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Zwiebach problem 12.4

  1. Oct 26, 2007 #1
    1. The problem statement, all variables and given/known data
    The first part of this problem makes no sense to me because n is a constant in gamma and when you multiply gamma by zeta the n is somehow supposed to get inside the zeta function...


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 27, 2007 #2

    It's pretty straight forward. What does equation (1) look like after the replacement t -> nt?
     
  4. Oct 27, 2007 #3
    I understand what he wants:

    [tex] \Gamma (s) \zeta (s) = \sum_{n=1}^{\infty}\Gamma (s) n^{-s} [/tex]

    and it is easy from there.

    But that equation is like saying that

    [tex](a x^2 + bx +n) (\sum_{n=1}^{\infty} n^2) = \sum_{n=1}^{\infty} (a x^2 + bx +n) n^2[/tex]

    My point is that there is very poor notation in this problem.

    Zwiebach uses an n outside the summation which simply does not make sense.
     
  5. Oct 27, 2007 #4
    My bad. I should have written:

    Replace t with nt in

    [tex]\Gamma(s) = \int_0^{\infty}dte^{-t}t^{s-1}[/tex]

    after you put the gamma inside the sum.
     
    Last edited: Oct 27, 2007
  6. Oct 27, 2007 #5
    I see. I was stupidly replacing t with nt before I put the gamma inside the sum. I take back what I said about the poor notation.
     
  7. Oct 28, 2007 #6
    How do you justify the last inequality?
     

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  8. Oct 31, 2007 #7
    Sorry. I mean equality.
     
  9. Oct 31, 2007 #8

    Dick

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    For the last equality, they are expanding 1/(1+f(t))=1-f(t)+f(t)^2-f(t)^3+... and keeping only the first few terms.
     
  10. Nov 1, 2007 #9
    Cool. I knew that 1/(1+t) = 1 -t + t^2 -t^3 +t^4. I didn't know it was true when you replaced t with an arbitrary function of t.

    Is there a quick way to prove that?
     
  11. Nov 1, 2007 #10
    This is not a statement about the function f, it is a statement about the number f(t).
     
    Last edited: Nov 1, 2007
  12. Nov 1, 2007 #11

    Dick

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    You can expand 1/(1+whatever) that way as long as |whatever|<1.
     
  13. Nov 1, 2007 #12
    I see. Thanks.
     
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