User:Fady adel nagy
The Proof of Riemann Hypothesis
Inventor engineer / Adel Nagy Asham Mena,
Fady Adel Nagy asham Mena,
Marian Adel Nagy asham Mena , ashammena@gmail.com , lecce,Italy Fadymena1999@gmail.com , lecce,Italy Marianadelnagy@gmail.com , lecce,Italy
- Corresponding author: ashammena@gmail.com
(Egypt)
ABSTRACT: In mathematics, the Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and the complex numbers with real part (1/2). It was proposed by Bernhard Riemann (1859). The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics . The Riemann hypothesis, along with Goldbach's conjecture, is part of Hilbert's eighth problem in David Hilbert's list of (23) unsolved problems; it is also one of the Clay Mathematics Institute's Millennium Prize Problems.The Riemann hypothesis proof is demonstrated in a simple way. In these papers zeta function connection with the distribution of prime numbers. These papers present laws and simple ways to find all the prime numbers. Finding of prime factors for a semi-prime.
Keywords: Riemann zeta function ,Primes ,Euler’s equation ,Complex number , Graphs of trig functions .
Introduction: http://www.claymath.org/millennium-problems/riemann-hypothesis The Riemann hypothesis asserts that all interesting solutions of the equation ζ(s) = 0 lie on a certain straight line The Riemann zeta function ζ(s) = 0 when s is one of (−2, −4, −6, .... )These are called its trivial zeros. However, the negative even integers are not the only values for which the zeta function is zero. The other ones are called non-trivial zeros. The Riemann hypothesis is concerned with the locations of these non-trivial zeros, and states that:The real part of every non-trivial zero of the Riemann zeta function is( ½).Thus, if the hypothesis is correct, all the non-trivial zeros lie on the critical line consisting of the complex numbers (1/2 + i t), where t is a real number and i is the imaginary unit
Definition of zeta. the zeta function satisfies the functional equation.
ζ(s) = 2^s π^(s-1) cos π(1-s)/2 Γ(1-s) ζ (1-s)
where: Γ(s) Γ(1-s) = π /sin πs is general gamma function. The functional definition shows that the Riemann zeta function have the infinitely zeros, called the trivial zeros, at the negative even integers .
1)The equation e^π = -1 , cos πj = -1
Euler's formula states that, for any real number x:
e^jx= cos x + j sin x
where e is the base of the natural logarithm, j is the imaginary unit, and cos and sin are the trigonometric functions cosine and sine respectively, with the argument x given in radians. Euler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation "our jewel" and "the most remarkable formula in mathematics.
e^jπ = -1
Euler's formula states that, for any real number x: e^jx= cos x + j sin x ---(1)
e^jπ = -1 ---(2)
the all complex numbers can be expressed in polar coordinates. Therefore, for some r and φ depending on x,This formula can be interpreted as saying that the function e^jφ is a unit complex number, j.e., traces out the unit circle in the complex plane as φ ranges through the real numbers. Here, φ is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counterclockwise and in radians. r =1
