imag94
Jul5-09, 08:17 AM
Accurate Proof verification of Riemann’s Hypothesis
Riemann Hypothesis states that \int \frac{1}{ln (x)} has a root at \frac{1}{2} when s=2
The time series expansion of Log function is,
[tex] \ln(x) = \frac {[x-1}{[x-2}+ \frac{1){3} \frac{x-3}{x-4} + \frac{1}{5}\frac{x-5}{x-6}+……. [\tex]
Let it be equal to [tex] mx + c [\tex] because of the Linear nature of Log function.
Now,
\[tex]int \frac{1}{ln(x)}=\int\frac{1}{mx+c}[\tex]
If we take x=2 from Log function we can deduce m=0.35 and c= 0.0007
Riemann stipulates that for any value x \int \frac{1}{ln (x) will have to be taken between limits 2 and x
So,
[tex] \int^2 _1/2 \frac{1}{ln(x)} [\tex] is done using Cauchys principal number taken between 2 to 1 and from 1 to ½
Which is,
[tex]{\ln(035x1+0.0007)-\\ln(0.5x0.35+0.0007)} + {\ln(0.35x2 +0.0007)-\ln(0.35x1+0.0007)}[\tex]
Taking the proper order of integrations and signs we get
[tex][-1.04782 + 0.35568]+[-1.04782+1.73897]=-0.69+0.69 =0[\tex]
Which proves that the root of Riemann’s ξ function is 1/2 when s=2
Mathew Cherian
Riemann Hypothesis states that \int \frac{1}{ln (x)} has a root at \frac{1}{2} when s=2
The time series expansion of Log function is,
[tex] \ln(x) = \frac {[x-1}{[x-2}+ \frac{1){3} \frac{x-3}{x-4} + \frac{1}{5}\frac{x-5}{x-6}+……. [\tex]
Let it be equal to [tex] mx + c [\tex] because of the Linear nature of Log function.
Now,
\[tex]int \frac{1}{ln(x)}=\int\frac{1}{mx+c}[\tex]
If we take x=2 from Log function we can deduce m=0.35 and c= 0.0007
Riemann stipulates that for any value x \int \frac{1}{ln (x) will have to be taken between limits 2 and x
So,
[tex] \int^2 _1/2 \frac{1}{ln(x)} [\tex] is done using Cauchys principal number taken between 2 to 1 and from 1 to ½
Which is,
[tex]{\ln(035x1+0.0007)-\\ln(0.5x0.35+0.0007)} + {\ln(0.35x2 +0.0007)-\ln(0.35x1+0.0007)}[\tex]
Taking the proper order of integrations and signs we get
[tex][-1.04782 + 0.35568]+[-1.04782+1.73897]=-0.69+0.69 =0[\tex]
Which proves that the root of Riemann’s ξ function is 1/2 when s=2
Mathew Cherian