take the particular selection of times, $s_n = n$. To initiate the induction, we will choose $\alpha_0 = 0$, and $u_0(s)$ the corresponding solution (clearly, $u_0(s_0) = u_0(0) = \alpha_0 = 0$). We describe now the $n \rightarrow n+1$ step of the iteration. We now assume inductively that there exists a choice of $\alpha_n$ so that $u_{\alpha_n}(s_{n}) =  0$ and describe the choice of $\alpha_{n+1}$, which is achieved through the Implicit Function Theorem.

We define now the map $\mathcal{T}_n$ given by $\mathcal{T}_n(\alpha) := u_{\alpha}(s_{n+1})$. We now seek an $\alpha_{n+1}$ in a small neighborhood, $\mathcal{B}_n$, of $\alpha_n$ so that $\mathcal{T}_{n}(\alpha_{n+1}) = 0$. According to a Taylor expansion of $\mathcal{T}_n$ in $\alpha$, we obtain for some $\alpha_\ast$ satisfying $|\alpha_\ast - \alpha_n| \le |\alpha_n - \alpha_{n+1}|$, $$\begin{aligned}
\mathcal{T}_n(\alpha_{n+1}) = \mathcal{T}_n(\alpha_n) + (\alpha_{n+1} - \alpha_n) \frac{{\ensuremath{\partial}}\mathcal{T}_n}{{\ensuremath{\partial}}\alpha}(\alpha_n) + \frac 1 2 (\alpha_{n+1} - \alpha_n)^2 \frac{{\ensuremath{\partial}}^2 \mathcal{T}_n}{{\ensuremath{\partial}}\alpha^2}(\alpha_\ast)\,.\end{aligned}$$ Accordingly, we may apply the Implicit Function Theorem to identify a $\alpha_{n+1}$ so that the left-hand side is zero if we can obtain three estimates: an upper bound on $| \mathcal{T}_n(\alpha_n)|$, a lower bound on $ \frac{{\ensuremath{\partial}}\mathcal{T}_n}{{\ensuremath{\partial}}\alpha}(\alpha_n)$, and an upper bound over $\sup_{\alpha_\ast \in \mathcal{B}_n} |\frac{{\ensuremath{\partial}}^2 \mathcal{T}_n}{{\ensuremath{\partial}}\alpha^2}|$.

We thus define the *error at the next time scale* created by 