Produs eulerian de numere prime
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Cezar Lupu
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Produs eulerian de numere prime
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.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Dragos Fratila
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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... )
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... )