# Nasty summation + derivative help

1. Jun 11, 2012

### exmachina

Edit: LOTS OF TYPOS (sorry guys)

Let:

$$f(r) = e^{-(a-r)^2}$$
$$g(r) = r e^{-(a-r)^2}$$

Where a is some constant

Can:

$$\dfrac{ \sum\limits^{r=\infty}_{r=-\infty} g(r) } {\sum\limits^{r=\infty}_{r=-\infty} f(r) }$$

Be simplified?

Last edited: Jun 11, 2012
2. Jun 11, 2012

### Millennial

Isn't it obvious?

3. Jun 11, 2012

### genericusrnme

Yes, look at the differences between f(r) and f'(r)

4. Jun 11, 2012

### HallsofIvy

Staff Emeritus
This is incorrect. If $f(r)= e^{-(a- r)^2}$ then $f'(r)= e^{-(a-r)^2}[-2(a- r)(-1)]= 2(a-r)e^{-(a-r)^2}$.

In order that $f'(r)$ exist, r must be a continuous variable. If r is not discrete, what does "$\sum_{r=-\infty}^{r=\infty}f(r)$" mean?

5. Jun 11, 2012

### exmachina

Sorry I was way too sloppy in my original post, I have since updated it, I had forgotten an r term.

Last edited: Jun 11, 2012
6. Jun 11, 2012

### exmachina

double post sorry

Last edited: Jun 11, 2012