Geometric sum using complex numbers

[fatpotato]

Homework Statement

Calculate the sums $S_5 = \sum_{k=0}^{n} \frac{\cos(kx)}{\cos^k(x)}$ and $S_6 = \sum_{k=0}^{n} \frac{\sin(kx)}{\sin^k(x)}$.

Relevant Equations

From $S = \sum_{k=0}^{n} e^{i(a+kb)}$ it is possible to derive expressions for the finite sum of sine and cosine where the argument of those functions are of the form$(a+kb)$ where $k$ is an integer : $$S_R = \sum_{k=0}^{n} \cos(a + bk) = \cos(a + \frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})}$$

And $$S_I = \sum_{k=0}^{n} \sin(a + bk) = \sin(a + \frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})}$$

Solution to the problem tells us that $S_5 + i S_6$ is the sum of the terms of a geometric sequence and thus the solutions should be :

$$S_5 = \frac{\sin( (n+1) x)}{\cos^n(x) \sin(x)},\,\,\,\, S_6 = \frac{\cos^{n+1}(x) - \cos((n+1)x)}{\cos^n(x) \sin(x)} , x \notin \frac{\pi}{2} \mathbb{Z}$$

I'm failing to see how they got to the answer. So far, I have written the terms of $S_5 + i S_6$ in both cartesian and polar form to try and find the common ratio $r$ but this method yields nothing.

The other method I fiddled with was taking the result from $S_R$, taking the special case where $a=0$ and dividing by $\cos(x)$ to try to get a new expression to apply the formula of geometric sums :

$$\frac{S_{R}}{\cos^k(x)} = \sum_{k=0}^{n} \frac{\cos(bk)}{\cos^k(x)} = \cos(\frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})} \cdot a_0 \frac{1 - r^{n+1}}{1 - r}$$

Where the first term is $a_0 = 1$ and the common ratio is $r = \cos^{-1}(x)$, but again, this leads me nowhere. Furthermore, computing series is not my forte and I'm almost certain the last step with the division isn't correct, i.e., it is wrong to embed an expression in a series.

Any help would be appreciated. Don't hesitate to point my mistakes, thank you.

Edit: Latex formatting.

 

[Antarres]

You're right in that your last attempt where you divided and then used formula for power series combined with this formula for cosine series, is not good, since a series of products is not a product of series, hence that formula doesn't apply.

It is often good to start with the guideline form the solution, so if we form $S_5 + iS_6$, we get:
$$S_5 + iS_6 = \sum_{k=0}^n \frac{\cos(kx)}{\cos^k(x)} + i \frac{\sin(kx)}{\sin^k(x)} = \sum_{k=0}^n \frac{\cos(kx)\sin^k(x) + i \sin(kx)\cos^k(x)}{(\cos(x)\sin(x))^k} $$
Now if we look at the sum we got, the denominator looks like geometric series(like they say in the solution of yours), so only the numerator is the problem. Like in your example, we would like to put numerator into exponential form, so that the general term of the series would be geometric, and then we can solve it. But looking at the formula in the numerator doesn't look like we can straightforward do it.

I would suggest that you use the formulas here that are directly derived from the Euler formula for exponential form of complex number:
$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2} \qquad \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

You might need to work some algebra to finally simplify it, but that would be the most direct way, I think. You need to get the numerator either into one exponent(where $k$ is in exponent so you can pull it out), or a sum of such terms(this is more likely to happen with this approach).

I'm not sure if there is an elegant solution, haven't worked on these kinds of problems in a while, but I hope I gave you some ideas.

 

[Davros_Kaled]

Should you have $\cos^k x$ in second summation ?

If so, the required sums can be constructed from the GPs $\text{ } \sum^{n}_{k=0} \left(\dfrac {e^{\pm ix}} {\cos{x}} \right) ^k$

These sums are $A = \dfrac{i}{\sin{x}} \left( \cos{x}- \dfrac{e^{ix(n+1)}}{\cos^n x}\right)$ and $B =\dfrac{-i}{\sin{x}} \left( \cos{x}- \dfrac{e^{-ix(n+1)}}{\cos^n x}\right)$
$A+B = 2S_5 = \dfrac{i}{\sin{x}\cos^n x} \left( -e^{ix(n+1)}+e^{-ix(n+1)} \right) = 2 \dfrac { \sin{(n+1)x}}{\sin{x}\cos^n x}$
$A-B = 2iS_6 = \dfrac{i}{\sin{x}} \left( 2\cos{x}- \dfrac{(e^{ix(n+1)}+e^{-ix(n+1))}}{\cos^nx} \right)$
$A-B = 2i \dfrac { \cos^{n+1}{x}-\cos{(n+1)x}}{\sin{x}\cos^n x}$

 

[fatpotato]

You're right in that your last attempt where you divided and then used formula for power series combined with this formula for cosine series, is not good, since a series of products is not a product of series, hence that formula doesn't apply.

Thank you for the reminder. I really needed to read "a series of products is not a product of series" to remember it! I tried to simplify the expression but got nowhere.

Should you have coskxcosk⁡x \cos^k x in second summation?

I am not sure about what you are referring to. Just to be sure, I double-checked and I wrote correctly the answers from my textbook.

What I'm really curious about, is how you managed to get this result :

These sums are A=isinx(cosx−eix(n+1)cosnx)A=isin⁡x(cos⁡x−eix(n+1)cosn⁡x) A = \dfrac{i}{\sin{x}} \left( \cos{x}- \dfrac{e^{ix(n+1)}}{\cos^n x}\right) and B=−isinx(cosx−e−ix(n+1)cosnx)B=−isin⁡x(cos⁡x−e−ix(n+1)cosn⁡x) B =\dfrac{-i}{\sin{x}} \left( \cos{x}- \dfrac{e^{-ix(n+1)}}{\cos^n x}\right)

I believe I understand why it is correct, but how did you find this? How did you define #A# and #B# in the first place?

 

[Davros_Kaled]

A is sum of GP with first term = 1 and common ratio $e^{ix}/\cos{x}$
B is sum of GP with first term = 1 and common ratio $e^{-ix}/\cos{x}$

If you run a specific example of your sums through Mathematica the second one is incorrect.
If you change $sin^k{x}$ to $cos^k{x}$ it is correct. Suspect answer from book.