Discrete Trigonometry

Since $s_0 = 0$ and $c_{n + 1} = s_n$ , $c_1 = 0$ .

Since $c_{n + 1} = s_n$ , $c_{2n + 1} = s_{2n}$ . Therefore, $s_{2n} = 2s_nc_n$ becomes $c_{2n + 1} = 2c_{n + 1}c_n$ after substitution.

Since $c_{2n + 1} = 2c_{n + 1}c_n$ , $c_{2(n + 1) + 1} = 2c_{(n + 1) + 1}c_{(n + 1)}$ , or $c_{2n + 3} = 2c_{n + 2}c_{n + 1}$ . Substituting $c_{n + 1} = \frac{c_{2n + 1}}{c_n}$ from $c_{2n + 1} = 2c_{n + 1}c_n$ gives $c_{2n + 3} = 2c_{n + 2}\frac{c_{2n + 1}}{c_n}$ , which means $c_{2n + 1}$ is a factor of $c_{2n + 3}$ . (In other words, an odd term of $c$ is a factor on the next odd term of $c$ .) Since $c_1 = 0$ , all odd terms of $c$ have a factor of $0$ , meaning all odd terms of $c$ are $0$ , so $c_{2n + 1} = 0$ .

Since $c_{2n} = 2c_n^2 - 1$ , and $c_{2n + 1} = 0$ , then all terms of $c$ with exactly one factor of $2$ are $2(0)^2 - 1 = -1$ , so $c_{4n + 2} = -1$ .

Since $c_{2n} = 2c_n^2 - 1$ , and $c_{2n + 1} = 0$ , then all terms of $c$ with exactly two factors of $2$ are $2(-1)^2 - 1 = 1$ , and all terms with more than two factors of $2$ are also $2(1)^2 - 1 = 1$ , so $c_{4n} = 1$ .

Therefore, the terms of $c$ cycle through the terms $(0, -1, 0, 1)$ for a $2p$ period of $2p = 4$ , so $p = 2$ .

Since $c_{n + 1} = s_n$ and the terms of $c$ cycle through the terms $(0, -1, 0, 1)$ , the terms of $s$ cycle through the terms $(-1, 0, 1, 0)$ .

The sum $S = \sum_{n = 0}^{\infty} \frac{c_{np} + s_{np}}{p^n}$ $=$ $\frac{1 + 0}{1} + \frac{-1 + 0}{2} + \frac{1 + 0}{4} + \frac{-1 + 0}{8} + \dots$ $=$ $1 - \frac{1}{2} + \frac{1}{4} -\frac{1}{8} + \dots$ $=$ $1 - \frac{1}{2}(1 - \frac{1}{2}(1 - \frac{1}{2}(1 \dots$

Substituting $S$ into itself gives $S = 1 - \frac{1}{2}S$ , and solving for $S$ gives $S = \frac{2}{3}$ . This means $a = 2$ and $b = 3$ , and $a + b = 2 + 3 = \boxed{5}$ .

The first few values of $c_n$ and $s_n$ seems to imply that $c_n = \Re \left(e^{-\frac {n\pi} 2 i}\right) = \cos \left(-\frac {n\pi} 2\right)$ and $s_n = \Im \left(e^{-\frac {n\pi} 2 i}\right) = \sin \left(-\frac {n\pi} 2\right)$ . Let us prove the claim satisfies the three equation given.

Equation 1: $c_{n+1} = \cos \left(-\frac {(n+1)\pi}2\right) = \cos \left(\frac {(n+1)\pi}2\right) = \cos \left(\frac \pi 2 + \frac {n\pi}2\right) = \sin \left(-\frac {n \pi}2\right) = s_n$

Equation 2: $c_{2n} = \cos \left(-(n+1)\pi \right) = 2 \cos^2 \left(-\frac {(n+1)\pi}2\right) - 1 = 2c_n^2 - 1$

Equation 3: $s_{2n} = \sin \left(-(n+1)\pi \right) = 2 \sin \left(-\frac {(n+1)\pi}2\right)\cos \left(-\frac {(n+1)\pi}2\right) = 2s_nc_n$

1. Since $c_n = \cos \left(-\frac {n\pi} 2\right)$ and $s_n = \sin \left(-\frac {n\pi} 2\right)$ , it means that $c_n$ and $s_n$ have a period of $\frac {n\pi}2 = 2\pi \implies n = 4$ . Therefore, $2p=4 \implies p=2$ .
2. And $\displaystyle \sum_{n=0}^\infty \dfrac {c_{2n}+s_{2n}}{2^n} = \sum_{n=0}^\infty \dfrac {\cos (- n\pi)+\sin (-n\pi)}{2^n} = \sum_{n=0}^\infty \dfrac {(-1)^n+0}{2^n} = \sum_{n=0}^\infty \left(-\frac 12\right)^n = \frac 1{1+\frac 12} = \frac 23$ . Therefore, $a+b=2+3=\boxed{5}$ .