Reed,Simon Theorem XII.1: Use of recursive substitution in the proof We have a function $F(\beta,\lambda)$ (polynomial of degree $n$) which is analytic near $\beta_0$ and $\lambda_0$. So we can write
$$F(\beta,\lambda)=\sum_{m=0}^n(\lambda-\lambda_0)^mf_m(\beta)$$
where $f_0(\beta_0)=F(\beta_0,\lambda_0)=0$ and $f_1(\beta_0)=\frac{\partial F}{\partial\lambda}(\beta_0,\lambda_0)\neq0$. We want to solve an equation $F(\beta,\lambda)=0$, which is equivalent to
\begin{equation}
\lambda=\lambda_0-\frac{f_0(\beta)}{f_1(\beta)}-\sum_{m=2}^n(\lambda-\lambda_0)^m\frac{f_m(\beta)}{f_1(\beta)}
\end{equation} 
We try to solve this equation with solution in the from
$$\lambda(\beta)=\sum_{k=1}^\infty\alpha_k(\beta-\beta_0)^k$$
By the use of recursive substitution into latter equation we can compute the $\alpha_k$'s
$$\alpha_1=-\left[\frac{f_0(\beta)}{f_1(\beta)}\right]'\bigg|_{\beta=\beta_0}\ ,\quad \alpha_2=-\frac{1}{2}\left[\frac{f_0(\beta)}{f_1(\beta)}\right]''\bigg|_{\beta=\beta_0}-\alpha_1^2\frac{f_2(\beta_0)}{f_1(\beta_0)}$$
I really don't see, how we used the recursive substition and where the formulas for $\alpha_1$ and $\alpha_2$ came from. When I compare the two equations and Taylor expand $f_0(\beta)$ and $f_1(\beta)$ near $\beta_0$, I end up with all the higher derivatives from the expansion. Any help would be appreciated.
 A: For brevity, write $z = \beta - \beta_0$, $w = \lambda - \lambda_0$, and $h_m(z) = -f_m(z+\beta_0)/f_1(z+\beta_0)$ for $m \neq 1$ and $h_1(z) = 0$ so that the equation becomes 
$$w = \sum_{m=0}^n h_m w^m \tag{1}$$
Assume that an analytic function $w(z) = \sum_{k \ge 0} \alpha_k z^k$ satisfies (1) in a neighborhood of $0$. Express $h_m(z) = \sum_{k\ge 0} \beta_k^m z^k$ and plug this expression into $(1)$, expand, and match the coefficients of the $z^k$'s to get:
$$\alpha_k = \sum_{m=0}^n \sum_{|\gamma| = k} \prod_{i=1}^m \alpha_{\gamma_i} \beta_{\gamma_{m+1}}^m \tag{2}$$
Where $|\gamma| = \gamma_1 + \gamma_2 +\cdots + \gamma_{m+1}$. Notice that no $\alpha_k$ appears in $(2)$ because that would imply some $\gamma_i = k$, so every other $\gamma_j=0$ for $j \neq i$, forcing multiplication by $\alpha_0=0$. This shows that $(2)$ works as a recursive definition for the $\alpha_k$'s and that every $\gamma_i \ge 1$. 
Now pick $z$ in the intersection of the discs of convergence of each $h_m$ so that, for all $m$, absolute convergence gives:
$$C = \max_m \left\{\sum_{k\ge 0} |\beta_k^m| |z|^k\right\}< \infty$$
It follows that for any $r \ge 1/|z|$ we must have $|\beta_k^m|\le C r^k$.
Finally, define $K = \max\{C, 2n\}$, $A = \frac 1{2K}$ and $R = \max\{1, r, |\alpha_1|, |\alpha_2|\}/A$ so that $|\alpha_1|\le AR$, $|\alpha_2|\le AR^2$, and $q := r/R \le A\Rightarrow 2A - q \ge A \Rightarrow (1-q/2A)^{-1}\le 2$. For $k\ge 3$, apply the Triangle Ineq. and a strong induction hypothesis on $(2)$ to obtain: 
\begin{align}
|\alpha_k| &\le \sum_{m=0}^n \sum_{|\gamma| = k} \prod_{i=1}^m |\alpha_{\gamma_i}| |\beta_{\gamma_{m+1}}^m| \le  \sum_{m=0}^n \sum_{|\gamma| = k} (AR)^{\gamma_1 + \cdots +\gamma_m} Cr^{\gamma_{m+1}} = C\sum_{m=0}^n \sum_{|\gamma| = k} (AR)^{k - \gamma_{m+1}} r^{\gamma_{m+1}} \\ & = CA^kR^k\sum_{m=0}^n \sum_{|\gamma| = k} (q/A)^{\gamma_{m+1}} = CA^kR^k\sum_{m=0}^n \sum_{i=0}^k \sum_{\gamma_1 + \cdots + \gamma_m = k - i}(q/A)^{i}
\end{align}
From the remark after $(2)$, we know that $\sum_{\gamma_1 + \cdots + \gamma_m = k - i}(q/A)^{i}$ contains $\binom{k-i-1}{m-1} \le 2^{k-i-1}$ terms (see Theorem 1 from here). This, combined with the inequalities given after the definition of $K$, gives:
\begin{align}
|\alpha_k|&\le AR^k\left[C(2A)^{k-1}n \sum_{i=0}^k (q/2A)^{i}\right] \le AR^k\left[C(2A)^{k-1}n (1-q/2A)^{-1}\right] \\ &\le AR^k\left[2nK^{2-k}\right] \le AR^k\left[2n/K\right] \le AR^k
\end{align}
