- #36
eljose
- 492
- 0
so you seem to be very clever to find "gaps" and fails in my works now i have a proof for you that is to proof that the set of functional equation...
[tex]k^{2}\chi(1/2+b-it)=\chi(1/2-b-it) [/tex]
[tex] \chi(1/2-b+it)\chi^{*}(1/2-b+it)=1 [/tex]
[tex]k^2=[\chi(1/2-b+it) [/tex] with []=modulus of the complex number
have a fixed b as it solution...and more complicate to prove that the only solutions to it for b and k are k=b+1=1...i am waiting for your respones oh masterminds of mathematics...
[tex]k^{2}\chi(1/2+b-it)=\chi(1/2-b-it) [/tex]
[tex] \chi(1/2-b+it)\chi^{*}(1/2-b+it)=1 [/tex]
[tex]k^2=[\chi(1/2-b+it) [/tex] with []=modulus of the complex number
have a fixed b as it solution...and more complicate to prove that the only solutions to it for b and k are k=b+1=1...i am waiting for your respones oh masterminds of mathematics...