Sum of a geometric series of complex numbers

[whatisreality]

Homework Statement

Given an integer n and an angle θ let
S_n(θ) = ∑(e^ikθ) from k=-n to k=n

And show that this sum = sinα / sinβ

Homework Equations

Sum from 0 to n of x^k is (x^k+1-1)/(x-1)

The Attempt at a Solution

The series can be rewritten by taking out a factor of e^-iθ as
e^-iθ∑(e^iθ)^k from 0 to 2n.

= (e^inθ-1)/(e^iθ-1)
Subsequent attempts at rearrangement, e.g. trying to realize the denominator by multiplying by e^-iθ-1 have not got me any closer to the answer. Not sure what to do from here...

 

[mhsd91]

Are you absolutely sure you printed the problem correctly? I ask since I get the following,

$S_n (\theta) = \sum \limits _{k=-n}^{n} e^{ik\theta}$

$= 1 + \sum \limits _{k=1}^{n} \left( e^{ik\theta} + e^{-ik\theta} \right)$

$= 1 + 2 \sum \limits _{k=1}^{n} \cos (k \theta )$

Where I see no possible way for rewriting in terms of what you have there. Are $\alpha, \beta$ constants?

 

[AdityaDev]

$e^{ik\theta}=cosk\theta+isink\theta$
Split the sumation from -n to -1, then from 1 to n. For k=0, you can simply substitute.
now again split the summation. On for summing cos terms and other for sin terms.
do you notice that the angle inside the cosine and sine terms are in AP?
That is:Σcoskθ from -n to -1 = $cos(-n\theta )+cos(-n\theta +\theta )+...+cos(-n\theta + (n-1)\theta )$
You have a simple formula for evaluating its sum:
Sin(a)+sin(a+b)+...+sin(a+(n-1)b)= try deriving it...

 

[Fredrik]

The series can be rewritten by taking out a factor of e^-iθ as
e^-iθ∑(e^iθ)^k from 0 to 2n.

Not sure what you did here.
$$(e^{i\theta})^k =(e^{-i\theta}e^{2i\theta})^k = e^{-ik\theta} e^{2ik\theta}.$$

 

[whatisreality]

Not sure what you did here.
$$(e^{i\theta})^k =(e^{-i\theta}e^{2i\theta})^k = e^{-ik\theta} e^{2ik\theta}$$

Reasoning as follows:
Series is z^-n+z^-n+1...+zⁿ
So if I take out a factor of z^-n, as in divide by it, then that becomes the powers are from 0 to 2n, and the geometric sum to n formula can be used.

Although come to think of it, it isn't entirely clear that that's allowed... does it break the rules?

 

[Fredrik]

Reasoning as follows:
Series is z^-n+z^-n+1...+zⁿ
So if I take out a factor of z^-n, as in divide by it, then that becomes the powers are from 0 to 2n, and the geometric sum to n formula can be used.

Although come to think of it, it isn't entirely clear that that's allowed... does it break the rules?

Thanks for the clarification. Yes, this you can do. I thought you were taking something that depends on k outside of the sum, but I see now that you're not.

 

[whatisreality]

$e^{ik\theta}=cosk\theta+isink\theta$
Split the sumation from -n to -1, then from 1 to n. For k=0, you can simply substitute.
now again split the summation. On for summing cos terms and other for sin terms.
do you notice that the angle inside the cosine and sine terms are in AP?
That is:Σcoskθ from -n to -1 = $cos(-n\theta )+cos(-n\theta +\theta )+...+cos(-n\theta + (-n-1)\theta )$
You have a simple formula for evaluating its sum:
Sin(a)+sin(a+b)+...+sin(a+(n-1)b)= try deriving it...

So evaluate 4 summations? When writing out the coskθ expansion, don't think I follow. Why isn't there a cos(-θ) term? That sounds like it has a really obvious answer... Sorry if it's a silly question!

 

[whatisreality]

Are you absolutely sure you printed the problem correctly? I ask since I get the following,

$S_n (\theta) = \sum \limits _{k=-n}^{n} e^{ik\theta}$

$= 1 + \sum \limits _{k=1}^{n} \left( e^{ik\theta} + e^{-ik\theta} \right)$

$= 1 + 2 \sum \limits _{k=1}^{n} \cos (k \theta )$

Where I see no possible way for rewriting in terms of what you have there. Are $\alpha, \beta$ constants?

They look completely different. But I can't find why mine is wrong, so maybe they're the same? But where are the sin terms going to come from?Oh, and I don't think alpha and beta are constant, they depend on n and/or theta according to the question!
I'll give it a go with this expression, see if I can get there. Thanks :)

Edit: definitely different. What kind of series is that, anyway?!

 

[AdityaDev]

So evaluate 4 summations? When writing out the coskθ expansion, don't think I follow. Why isn't there a cos(-θ) term? That sounds like it has a really obvious answer... Sorry if it's a silly question!

It was a typo. Its edited now

 

[Ray Vickson]

Homework Statement

Given an integer n and an angle θ let
S_n(θ) = ∑(e^ikθ) from k=-n to k=n

And show that this sum = sinα / sinβ

Homework Equations

Sum from 0 to n of x^k is (x^k+1-1)/(x-1)

The Attempt at a Solution

The series can be rewritten by taking out a factor of e^-iθ as
e^-iθ∑(e^iθ)^k from 0 to 2n.

= (e^inθ-1)/(e^iθ-1)
Subsequent attempts at rearrangement, e.g. trying to realize the denominator by multiplying by e^-iθ-1 have not got me any closer to the answer. Not sure what to do from here...

