}+it)=1[/itex] which isn't true for all t.

And I don't quite see why

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

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

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Dschumanji

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Ashwin_Kumar

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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)

[itex]\Gamma(\frac{1}{2}+it)=...[/itex]

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.

When you write( i have put three dots instead because i don't wish to copy down the whole thing)

[itex]\Gamma(\frac{1}{2}+it)=...[/itex]

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.

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dimension10

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Ashwin_Kumar said: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)

[itex]\Gamma(\frac{1}{2}+it)=...[/itex]

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

Ashwin_Kumar said: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?

Ashwin_Kumar said: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.

Ashwin_Kumar said: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.

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dimension10

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Dschumanji said:

Even taking it as a limit, you get infinite.

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dimension10

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micromass said: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 [itex]\Gamma(\frac{1}{2}+it)=1[/itex] 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.

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dimension10

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micromass said: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.

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dimension10 said: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

[tex]\Gamma(\frac{1}{2}+it)=1[/tex]

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

What you could perhaps say is that

[tex]e^{i\infty}=\lim_{x\rightarrow}{e^{ix}}[/tex]

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

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dimension10 said:Ok. I understood. I'll try to do that. Thanks.

Don't bother, the series converges.

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Dschumanji

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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.dimension10 said:Even taking it as a limit, you get infinite.

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:

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

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:

[itex]0\cdot\infty[/itex]

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

[itex]\infty = \frac{c}{0}[/itex]

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).

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[tex]\frac{c}{0}=\infty[/tex]

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

[tex]f(x)=\frac{c}{x}[/tex]

is meromorphic and takes on the value [itex]\infty[/itex] in 0.

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

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Dschumanji

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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 said:

[tex]\frac{c}{0}=\infty[/tex]

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

[tex]f(x)=\frac{c}{x}[/tex]

is meromorphic and takes on the value [itex]\infty[/itex] in 0.

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

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Dschumanji said: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 [itex]0\cdot \infty[/itex], [itex]\infty+\infty[/itex], [itex]\infty\cdot \infty[/itex] and [itex]0/0[/itex] will remain undefined. See http://en.wikipedia.org/wiki/Riemann_sphere

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Dschumanji

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Ah, thank you, Micromass!micromass said:Things like [itex]0\cdot \infty[/itex], [itex]\infty+\infty[/itex], [itex]\infty\cdot \infty[/itex] and [itex]0/0[/itex] will remain undefined. See http://en.wikipedia.org/wiki/Riemann_sphere

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dimension10

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

Isn't [tex]\infty\times \infty=\infty[/tex] and [itex]\infty+\infty=\infty[/itex]

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disregardthat

Science Advisor

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dimension10 said:Isn't [tex]\infty\times \infty=\infty[/tex] and [itex]\infty+\infty=\infty[/itex]

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

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dimension10

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Nebuchadnezza

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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.

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dimension10 said:Isn't [tex]\infty\times \infty=\infty[/tex] and [itex]\infty+\infty=\infty[/itex]

Not in the complex numbers.

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3.1415926535

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Nebuchadnezza said:I also made a discovery ! This graph here shows that clearly [tex]\sum_{k=1}^{\infty }\frac{1}{k} = 0 [/tex]

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

No it doesn't. It shows that [tex]\lim_{x\rightarrow \infty}\frac{1}{x}=0[/tex]

Here is the graph for the sum

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

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3.1415926535 said:No it doesn't. It shows that [tex]\lim_{x\rightarrow \infty}\frac{1}{x}=0[/tex]

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

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disregardthat

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micromass said:I think you didn't pick up the sarcasm

You'll get warned for less!

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3.1415926535

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micromass said:I think you didn't pick up the sarcasm

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

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3.1415926535 said:Pardon me, I didn't know Nebuchadnezza was being sarcastic

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

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Ashwin_Kumar

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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)

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dimension10

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Ashwin_Kumar said:

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.

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3.1415926535

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dimension10 said: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

[tex]sin(\frac{\pi }{4}+\frac{\pi it}{2})=sin(\pi(\frac{2it+1}{4}))\neq sin(\pi(\frac{4it+1}{4}))[/tex]

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Ashwin_Kumar said: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 [itex]1/0=\infty[/tex] and we do that in the Riemann sphere. Taking different functions, like 1/x, 1/x

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Ashwin_Kumar

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dimension10

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3.1415926535 said:As I can see you have made a mistake in the 2nd line of the third page

[tex]sin(\frac{\pi }{4}+\frac{\pi it}{2})=sin(\pi(\frac{2it+1}{4}))\neq sin(\pi(\frac{4it+1}{4}))[/tex]

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

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

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Hurkyl

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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

- [itex]e^{x - \infty} =0 [/itex ]
- [itex]\log (x + i \infty) = \infty + i \frac{\pi}{2}[/itex]

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Ashwin_Kumar

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

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