Riemann proved three results: analytically continue ζ(s) over the whole complex plane s =σ + it with a pole s =1;(Theorem A) functional equation ξ(t) = G(s<sub>0</sub>)ζ (s<sub>0</sub>), s&...Riemann proved three results: analytically continue ζ(s) over the whole complex plane s =σ + it with a pole s =1;(Theorem A) functional equation ξ(t) = G(s<sub>0</sub>)ζ (s<sub>0</sub>), s<sub>0</sub> =1/2 + it and (Theorem B) product expression ξ<sub>1</sub>(t) by all roots of ξ(t). He stated Riemann conjecture (RC): All roots of ξ (t) are real. We find a mistake of Riemann: he used the same notation ξ(t) in two theorems. Theorem B must contain complex roots;it conflicts with RC. Thus theorem B can only be used by contradiction. Our research can be completed on s<sub>0</sub> =1/2 + it. Using all real roots r<sub>k</sub><sub> </sub>and (true) complex roots z<sub>j</sub> = t<sub>j</sub> + ia<sub>j</sub> of ξ (z), define product expressions w(t), w(0) =ξ(0) and Q(t) > 0, Q(0) =1 respectively, so ξ<sub>1</sub>(t) = w(t)Q(t). Define infinite point-set L(ω) = {t : t ≥10 and |ζ(s<sub>0</sub>)| =ω} for small ω > 0. If ξ(t) has complex roots, then ω =ωQ(t) on L(ω). Finally in a large interval of the first module |z<sub>1</sub>|>>1, we can find many points t ∈ L(ω) to make Q(t) . This contraction proves RC. In addition, Riemann hypothesis (RH) ζ for also holds, but it cannot be proved by ζ.展开更多
文摘Riemann proved three results: analytically continue ζ(s) over the whole complex plane s =σ + it with a pole s =1;(Theorem A) functional equation ξ(t) = G(s<sub>0</sub>)ζ (s<sub>0</sub>), s<sub>0</sub> =1/2 + it and (Theorem B) product expression ξ<sub>1</sub>(t) by all roots of ξ(t). He stated Riemann conjecture (RC): All roots of ξ (t) are real. We find a mistake of Riemann: he used the same notation ξ(t) in two theorems. Theorem B must contain complex roots;it conflicts with RC. Thus theorem B can only be used by contradiction. Our research can be completed on s<sub>0</sub> =1/2 + it. Using all real roots r<sub>k</sub><sub> </sub>and (true) complex roots z<sub>j</sub> = t<sub>j</sub> + ia<sub>j</sub> of ξ (z), define product expressions w(t), w(0) =ξ(0) and Q(t) > 0, Q(0) =1 respectively, so ξ<sub>1</sub>(t) = w(t)Q(t). Define infinite point-set L(ω) = {t : t ≥10 and |ζ(s<sub>0</sub>)| =ω} for small ω > 0. If ξ(t) has complex roots, then ω =ωQ(t) on L(ω). Finally in a large interval of the first module |z<sub>1</sub>|>>1, we can find many points t ∈ L(ω) to make Q(t) . This contraction proves RC. In addition, Riemann hypothesis (RH) ζ for also holds, but it cannot be proved by ζ.
