The proof by contradiction of the negation of Riemann Hypothesis

21 Pages

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

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
21 of 21 pages

Details

Title
The proof by contradiction of the negation of Riemann Hypothesis
Author
Year
2013
Pages
21
Catalog Number
V213699
ISBN (Book)
9783656422198
File size
1769 KB
Language
English
Notes
My approach to the most famous unsolved problems in mathematics so far.
Tags
riemann, hypothesis
Quote paper
Prof. Dr. med. John Bredakis (Author), 2013, The proof by contradiction of the negation of Riemann Hypothesis, Munich, GRIN Verlag, https://www.grin.com/document/213699 