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

In summary, the conversation discusses finding a compact expression for the infinite sum of e^(i*n*x). The person is given a hint to use the fact that S(x)-1= S(x)*e^(ix), but almost. They suggest converting the series into trigonometric terms but it doesn't seem to get them the solution. Another person suggests using the formula for the sum of a finite geometric series and the original person realizes their mistake and suggests the sum to be [1-e^(i*n*x)]/[1-e^(i*x)]. The conversation ends with a confirmation that this is likely the answer, with a minor correction to the formula.
  • #1
eschiesser
18
0

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!
 
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  • #2
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.
 
  • #3
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!
 
Last edited:
  • #4
eschiesser said:
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.
 
  • #5
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!
 
  • #6
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.
 
  • #7
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!
 
  • #8
Look up the formula again. You are summing n terms. Shouldn't it be n+1 in your formula?
 
  • #9
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.
 

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

1. What is an infinite sum of e^(n*i*x) terms?

An infinite sum of e^(n*i*x) terms is a mathematical series that involves the constant e raised to the power of a complex number (n*i*x) multiplied by a variable n that starts at 0 and increases by 1 with each term. This series is also known as a Fourier series and is commonly used in mathematics and physics.

2. What is the purpose of an infinite sum of e^(n*i*x) terms?

The purpose of an infinite sum of e^(n*i*x) terms is to represent a periodic function using a combination of sine and cosine functions. This allows for easier analysis and manipulation of the function, as well as providing a way to approximate any continuous function using a finite number of terms.

3. How is an infinite sum of e^(n*i*x) terms calculated?

To calculate an infinite sum of e^(n*i*x) terms, one must use the formula a_n = (1/2π) * ∫f(x)e^(-n*i*x)dx, where a_n is the coefficient for each term in the series and f(x) is the periodic function being represented. This formula is derived from the Euler's formula, e^(ix) = cos(x) + i*sin(x), and can be used to find the coefficients for each term in the series.

4. What is the convergence of an infinite sum of e^(n*i*x) terms?

The convergence of an infinite sum of e^(n*i*x) terms depends on the function being represented. In some cases, the series may converge to the original function, while in others it may only converge to an approximation. The convergence also depends on the values of x and n, and certain values may result in divergence.

5. What are the applications of an infinite sum of e^(n*i*x) terms?

An infinite sum of e^(n*i*x) terms has many applications in mathematics and physics. It is commonly used in Fourier analysis to represent periodic functions, as well as in signal processing, differential equations, and quantum mechanics. It is also a fundamental concept in the study of harmonic functions and complex analysis.

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