# Riemann Zeta function on positive odd

Posted 1 month ago
249 Views
|
0 Replies
|
0 Total Likes
|
 Hello members, I am interesting on Zeta Function, Riemann Hypothesis and, linear independence and formulas of zeta on positive odd. I demonstrated analytically the formula for Zeta[4n-1]; n=2,3,4, ...Example: Zeta_[4n-1]=For[n=2,n<=11,n++,Print[N[(1/(2n-1))Sum[Zeta[2p]*Zeta[4n-1-2p],{p,1,2n-2}],16]]] I did the comparison between this formula with the value given by Mathematica, Zeta[4n-1]; the absolute and relative errors are small. Me, I was waiting to see that, very very small. Somebody can help me on this calculation and give me the commentaries.L'erreur absolue; In[18]:= For[n=2,n<=11,n++,Print[Abs[N[(1/(2n-1)) (Sum[Zeta[2p]*Zeta[4n-1-2p],{p,1,2n-2}])-Zeta[4n-1],16]]]] L'erreur relative en pourcentage; In[20]:= For[n=2,n<=11,n++,Print[100Abs[N[ (Sum[Zeta[2p]*Zeta[4n-1-2p],{p,1,2n-2}]/(2n-1)-Zeta[4n-1])/Zeta[4n-1],16]]]] Sincerely yourMundankulu Kabongo