The proof by contradiction of the negation of Riemann Hypothesis
The proof by contradiction of the
negation of Riemann Hypothesis
Riemann Hypothesis s=(+i.t)
All the non trivial zeros of the Zeta function (s) in the critical strip
(0<Real s<1) are located on the critical line (Real s=1/2)
By my effort to solve:
The greatest unsolved problem in Mathematics
I found that:
The Riemann Hypothesis is not valid
Iff there exist p,q,t, (p#q) to match this equation
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
The above equation is correct:
0<1
2 . log()
For t>0 and M.t = t +p.q M = + (p.q).
log() .
2
Those distinct t's (Roots of t -M.t+p.q=0) exist , but cannot
be defined , since the exact value of (>1) is not known
http://Mathhighways.blogspot.com/
John Bredakis
Athens Greece 2013
1
A fact for the Zeta function (s) s=(+i.t)
2 2
n=+ -n ..x (x)-1 +oo -n ..x (x) 1
(x) = e = e =
n=- 2 n=1 (1/x)
x
-s/2 -(1-s)/2
.(s/2).(s) = .[(1-s)/2].(1-s)
+oo
-1 s/2 (1-s)/2 -1 (x)-1
= + x + x .x ..dx
s.(1-s) 1 2
Checking the Riemann Hypothesis in the critical strip
s=(p
i.t) or s=(qi.t) 0<p<1 0<q<1 q=1-p
I found a formula for A and B
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
And then I made a Working Hypothesis
ie: under what conditions A=B=O
expecting that this zero belongs to (s)
For the proof of the formulas for A and B
see text
2
For the proof of the formulas for A and B
see text
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
2
-t +p.q - i.(p-q).t
A =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
+ i. x - x .x .sinlog(x)...dx
1 2 2
-----
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q + i.(p-q).t
B =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
- i. x - x .x .sinlog(x)...dx
1 2 2
3
My Working Hypothesis
ie: under what conditions A=B=O
expecting that this zero belongs to (s)
Suppose that:
2 M = Icos
t +p.q 0<p<1
M = = Icos q=1-p
2 2 2 2
t +p.q + (p-q) .t 0<q<1
+oo
p/2 q/2 -1 t (x)-1
Icos = x + x .x .coslog(x)...dx
1 2 2
And suppose also that:
N = Isin
(p-q).t
N = = Isin
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
Isin = x - x .x .sinlog(x)...dx
1 2 2
----
+oo
p/2 q/2 -1 t (x)-1
Icos = x + x .x .coslog(x)...dx
1 2 2
2
(x)-1 +oo -n ..x
= e
2 n=1
+oo
p/2 q/2 -1 t (x)-1
Isin = x - x .x .sinlog(x)...dx
1 2 2
4
By my Working Hypothesis
A=B=0
2
(p-q).t [t + p.q]
Isin = and Icos = 2 2 2 2
=t +p.q +(p-q) .t
.Isin .Icos 0<p<1
ie: = = 1
(p-q).t 2 q=1-p
[t +p.q] 0<q<1
By my Working Hypothesis
The following integral vanishes (becomes zero)
t 2 2 2 2
log(x). = A = t +p.q + (p-q) .t
2
p/2 q/2 p/2 q/2
+oo x -x x +x
-1 (x)-1
..sin(A) - .cos(A).x ..dx
1 (p-q).t 2 2
t +p.q
By the Mean Value Theorem of Integral Calculus
there must be a >1 so that
p/2 q/2 p/2 q/2
- +
t t
.sinlog(). = .coslog().
