Simplifying a ratio of Gaussian summations

[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?

 

[Millennial]

Isn't it obvious?

 

[genericusrnme]

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

 

[HallsofIvy]

Let:

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

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

Where a is some constant

Can:

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

Be simplified?

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?

 

[exmachina]

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

 

[exmachina]

double post sorry