<s>[INST] System of recurrence relations with Taylor series expansion Find $a_n,b_n$ where $a_0=1,b_0=0$ for the following relations:
$a_{n+1}=2a_n+b_n$
$b_{n+1}=a_n+b_n$
Using generating functions, the system is:
$f(x)-a_0=2xf(x)+xg(x)$
$g(x)-b_0=xf(x)+xg(x)$
Solving for $f(x)$ and $g(x)$ gives:
$$f(x)=\frac{1-x}{x^2-3x+1}$$
$$g(x)=\frac{x}{x^2-3x+1}$$
Taylor series for $f(x)$ and $g(x)$,
$\sum\limits_{n=0}^{\infty}\frac{f^{(n)}(a_0)}{n!}(x-a_0)^n$
$\sum\limits_{n=0}^{\infty}\frac{g^{(n)}(b_0)}{n!}(x-b_0)^n$
$n^{th}$ derivatives of $f,g$ can be found using partial fractions:
$$f^{(n)}(x)=\frac{(-1)^n n!(\sqrt{5}-5)}{10\left(x-\frac{3-\sqrt{5}}{2}\right)^{n+1}}+\frac{(-1)^n n!(-\sqrt{5}-5)}{10\left(x-\frac{3+\sqrt{5}}{2}\right)^{n+1}}$$
$$g^{(n)}(x)=\frac{(-1)^n n!(3\sqrt{5}-5)}{10\left(x-\frac{3-\sqrt{5}}{2}\right)^{n+1}}+\frac{(-1)^n n!(-3\sqrt{5}-5)}{10\left(x-\frac{3+\sqrt{5}}{2}\right)^{n+1}}$$
$n^{th}$ derivatives at $a_0$ and $b_0$:
$$f^{(n)}(a_0)=\frac{(-1)^n n!(\sqrt{5}-5)}{10\left(\frac{\sqrt{5}-1}{2}\right)^{n+1}}+\frac{(-1)^n n!(-\sqrt{5}-5)}{10\left(\frac{-1-\sqrt{5}}{2}\right)^{n+1}}$$
$$g^{(n)}(b_0)=\frac{(-1)^n n!(3\sqrt{5}-5)}{10\left(-\frac{3-\sqrt{5}}{2}\right)^{n+1}}+\frac{(-1)^n n!(-3\sqrt{5}-5)}{10\left(-\frac{3+\sqrt{5}}{2}\right)^{n+1}}$$
How to derive $a_n$ and $b_n$ in Taylor series?
 [/INST] You seem to have done all the necessary work, but confused a couple things. The Taylor series should be evaluated about the point $x=0$, not about the first terms of the sequences $a_0=1$, $b_0=0$. In other words, the $n^{\rm th}$ derivatives of $f$ and $g$ that you computed with partial fractions should be evaluated at $0$, and then you can use Taylor's formula to obtain the terms of the sequences: $a_n=f^{(n)}(0)/n!$ and $b_n=g^{(n)}(0)/n!$. So, as long as your derivatives are correct,
$$a_n=\frac{5-\sqrt5}{10}\left(\frac2{3-\sqrt5}\right)^{n+1}+
\frac{5+\sqrt5}{10}\left(\frac2{3+\sqrt5}\right)^{n+1}$$
and
$$b_n=\frac{5-3\sqrt5}{10}\left(\frac2{3-\sqrt5}\right)^{n+1}+
\frac{5+3\sqrt5}{10}\left(\frac2{3+\sqrt5}\right)^{n+1}.$$
For that matter, it's possible to skip the differentiation altogether and find the terms of the sequences directly from the partial fractions decomposition of the functions.
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