Is there an efficient way to calculate the following power series?

Question

I want to find coefficients of a power series $K_p$ given by the equation: $$\frac{1-z^2}{1+z^2-2z\cos(\theta)} = \sum_{p=0}^{\infty}K_pz^p$$ where $\theta$ is a constant. I have checked that $K_0=1, K_1=2\cos(\theta), K_2 = 2\cos(2\theta), K_3 = 2\cos(3\theta)$. And I know that the answer is $K_p = 2\cos(p\theta)$. But is there an efficient way to derive the answer for an arbitrary $p$?

Answer 1

Hint

I guess that $z\in\mathbb R$. Then, take the real part of $$\sum_{p=0}^\infty (e^{i\theta }z)^p =\frac{1}{1-ze^{i\theta }}.$$

Answer 2

If we use the generating function of the Chebyshev polynomials of the second kind (see Abramowitz-Stegun 22.9.10) $$1/(1-2xz+z^2)=\sum_{n\ge 0} U_n(x)z^n,$$ the result should follow from a recurrence involving the transition $U_n \leftrightarrow U_{n-2}$ for the Chebyshev polynomials.

Answer 3

The coefficients are $2\cos(p\theta) $.

(Just doing again what has been done many times before.)

Writing $t$ for $\theta$,

$\dfrac{1-z^2}{1+z^2-2z\cos(t)} = \sum_{p=0}^{\infty}a_pz^p $ so

$\begin{array}\\ 1-z^2 &=(1+z^2-2z\cos(t))\sum_{p=0}^{\infty}a_pz^p\\ &=(1-cz+z^2)\sum_{p=0}^{\infty}a_pz^p \qquad c = 2\cos(t)\\ &=\sum_{p=0}^{\infty}a_pz^p-cz\sum_{p=0}^{\infty}a_pz^p+z^2\sum_{p=0}^{\infty}a_pz^p\\ &=\sum_{p=0}^{\infty}a_pz^p-\sum_{p=1}^{\infty}ca_{p-1}z^p+\sum_{p=2}^{\infty}a_{p-2}z^p\\ &=a_0+a_1z-ca_0z+\sum_{p=2}^{\infty}(a_p-ca_{p-1}+a_{p-2})z^p\\ \end{array} $

so

$a_0 = 1,\\ a_1-ca_0 = 0,\\ a_2-ca_1+a_0 = -1,\\ a_p-ca_{p-1}+a_{p-2}=0 \text{ for }p > 2. $

Therefore

$\begin{array}\\ a_1 &= c,\\ &= 2\cos(t),\\ a_2 &=-1+ca_1-a_0\\ &=-2+c^2\\ &=-2+4\cos^2(t)\\ &=-2+2(\cos(2t)+1)\\ &=2\cos(2t)\\ a_p &=2\cos(t)a_{p-1}-a_{p-2} \qquad\text{for } p \ge 3\\ \end{array} $

Therefore, if we can show that $2\cos(pt) =4\cos(t)\cos((p-1)t)-2\cos((p-2)t) $ for $p \ge 3$ we are done.

Since $\cos(a\pm b) =\cos(a)\cos(b)\mp\sin(a)\sin(b) $, $2\cos(a)\cos(b) =\cos(a+b)+\cos(a-b) $ so $2\cos(t)\cos((p-1)t) =\cos(pt)+\cos((p-2)t) $ and we are done.