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Apostol & floor function problem

  1. Feb 18, 2007 #1

    I'm working through some questions in Apostol's Calculus text (vol. 1). The problem is with ex. 7 on p.64. Here, he states that

    "by changing the index of summation, note that

    [tex]\sum_{n=a}^{b-1} [\frac{na}{b}] = \sum_{n=a}^{b-1} [\frac{a(b-n)}{b}][\tex] "

    I'm comfortable with the solution from this point on. I just don't see how these two are equivalent.

    Any help would be appreciated!

    p.s. Sorry about the LaTeX not displaying properly.
    Last edited: Feb 18, 2007
  2. jcsd
  3. Feb 18, 2007 #2


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    Are there any specific conditions on a,b and n?

    (I can't see the answer to your question; just thought I'd get the LaTex working for you!)
  4. Feb 18, 2007 #3


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    It works if you make the change of index n=b-m. Then,

    [tex]\sum_{n=a}^{b-1} [\frac{na}{b}] = \sum_{m=b-a}^{1} [\frac{a(b-m)}{b}][/tex]

    You can use Gauss's sum [itex]\sum_i^n i=n(n+1)/2[/itex] to show that the equality you wrote is not true unless a=1 or a=b.
    Last edited: Feb 18, 2007
  5. Feb 18, 2007 #4

    matt grime

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    You're just doing the sum in reverse order.
  6. Feb 18, 2007 #5
    Thanks for the responses. The conditions are that a and b are positive integers, and that they are coprime. Sorry for not stating that at the outset.
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