# Math Help - method of differnce

1. ## method of differnce

the 2 equations above are from my books, can i modify the equation to the fourth one, and use it?

2. ## Re: method of differnce

If you are asking "Can I go from $\sum [f(r)- f(r+1)]$ to $\sum [f(r- 1)- f(r)$?" the answer is "that depends".

It depends upon exactly what that sum is- what values of r is it over?

If you have $\sum_{r= a}^b [f(r)- f(r+ 1)$, then letting n= r+1, so that r= n- 1, f(r)- f(r+ 1) becomes f(n- 1)- f(n). But when r= a, n= a+ 1 and when r= b, n= b+ 1. So we have $\sum_{r= a}^b [f(r)- f(r+ 1)]= \sum_{n= a+1}^{b+1} [f(n)- f(n-1)]$. Of course we could change the "dummy index" at will so we can write $\sum_{r= a}^b [f(r)- f(r+ 1)]= \sum_{r= a+1}^{b+1} [f(r)- f(r-1)]$.
But notice that the limits on the sum have changed.

Of course, if the sum were from $-\infty$ to $\infty$ that wouldn't matter.