r e^jφ = r ( cos φ + j sin φ)
The original proof is based on the Taylor series expansions of the exponential function . Power series definition
e^x =1+ x +(x^2) /2! + (x^3) /3!+ (x^4) /4! + (x^5) /5! + ------(3)
Using Power series definition
x = πy
e^πy =1+ πy + (πy^2) /2! + (πy^3) /3! + (πy^4) /4! + (πy^5) /5! + ------(4)
Euler's formula provides a powerful connection between analysis and trigonometry, and provides an interpretation of the sine and cosine functions as weighted sums of the exponential function:
Cos x = (e^jx + e^(-jx)) / 2 ---(5)
sin x = (e^jx - e^(-jx)) / (2 j) ---(6)
Cos jx = (e^x + e^(-x)) / 2 ---(7)
sin jx = (e^(-x) - e^x) / (2 j) ---(8)
Cos jπy = (e^πy + e^(-πy)) / 2 ---(9)
sin jπy = (e^(-πy) - e^πy) / (2 j) ---(10)
Using Euler's formula,
e^jx = cos x + j sin x
Compensation in the equation for the value x = - jπy
e^πy = cos jπy - j sin jπy ---(11)
Using Euler's formula,
e^jx = cos x + j sin x
Compensation in the equation for the value x = jπy
e^(-πy) = cos jπy + j sin jπy ---(12)
Power series definition
e^x =1+ x + (x^2) / 2! + (x^3) / 3! + (x^4) / 4! + (x^5) / 5! + ------
X= πy
e^πy=1+ πy + (πy)^2 /2! + (πy)^3 /3! + (πy)^4 /4! + (πy)^5 /5! + --- ---(13)
Cos x = 1- (x^2)/ 2! + (x^4)/ 4! – (x^6)/ 6! + ---
X= jπy
Cos jπy = 1- (jπy)^2 / 2! + (jπy)^4 / 4! – (jπy)^6 / 6! +---
Cos jπy = 1+ (πy)^2 / 2! + (πy)^4 / 4! + (πy)^6 / 6! +--- ---(14)
Sin x = x – (x^3) / 3! + (x^5) / 5! – (x^7) / 7! +---
X= jπy
Sin jπy = jπy – (jπy)^3 / 3! + (jπy)^5 / 5! – (jπy)^7 / 7! +---
Sin jπy = jπy + j (πy)^3 / 3! + j (πy)^5 / 5! + j (πy)^7 / 7! +---
( j / j )Sin jπy = j[ πy + (πy)^3 / 3! + (πy)^5 / 5! + (πy)^7 / 7! +--- ]
- j Sin jπy = πy + (πy)^3 / 3!+(πy)^5 / 5! + (πy)^7 / 7! +--- ---(15)
Compensation in the equation N0. 13 ,14,15
e^πy = cos jπy - j sin jπy ---(16)
e^(-πy)= Cos jπy+j Sin jπy ---(17)
1-1) complex circle
complex circle * j = complex circle , φ*j = φj
In the figure of complex circle when φ = π
We conclude that cos jπ = -1 , j sin jπ = 0
compensation in equation No:11 OR No: 16 , Y=1
e^πy = cos jπy - jsin jπy
e^π = cos jπ - j sin jπ
e^π = -1 - 0j
compensation in equation No:12 , OR No: 17 , Y=1
e^(-πy)= cos jπy + jsin jπy
e^(-π) = cos jπ + j sin jπ
e^(-π) = -1 + 0j
e^(±πy) = cos jπy = + 1
when y = ± (2,4,6,---)
e^(±πy) = cos jπy = -1
when y = ± (1,3,5,---)
2) Another The equation e^π = -1 , cos πj = -1
In mathematics, de Moivre's formula , (cos x + j sinx)^n = cos(nx) + j sin(nx) where i is the imaginary unit ( j )^2 = −1 The formula is important because it connects complex numbers and trigonometry. By expanding the left hand side and then comparing the real and imaginary parts under the assumption that x is real, it is possible to derive useful expressions for cos(nx) and sin(nx) in terms of cos(x) and sin(x). the formula is not valid for non-integer powers n. However, there are generalizations of this formula valid for other exponents. These can be used to give explicit expressions for the nth roots of unity.
