Subsection4.1.2Constructing the conjugation as a power series
The idea behind our next proof is to assume that the conjugation can be written as a power series. Like our previous proof, we should prove that the process converges. We'll skip that step here as we already know that the conjugation exists. Thus, this construction could be viewed as another way to generate an approximation.
Since \(f\) is analytic at zero, it can be expanded in a Taylor series. The conditions \(f(0)=0\) and \(f'(0)=\lambda\) give us a bit of information on the coefficients, but they're mostly arbitrary:
\begin{equation*} f(z)=\lambda z + \sum _{k=2}^{\infty } a_kz^k. \end{equation*}Now, we seek an analytic function \(\varphi\) as the conjugacy, thus we assume that it's given by a power series:
\begin{equation*} \varphi (z)=\sum _{k=0}^{\infty } c_kz^k. \end{equation*}We hope to choose the coefficients \(c_k\) such that
\begin{equation*} \varphi \circ f(z)=L\circ \varphi (z). \end{equation*}Now, the right hand side is pretty simple:
\begin{equation*} L(\varphi (z)) = \lambda \left(c_0+c_1z+c_2z^2+c_3z^3+\cdots \right)= \lambda c_0+\lambda c_1z + \lambda c_2z^2+\lambda c_3z^3+\cdots . \end{equation*}The expression on the left is a bit trickier:
\begin{equation*} \begin{array}{ll} \varphi (f(z)) = c_0 & + \text{ } c_1\left(\lambda z + a_2z^2+a_3z^3+ a_4z^4+\cdots \right) \\ & + \text{ } c_2\left(\lambda z + a_2z^2+a_3z^3+ a_4z^4+\cdots \right){}^2 \\ & + \text{ } c_3\left(\lambda z + a_2z^2+a_3z^3+ a_4z^4+\cdots \right){}^3 \\ & + \text{ } c_4\left(\lambda z + a_2z^2+a_3z^3+ a_4z^4+\cdots \right){}^4+\cdots \\ \end{array} \end{equation*}Nonetheless, we can compare coefficients of powers of \(z\) to see what the \(c_k\)s must be in order to make the conjugacy work. The constant term on the right is \(\lambda c_0\text{,}\) while on the left it's just \(c_0\text{.}\) The only way that can work is if \(c_0=0\text{.}\) The coefficient of \(z\) on both sides is \(\lambda c_1\text{.}\) I guess that means that \(c_1\)can be any non-zero value. This just means that there are many possible choices for the conjugacy parametrized by the choice of \(c_1\text{.}\) We'll make the simplest possible choice for \(c_1\text{,}\) namely \(c_1=1\text{.}\)
The coefficient of \(z^2\) on the right is \(\lambda c_2\) while on the left it's \(c_1a_2+c_2\lambda ^2\text{.}\) Since we already know that \(c_1=1\text{,}\) this yields \begin{equation*} c_2=\frac{a_2}{\lambda -\lambda ^2}. \end{equation*}In general, when we compare the coefficients of \(z^n\text{,}\) we obtain an equation involving \(c_0,c_1,\ldots ,c_n\) as well as some of the \(a_k\)s, which are known. Since \(c_0,c_1,\ldots ,c_{n-1}\) are already known, we can solve for \(c_n\) to obtain a recursive procedure to find all the \(c_k\)s. Pushing the computations we have so far one step further to illustrate the point, we find that the coefficient of \(z^3\) on the right \(\lambda c_3\) while on the left it's \(c_1a_3+2c_2\lambda a_2+c_3\lambda ^3\text{.}\) Using the known values for \(c_1\) and \(c_2\text{,}\) we get
\begin{equation*} c_3=\left(a_3+2\lambda \frac{a_2^2}{\lambda -\lambda ^2}\right)/\left(\lambda -\lambda ^3\right). \end{equation*}In general, each \(c_k\) is given by
\begin{equation*} c_k = \frac{P_k(a_2,\ldots,a_k,c_2,\ldots,c_{k-1})}{\lambda^k-\lambda}, \end{equation*}where each \(P_k\) is a polynomial in the variables \(a_j\) and \(c_j\) with positive coefficients.