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Simplifying finite geometric series expression

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    I've come across the type of sum in several places/problems but seem to be making no progress in trying to simplifying it further.

    We have a finite series of some exponential function.

    [tex] \sum_{n=0}^{N}e^{-na} [/tex]

    Where a is some constant, a quantum of energy or a phase factor etc.

    Now I know that this is actually a geometric series with the common ratio [itex]e^{-a}[/itex] and the first term = 1. So the using the sum of N terms of a geometric series we have.


    [tex] \sum_{n=0}^{N}e^{-na} = \frac{1-e^{-Na}}{1-e^{-a}} [/tex]

    If I take out a factor of [itex] e^{-Na/2} [/itex] from the numerator I can rewrite it as an hyperbolic sine function. I can do something similar for the denominator using a factor of [itex]e^{-a/2}[/itex], this gives:

    [tex]\sum_{n=0}^{N}e^{-na} = \frac{e^{-Na/2}}{e^{a/2}}\frac{sinh{\frac{Na}{2}}}{sinh{\frac{a}{2}}}[/tex]

    Here's where I get stuck. I can't think how to reduce this any further or clean it up. Can anyone suggest anything or point out a mistake I've made.

    Thanks

    BM
     
  2. jcsd
  3. Aug 28, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey Beer-Monster.

    The expression you've given in the second last equation line is very simple: Did you have a reason why you wanted to get it in a particular form (like a hyperbolic trig function)?
     
  4. Aug 29, 2012 #3
    One example where I felt something like that was needed involved interference from multiple wave sources. Each source successive source considered would add a phase differences [itex]\delta[/itex]. I managed to treat the waves as complex exponentials, with led to a similar sum but with a complex exponential. I was asked to determine the intensity as a function of angle from the sources. This seemed to suggest an answer as a trigonometric function of [itex]\theta[/itex].
     
  5. Sep 4, 2012 #4
    You can simplify the first ratio to exp (-(N+1)a/2). I don't think this helps much.
     
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