# Apostol & floor function problem

1. Feb 18, 2007

### HamishMc

Hi,

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

$$\sum_{n=a}^{b-1} [\frac{na}{b}] = \sum_{n=a}^{b-1} [\frac{a(b-n)}{b}][\tex] " I'm comfortable with the solution from this point on. I just don't see how these two are equivalent. Any help would be appreciated! p.s. Sorry about the LaTeX not displaying properly. Last edited: Feb 18, 2007 2. Feb 18, 2007 ### cristo Staff Emeritus Are there any specific conditions on a,b and n? (I can't see the answer to your question; just thought I'd get the LaTex working for you!) 3. Feb 18, 2007 ### quasar987 It works if you make the change of index n=b-m. Then, [tex]\sum_{n=a}^{b-1} [\frac{na}{b}] = \sum_{m=b-a}^{1} [\frac{a(b-m)}{b}]$$

You can use Gauss's sum $\sum_i^n i=n(n+1)/2$ to show that the equality you wrote is not true unless a=1 or a=b.

Last edited: Feb 18, 2007
4. Feb 18, 2007

### matt grime

You're just doing the sum in reverse order.

5. Feb 18, 2007

### HamishMc

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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