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Taking the logarithm of the euler product

  1. Dec 18, 2003 #1
    can some one explain to me how is taking the logarithm of euler product gives you -sum(p)[log(1-p^s)]+log(s-1)=log[(s-1)z(s)]?

    my question is coming after encoutering this equation in this text in page number 2: http://claymath.org/Millennium_Priz...sis/_objects/Official_Problem_Description.pdf

    btw, does taking a logarithm out of a product gives you the summation or what?

    thanks in advance.
  2. jcsd
  3. Dec 18, 2003 #2
    I don't know if this is of any use to you... But the logarithm of a product is the sum of the logarithms of the factors. That is, log(ab) = log(a) + log(b).
  4. Dec 18, 2003 #3
    yes, i know this but how can you interpret the euler product like this.
    lets see it as an example the product of p is 1*2*3...*n so you take the logarithm and you get log1*2*3=log1+log2+log3 which is the summation on this i understand but why adding the term log(s-1)?
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