e^-πy = cos jπy + j sin jπy
m = jy
cos(π m) + j sin (π m) = ( cos π + jsinπ )^m
sin π = 0 , cos π = -1
cos(π m) + j sin (π m) = (cos π + jsinπ)^m
cos(π m) + j sin (π m) = (-1 + j 0 )^m
cos(π m) + j sin (π m) = (-1)^m
cos(π m) + j sin (π m) = -1
cos jπy + j sin jπy = -1
Y=(±(1,3,5,7,---))
cos jπy + j sin jπy = 1
Y=(±(2,4,6,8,---))
cos jπy = ±1 ,
Y= (±(1,2,3,4,---))
3) The definition of genius Riemann Hypothesis :
My equation: Anesti (s) = Æ (S) = ( Sin πS)/2 Rad
sin πs = 0 Rad , when s = ±(1,2,3,4,----)
cos π/2 = 0
Sin π s = cos(π/2 - π s)
Æ (S) = ½ sin πS
Æ (S) = ½[0+sin πS ]
Æ (S) = ½[cos π/2 + cos(π/2 - πS) ]
Æ (S) = ½[cos π/2 + cos(π/2 - 2πs /2)]
Æ (S) = ½[cos π/2 + cos(π/2 - π s/2 - πs/2) ]
Æ (S) = ½[cos (π-π s + π s)/2 + cos((π-π s)/2 - πs/2) ]
Æ(S) = ½[cos((π(1-s)/2)+(π s)/2)+ cos((π(1-s)/2)-(πs)/2)]
Cos a Cos b = 1/2 [cos( a+b)+cos(a-b)]
Æ (S) = cos π(1-s)/2 * cos πs/2 = 0 - - - (18)
(Æ (S) = ( Sin πS)/2)
(Equation No:18)*( 2/sin π s) Æ(S)
Æ (S) = (cos π(1-s)/2) (cos πs/2) ( 2/sin πS) Æ(S)
* ( 2^-s 2^s π π^-1 π^-s π^s )
Æ (S) = (cos π(1-s)/2) (cos πs/2) (2/sin πS) Æ (S)(2^-s)(2^s)(π)(π^-1)(π^-s)(π^s)
Æ (S) = 2 (π^-s) ( 2^-S)(cos πs/2)(cos π(1-s)/2) (2^s)(π^-1)(π^s)(π/sin πS) Æ (S)
(General : Γ(S)Γ(1-S)= π/sin πs )
Æ (S)= 2 π^-s 2^-S cos πs/2 2^s π^-1 π^s cos π(1-s)/2 Γ(S)Γ(1-S) Æ (S)
Æ (S)= 2^(1-s) π^-s cos πs/2 Γ(S) 2^s π^(s-1) cos π(1-s)/2 Γ(1-S) Æ (S) ----(19)
Æ (S) = ( Sin π S)/2 = 0 when s = ±(1,2,3,4,----)
2^(1-s) π^-s cos πs/2 Γ(S) 2^s π^(s-1) cos π(1-s)/2 Γ(1-S) Æ (S)= 0 ---(20)
2^(1-s) π^-s cos πs/2 Γ(S) Æ (S) = 0/ [2^s π^(s-1) cos π(1-s)/2 Γ(1-S) ] ---(21)
T(s) =2^s π^(s-1) cos π(1-s)/2 Γ(1-S)
½ sin πS = cos π(1-s)/2 * cos πs/2
Γ(S)Γ(1-S)= π/sin πs
2^(1-s) π^(-s) ½ [Sin πs/2 / cos π(1-S)/2 ] [ π /(Γ(1-S) sin π s )] Æ (S) = 0 / T(s)
Æ (S) = [ 0 * (2^(s-1) π^(s-1) 2 cos π(1-S)/2 Γ(1-S) ) ] / T(s)
Æ (S) = 0 / T(s)
Æ (1-S) = Æ (S) = ( Sin πS)/2 = 0
when S = ± (1,2,3,----)
Æ (1-S) = 0 / T(s)
Æ (S) = 0 = ( Sin πS)/2
when S = ± (1,2,3,----)
Æ (1-S) = Æ (S) / T(s)
Æ (1-S) = Æ (S) / [ 2^s π^(s-1) cos π(1-s)/2 Γ(1-S) ]
Æ (S) = 2^s π^(s-1) cos π(1-s)/2 Γ(1-S) Æ (1-S)
Æ = ζ
ζ(S) = 2^s π^(s-1) cos π(1-s)/2 Γ(1-S) ζ (1-S)
4)The proof of genius Riemann Hypothesis .
cos π/2 = 0
sin π/2=1
ζ(S)=0 = Æ (S) = ½ Sin π S Rad
Æ(s) = ½ Sin πS
Compensation in the equation for the value S= ( ½ ± Yj)
Æ(½ ±Yj) = ½ Sin π (½ ±Yj)
Æ(½ ±Yj) = ½ [Sin π/2 cos πYj – cos π/2 Sin πYj ]
Æ(½ ±Yj) = ½ [ 1 * cos πYj – 0 * Sin πYj ]
Æ(½ ±Yj) = ½ [ cos πYj ]
Cos πYj = ±1 , Y=(±(1,2,3,---))
Æ(½ ±Yj) = ± ½
Æ = ζ
ζ (½±Yj) = ± ½
The real part of every non-trivial zero of genius Riemann zeta function is 1/2.
