Finding the sum of a trigonometric series

[Skyrior]

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!

 

[SammyS]

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 .

 

[Skyrior]

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]

Complex numbers can be written in exponential form: http://mathworld.wolfram.com/ComplexNumber.html

z=r(cosθ+isinθ)=r e^iθ.

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

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

ehild

 

[Skyrior]

Complex numbers can be written in exponential form: http://mathworld.wolfram.com/ComplexNumber.html

z=r(cosθ+isinθ)=r e^iθ.

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

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

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...

 

[ehild]

Multoply both the numerator and denominator by the conjugate of the denominator.

ehild

 

[SammyS]

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.

$\displaystyle U= \sum_{k=0}^{n} \left(a\left(\cos(\theta)+i\sin(\theta)\right)\right)^k$

$\displaystyle \quad\ \ =1+a\left(\cos(\theta)+i\sin(\theta)\right)+ a^2\left(\cos(\theta)+i\sin(\theta)\right)^2+\ \dots\ +a^{n}\left(\cos(\theta)+i\sin(\theta)\right)^{n}$​

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?

 

[csleong]

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?