There is a reccurent equation: $a_{n+2}-2\cos(\phi)a_{n+1}+a_n=0$ and I must to solve it. I found generating function: $$A(t) = \dfrac{1 - t\cos(\phi)}{1 - 2t\cos(\phi) + t^2},$$ but I can not extend it into series to solve this equation. Please help me.
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set $a_n=q^n$ in your equation – Dr. Sonnhard Graubner Oct 23 '14 at 16:59
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To pursue the GF further, note that the denominator has roots $e^{\pm i\phi}$. So you can factor the bottom, do partial fractions, and expand the terms in two geometric series. – Semiclassical Oct 23 '14 at 17:16
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Also, in writing that GF you've assumed a certain pair of initial values (i.e. $a_0,a_1$). So that should be in the question as well. – Semiclassical Oct 23 '14 at 17:39
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Set $z=te^{i\phi}$ then your generating function can be written as $$A(t)=\frac{1-t\cos(\phi)}{1-2t\cos(\phi)+t^2}=\frac{1-\frac{1}{2}(z+\bar{z})}{(z-1)(\bar{z}-1)}=\frac{(1-z)+(1-\bar{z})}{2(z-1)(\bar{z}-1)}=\frac{1}{2}(\frac{1}{1-z}+\frac{1}{1-\bar{z}})$$ For $|z|<1$ we have the following series expansion $$A(t)=\frac{1}{2}(\sum^{\infty}_{k=0}z^k+\sum^{\infty}_{k=0}\bar{z}^k)=\frac{1}{2}(\sum^{\infty}_{k=0}2t^k\cos(k\phi))=\sum^{\infty}_{k=0}t^k\cos(k\phi)$$ So you have $a_k=\cos(k\phi)$. Now notice that \begin{align}a_{k+2}-2\cos(\phi)a_{k+1}+a_k&=\cos((k+2)\phi)-2\cos(\phi)\cos((k+1)\phi)+\cos(k\phi)\\& =\cos((k+2)\phi)-2(\frac{\cos((k+2)\phi)+\cos(k\phi)}{2})+\cos(k\phi)\\&= \cos((k+2)\phi)-\cos((k+2)\phi)\\&= 0 \end{align} so $a_k=\cos(k\phi)$ is a solution.
Arian
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HINT:
$(e^{i n \theta})$ , $(e^{-i n \theta})$ or if you want, $(\cos(n \theta))$, $(\sin(n \theta))$ are solutions.
orangeskid
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