$a\in\Delta_{n-1}$. Then, for all $c\in C$ and $k \leq n$ we have $$\frac{1}{2}\leq\frac{\left|\tau_k(a)\right|}
{\left|(f_a^k)'c_0(a)\right|}\leq 2.$$

We have $$\label{f1-s3.2A}
\tau_{k}(a)=1+f_a'c_{k-1}(a)\cdot\tau_{k-1}(a).$$ Using this inductively and then dividing the result by $(f_a^{k})'c_0(a)$, which is nonzero by (G2$)_{n-1}$ we obtain $$\frac{\tau_{k}(a)}{(f_a^{k})'c_0(a)}=1+\sum_{i=1}^{k}\frac{1}
{(f^i)'c_0(a)}.$$ This lemma then follows from applying (G2)$_{n-1}$.

For $a_*\in[0,1)$, $c \in C$ and $c_0=fc$, define $$I_n(a_*,c)=[a_*-D_{n}(c_0), a_*+D_{n}(c_0)]$$ where $
D_n(c_0)$ is the same as the one in (\[Theta\]) with $f = f_{a_*}$.

\[samp\] Let $a_*\in\Delta_{n-1}$. For all $c\in C$, $a \in I_{n}(a_*,c)$ and $k \leq n$ we have $$\frac{1}{2} < \frac{|\tau_{k}(a)|}{|\tau_{k}(a_*)|}\leq 2.$$

Let $k \leq n$. To prove this lemma we inductively assume that, for all $j < k$, $$\label{samp-induct}
\frac{|\tau_j(a)|}{|\tau_j(a_*)|}\leq 2, \ \ \ \text{for all} \ a
\in I_{n}(a_*, c).$$ We then prove the same estimate for $j = k$. Write $I_{n}$ for $I_{n}(a_*,c)$. For all $a \in I_{n}$ we have $$\begin{aligned}
\left|\log\frac{|\tau_{j+1}(a)|}{|\tau_{j+1}(a_*)|}-
\log\frac{|\tau_{j}(a)|}{|\tau_{j}(a_*)|}\right| \leq&
\left|\log\frac{|\tau_{j+1}(a)|}{|\tau_j(a)|}-
\log|f_{a_*}'c_j(a_*)|\right|\\
+\left|\log\frac{|\tau_{j+1}(a_*)|}{|\tau_j(a_*)|}-
\log|f_{a_*}'c_j(a_*)|\right|&\leq (I)_a +(I)_{a_*} + (I\!I),\end{aligned}$$ where $$(I)_a = \left|\log\frac{\tau_{j+1}(a)} {\tau_j(a)}-
\log|(f_{a})'c_j(a)|\right|, \ \ \ (I\!I) = \left|\log
\frac{(f_{a})'c_j(a)}{(f_{a_*})'c_j(a_*)} \right|.$$ We claim that $$\label{twoa}
|(I\!I)|\leq 2L^{-\frac{1}{3}}\cdot d_j^{-1}\left[\sum_{i=0}^{n-1}d_i^{-1}\right]^{-1},$$ where $d_i$ is the same as the one in (\[Theta\]) with $f=f_{a_*}$.

To prove (\[twoa\]), first we use (\[samp-induct\]) and Lemma \[trans\] to obtain $$\begin{aligned}
\label{R1add}
