Affirmative Proof of Riemann Hypothesis Surajit Ghosh, India,[email protected] 31’st January, 2018 Abstract Real part of all the zeros of zeta function lies on z = 12 ± iy in complex plane as Riemann Conjectured. Proof is arrived by using Eulers product form of zeta function and connecting it with Eulers identity eiπ + 1 = 0 dealing complex numbers in exponential form in the unit circle.

Keywords:Zeta function; Riemann Hypothesis[MSC2010 Code:11M26]

1

1

Introduction to Zeta Function

In 1737, Euler published a paper where he derived a tricky formula that pointed to a wonderful connection between the sum of the reciprocals of all natural integers (zeta function in its simplest form) and all prime numbers. 1+

1 1 1 1 2.3.5.7.11.... + + + + ... = 2 3 4 5 1.2.4.6.8....

Left hand side is a harmonic series and diverges. Whereas the right hand side is the product of sums of geometric series of reciprocal of all primes as shown below: ! !2 !3 !4 2 1 1 1 1 1+ + + ... = + 2 2 2 2 1 1+

1 3

! +

1 3

!2

1 3

+

!3 +

1 3

!4 ... =

3 2

Zeta Function in Euler Product form is as follows: ∞ X Y 1 1 = ........(1) s n 1 − P −s p n=1

Which is equivalent to: ! ∞ X Y 1 1 1 1 1 = 1 + s + 2 + 3 + 4 ... ........(2) ns p p s p s p s p n=1

2 2.1

Unit Circle and projection to Unit Sphere Unit Circle

Eulers exponential form of complex numbers can be explained as representation of all numbers (real as well as imaginary) as e(base of natural logarithm) raised to an imaginary power iθ.In complex plane points on the unit circle represents all possible expressions of unity in the form of eiθ . A complex number having modulus not equaling 1 differs from unity by its radius only. Two equation of complex circles Z1 = r1 eix and Z2 = r2 eix never intersect although identical equations may coincide. The exponential form of complex numbers has advantage of ease of doing exponentiation of complex numbers over its regular Cartesian form. To multiply a complex number with itself or to find the square of a complex number we just need to double its θ and square its radius. Similarly higher exponents of a complex number can also be found.eiπ + 1 = 0 or eiπ = −1 is a special case when θ rotated to π completes the unit circle.The beauty of the equation is, it encompasses 5 mathematical constants e(base of natural logarithm), i(square root of -1 being unit complex number), π(the ratio

2

of a circle’s circumference to its diameter), 1(result of multiplicative identity), 0(result of additive identity). All the complex numbers lying on all the rays upto infinity in complex plane cut the unit circle in an angle equal to its θ. This way eiπ + 1 = 0 is a manifestation of infinity itself.Stereographic projection is a particular mapping that projects a plane to a sphere and vice versa. The projection is defined on the entire sphere, except at one point: the projection point. When Unit circle in complex plane is projected to unit sphere the points within the area of unit circle gets mapped to southern hemisphere, the points on the unit circle gets mapped to equatorial plane, the points outside the unit circle gets mapped to northern hemisphere. South pole represents the origin, North pole is an extension of infinity.

2.2

Half radius unit Circle

We shall derive another unit circle in Eulers exponential form z = reix . When x = π3 in polar from of z = r(cos(x) + i.sin(x)) = 1 and r = 21 , in exponential z form it can be written as zb = 12 eix = 1. zb = eln(z)−ln2 = eln 2 = z2 . Which means logarithm of any complex number become a continuous function and straightway equals half of the original complex number without needing any branch cuts. Multiple branches in surfaces meets in the unit sphere to give unique principal value obeying the logarithmic rules.

3

Riemann Hypothesis

Riemann showed that zeta function have the property of analytic continuation in the whole complex plane except for s=1 where the zeta function has its pole. Riemann Hypothesis is all about non trivial zeros of zeta function. There are trivial zeros which occurs at every negative even integers.There are no zeros for s > 1. All other zeros lies at a critical strip 0 < s < 1. In this critical strip he conjectured that all other zeros lies on a critical complex line of the form of z = 12 ± iy i.e. the real part of all those complex numbers equals 21 .In the strip 0 < Re(s) < 1 the zeta function satisfies the functional equation: ! πs ζ(s) = 2s π (s−1) sin Γ(1 − s)ζ(1 − s).....(3) 2

4

Zeta function in Riemann Sphere

If we want to find the non trivial zeros of zeta function from the above functional equation we can say ζ(s) will be zero when ζ(1−s) will be zero. But ζ(1−s) will be zero when ζ(1 − 1 + s) i.e. ζ(s) will be zero. This is paradoxical. We fall into this paradox due to the fact ζ(1) is not defined. Can we define it analytically? Lets Try. Lets map zeta Function in 0 < Re(s) < 1 to Riemann sphere and see whether this paradox is solved. The idea is, we shall use the extended domain of Riemann sphere (as it encloses infinity) to limit the principal value of ζ(1) which 3

