Simplifying finite geometric series expression

[Beer-monster]

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.

\sum_{n=0}^{N}e^{-na}

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 e^{-a} and the first term = 1. So the using the sum of N terms of a geometric series we have.


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

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

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

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

 

[chiro]

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)?

 

[Beer-monster]

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 \delta. 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 \theta.

 

[Chestermiller]

You can simplify the first ratio to exp (-(N+1)a/2). I don't think this helps much.