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Nasty summation + derivative help

  1. Jun 11, 2012 #1
    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?
     
    Last edited: Jun 11, 2012
  2. jcsd
  3. Jun 11, 2012 #2
    Isn't it obvious?
     
  4. Jun 11, 2012 #3
    Yes, look at the differences between f(r) and f'(r)
     
  5. Jun 11, 2012 #4

    HallsofIvy

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    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].

    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?
     
  6. Jun 11, 2012 #5
    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
  7. Jun 11, 2012 #6
    double post sorry
     
    Last edited: Jun 11, 2012
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