Show that $\sum_{k=0}^{\infty} p^k\cos(kx) = \frac{1-p\cos(x)}{1-2p\cos(x)+p^2}$ using complex numbers.

Question

I have this equality that I'm trying to show. I have tried many times but I can't get it to work.

Could you please help me?

I would like to first make this finite sum equal something.

$$\sum_{k=0}^np^k\cos(kx) = S_{n}(p, x), -1 < p < 1 $$

And then use it to calculate the infinite sum, which should be the right side.

$$\lim_{n \to {\infty}} S_n(p, x)=\sum_{k=0}^{\infty}p^k\cos(kx) = \frac{1-p\cos(x)}{1-2p\cos(x)+p^2}$$

I would like to use complex numbers to show this.

EDIT: I know that I have to use Euler's identity, I just can't get the algebra to work.

I have gotten so far but I don't know how to continue.

$$\frac{(pe^{ix})^{(n+1)/2}((pe^{ix})^{-(n+1)/2}-(pe^{ix})^{(n+1)/2})}{(pe^{ix})^{1/2}((pe^{ix})^{-1/2}-(pe^{ix})^{1/2})}$$

Thank you in advance!

Answer 1

If $p$ is a real number then $ p^k \cos(kx)$ is the real part of the complex number $$ p^k (\cos(kx) + i \sin(kx)) = p^k e^{ikx} = (pe^{ix})^k \, . $$ So you can compute the geometric sum $$ \sum_{k=0}^n (pe^{ix})^k = \frac{1-(pe^{ix})^{n+1}}{1-pe^{ix}} $$ and determine its real part. That is done by expanding the fraction with the conjugate of the denominator: $$ \frac{(1-(pe^{ix})^{n+1})(1-pe^{-ix})}{(1-pe^{ix})(1-pe^{-ix})} = \frac{1 -p^{n+1} e^{i(n+1)x} -pe^{-ix} + p^{n+2}e^{inx} }{1-2p \cos(x) + p^2} \, . $$ It follows that $$ \sum_{k=0}^np^k\cos(kx) = \frac{1 - p^{n+1}\cos((n+1)x) - p \cos(x) + p^{n+2} \cos(nx)}{1-2p \cos(x) + p^2} \, . $$

Answer 2

Hint:

$p^k\cos(kx)=\operatorname{Re}\bigl(p^k\mathrm e^{ikx}\bigr)=\operatorname{Re}\bigl(p\mathrm e^{ix}\bigr)^k$, so you just have to calculate the sum of a geometric series, and take its real part.

Answer 3

This is an attempt to verify this result directly.

$\begin{array}\\ S(n, p, x) &=(1-2p \cos(x) + p^2)\sum_{k=0}^np^k\cos(kx)\\ &=\sum_{k=0}^np^k\cos(kx)-2p \cos(x)\sum_{k=0}^np^k\cos(kx)+p^2\sum_{k=0}^np^k\cos(kx)\\ &=\sum_{k=0}^np^k\cos(kx)-\sum_{k=0}^n2p^{k+1} \cos(x)\cos(kx)+\sum_{k=0}^np^{k+2}\cos(kx)\\ &=\sum_{k=0}^np^k\cos(kx)-\sum_{k=0}^np^{k+1}(\cos((k+1)x)\\ &\qquad+\cos((k-1)x)+\sum_{k=0}^np^{k+2}\cos(kx)\\ &=\sum_{k=0}^np^k\cos(kx)-\sum_{k=0}^np^{k+1}\cos((k+1)x)\\ &\qquad-\sum_{k=0}^np^{k+1}\cos((k-1)x)+\sum_{k=0}^np^{k+2}\cos(kx)\\ &=\sum_{k=0}^np^k\cos(kx)-\sum_{k=1}^{n+1}p^{k}\cos(kx)-p\cos(-x)\\ &\qquad-\sum_{k=0}^{n-1}p^{k+2}\cos(kx)+\sum_{k=0}^np^{k+2}\cos(kx)\\ &=1-p^{n+1}\cos((n+1)x)-p\cos(x)+p^{n+2}\cos(nx)\\ \end{array} $

And it works!