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Old 2008-06-11, 07:26   #1
Carl Fischbach
 
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Oct 2007

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Default Riemann's hypothesis is incorrect a proof

I would be greatful for any feedback on this proof so it is free of mistakes
before I take it any further.


from Euler's formula e^ipi = -1

let 0 + i0 = 1/1^s +1/2^s + 1/3^s +1/4^s . . .

now you choose any term except 1/1^s and move it to the left side of
the equation I've choosen 1/2^s

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

from Euler's formula you now have

e^ipi/2^s = 1/1^s +1/3^s +1/4^s. . .

substituting you get

0 +i0 = 1/2^s +e^ipi/2^s

multiplying both sides by 2^2s you get

0+ i0 = 2^s + 2^s*e^ipi

next

0 +i0 = 2^(a +iq)+2^(a+iq)*e^ipi

then

0 +i0 = 2^a*2^iq + 2^a*2^iq*e^ipi

let 2^q = e^pi

therefore

0 +i0 = 2^a*e^ipi + 2^a*e^ipi*e^ipi

let -1 = e^ipi

0 + i0 = 2^a*-1 + 2^a*-1*-1

add i0 to both sides

0 +i0 = -2^a + 2^a +i0

therefore

0 +i0 = 0 +i0

Note (a) can be any real value not just 1/2 therefore Riemann's
hypothesis is incorrect.
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Old 2008-06-11, 12:01   #2
cheesehead
 
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I looked for a smilie that expressed my reaction, but the best I can do is

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Old 2008-06-11, 12:06   #3
Mini-Geek
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Quote:
Originally Posted by cheesehead View Post
I looked for a smilie that expressed my reaction, but the best I can do is

I've got a few:








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Old 2008-06-11, 12:19   #4
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by Carl Fischbach View Post
I would be greatful for any feedback on this proof so it is free of mistakes
before I take it any further.


from Euler's formula e^ipi = -1

let 0 + i0 = 1/1^s +1/2^s + 1/3^s +1/4^s . . .

now you choose any term except 1/1^s and move it to the left side of
the equation I've choosen 1/2^s

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

from Euler's formula you now have

e^ipi/2^s = 1/1^s +1/3^s +1/4^s. . .

substituting you get

0 +i0 = 1/2^s +e^ipi/2^s

multiplying both sides by 2^2s you get

0+ i0 = 2^s + 2^s*e^ipi

next

0 +i0 = 2^(a +iq)+2^(a+iq)*e^ipi

then

0 +i0 = 2^a*2^iq + 2^a*2^iq*e^ipi

let 2^q = e^pi

therefore

0 +i0 = 2^a*e^ipi + 2^a*e^ipi*e^ipi

let -1 = e^ipi

0 + i0 = 2^a*-1 + 2^a*-1*-1

add i0 to both sides

0 +i0 = -2^a + 2^a +i0

therefore

0 +i0 = 0 +i0

Note (a) can be any real value not just 1/2 therefore Riemann's
hypothesis is incorrect.
Your first mistake is that you are unaware that you have no
understanding of mathematics.
Your second mistake is total ignorance about the Zeta function.
Your third mistake is not realizing that the representation that you
are using for the Zeta function ONLY CONVERGES for real(s) > 1.
Your fourth mistake is not realizing that your so-called argument is
nonsense.
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Old 2008-06-11, 14:18   #5
wblipp
 
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May 2003
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Quote:
Originally Posted by R.D. Silverman View Post
Your first mistake is that you are unaware that you have no understanding of mathematics.

Your second mistake is total ignorance about the Zeta function.

Your third mistake is not realizing that the representation that you
are using for the Zeta function ONLY CONVERGES for real(s) > 1.

Your fourth mistake is not realizing that your so-called argument is
nonsense.
Who are you and what have you done with the real R.D. Silverman?
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Old 2008-06-11, 14:55   #6
Carl Fischbach
 
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Oct 2007

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Default an interesting twist

I am not really sure of myself with complex math but I thought I would
give Riemann's hypothesis a shot, the feedback is useful. But I am sure of
this get to the bottom of the year 2012 your really going to need a
life boat then.
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Old 2008-06-11, 15:07   #7
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by Carl Fischbach View Post
I am not really sure of myself with complex math but I thought I would
give Riemann's hypothesis a shot, the feedback is useful. But I am sure of
this get to the bottom of the year 2012 your really going to need a
life boat then.
I am curious. You admit that you are unsure of yourself.
Why then would you even try something like this? Don't you
realize it makes you look foolish? I am sure that you are equally
ignorant about (say) neurosurgery. But you wouldn't presume to
suggest a new surgical technique would you?

What is it about mathematics that people who are totally ignorant
(and who know they are ignorant) think they can magically solve
problems that have eluded PhD's for centuries??? We see this phenomenon
constantly on the Internet.

I also suggest that you need to go back to school to learn how
to write cogent English. Not only is your mathematics bad, but so is
your spelling, grammar, punctuation, and sentence structure. Indeed, your
use of the phrase "complex math" doesn't even mean what you
clearly intended. "complex math" is not the same as "mathematics of
complex valued functions". Your public writing makes you appear to be a
total cretin. Do you like to appear foolish???
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Old 2008-06-11, 15:11   #8
retina
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We are all fools at some time in our lives. Nothing wrong with that as long as we learn from it and try to improve.
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Old 2008-06-11, 15:35   #9
bsquared
 
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Feb 2007

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Quote:
Originally Posted by retina View Post
We are all fools at some time in our lives. Nothing wrong with that as long as we learn from it and try to improve.
Evidence of improvement? Compare to, say, this thread: http://www.mersenneforum.org/showthread.php?t=9449
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Old 2008-06-11, 15:38   #10
Carl Fischbach
 
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Oct 2007

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Default nothing ventured nothing gained

I may be an amateur and it may be along shot, but amatuers have
cracked problems before and I don't care.
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Old 2008-06-11, 16:22   #11
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by Carl Fischbach View Post
I may be an amateur and it may be along shot, but amatuers have
cracked problems before and I don't care.

No, you are not an "amateur". An amateur is someone who knows
at least a little about the subject under discussion. You tried to disprove
R.H. without even knowing what the Zeta function *is*.

You certainly have the right not to care about looking foolish.

Also, your English still hasn't improved. Someone who is deliberately
ignorant and who refuses to learn from clear prior mistakes is a crank.
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