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Simplifying finite geometric series expression

  • #1
296
0

Homework Statement



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
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
131
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)?
 
  • #3
296
0
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].
 
  • #4
20,143
4,212
You can simplify the first ratio to exp (-(N+1)a/2). I don't think this helps much.
 

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