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Bear with me, I know nothing.
Eons ego @micromass told me about this beautiful formula:
\frac {\pi^2} 6 = \prod\limits_{P}\left( 1-\frac 1 {p^2}\right) ^{-1}
where p are primes. Just a few minutes ago I have learned about the Basel problem and its solution:
\sum \limits_{n=1}^{\infty} \frac 1 {n^2} = \frac {\pi^2} 6
What struck me was that it is the same π²/6 in both cases.
Somehow I feel like it can be actually the same formula - just the one based on prime numbers takes into account fact that repetitions of prime factors cancel out when we try to sum the fractions finding the common denominator. Am I right, or am I completely off, as usual?
Eons ego @micromass told me about this beautiful formula:
\frac {\pi^2} 6 = \prod\limits_{P}\left( 1-\frac 1 {p^2}\right) ^{-1}
where p are primes. Just a few minutes ago I have learned about the Basel problem and its solution:
\sum \limits_{n=1}^{\infty} \frac 1 {n^2} = \frac {\pi^2} 6
What struck me was that it is the same π²/6 in both cases.
Somehow I feel like it can be actually the same formula - just the one based on prime numbers takes into account fact that repetitions of prime factors cancel out when we try to sum the fractions finding the common denominator. Am I right, or am I completely off, as usual?
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