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Euler's solution to the Basel problem

  1. May 6, 2010 #1
    Can someone find a good explanation of how Euler did it? I can't seem to find anything article or whatnot that carefully explains what he did to solve the problem. From what I can gather, he seems to follow a method of writing out sin(x) as an infinite series(taylor polynomial) and divides by x and from there I get lost in the mishmash. Why does he even choose to use sin(x) and not cos(x) or tan(x) for that matter?
     
  2. jcsd
  3. May 6, 2010 #2
    The function

    sin(pi x)/(pi x)

    is zero at the nonzero integers, so it should be equal to:

    sin(pi x)/(pi x) = product over n of (1-x^2/n^2)

    The right hand side is a product that converges and it is zero when x isa nonzero integer. The normalization is correct, because for x = 1 it is 1 while the limit for x to 1 of the left hand side is also 1. So, the above indentity looks correct (but you can't rigorously prove that, Euler only conjectured the identity).

    The coefficent of x^2 of the left hand side is:

    -pi^2/6

    And from the right hand side it is minus the sum of 1/n^2 from n = 1 to infinity. To get an x^2 term, you need to take it from one factor of the infinite product, say the nth, and then you need to take the 1 from all other factors. You then get -1/n^2, and all n from 1 to infinity contribute.

    Now, you can just as well do this using cos(pi x). The zeroes are at x = (n+1/2), so you would conjecture that:

    cos(pi x) = Product over n of [1-x^2/(n+1/2)^2]

    Normalization is correct as can be seen from putting x = 1 on bith sides. Extracting the coefficient of x^2 gives:

    -pi^2/2 = -Sum over n from n= 0 to infinity of 1/(n+1/2)^2

    You can rewrite this as:

    sum over n 1/(2n+1)^2 = pi^2/8

    You then use the following trick. If we put:

    Zeta(2) = sum from n = 1 to infinity of 1/n^2

    Then clearly the sum of the inverse squares of the even numbers only is:

    sum from n = 1 to infinity of 1/(2n)^2 = 1/4 Zeta(2)

    So, the sum over only the inverse squares if the odd numbers must be
    3/4 Zeta(2). So we have:

    pi^2/8 = 3/4 Zeta(2) -------->

    Zeta(2) = pi^2/6.
     
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