This was the proof of genius Riemann Hypothesis
5)Another proof of Riemann Hypothesis definition :
Γ(s) Γ(1-s) = π / sin πs *Æ (s)
Æ (s) Γ(S)Γ(1-s) = (π/(sin πs)) Æ (s)
Æ (s) Γ(S)Γ(1-s) sin πs π^-1 = Æ (s)
sin πs = 2 cos π (1-s)/2 cos π S/2
Æ (s) = Γ(S)Γ(1-S) 2 cos π(1-S)/2 cos π(S/2) π^-1 Æ (s)
*(2^-s 2^s π^-s π^s )
Æ (s) = Γ(S)Γ(1-S) 2 cos π (1-s)/2 cos π S/2 π^-1 Æ (s) 2^-s 2^s π^-s π^s
Æ(S) = 2^(1-s) π^-s cos πs/2 Γ(S) 2^s π^(s-1) cos π(1-s)/2 Γ(1-S) Æ (S)
Look at equation No. 19
6)Using equation No. 18
ζ(S) = cos π(1-S)/2 * cos πs/2 = 0
Either ( cos π(1-S)/2 ) OR ( cos πs/2 ) equals Zero , when compensation of (S) by all integer numbers, we will know with which value the equation equals to zero .
ζ1(S) = cos πs/2 Rad -------(22)
ζ2(S) = cos π(1-s)/2 Rad -------(23)
6-1) Compensation in the equation No. 22
ζ1(S) = cos πs/2 Rad
Changing from radian to degrees
ζ1(S) = cos 180πs/π2
ζ1(S) = cos 90s
Compensation in the equation for the value
S = 0,4,8,12,-------- ∞( increase+4 )
ζ1(4) = cos 90(4) = cos 360 =1
ζ1(0,4,8,12,---∞) = 1
Compensation in the equation for the value
S = 2,6,10,14,---- ∞(increase +4 )
ζ1(2) = cos 90(2) = cos 180 = -1
ζ1(2,6,10,14,---∞) = -1
Compensation in the equation for the value
S=1,5,9,13,---- ∞(increase +4 )
ζ1(1) = cos 90(1) = cos 90 = 0
ζ1(1,5,9,13,---∞) = 0
Compensation in the equation for the value
S=3,7,11,15,----- ∞(increase +4 )
ζ1(3) = cos 90(3) = cos 270 = 0
ζ1(3,7,11,15,---∞) = 0
Therefore the difference is four,then the result always the same The circle is divided into four parts, starting from zero and the numbers are distributed as the following : All even numbers are located on the axis X , All odd numbers are located on the axis Y .
6-2) The second proof of genius Riemann Hypothesis .
