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I Basel problem, primes and π²/6

  1. Dec 15, 2015 #1

    Borek

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    Bear with me, I know nothing.

    Eons ego @micromass told me about this beautiful formula:

    [tex]\frac {\pi^2} 6 = \prod\limits_{P}\left( 1-\frac 1 {p^2}\right) ^{-1}[/tex]

    where p are primes. Just a few minutes ago I have learned about the Basel problem and its solution:

    [tex]\sum \limits_{n=1}^{\infty} \frac 1 {n^2} = \frac {\pi^2} 6 [/tex]

    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?
     
    Last edited by a moderator: Jun 19, 2016
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  3. Dec 15, 2015 #2

    micromass

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  4. Jun 1, 2016 #3

    Borek

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    Here comes a follow up question:

    At first sight both formulas listed in my first post look differently, but it turns out it is actually the same thing. There are many formulas allowing calculation of π - or some value related to π. Do we know anything about how many of them are independent?
     
  5. Jun 1, 2016 #4

    micromass

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    What do you mean with independent? In a strict sense, none of the formulas are independent in the sense that we can accept one formula as the definition of ##\pi## and derive all other formulas from it.
     
  6. Jun 1, 2016 #5

    Borek

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    Good point, perhaps my thinking is flawed somehow.

    In the above case we don't need to know that both formulas produce the same result to prove they are equivalent. It is enough to rearrange them, so they are both describing exactly the same calculation. In this sense I don't consider them to be independent.
     
  7. Jun 19, 2016 #6

    Borek

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    Actually we started with stating "it is the same formula" (it can be obtained just by rearranging). So what if by "independent" I mean "not the same"?
     
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