Finding the sum of a trigonometric series

AI Thread Summary
The discussion centers around finding the sum of the series 1 + a cos θ + a² cos 2θ + a³ cos 3θ + ... + aⁿ cos nθ. The proposed solution involves using De Moivre's theorem and the properties of geometric series, but the original poster struggles to apply these concepts correctly. They express confusion over identifying the correct ratio for the geometric series and how to isolate the real part of the complex representation. Several participants suggest using De Moivre's theorem directly to simplify the problem, emphasizing the importance of understanding the geometric series formula. Ultimately, the original poster feels overwhelmed and considers abandoning the problem due to the complexity involved.
Skyrior
Messages
8
Reaction score
0
1. The problem statement

Find the sum of the series:
a. 1 + a cos θ + a^{2} cos 2θ + a^{3} cos 3θ + ... + a^{n} cos nθ

Apparently, the answer is:

\frac{a^{n+1}(a cos nθ - cos(n+1)θ) - a cos θ + 1)}{a^{2} - 2a cos θ + 1}

2. The attempt at a solution

= The real part of z^{0} + z^{1} + ... + z^{n}
= 1 + a cos θ + a^{2} (cos^{2} θ - sin^{2} θ) + a^{3}(cos^{3} θ - 3 cos θ sin^{2} θ) + ... + a^{n}(cos nθ)
(Basically, binomial expansion)

I continued on, but I got something I couldn't understand:

= 1 (1 - a^{2} + a^{4} - a^{6} + ...) + cos θ (a - 3a^{3} + 5a^{5} + ...) + cos^{2} θ (a^{2} - 8a^{4} + 18a^{6} + ...) + cos^{3} θ (4a^{3} - 20a^{5} + 56a^{7} + ...) + ... + cos^{n} θ (2^{n-1} a^{n} - ?)

I think the right way to the solution is much simpler than this, but I just couldn't figure it out...I'm stuck on it for hours...

3. Relevant Equations
n.a.
We are supposed to know binomial expansion, binomial expansion of complex numbers, De Movire's Theorem, roots of complex numbers, double angle identity, trigonometric addition formula.

We didn't learn: multiple angle formula, Hyperbolic Functions.

There was a previous question that asks you to:

"Use complex number methods to show that ..."
So I presume that the question I asked is also about complex numbers...

btw, I'm currently in high school, specifically taking IB Math HL, I'm not familiar with the multiple angle formula: http://mathworld.wolfram.com/Multiple-AngleFormulas.html (I don't think we learned that). We learned De Movire's Theorem though. I think I might be using the theorem in a wrong way. Thanks for any help!
 
Last edited:
Physics news on Phys.org
Skyrior said:
1. The problem statement

Find the sum of the series:
a. 1 + a cos θ + a^{2} cos 2θ + a^{3} cos 3θ + ... + a^{n} cos nθ

Apparently, the answer is:

\frac{a^{n+1}(a cos nθ - cos(n+1)θ) - a cos θ + 1)}{a^{2} - 2a cos θ + 1}

2. The attempt at a solution

= The real part of z^{0} + z^{1} + ... + z^{n}
= 1 + a cos θ + a^{2} (cos^{2} θ - sin^{2} θ) + a^{3}(cos^{3} θ - 3 cos θ sin^{2} θ) + ... + a^{n}(cos nθ)
(Basically, binomial expansion)

I continued on, but I got something I couldn't understand:
= 1 (1 - a^{2} + a^{4} - a^{6} + ...) + cos θ (a - 3a^{3} + 5a^{5} + ...) + cos^{2} θ (a^{2} - 8a^{4} + 18a^{6} + ...) + cos^{3} θ (4a^{3} - 20a^{5} + 56a^{7} + ...) + \dots\\ \dots\ + cos^{n} θ (2^{n-1} a^{n} - ?)

I think the right way to the solution is much simpler than this, but I just couldn't figure it out...I'm stuck on it for hours...

btw, I'm currently in high school, specifically taking IB Math HL, I'm not familiar with the multiple angle formula: http://mathworld.wolfram.com/Multiple-AngleFormulas.html (I don't think we learned that). We learned De Movire's Theorem though. I think I might be using the theorem in a wrong way. Thanks for any help!
De Moivre's theorem should do it. (Euler's theorem works as well or better.)

Do you know the sum for a finite geometric series? It's not difficult to derive.

Let \displaystyle \ S=\sum_{k=0}^{n}r^k=1+r+r^2+\dots+r^{n-1}+r^n \ .\

Then \displaystyle \ r\cdot S=r\sum_{k=0}^{n}r^k=r+r^2+\dots+r^{n-1}+r^n+r^{n+1} \ .\

Look at rS - S and solve for S .
 
SammyS said:
De Moivre's theorem should do it. (Euler's theorem works as well or better.)

Do you know the sum for a finite geometric series? It's not difficult to derive.

Let \displaystyle \ S=\sum_{k=0}^{n}r^k=1+r+r^2+\dots+r^{n-1}+r^n \ .\

Then \displaystyle \ r\cdot S=r\sum_{k=0}^{n}r^k=r+r^2+\dots+r^{n-1}+r^n+r^{n+1} \ .\

Look at rS - S and solve for S .

We didn't learn Euler's Theorem but we learned sum for a finite geometric series.

Sorry, but I still don't think I get it. I'm sorry for not stating it, but I did attempt to use geometric series, but I don't know what is r. I think r should be something like a (cos 2θ/cosθ) but (cos 3θ/cos 2θ) seems to be different from cos 2θ/cos θ.
 
ehild said:
Complex numbers can be written in exponential form: http://mathworld.wolfram.com/ComplexNumber.html

z=r(cosθ+isinθ)=r e.

The series is the real part of z0+z1+z2+z3...

What do you get if you replace z by a*e?

ehild

erm...I'm quite dumb...I still don't get it.

Sum of series of real+imaginary would then be: (1-(a(e^{iθ}))^{n})/(1-(a(e^{iθ})))?

Uh...

Well, I'm really confused. I know how to get to this part, but I never know how I could remove the imaginary part.

The real part of the series is
= [1-a^{n}cos(nθ)]/[1-a cos(θ)]? It just doesn't make sense to me...

erm...Is it OK to do so? Because I don't think it is right anyways

I thought again, and I think a^{2} - 2a cos θ + 1, the denominator of the answer, is the expansion of (x-y)^{2}. But I don't know how to get it nor what should be x and y.

Thank you for both of your help, but I don't think I would want to continue on this question, it has already wasted half of my day and within my brain capacity I will never be able to comprehend it anyways...
 
Multoply both the numerator and denominator by the conjugate of the denominator.

ehild
 
Let's go back & see if we can solve this using de Moivre's formula pretty much straight up.

Let U be the series defined as follows.
\displaystyle U=<br /> \sum_{k=0}^{n} \left(a\left(\cos(\theta)+i\sin(\theta)\right)\right)^k<br />

\displaystyle \quad\ \ =1+a\left(\cos(\theta)+i\sin(\theta)\right)+<br /> a^2\left(\cos(\theta)+i\sin(\theta)\right)^2+\ \dots\ <br /> +a^{n}\left(\cos(\theta)+i\sin(\theta)\right)^{n}<br />​
Applying de Moivre's formula to the above result, can you see that the real part of U is the same as the series you're to evaluate?
 
Please note the different between cos(2x) and (cosx)^2 (sorry I'm lazy to find the delta symbol).

so we use alternative form of expression for cosx = 0.5[e^(-ix)+e^(ix)] and your equation can simply solve with finite geometric series?
 
Back
Top