Counting preimages Algebraically (7)

Vasco,

That addition to the document is a great help. In the following, I have tried to gain a better understanding of the reasoning and the meaning of $\Omega$.

For the case of the analytic function where $a$ is not a critical point, (f(z) - p) is the Taylor series (T[f]) of $f(z)$ minus $f(a)$ as it converges in the neighborhood of a. The first derivative dominates.

$\hspace{5em}f(z) - p = T[f](a) - f(a)$                      (1)

For the case of the analytic function where $a$ is a critical point, $\Omega$ is the Taylor series of $f^{(n)}(z)$ at (a), where $f^{(n)}$ is the first non-vanishing derivative. The first term dominates. Note that $\Delta = (z-a)$ (from p. 346, bottom)

$\hspace{5em}f(z) - p = T[f^{(n)}](a) \Delta^n$                 (2)

For a polynomial P with factors $(z-a)^n$, we know that $a$ is a critical point for all derivatives from $P^{(1)}$ to $P^{(n-1)}$. And we know that $P(z) = \Omega(z) A^n$. This looks very much like equation (2). What is $\Omega$? Is it the Taylor series of the first non-vanishing derivative $P^{(n)}$? Note that $A = (z-a)$ (from p. 346, paragraph 2)

$\hspace{5em}? P(z) = T[P^n(z)]A^n$                      (3)

Gary

Vasco,

Yes, I guess that should have been obvious to me. Needham defined it clearly enough. But I got wrapped up in the Taylor series and trying to see if $\Omega$ had anything to do with it. Can we say that in both cases, $\Omega$ is what is left when we factor out the difference term ($A^n$ or $\Delta^n$) whose power reveals the multiplier? We might translate $\Omega$ as "the rest of it". I feel the need for some diagrams that illustrate the multipliers of the general analytic function, the analytic function with a critical point, and the polynomial. But perhaps there is not time enough for everything one might want.

Gary