Nasty summation + derivative help

[exmachina]

Edit: LOTS OF TYPOS (sorry guys)

Let:

[tex] f(r) = e^{-(a-r)^2} [/tex]
[tex] g(r) = r e^{-(a-r)^2} [/tex]

Where a is some constant

Can:

[tex]
\dfrac{ \sum\limits^{r=\infty}_{r=-\infty} g(r) } {\sum\limits^{r=\infty}_{r=-\infty} f(r) }
[/tex]

Be simplified?

 

[Millennial]

Isn't it obvious?

 

[genericusrnme]

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

 

[HallsofIvy]

Let:

[tex] f(r) = e^{-(a-r)^2} [/tex]
[tex] f'(r) = 2e^{-(a-r)^2} [/tex]

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

Where a is some constant

Can:

[tex]
\dfrac{ \sum\limits^{r=\infty}_{r=-\infty} f'(r) } { \sum\limits^{r=\infty}_{r=-\infty} f(r) }
[/tex]

Be simplified?

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