may not be clear in two dimensional complex plane.Zeta Function is analytic in entire complex plane except at z = 1 ± iy where it blows to infinity due to harmonic nature of ζ(1). When we try to project zeta Function to unit sphere we get two infinity, one at true infinity and another at unity. Both This infinity wants to get mapped conventionally to one point i.e. the north pole. We know ζ(∞) limits to 1. So what if we don’t bother too much about ζ(∞) and map ζ(∞) to (1,0,0).Now the North Pole becomes a hole. But conventionally North pole is the projection point, how can a function known to be analytic be left with a hole? Some point has to fill the hole to preserve the analytic character of zeta function. If we deploy modern day super computer to map rest of the points we shall see after infinite amount of time that all the pixels of Riemann sphere are mapped except the north pole. As all other points has already been exhaustively mapped we are left only with ζ(1) which is the only candidate to sit on the north pole. Now C ∪ 1 being the point of compactification of the complex plane C the only possible singularity for entire zeta functions is now defined on C ⊂ C ∪ 1 , is the point 1 and thus zeta function now become an entire function in the region 0 < Re(s) < 1 in literal meaning. Also in compact zeta function there will be a point smax ∈ M where |ζ(s)| attains its maximum. The minimum of |ζ(s)| in unit circle is therefore achieved at some point smin in the interior of unit circle, but not at any point of its boundary.In complex analysis, Liouville’s theorem states that every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f (z)| ≤ M for all z in C is constant.We can find the mean modulus applying Gauss’s Mean Value Theorem as follows: Z 2π 1 |ζ(a + eiθ )dθ| |ζ(a)| = 2π 0 Gauss’s Mean Value Theorem further requires that in case of a function having constant value in some neighborhood, smax = smin = savg all should coincide at the center (0,0) of the unit circle drawn on the neighborhood . |ζ(0)| = 21 shall be the constant modulus or radius of compactified zeta function. Points on half radius unit circle (as shown above) gives the Maximum modulus. The Maximum modulus principle implies then that ζ( smax = smin = savg ) = 0 will be the zeros of the zeta function which all have half as real part in the the half radius unit circle. Replacing s with (1 − s) and sin( πs 2 ) with its versine 2 1 πs 2 versin( ) = sin πs = 1 − cos πs in the equation (3) we get equation of zeta 2 2 function in unit sphere: ζ(1 − s) = 2s+1 π (s−1) (1 − cos2 πs)Γ(1 − s)ζ(s).....(4) Putting s=0 in the equation (4) we can set the limiting value of ζ(1) = 0 .

4

5

Solving the zeros of Zeta function algebraically

5.0.1

Trivial Zeros

In eulers exponential form of representing complex numbers in unit circle we can rewrite the Equation (2) as follows: ! ∞ Y X 1 iθ 2 i2θ 3 i3θ = 1 + re + r e + r e ... ........(5) ns p n=1 ! Now any such factor

1 + reiθ + r2 ei2θ + r3 ei3θ ...

will be zero if

! reiθ + r2 ei2θ + r3 ei3θ ...

= −1 = eiπ .....(6)

As we are into the unit circle and infinite number of terms of such factors incrementally adds upto unity we can say analytically that θ + 2θ + 3θ + 4θ... = π r + r2 + r3 + r4 .... = 1 We can solve θ and r as follows: θ + 2θ + 3θ + 4θ... =

π

θ(1 + 2 + 3 + 4...) =

π

θ.ζ(−1) = π −1 θ. = π 12 θ = −12π

r + r2 + r3 + r4 .... = 2

3

1

4

r(1 + r + r + r + r ....) = 1 1 r = 1 1−r r = 1−r 1 r= 2

Now Plugging this general values of r and θ in r cos θ we can calculate the general value of the real part of all roots of zeta functions as follows: r cos θ =

1 1 cos(−12π) = 2 2

Notable to see that ζ(−1) is responsible for trivial zeros of zeta function at all even negative integers. If we trace them at complex plane they will be lying on the half radius unit circle and in Riemann sphere they will be projected in the Southern Hemisphere.

5

5.0.2

Non Trivial Zeros

Zeta Function in Eulers product form of equation (1) can be rewritten as follows: ! ! ∞ Y X Y 1 1 reiθ = .......(7) ... = ns reiθ − 1 1 − re1iθ p p n=1 Multiplying both numerator and denominator with reiθ + 1 we get: ! ∞ Y X reiθ (reiθ + 1) 1 = ........(8) ns (reiθ − 1)(reiθ + 1) p n=1 Now any such factor of equation (8) will be zero if the numerator reiθ (reiθ +1) can be shown to be zero : reiθ (reiθ + 1)

=

0



re (re − e )

=

0

2 i2θ

= 0 = rei(π−θ)

r e





i(π−θ)

− re r2 ei2θ

We can solve θ and r as follows: 2θ = 3θ = θ=

(π − θ) π π 3

r2 = 2

r = r r=

r r r 1

Now Plugging this general values of r and θ in r cos θ we can calculate the general value of the real part of all roots of zeta functions as follows: π 1 r cos θ = 1. cos( ) = 3 2 Notable to see that ζ(1) is responsible for non trivial zeros of zeta function at z = 12 ± iy. If we trace them at complex plane they will be lying on the half radius unit circle and in Riemann sphere they also will be projected in the Southern Hemisphere.

6

Conclusion

Hence we conclude with affirmative confirmation of Riemann Hypothesis. Together with the trivial zeros, non trivial zeros completes a half radius unit circle in the complex plane and in Riemann sphere they are projected at an zenith angle of π3 i.e. latitude of 60◦ South.

6

Affirmative Proof of Riemann Hypothesis -

form of zeta function and connecting it with Eulers identity eiπ +1=0 ..... Now any such factor of equation (8) will be zero if the numerator reiθ(reiθ+1).

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