Apostol & floor function problem

[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

$$\sum_{n=a}^{b-1} [\frac{na}{b}] = \sum_{n=a}^{b-1} [\frac{a(b-n)}{b}][\tex] "<br /> <br /> I'm comfortable with the solution from this point on. I just don't see how these two are equivalent.<br /> <br /> Any help would be appreciated!<br /> <br /> p.s. Sorry about the LaTeX not displaying properly.$$

 

[cristo]

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

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

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.

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!)

 

[quasar987]

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

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

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.

 

[matt grime]

You're just doing the sum in reverse order.

 

[HamishMc]

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

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.