Compensation in the equation No. 23
ζ2(S) = cos π(1-S)/2 (Rad)
Changing from radian to degrees
ζ2(S) = cos 90(1-S)
Compensation in the equation for the value S = -3,-7,--∞(increase +4 )
ζ2(-3) = cos 90(1-(-3))= cos 360 = 1
ζ2(-7) = cos 90(1-(-7))= cos 720 = 1
Compensation in the equation for the value S = -1,-5,--∞(increase +4 )
ζ2(-1) = cos 90(1-(-1))= cos 180 = -1
Compensation in the equation for the value S = 0,-2,-4,-6, ---- ∞
ζ2(-2) = cos 90(1-(-2))= cos 270 = 0
ζ2(-4) = cos 90(1-(-4))= cos 450 = 0
Therefore the difference is four, then the result always the same
8) Using Riemann formula,
Æ (S) = 2^s π^(s-1) cos π(1-s)/2 Γ(1-s) Æ (1-S)
Æ (S) =1/2 sin πS = 0 when S = ± (1,2,3,4,----)
Æ (1-S) =1/2 sin πS= Æ(S) = ζ(s)
Æ (S) = 2^s π^(s-1) cos π(1-s)/2 Γ(1-s) Æ (S)
Æ (S) Γ(1-S) = Æ (S) (2^(-S)) ( π^(1-S)) / cos(π(1-S)/2)
Γ(1-S) = (2^(-S)) ( π^(1-S)) / cos(π(1-S)/2 ---(24)
Sin πs = 2(cos(π (1-S))/2) (cos (π S)/2 )
Γ(1-S) 1/2 Sin π S = 2^(-S) π^(1-S) cos(π S)/2
ζ(S) =1/2 sin πS
Γ(1-S) ζ(S) = 2^(-S) π^(1-S) cos(π S)/2 ---(25)
Compensation in the equation No.24
Sin πs = 2(cos(π (1-S))/2) (cos (π S)/2 )
Γ(1-S) = π / (sin π s * Γ(S) )
π / (sin π s * Γ(S) ) = (2^(1-S) π^(1-S) cos (πS/2) / (Sin π S)
Γ(S) = (2^(S-1) π^(S )) / cos(πS/2) ---(26)
Sin πs = 2(cos(π (1-S))/2) (cos (π S)/2 )
Γ(S) = (2^S π^S cos π(1-s)/2) / Sin π S
Γ(S) 1/2 Sin π S = 2^S π^S 2^(-1) cos π (1-s)/2
ζ(S) =1/2 sin πS
Γ(S) ζ (S) = 2^(S-1) π^S cos π(1-s)/2 ---(27)
Æ (S) = ζ(S) = cos π(1-S)/2 cos πS/2 = (2^(-1) π ) / Γ(S)Γ(1-S)
Γ(S) / Γ(1-S) = 2^(2s-1) π^(2s-1) cos π (1-s)/2 / cos πs/2
References: [1] https://www.youtube.com/watch?v=UGj6mfCSZfY [2] Official Problem Description With Links to Riemann’s paper by Clay Mathematics Institute .
http://WWW.claymath.org/millenium problems/Riemann hypohesis
[3] Brent, Richard P. "On the Zeros of the Riemann Zeta Function in the Critical Strip."
Mathematics of Computation 33.148 (1979): 1361. Print.
[4] Edwards, Harold M Riemann's Zeta Function. Academic Press, New York. 1974, ISBN: 978-0486417400 [5] Hutchinson, J. I. "On the Roots of the Riemann Zeta Function." Transactions of the
American Mathematical Society 27.1 (1925): 49. Print.
[6] Lehmer, D. H. "Extended Computation of the Riemann Zeta-function." Mathematika 3.02 (1956): 102. Print. [7] https://en.wikipedia.org/wiki/Riemann_hypothesis. [8] Carlson, James, Arthur Jaffe and Andrew Wiles, eds. The Millennium Prize problems American Mathematical Society, Providence. 2006. ISBN: 978- 0821836798 [9] Ivic, Aleksander. The Riemann Zeta-Eunction-The Theory of the Riemann Zeta-Function with and Applications. John Wiley New York, 1985, ISBN: 978-0471806349 [10] Titchmarsh, E. C. The Theory of the Riemann Zeta Function. 2nd ed. Revised by R. D. Heath Brown. Oxford University Press, Oxford. 1986. ISBN: 978- 0198533696 [11] Derbyshire, John. Prime Obsession. Joseph Henry Press, Washington, DC. 2003. ISBN: 9780309085496. [12] Devlin, Keith. The Millennium Problems. Basic Books, New York. 2003. ISBN: 978-0465017300 [13] Sabbagh, Karl. The Riemann Hypothesis. Farrar, Straus, and Giroux, New York. 2003. ISBN: 97803742500 72. [14] du Sautoy, Marcus. The Music of the Primes. Harper, New York. 2003. ISBN: 978-0066210704. [15] Stewart, lan. Visions of Infinity. Basic Books, New York. 2013 ISBN: 978-0465022403