If $S_0 = \sum_{k=1}^n e^{i k \theta}$, then $S = 1 + 2 \text{Re}(S_0)$, where $\text{Re}$ denotes the "real part". Use the geometric sum formula to get $S_0$, then re-write it to extract the real and imaginary parts.

 

[whatisreality]

If $S_0 = \sum_{k=1}^n e^{i k \theta}$, then $S = 1 + 2 \text{Re}(S_0)$, where $\text{Re}$ denotes the "real part". Use the geometric sum formula to get $S_0$, then re-write it to extract the real and imaginary parts.

What does S denote? I am a little bit lost. And should one part, either real or imaginary, be zero? Because the whole series is meant to sum to just one term?

 

[Ray Vickson]

What does S denote? I am a little bit lost. And should one part, either real or imaginary, be zero? Because the whole series is meant to sum to just one term?

My $S$ is your $S_n(\theta)$.

In your original sum, for every $+k$ term there is also a corresponding $-k$ term, so the imaginary parts cancel out when you perform the whole sum. The reason this happens is that the $+k$ and $-k$ terms taken together produce
$$e^{ik\theta} + e^{-ik\theta} = 2 \cos(k \theta) = 2\, \text{Re}(e^{i k \theta}).$$
Therefore, it suffices to sum the real parts for $k$ from 1 to $n$. What makes this do-able is the fact that the sum of the real parts equals the real part of the sum. Thus, you can sum first, then take the real part; that will allow you to perform calculations like $\sum_{k=1}^n \cos(k \theta)$ by using geometric sums.

 

[mhsd91]

I have another approach, which almost solved it. I think something is wrong with problem itself, or perhaps som eother information is missing. However, will give my seconds try, and perhaps you'll find, some mistake I made and solve it... Anyways,
$S_n(\theta) = \sum \limits _{k=-n}^n e^{ik\theta}$

adding $e^0 - 1 = 0$ to the sum enables us to write,

$= \left( \sum \limits _{k=0}^n e^{+ik\theta}\right) + \left( \sum \limits _{k=0}^n e^{-ik\theta}\right) - 1$

Then using the final sum formula for a geometrical series as you yourself posted,

$= \frac{e^{+i(n+1)\theta}-1}{e^{+i\theta}-1} + \frac{e^{-i(n+1)\theta}-1}{e^{-i\theta}-1} - 1$

Make all fractions have the same denominator

$= -\frac{e^{+i(n+1)\theta}-1}{1-e^{i\theta}} + \frac{e^{-i(n+1)\theta + i\theta}-e^{i\theta}}{1-e^{i\theta}} - \frac{1-e^{i\theta}}{1-e^{i\theta}}$

Rearrange,

$= \frac{1}{1-e^{i\theta}} \left( -e^{+i(n+1)\theta} + 1 + e^{-i(n+1)\theta + i\theta}-e^{i\theta} - 1+e^{i\theta}\right)$

Some stuff goes away,

$= \frac{1}{1-e^{i\theta}} \left( -e^{+i(n+1)\theta}+ e^{-i(n+1)\theta}\cdot e^{i\theta}\right)$

In which WolframAlpha.com simplifies to,

$= \cos(n \theta) + \frac{\cos(\theta / 2)}{\sin(\theta / 2)}\cdot \sin(n \theta)$As said, this is as far as I come. I suggest skim over my calculations and se if you find any faults.. If not, there is something wrong with the problem, or with the geometric series sum formula..

Good luck!

 

[whatisreality]

I have another approach, which almost solved it. I think something is wrong with problem itself, or perhaps som eother information is missing. However, will give my seconds try, and perhaps you'll find, some mistake I made and solve it... Anyways,
$S_n(\theta) = \sum \limits _{k=-n}^n e^{ik\theta}$

adding $e^0 - 1 = 0$ to the sum enables us to write,

$= \left( \sum \limits _{k=0}^n e^{+ik\theta}\right) + \left( \sum \limits _{k=0}^n e^{-ik\theta}\right) - 1$

Then using the final sum formula for a geometrical series as you yourself posted,

$= \frac{e^{+i(n+1)\theta}-1}{e^{+i\theta}-1} + \frac{e^{-i(n+1)\theta}-1}{e^{-i\theta}-1} - 1$

Make all fractions have the same denominator

$= -\frac{e^{+i(n+1)\theta}-1}{1-e^{i\theta}} + \frac{e^{-i(n+1)\theta + i\theta}-e^{i\theta}}{1-e^{i\theta}} - \frac{1-e^{i\theta}}{1-e^{i\theta}}$

Rearrange,

$= \frac{1}{1-e^{i\theta}} \left( -e^{+i(n+1)\theta} + 1 + e^{-i(n+1)\theta + i\theta}-e^{i\theta} - 1+e^{i\theta}\right)$

Some stuff goes away,

$= \frac{1}{1-e^{i\theta}} \left( -e^{+i(n+1)\theta}+ e^{-i(n+1)\theta}\cdot e^{i\theta}\right)$

In which WolframAlpha.com simplifies to,

$= \cos(n \theta) + \frac{\cos(\theta / 2)}{\sin(\theta / 2)}\cdot \sin(n \theta)$As said, this is as far as I come. I suggest skim over my calculations and se if you find any faults.. If not, there is something wrong with the problem, or with the geometric series sum formula..

Good luck!

Pretty sure all that is right! If you keep going, make a common denominator, then it simplifies down to sin/sin! What a horrible question. Thanks for your help :)