Starting to prove by De Moivre's Theorem

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In summary, the problem is to find a geometric progression with common ratio exp(i*theta)/2, and to factor it.
  • #1

rock.freak667

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

Homework Statement



Use de Moivre's theorem to show that

[tex]\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)}[/tex]

Homework Equations



[tex](cos\theta + isin\theta)^n=cosn\theta + isinn\theta[/tex]

The Attempt at a Solution



Just need a little help on how to start this question. Where would I get the [itex]\frac{sinn\theta}{2^n}[/itex] from?
 
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  • #2
Just need a little help on how to start this question. Where would I get the [itex]\frac{sinn\theta}{2^n}[/itex] from?
As the imaginary part of [itex](cos\theta + isin\theta)^n[/itex], presumably. Oh, I suppose that doesn't work literally.
 
  • #3
Hurkyl said:
As the imaginary part of [itex](cos\theta + isin\theta)^n[/itex], presumably. Oh, I suppose that doesn't work literally.

(btw, the ∑ is to N, not to ∞)

Yes it does … sum [itex]\left(\frac{1}{2}(cos\theta + isin\theta)\right)^n[/itex] from 1 to N, get rid of the complex denominator in the usual way, and take the imaginary part. :smile:
 
  • #4
But if I expand [itex](cos\theta +isin\theta)^n[/itex] how do I know when to stop since n varies?


[tex](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 +...[/tex]
 
  • #5
rock.freak667 said:
But if I expand [itex](cos\theta +isin\theta)^n[/itex] how do I know when to stop since n varies?


[tex](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 +...[/tex]

Don't expand it! Use deMoivre!
 
  • #6
Dick said:
Don't expand it! Use deMoivre!

But then that would just give me that


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

the left has what I need, the right doesn't seem to be giving me any ideas.
 
  • #7
rock.freak667 said:
But then that would just give me that


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

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?
 
  • #8
Forgive me if I am slow with this, I am very bad with these problems...But back to it. [itex]sin n\theta = Im[(cos\theta +i sin\theta)^n][/itex]

putting that into

[tex]\sum_{n=1} ^{N} \frac{sinn\theta}{2^n} [/tex]

I get

[tex]
\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[/tex]

[tex]\sum_{n=1} ^{N} (\frac{sin\theta}{2})^n[/tex]Which is a GP with first term,a=[itex]\frac{sin\theta}{2}[/itex] and common ratio,r=[itex]\frac{sin\theta}{2}[/itex] and |r|<1 ?
 
  • #9
rock.freak667 said:
Forgive me if I am slow with this, I am very bad with these problems...But back to it. [itex]sin n\theta = Im[(cos\theta +i sin\theta)^n][/itex]

putting that into

[tex]\sum_{n=1} ^{N} \frac{sinn\theta}{2^n} [/tex]

I get

[tex]
\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[/tex]

[tex]\sum_{n=1} ^{N} (\frac{sin\theta}{2})^n[/tex]Which is a GP with first term,a=[itex]\frac{sin\theta}{2}[/itex] and common ratio,r=[itex]\frac{sin\theta}{2}[/itex] 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.
 
  • #10
Dick said:
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 [itex]e^{i\theta}=cos\theta +i sin\theta[/itex] 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?
 
  • #11
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 [itex]2^{N+1}sin\theta[/itex], I have [itex]2^Nsin\theta[/itex]. Shall recheck my algebra in a while.
 
  • #12
rock.freak667 said:
Not too clear here.

I know that [itex]e^{i\theta}=cos\theta +i sin\theta[/itex] 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.
 
  • #13


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
 
  • #14


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!
 

1. What is De Moivre's Theorem?

De Moivre's Theorem is a mathematical concept that states that for any complex number z and any positive integer n, the nth power of z can be expressed as the product of the nth power of the absolute value of z and the nth power of the angle of z in polar coordinates.

2. How is De Moivre's Theorem used in proofs?

De Moivre's Theorem is often used in proofs involving complex numbers and trigonometric functions. It allows for simplification and manipulation of complex expressions, making it a powerful tool in mathematical proofs.

3. What are the key steps in proving by De Moivre's Theorem?

The key steps in proving by De Moivre's Theorem are: 1) expressing the complex number in polar form, 2) applying De Moivre's Theorem to simplify the expression, and 3) converting the simplified expression back to rectangular form if needed.

4. Are there any limitations to using De Moivre's Theorem in proofs?

De Moivre's Theorem can only be applied to complex numbers and does not work for real numbers. Additionally, it is only valid for positive integer powers, so it cannot be used for fractional or negative powers.

5. Can De Moivre's Theorem be used in other areas of mathematics?

Yes, De Moivre's Theorem has applications in various areas of mathematics, such as geometry, trigonometry, and calculus. It is also used in fields like physics and engineering for solving complex equations and analyzing systems.

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