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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,
\ln(x) = \frac {[x-1}{[x-2}+ \frac{1){3} \frac{x-3}{x-4} + \frac{1}{5}\frac{x-5}{x-6}+……. [\tex]<br /> Let it be equal to mx + c [\tex] because of the Linear nature of Log function.<br /> <br /> Now,<br /> \int \frac{1}{ln(x)}=\int\frac{1}{mx+c}[\tex]&lt;br /&gt; If we take x=2 from Log function we can deduce m=0.35 and c= 0.0007&lt;br /&gt; Riemann stipulates that for any value x \int \frac{1}{ln (x) will have to be taken between limits 2 and x&lt;br /&gt; So,&lt;br /&gt; \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&amp;lt;br /&amp;gt; Which is, &amp;lt;br /&amp;gt; {\ln(035x1+0.0007)-\\ln(0.5x0.35+0.0007)} + {\ln(0.35x2 +0.0007)-\ln(0.35x1+0.0007)}[\tex]&amp;amp;lt;br /&amp;amp;gt; Taking the proper order of integrations and signs we get&amp;amp;lt;br /&amp;amp;gt; [-1.04782 + 0.35568]+[-1.04782+1.73897]=-0.69+0.69 =0[\tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Which proves that the root of Riemann’s ξ function is 1/2 when s=2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; 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,
\ln(x) = \frac {[x-1}{[x-2}+ \frac{1){3} \frac{x-3}{x-4} + \frac{1}{5}\frac{x-5}{x-6}+……. [\tex]<br /> Let it be equal to mx + c [\tex] because of the Linear nature of Log function.<br /> <br /> Now,<br /> \int \frac{1}{ln(x)}=\int\frac{1}{mx+c}[\tex]&lt;br /&gt; If we take x=2 from Log function we can deduce m=0.35 and c= 0.0007&lt;br /&gt; Riemann stipulates that for any value x \int \frac{1}{ln (x) will have to be taken between limits 2 and x&lt;br /&gt; So,&lt;br /&gt; \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&amp;lt;br /&amp;gt; Which is, &amp;lt;br /&amp;gt; {\ln(035x1+0.0007)-\\ln(0.5x0.35+0.0007)} + {\ln(0.35x2 +0.0007)-\ln(0.35x1+0.0007)}[\tex]&amp;amp;lt;br /&amp;amp;gt; Taking the proper order of integrations and signs we get&amp;amp;lt;br /&amp;amp;gt; [-1.04782 + 0.35568]+[-1.04782+1.73897]=-0.69+0.69 =0[\tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Which proves that the root of Riemann’s ξ function is 1/2 when s=2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Mathew Cherian
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