|Dec18-03, 05:56 AM||#1|
taking the logarithm of the euler product
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_Prize...escription.pdf
btw, does taking a logarithm out of a product gives you the summation or what?
thanks in advance.
|Dec18-03, 08:32 AM||#2|
|Dec18-03, 12:28 PM||#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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