Proof Of Riemann Hypothesis

Proof for Riemann Hypothesis. Abstract: Proof of Riemann’s hypothesis that the real part of the solution of Zeta function is ½ is proved. Historical development of this area of Mathematics from Gauss, Legrange, Euler, Riemann to Hilbert is discussed. Initially a surrogate for zeta function is derived and using Cauchy principal number of integration between bounds it is proved that the real part of Riemann zeta function is ½. Also Riemann’s ξ =∫ 1/Ln(x) gives the number of Prime numbers between 0 and that number. I choose the original ξ function of Riemann to prove it. This is given in the book “God created Integers” by Prof. Stephen Hawkings published by Penguin which is a good reference for this. Many late attempters use the modified version of 1/x2 instead of 1/Ln(x) to prove. The problem with that it won’t be accurate. A comprehensive exposition of Riemann Zeta function is given in the book “Mysteries of Mathematics” Prof. Calvin Clawson. Subscription of Proof Subscription: Legrange and Gauss conjured that п(x) the function counting all the primes less than x asymptotically approaches Li(x) meaning п(x)/Li(x) tend to 1, where, n Li(x) = ∫ dx/ln(x) 2 Euler created a time series solution to the function Li(x) and Riemann named it the ξ function adding his own solution to Euler’s work. In Riemann’s words “a value x is the root of a function f(x) if f(x)=0. A root of the function ξ(x) is real if and only if the root of the zeta function is complex number with real part equal to ½”. Proving the real part to be ½ was left undone by Riemann. Hilbert later on added, finding the proof for Riemann hypothesis as one of the problem that remain un resolved in Mathematics.

I start where Riemann left his hypothesis without the proof. n As said earlier Li(x) = ∫ dx/Ln(x) 2 Riemann conjured that the function ξ = ∫ 1/ln(x) has the real part of the root at ½ when s=2 . Consider the Log function time series expansion, Log (x)= (x-1/x-2)+1/3(x-3/x-4)+1/5(x-4/x-5)+……..

It’s differential equation can be written as a f ”(x)+ b f(x) + c = ln(x) Applying Taylor series to the time series whose 3rd and 2nd term forms the first two terms of the differential equation and the rest of the terms forming c we have, 2. f “(x)/2!+ 1. f’(x)+c. Here we see a = 1 and b =1 making the log function Homogeneous so Eliptical function per se and so the roots must be Imaginary or Complex roots, which means the Integral of it’s reciprocal also must be Eliptical and so with complex roots. System Equations that generate sustained oscillations are of the Eliptical Nature. Instead of transforming this function to Li(x) or Riemann ξ function, the same differential equation is equated to Li(x) with the intention of finding a suitable “c” formula so that there won’t be any contradiction to the new equation. Through this approach we can get any number of “c” and if and b are made equal to this we have a solution to Riemann Hypothesis without lose in generality. So we write the equation as, f”(x)+f’(x)+c = ∫ 1/Ln(x) Taking f(x)=Ln(x) the above formula simplifies to. (-1/x)+(1/x)+c = ∫1/Ln(x) Where c = [x/Ln(x)]e0,25 The Table is the values of Riemann Li(x) calculated from above equation along with actual values of ξ function found from inspection from numbers selected randomly between 0.5 and 90. X 0.5 1.1 2 5 10 20 30 80 90

Actual Primes

5 6 9 10 22 27

Riemann Li(x) calculated -2.526 14.819 4.03 4.14 5.666 8.6175 11.3522 23.4523 25.6904

The numbers come very close. To actual numbers 1 is added to include 1 as a Prime which some may not agree. If that is subtracted too the numbers remain close. The numbers between .5 and 2 is used to calculate the area between them. The idea is to get a viable “c” so that “a” and “b” chosen equal to that can prove the Riemann hypothesis that all roots of Riemann ξ function lie on the real plane of “1/2”. The Eliptical nature of the equation for it is proved above having equal “a” and “b”.

Now the area between 2 and ½ is found out as, [((x/Ln(x))e

0.25

1.1 2 0.25 )] - [((x/Ln(x))e ] =0 ½ 1.1

[14.819+2.926]-[14.819-4.03]=0 In this we see the negative region overlapping into the positive region and this is done intentionally to avoid division by 0 if 1 is taken as intermediate limit between 2 and ½ where division by ln(1) occurs which lead to indeterminacy. To avoid this a close range number 1.1 is taken as the intermediate limit and error discounted as due to the particular nature of number system though the result of the nature of the Riemann Hypothesis is not surrendered to this error. So the minute error can be attributed to this overlap if corrected or taken into consideration can lead to exact area being equal to 0. So it is affirmed that the area between 2 and ½ is 0. Now the Riemann Hypothesis Proof. Since a=1 and b =1 the roots will be (b+/-√b2-4ac)/2a. So it is (1+/-√1-4c)/2= 1/2 +/- √1-4c/2 where we see that the roots are complex and if we vary c from what is calculated to any number(verified only till 90) and changing ‘a’ and ‘b’ equal to “c” or any equal values less than calculated “c” for any value “x” we get infinite number of complex roots with real part at ½. ½ is seen to be remaining constant for all roots only complex part varying. Roots with “-i” is discarded by number Theorists. So all the roots of Riemann ξ function lie on the complex plane at ½ from 1 through any large number(till 90 already proved above). We already proved with Homogeneous Differential Equation ½ is a root of Riemann ξ function (area between 2 and ½) and the differential equation compliments it with real part of it’s root at ½ for many “c” and equal “a” and “b”