# How can I get the sum inside the exponential?

1. Jun 6, 2012

### matlabber

1. The problem statement, all variables and given/known data
Show if true:
$$\sum_{i=1}^{n-1}e^{2 \pi ift}=\frac{e^{2 \pi ifn}-1}{e^{2 \pi if}-1}=e^{\pi if(n-1)}\frac{e^{\pi ifn}-e^{-\pi i f n}}{e^{ \pi if}-e^{-\pi if}}\\\\\\\\$$

2. Relevant equations

I'm really stuck here, just looking for a suggestion as to what equation to use...
Can I take the log? How can I get the sum inside the exponential?

3. The attempt at a solution

Last edited: Jun 6, 2012
2. Jun 6, 2012

### vela

Staff Emeritus
Re: Euler

Hint: Geometric series.

3. Jun 6, 2012

### D H

Staff Emeritus
Re: Euler

This does not look right, no matter how I read your sum. The expression is incorrect if the sum is over i; you should have a t somewhere on the right hand side. The expression is also incorrect if the sum is over t rather than i.

$$\sum_{i=1}^{n-1} e^{2 \pi i f t} = \frac{e^{2\pi f n t} - e^{2\pi f t}} {e^{2\pi f t} - 1} = e^{2\pi f t} \frac{e^{2\pi f (n-1) t} - 1} {e^{2\pi f t} - 1}$$

and

$$\sum_{t=1}^{n-1} e^{2 \pi i f t} = \frac{e^{2\pi i f n} - e^{2\pi i f}} {e^{2\pi i f} - 1} = e^{2\pi i f} \frac{e^{2\pi i f (n-1)} - 1} {e^{2\pi i f} - 1}$$

Since you mentioned Euler in the title, I suspect it is the latter that you are supposed to prove.

Or perhaps it is

$$\sum_{t=0}^{n-1} e^{2 \pi i f t} = \frac{e^{2\pi i f n} - 1} {e^{2\pi i f} - 1}$$

4. Jun 6, 2012

### matlabber

Re: Euler

Hi,

You are correct my mistake.

$$\sum_{t=0}^{n-1}e^{2 \pi ift}=\frac{e^{2 \pi ifn}-1}{e^{2 \pi if}-1}=e^{\pi if(n-1)}\frac{e^{\pi ifn}-e^{-\pi i f n}}{e^{ \pi if}-e^{-\pi if}}\\\\\\\\$$

Last edited: Jun 6, 2012
5. Jun 6, 2012

### matlabber

Re: Euler

Using Geometric Series I get...

$$\frac{1-e^{2 \pi ifn}}{1-e^{2 \pi if}} =\frac{e^{2 \pi ifn}-1}{e^{2 \pi if}-1}$$

Do I want to split up the pieces in the numerator? or Factor?

Last edited: Jun 6, 2012
6. Jun 13, 2012

### joseirs

Re: Euler

You have:

$\frac{e^{2nx}-1}{e^{2x}-1}=\frac{e^{nx}}{e^{x}}\;\frac{e^{nx}-e^{-nx}}{e^x-e^{-x}}$

and the result follows simplifying the first factor in the right had side of the equality.

7. Jun 13, 2012

### dimension10

Re: Euler

How is this even related to Euler's formula, since i is an index? But, as vela said, geometric series makes it obvious here. But first you really have to simplify that mess. Replace 2πft with something else, like $A$, or even $x$ since it isn't used so far.