(p-q).t 2 2 2
t +p.q
r
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
5
So far we have that iff the formulas of A and B are correct
Then the following is also correct
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
The next step is to select a t to satisfy the above
p/2 q/2
2 +
M ± M - 4.(p.q) -1
tanlog(). = .(p-q).M
4 p/2 q/2
-
>1
0<p<1 0<q<1 [2p-1]/2
+ 1
q=1-p -1
Or tan() = .(p-q).M
[2p-1]/2
- 1
Working on the left part of the above equation
>1 0<1
2
log() = 2./+ -4.(p.q) Or log() = 2./ -
-----=-----
2./ + 2./ -
= e r = e
6
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
Working on the right part of the above equation
0<p<1 0<q<1 q=1-p >1
[2p-1]/2 S+1 M.R M.R
= = = S
S-1 (p-q) (2p-1)
S+1 2/[2p-1]
=
S-1
>1 S+1 2/[2p-1]
S+1 2/[2p-1] log
= = e S-1
S-1
>1 0<1
2./ + 2./ -
= e r = e
-----=log()------
±
S+1 2/[2p-1] M.R R=tan(.)
= .log S = 2
2. S-1 (2p-1)
0<1
2
= - 4.(p.q) 0<p<1 0<q<1 q=1-p
0<1
2 . log()
For t>0 and M.t = t +p.q M = + (p.q).
log() .
2
Those distinct t's (Roots of t -M.t+p.q=0) exist , but cannot
be defined , since the exact value of (>1) is not known
Those distinct roots are: t1,2 = (M
T)/2
7
Regardless of whether p>q or p<q 0<p<1 0<q<1 q=1-p >1
[p-q]=[2p-1] [p-q]=[2p-1]
[p-q]/2 S+1 M.R+(p-q) S+1 2/[2p-1]
= = =
S-1 M.R-(p-q) S-1
-----------------------------------------------------------
[q-p]/2 S+1 -1 M.R+(p-q) -1 M.R+(q-p)
= = =
S-1 M.R-(p-q) M.R-(q-p)
[q-p]/2 M.R+(q-p) M.R+(q-p) 2/[2q-1]
= =
M.R-(q-p) M.R-(q-p) [2q-1]=[q-p]
Therefore in any case >1 and close (to very close) to 1
As expected
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
0<1
2 . log()
For t>0 and M.t = t +p.q M = + (p.q).
log() .
2
Those distinct t's (Roots of t -M.t+p.q=0) exist , but cannot
be defined , since the exact value of (>1) is not known
Those distinct roots are: t1,2 = (M
T)/2
8
The proof of the formulas for A and B
Checking the Riemann Hypothesis in the critical strip
-s/2 -(1-s)/2
.(s/2).(s) = .[(1-s)/2].(1-s)
+oo
-1 s/2 (1-s)/2 -1 (x)-1
= + x + x .x ..dx
s.(1-s) 1 2
2 2
n=+ -n ..x (x)-1 +oo -n ..x (x) 1
(x) = e = e =
n=- 2 n=1 (1/x)
x
A.
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2
q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
---------------------------------------------------------------
2
-t +p.q-i.(p-q).t
-1 -1
= =
[p+i.t].[q-i.t] 2 2 2 2 2
t +p.q-i.(p-q).t t +p.q +(p-q) .t
=
+oo
(p+i.t)/2 (q-i.t)/2 -1 (x)-1
+ x + x .x ..dx
1 2
9
Checking the Riemann Hypothesis in the critical strip
-s/2 -(1-s)/2
.(s/2).(s) = .[(1-s)/2].(1-s)
+oo
-1 s/2 (1-s)/2 -1 (x)-1
= + x + x .x ..dx
s.(1-s) 1 2
2 2
n=+ -n ..x (x)-1 +oo -n ..x (x) 1
(x) = e = e =
n=- 2 n=1 (1/x)
x
B.
