Starting to prove by De Moivre's Theorem

[rock.freak667]

[SOLVED] Starting to prove by De Moivre's Theorem

Homework Statement

Use de Moivre's theorem to show that

\sum_{n=1} ^{\infty} \frac{sinn\theta}{2^n}=\frac{2^{N+1}sin\theta+sinN\theta-2sin(N+1)\theta}{2^N(5-4cos\theta)}

Homework Equations

(cos\theta + isin\theta)^n=cosn\theta + isinn\theta

The Attempt at a Solution

Just need a little help on how to start this question. Where would I get the \frac{sinn\theta}{2^n} from?

 

[Hurkyl]

Just need a little help on how to start this question. Where would I get the \frac{sinn\theta}{2^n} from?

As the imaginary part of (cos\theta + isin\theta)^n, presumably. Oh, I suppose that doesn't work literally.

 

[tiny-tim]

As the imaginary part of (cos\theta + isin\theta)^n, presumably. Oh, I suppose that doesn't work literally.

(btw, the ∑ is to N, not to ∞)

Yes it does … sum \left(\frac{1}{2}(cos\theta + isin\theta)\right)^n from 1 to N, get rid of the complex denominator in the usual way, and take the imaginary part.

 

[rock.freak667]

But if I expand (cos\theta +isin\theta)^n how do I know when to stop since n varies?


(cos\theta +isin\theta)^n=(isin\theta)^n + ^nC_1(isin\theta)^{n-1}cos\theta +^nC_2(isin\theta)^{n-2}cos^2\theta +...

 

[Dick]

But if I expand (cos\theta +isin\theta)^n how do I know when to stop since n varies?


(cos\theta +isin\theta)^n=(isin\theta)^n + ^nC_1(isin\theta)^{n-1}cos\theta +^nC_2(isin\theta)^{n-2}cos^2\theta +...

Don't expand it! Use deMoivre!

 

[rock.freak667]

Don't expand it! Use deMoivre!

But then that would just give me that


sin n\theta = Im[(cos\theta +i sin\theta)^n

the left has what I need, the right doesn't seem to be giving me any ideas.

 

[Hurkyl]

But then that would just give me that


sin n\theta = Im[(cos\theta +i sin\theta)^n

the left has what I need, the right doesn't seem to be giving me any ideas.

Not even after plugging it into the sum and simplifying?

 

[rock.freak667]

Forgive me if I am slow with this, I am very bad with these problems...But back to it. sin n\theta = Im[(cos\theta +i sin\theta)^n]

putting that into

\sum_{n=1} ^{N} \frac{sinn\theta}{2^n}

I get

<br /> \sum_{n=1} ^{N} \frac{Im[(cos\theta +i sin\theta)^n]}{2^n} =\sum_{n=1} ^{N} (\frac{Im[(cos\theta +i sin\theta)}{2})^n

\sum_{n=1} ^{N} (\frac{sin\theta}{2})^nWhich is a GP with first term,a=\frac{sin\theta}{2} and common ratio,r=\frac{sin\theta}{2} and |r|<1 ?

 

[Dick]

Forgive me if I am slow with this, I am very bad with these problems...But back to it. sin n\theta = Im[(cos\theta +i sin\theta)^n]

putting that into

\sum_{n=1} ^{N} \frac{sinn\theta}{2^n}

I get

<br /> \sum_{n=1} ^{N} \frac{Im[(cos\theta +i sin\theta)^n]}{2^n} =\sum_{n=1} ^{N} (\frac{Im[(cos\theta +i sin\theta)}{2})^n

\sum_{n=1} ^{N} (\frac{sin\theta}{2})^nWhich is a GP with first term,a=\frac{sin\theta}{2} and common ratio,r=\frac{sin\theta}{2} and |r|<1 ?

You can't factor the nth power outside the Im, and you don't want to. The left side is a geometric progression with common ratio exp(i*theta)/2. Sum it first, then take the imaginary part of both sides.

 

[rock.freak667]

You can't factor the nth power outside the Im, and you don't want to. The left side is a geometric progression with common ratio exp(i*theta)/2. Sum it first, then take the imaginary part of both sides.

Not too clear here.

I know that e^{i\theta}=cos\theta +i sin\theta so the imaginary part would just be the sine. But if I can't factor the n outside of the Im, how do I get it as a GP?

 

[rock.freak667]

ah I now got what you meant.

I was to consider the sum from 1 to N of [exp(i*theta)/2]^n and then take the I am part. Well I got everything in the Denominator and two of the terms in the numerator. I am just missing the 2^{N+1}sin\theta, I have 2^Nsin\theta. Shall recheck my algebra in a while.

 

[Dick]

Not too clear here.

I know that e^{i\theta}=cos\theta +i sin\theta so the imaginary part would just be the sine. But if I can't factor the n outside of the Im, how do I get it as a GP?

Forget the original series. Sum (exp(i*theta)/2)^n. Then take the imaginary part. The original series will appear on the left side. It will be a surprise.

 

[usman94]

Can some1 post the whole solution? How to get the 2^N[5-4Cos(theta)]


How do we get cos? If we are taking the imaginry part? Please solve it and post it... thanks in advance

 

[usman94]

We want to show that
Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)].
-------------------------
Note that e^(iθ) = cos θ + i sin θ.
So, De Moivre's Theorem yields for any positive integer n
e^(inθ) = (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ).

Next, note that
Σ(n=1 to N) sin(nθ) / 2^n
= I am [Σ(n=1 to N) e^(inθ) / 2^n], by the first observation
= I am {Σ(n=1 to N) [e^(iθ)/2]^n}.

Thus, we need to compute the imaginary part of the finite geometric series
Σ(n=1 to N) [e^(iθ)/2]^n
= (e^(iθ)/2) Σ(k=0 to N-1) [e^(iθ)/2]^k, by re-indexing the sum
= (e^(iθ)/2) (1 - (e^(iθ)/2)^N) / (1 - e^(iθ)/2)
= e^(iθ) (1 - (e^(iθ)/2)^N) / (2 - e^(iθ))
= e^(iθ) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))]

Multiply numerator and denominator by the conjugate (2 - e^(-iθ)):
==> e^(iθ)(2 - e^(-iθ)) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))(2 - e^(-iθ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (e^(iθ) + e^(-iθ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (2 cos θ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 4 cos θ)].

Now, it's a matter of simplifying the numerator.
(2e^(iθ) - 1) (2^N - e^(iNθ))
= 2^(N+1) e^(iθ) - 2^N - 2e^(i(N+1)θ) + e^(iNθ)
= 2^(N+1) (cos θ + i sin θ) - 2^N - 2 (cos(N+1)θ + i sin(N+1)θ) + (cos(Nθ) + i sin(Nθ))
= [2^(N+1) cos θ - 2^N - 2 cos(N+1)θ + cos(Nθ)] + i [2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ)]

Its imaginary part is 2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ).

Therefore, we have obtained
Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)]
as required.

I hope this helps!