Let $f$ and $g$ be real analytic functions, both of which are defined on an open interval $I$. Assume that $g$ does not vanish on $I$. Then the function
\begin{align*} h(x) &= \frac{f(x)}{g(x)} \\ \end{align*}
is real analytic on $I$. Moreover, if $I$ is centered at the point $c$ and if:
\begin{align*} f(x) &= \sum\limits_{j=0}^\infty a_j (x - c)^j \\ g(x) &= \sum\limits_{j=0}^\infty b_j (x - c)^j \\ \end{align*}
then the power series expansion of $h$ about $c$ may be obtained by formal long division of the latter series into the former. That is, the zeroeth coefficient $c_0$ is $a_0/b_0$ and the order one coefficient is
\begin{align*} c_1 &= \frac{1}{b_0} \left( a_1 - \frac{a_0 b_1}{b_0} \right) \\ \end{align*}
Letting $w = x-c$, here is my work for calculating $c_1$:
\begin{align*} \left( \frac{a_0}{b_0} + c_1 w \right) \left( b_0 + b_1 w \right) = a_0 + a_1 w \\ a_0 + \frac{a_0 b_1 w}{b_0} + b_0 c_1 w + b_1 c_1 w^2 - a_0 - a_1 w = 0 \\ \frac{a_0 b_1}{b_0} + b_0 c_1 + b_1 c_1 w - a_1 = 0 \\ c_1 = \frac{1}{b_0 + b_1 w} \left( a_1 - \frac{a_0 b_1}{b_0} \right) \\ \end{align*}
Can someone explain the differing $b_1 w$ term in my answer?