Infinite Sum of e^(n*i*x) terms n=0,1,2

[eschiesser]

Homework Statement

I am asked to find a "compact expression" for the infinite sum:

S(x) = 1 + e^(ix) + e^(2ix) + e^(3ix) +...+ e^(i*n*x)

I am given a hint: "Note that it isn't true that S(x)-1= S(x)*e^(ix), but almost. Use this fact."

Homework Equations

e^(ix)=cos(x) + isin(x), the famous Euler's formula, is all I can think of that would be helpful in solving this.

The Attempt at a Solution

Thus far, the only thing I have managed to do is convert the series into trigonometric terms:

1+(cosx+isinx)+(cos2x+isin2x)+... etc. I have a feeling this is not going to get me the solution though. Any insight would be appreciated. Thanks!

 

[Dick]

If x is real, the infinite sum doesn't converge. You can certainly write a compact expression for the finite sum. It's a geometric series.

 

[eschiesser]

I had that thought, but for some reason I dismissed it because it doesn't converge. Would the sum be something along these lines:

1/[1-e^(ix)]

This seems too simple to be the "compact expression" the problem is looking for. Though it is quite compact.

Thanks for you help!

edit: I thought of another possible solution:

[1-e^(i*n*x)]/[1-e^(i*x)]

Is it within the parameters of the problem to have the integer "n" included in the answer?

Thanks again!

 

[Dick]

I had that thought, but for some reason I dismissed it because it doesn't converge. Would the sum be something along these lines:

1/[1-e^(ix)]

This seems too simple to be the "compact expression" the problem is looking for. Though it is quite compact.

The clue told you that was wrong. The INFINITE series doesn't converge. Use the formula for the sum of a FINITE geometric series. It has an extra term in the numerator.

 

[eschiesser]

I thought of another possible solution:

[1-e^(i*n*x)]/[1-e^(i*x)]

Is it within the parameters of the problem to have the integer "n" included in the answer?

I was typing this as you posted ha. This would be the sum of a finite geometric series, no?

I now realize what the clue was telling me. This answer makes the most sense to me. Do you think this is what the problem is looking for? Thanks again!
Thanks again!

 

[Dick]

That's all I can think that the problem might be asking for. BTW I don't think the 'n' in your term in the numerator term is quite right.

 

[eschiesser]

I had the same thought. Should it be defined to mean the number of "n" terms? Maybe a large N? Or are you saying there is something more fundamentally wrong with it. Thanks!

 

[Dick]

Look up the formula again. You are summing n terms. Shouldn't it be n+1 in your formula?

 

[stormazrael]

i think the hint should be S(x)-1= S(x)*e^(ix) - e^(n+1)ix
for instance, if n is 3
S(x)= 1 + e^(ix) + e^(2ix) + e^(3ix)
S(x)*e^(ix) = e^(ix) + e^(2ix) + e^(3ix) + e^(4ix)
by my hint if z=e^(ix)
S(x) - 1 = S(x)*z+z^(n+1)
so S(x) = [(1-z^(n+1)]/(1-z)

This answer should be right , because I got almost the same homework. My prof gave us the answer but hint.