Riemann's ζ\zeta Function and Prime Numbers

We know that π26=112+122+132+122×π26=122×112+122×122+122×132+122×π26=122+142+162+(1122)π26=112+132+152+34×π26=112+132+152+π28=112+132+152+π224=122+142+162+(18124)π2=112122+132π212=112122+132\begin{aligned} \because \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2} \times \dfrac{1}{1^2}+\dfrac{1}{2^2} \times \dfrac{1}{2^2}+\dfrac{1}{2^2} \times \dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore (1-\dfrac{1}{2^2})\dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \dfrac{3}{4} \times \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \therefore \dfrac{\pi^2}{8} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \dfrac{\pi^2}{24} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore (\dfrac{1}{8}-\dfrac{1}{24})\pi^2 &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \therefore \dfrac{\pi^2}{12} &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \end{aligned} As for π26=112+122+132+\dfrac{\pi^2}{6}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots, let zz takes 22's place, we can get the Riemann's ζ\zeta function. And if we continue, we can get ζ(z)=11z+12z+13z+(112z)ζ(z)=11z+13z+15z+13z×(1122)ζ(z)=13z×11z+13z×13z+13z×15z+(113z)(112z)ζ(z)=11z+15z+17z+(115z)(113z)(112z)ζ(z)=11z+17z+111z+1=ζ(z)(112z)(113z)(115z)ζ(z)=1(112z)(113z)(115z)(117z)\begin{aligned} \zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{2^z}+\dfrac{1}{3^z}+\cdots \\ (1-\dfrac{1}{2^z})\zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{3^z}+\dfrac{1}{5^z}+\cdots \\ \therefore \dfrac{1}{3^z} \times (1-\dfrac{1}{2^2})\zeta(z) &=\dfrac{1}{3^z} \times \dfrac{1}{1^z}+\dfrac{1}{3^z} \times \dfrac{1}{3^z}+\dfrac{1}{3^z} \times \dfrac{1}{5^z}+\cdots \\ \therefore (1- \dfrac{1}{3^z}) (1-\dfrac{1}{2^z})\zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{5^z}+\dfrac{1}{7^z}+\cdots \\ \therefore (1-\dfrac{1}{5^z})(1- \dfrac{1}{3^z}) (1-\dfrac{1}{2^z})\zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{7^z}+\dfrac{1}{11^z}+\cdots \\ \therefore 1 &=\zeta(z)(1-\dfrac{1}{2^z})(1- \dfrac{1}{3^z})(1-\dfrac{1}{5^z})\cdots \\ \therefore \zeta(z) &=\dfrac{1}{(1-\dfrac{1}{2^z})(1- \dfrac{1}{3^z})(1-\dfrac{1}{5^z})(1-\dfrac{1}{7^z})\cdots} \end{aligned}

Note by Edward Christian
4 weeks ago

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