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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 |
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September 23rd, 2012, 11:34 AM | #2 | |
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6 | Re: Is Riemanns hypothesis proven wrong? Quote:
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 | |
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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...
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September 24th, 2012, 04:11 AM | #4 | |
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory | Re: Is Riemanns hypothesis proven wrong? Quote:
The definition you wrote, But riemann conjectured that all the zeros of zeta lies in the region | |
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October 2nd, 2012, 04:16 AM | #5 | |
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond | Re: Is Riemanns hypothesis proven wrong? Quote:
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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.
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October 2nd, 2012, 10:54 AM | #7 | |
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory | Re: Is Riemanns hypothesis proven wrong? Quote:
This is very important in calculating the values of Let us calculate we know that See, how useful it is? . | |
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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 |
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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 |
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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 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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