Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Taking the logarithm of the euler product
  1. Euler constants (Replies: 2)

Loading...