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Produs eulerian de numere prime
Posted: Tue Oct 09, 2007 12:45 am
by Cezar Lupu
Sa se arate ca daca \( p \) este numar prim atunci \( \prod_{p}\left(1+\frac{1}{p(p+1)}\right)=\frac{\zeta(2)}{\zeta(3)} \), unde \( \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}} \) repreznta functia zeta a lui Riemann.
Posted: Tue Oct 09, 2007 8:41 am
by Dragos Fratila
Folosind
\( \zeta(s)=\prod_p \left(1-\frac{1}{p^s}\right)=1 \)
Avem de aratat ca:
\( \prod_p\left(1+\frac1{p(p+1)}\right)\prod_p\left(1-\frac{1}{p^2}\right)= \prod_p\left(1-\frac1{p^3}\right)\Leftrightarrow \)
\( \prod_p\left(1-\frac1{p^2}+\frac1{p(p+1)}-\frac1{p^3(p+1)}\right)= \prod_p\left(1-\frac1{p^3}\right) \)
cu un calcul simplu in stanga (aducere la acelasi numitor ...) se termina problema.
... nu stiu daca am voie sa schimb factorii daca nu sunt toti la fel fata de 1. (cred ca nu totusi... )
