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 September 23rd, 2012, 09:17 AM #1 Newbie   Joined: Sep 2012 Posts: 12 Thanks: 0 Is Riemanns hypothesis proven wrong? This proof proves Riemann's hypothesis wrong, I've posted it elsewhere but I never got an answer whats wrong with this proof. Maybe someone could shed some light on this for the benefit of my friend Semjase. From Euler's equation e^(i*pi) = -1 Let 0 + 0i = 1/1^s + 1/2^s + 1/3^s + 1/4^s . . . Moving 1/2^s to the left side of the equation you get -1/2^s = 1/1^s + 1/3^s + 1/4^s . . . substituting -1 = e^i*pi e^(i*pi)/2^s = 1/1^s + 1/3^s + 1/4^s . . . add 1/2^s to both sides of the equation 1/2^s + e^(i*pi)/2^s = 1/1^s + 1/2^s + 1/3^s + 1/4^s . . . which gives 0 + 0i = 1/2^s + e^(i*pi)/2^s multiply both sides 2^(2*s) 0 + 0i = 2^s + (2^s)*e^(pi*i) next 0 + 0i = (2^a)*2^(q*i) + (2^a)*2^(q*i)*e^(pi*i) let 2^q = e^pi 0 + 0i = (2^a)*e^(pi*i) + (2^a)*e^(pi*i)*e^(pi*i) let e^(pi*i) =-1 0 + 0i = (2^a)*-1 + (2^a)*-1*-1 0 + 0i = - 2^a + 2^a Now (a) can be any real number therefore Riemann's hypothesis is incorrect
September 23rd, 2012, 11:34 AM   #2
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Re: Is Riemanns hypothesis proven wrong?

Quote:
 Originally Posted by Semjase This proof proves Riemann's hypothesis wrong, I've posted it elsewhere but I never got an answer whats wrong with this proof. Maybe someone could shed some light on this for the benefit of my friend Semjase. From Euler's equation e^(i*pi) = -1 Let 0 + 0i = 1/1^s + 1/2^s + 1/3^s + 1/4^s . . .
This makes no sense at all. Do you mean "if" that

Moving 1/2^s to the left side of the equation you get

-1/2^s = 1/1^s + 1/3^s + 1/4^s . . .

substituting -1 = e^i*pi

e^(i*pi)/2^s = 1/1^s + 1/3^s + 1/4^s . . .

add 1/2^s to both sides of the equation

1/2^s + e^(i*pi)/2^s = 1/1^s + 1/2^s + 1/3^s + 1/4^s . . .

which gives

0 + 0i = 1/2^s + e^(i*pi)/2^s

multiply both sides 2^(2*s)

0 + 0i = 2^s + (2^s)*e^(pi*i)

next

0 + 0i = (2^a)*2^(q*i) + (2^a)*2^(q*i)*e^(pi*i)

let 2^q = e^pi

0 + 0i = (2^a)*e^(pi*i) + (2^a)*e^(pi*i)*e^(pi*i)

let e^(pi*i) =-1

0 + 0i = (2^a)*-1 + (2^a)*-1*-1

0 + 0i = - 2^a + 2^a

Now (a) can be any real number therefore Riemann's hypothesis is incorrect[/quote]
That's non-sense. What you have proved is that "0= 0". Do you not understand that if you manipulate an equation, multiplying both sides of an equation by something involving the variable "x", you can introduce "new" solutions, getting an equation that has solutions the original equation didn't? For example to "disprove" the statement "The only solution to x- 1= 0 is 1" do the following: mutiply both sides of the equation by x to get $x^2- x= 0$. That factors as $x(x- 1)= 0$ showing that both 1 and 0 satisfy the equation. You have simply manipulated the original equation to get a new equation that has all real numbers as solution. That says nothing about what numbers satisfy the original equation.

 September 23rd, 2012, 04:49 PM #3 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Is Riemanns hypothesis proven wrong? Yes and in any case your original equation is not even the expression of Zeta in the strip ]0,1[, it is only valid for Re(s)>1 and it is known that Zeta does not have zero in that region...
September 24th, 2012, 04:11 AM   #4
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Re: Is Riemanns hypothesis proven wrong?

Quote:
 Originally Posted by Semjase Let 0 + 0i = 1/1^s + 1/2^s + 1/3^s + 1/4^s . . .
This is not the riemann zeta. The formal definition of riemann zeta is the functional equation :

$\zeta(s)= 2^s \pi^{s-1} \sin$$\frac{\pi s}{2}$$ \Gamma(1-s) \zeta(1-s)$.

The definition you wrote,

$\zeta(s)= \sum_{k=1}^{\infty} \frac{1}{k^s}$ is only valid for $\Re(s) > 1$.

But riemann conjectured that all the zeros of zeta lies in the region $\Re(s)= \frac{1}{2}$. And so, your proof isn't valid for this case since 1/2<1

October 2nd, 2012, 04:16 AM   #5
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Re: Is Riemanns hypothesis proven wrong?

Quote:
 Originally Posted by mathbalarka The formal definition of riemann zeta is the functional equation : $\zeta(s)= 2^s \pi^{s-1} \sin$$\frac{\pi s}{2}$$ \Gamma(1-s) \zeta(1-s)$.
This appears to define the zeta in terms of itself.

 October 2nd, 2012, 10:00 AM #6 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Is Riemanns hypothesis proven wrong? This is the functional equation of Zeta, showing a symmetry of the function with respect to the critical line Re(s)=1/2. I don't know if it is the formal definition, but it I think it defines Zeta uniquely.
October 2nd, 2012, 10:54 AM   #7
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Re: Is Riemanns hypothesis proven wrong?

Quote:
 Originally Posted by greg1313 This appears to define the zeta in terms of itself.
Yes. Every functional equation defines the function used in the equation uniquely (maybe) in terms of that function.

This is very important in calculating the values of $\zeta(s)$ for Re[s] < 1.

Let us calculate $\zeta(-1)$ using this. We will get :

$\zeta(-1)= 2^{-1} \pi^{-2} \sin$$\frac{-\pi}{2}$$ \Gamma(2)\zeta(2)$

we know that $\zeta(2)= \pi^2/6$;

$\Rightarrow \zeta(-1)= 2^{-1} \pi^{-2} (-1) 1! \pi^2 6^{-1}$

$= - \frac{1}{12}$

See, how useful it is?

.

 October 2nd, 2012, 11:10 AM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Is Riemanns hypothesis proven wrong? Of course the functional equation only tells you how to find one side when you already know how to find the other. The usual way, I think, is to define it for Re(s) > 1 and then use the analytic continuation to define it for the rest of $s\ne 1.$
 October 2nd, 2012, 12:12 PM #9 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Is Riemanns hypothesis proven wrong? Yes Mathbalarka, that is what Ramanujan wrote in his letter to Hardy, namely $1+2+3+...=\frac{-1}{12}$. They thought this guy was crazy, but actually he knew this property of Zeta but did not know the right notation. I am not sure if he knew the Riemann paper or if he comes to it by himself..knowing how great of a mathematician he was, that would not surprised me!
 October 2nd, 2012, 02:26 PM #10 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Is Riemanns hypothesis proven wrong? If Dougy is right, that $\zeta(-1)= 1 + 2 + 3 + ... = -\frac{1}{12}$ why do we accept this result? To me it doesn't make sense, the sum on the left is not even close to the value on the right. Can somebody explain the underlying philosophy that would force us to conclude such an 'absurdity', IMHO

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