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2
q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
---------------------------------------------------------------
2
-t +p.q+i.(p-q).t
-1 -1
= =
[p-i.t].[q+i.t] 2 2 2 2 2
t +p.q+i.(p-q).t t +p.q +(p-q) .t
=
+oo
(p-i.t)/2 (q+i.t)/2 -1 (x)-1
+ x + x .x ..dx
1 2
10
Checking the Riemann Hypothesis in the critical strip
A+B
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
A+B -2.t +p.q +oo
(p+i.t)/2 (p-i.t)/2 -1 (x)-1
= + x +x .x ..dx
2 2 2 2 1 2
t +p.q+(p-q) .t
+oo
(q+i.t)/2 (q-i.t)/2 -1 (x)-1
+ x +x .x ..dx
1 2
2
A+B -2.t +p.q +oo
p/2 q/2 -1 t (x)-1
= + x +x .x .coslog(x)...dx
2 2 2 2 1 2 1
t +p.q+(p-q) .t
11
Checking the Riemann Hypothesis in the critical strip
A-B
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
A-B +oo
-2.i.(p-q).t (p+i.t)/2 (p-i.t)/2 -1 (x)-1
= + x -x .x ..dx
2 2 2 2 1 2
t +p.q+(p-q) .t
+oo
(q+i.t)/2 (q-i.t)/2 -1 (x)-1
- x -x .x ..dx
1 2
A-B +oo
-2.i.(p-q).t p/2 q/2 -1 t (x)-1
= + i. x -x .x .sinlog(x)...dx
2 2 2 2 1 2 1
t +p.q+(p-q) .t
12
Checking the Riemann Hypothesis in the critical strip
By Combination
A=
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q - i.(p-q).t
A =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
+ i. x - x .x .sinlog(x)...dx
1 2 2
13
Checking the Riemann Hypothesis in the critical strip
By Combination
B=
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q + i.(p-q).t
B =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
- i. x - x .x .sinlog(x)...dx
1 2 2
14
Checking the Riemann Hypothesis in the critical strip
By Combination Summary
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
----------------------------------------------------------------
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q - i.(p-q).t
A =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
+ i. x - x .x .sinlog(x)...dx
1 2 2
2
-t +p.q + i.(p-q).t
B =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
- i. x - x .x .sinlog(x)...dx
1 2 2
15
Some general remarks on the Zeta function (s)
And a reference to my pdf
Understanding the Zeta function and the Riemann Hypothesis
16
For further details see my pdf:
Understanding the Zeta function
and the Riemann Hypothesis
http://Mathhighways.blogspot.com/
John Bredakis
Athens Greece 2013
17
References
And a lot of personal work
------
I like to express my special thanks to Professor of Mathematics
Elias Kastanas
for his advise and check up
18
Additional
References From The Internet
-----
And my pdf:
Big Bang in Math,John Bredakis Method & the Gamma function
19
20
The reason to deal with the Zeta function
(s),
despite my shallow knowledge of complex analysis
-------
The common things between those pdf of mine
Big Bang in Math , John Bredakis method & the Gamma function
and
The proof by contradiction of the negation of Riemann Hypothesis
The Gamma function and the tranfer of i
from denominator to numerator
Trying to understand the difficult to comprehend topic of complex
analysis I ended up to the calculus of residues.
And then I realized that the complex analysis can be bypassed to
a large part by the classical advanced analysis , hence my pdf
An attempt to bypass the calculus of residues
Searching very carefully in the Internet , I found the one and only pdf
by Theodore Yoder (Introduction to Riemann Hypothesis)
unique in the sense of handling the
(s),with minimal complex analysis
and epecially with the introduction of the idea of analytic continuation
Thereafter I needed only few lines from the pdf by Lorentzo Menici
to comprehend the Riemann-Siegel formula , hence my pdf:
Understanding the Zeta function and the Riemann Hypothesis
This is the whole story
Handling Mathematics for many years
I must admitt that my mentor in Mathematics was my beloved
Uncle Fotis , a medical doctor , brilliant mathematical brain
and great teacher of Mathematics
Sincerelly:
John Bredakis MD
http://Mathhighways.blogspot.com/
Athens Greece 2013
21
Ende der Leseprobe aus 21 Seiten
- nach oben
- Arbeit zitieren
- Prof. Dr. med. John Bredakis (Autor:in), 2013, The proof by contradiction of the negation of Riemann Hypothesis, München, GRIN Verlag, https://www.grin.com/document/213699
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