Asymptotic powerseries development - Revision history

Bo198214: sign correction

2011-06-07T14:17:40Z

sign correction

Bo198214: typo

2011-06-07T14:16:53Z

typo

Bo198214: better wording

2011-06-07T14:16:12Z

better wording

Bo198214: initial gathering

2011-06-07T14:12:24Z

initial gathering

New page

Let $f$ be holomorphic on $D$ and let $a$ be a point on the boundary $\partial D$.
We say that $f$ has an asymptotic powerseries (development) at $a$, iff there exist $(c_n)_{n\in\N_0}$ such that
$$\lim_{z\to a} \frac{1}{(z-a)^N}\left(f(z)-\sum_{n=0}^N c_n (z-a)^n\right) = 0$$
for all $N\in\N_0$.

If $f$

Every function can have at most one asymptotic powerseries at $a$ because
the coefficients $c_n$ are determined by:
$$c_0=\lim{z\to a} f(z),\quad c_{N+1}=\lim_{z\to a} \frac{1}{(z-a)^{N+1}}\left(f(z)-\sum_{n=0}^{N} c_n (z-a)^n\right)$$

However there may be different functions with the same asymptotic powerseries: for example $e^{1/x}$ and $f(x)=0$ both have the asymptotic powerseries $c_n=0$ at 0.

The development (the coefficients) may depend on the region $D$.

'''Theorem'''. Let $D$ be of the following simplicity: for each $z\in D$ there is a sequence $a_n\to a$, such that $[a_n,a]\subset D$. If all the limits $c_n=\lim_{z\to a} \frac{f^{(n)}(z)}{n!}$ exist then $f$ has the asymptotic development with coefficients $c_n$ in $a$.

'''Theorem of Ritt'''. For any proper sector $S$ at $a$ and for every given sequence of $c_n$ there exists a holomorphic function $f$ on $S$ having the asymptotic powerseries $\sum_{n=0}^\infty c_n (z-a)^n$ at $a$.

@@ Line 8: / Line 8: @@
 $$c_0=\lim_{z\to a} f(z),\quad c_{N+1}=\lim_{z\to a} \frac{1}{(z-a)^{N+1}}\left(f(z)-\sum_{n=0}^{N} c_n (z-a)^n\right)$$
-However there may be different functions with the same asymptotic powerseries: for example $e^{1/x}$ and $f(x)=0$ both have the asymptotic powerseries $c_n=0$ at 0.
+However there may be different functions with the same asymptotic powerseries: for example $e^{-1/x}$ and $f(x)=0$ both have the asymptotic powerseries $c_n=0$ at 0.
 The development (the coefficients) may depend on the region $D$.

@@ Line 6: / Line 6: @@
 Every function can have at most one asymptotic powerseries at $a$ because
 the coefficients $c_n$ are determined by:
-$$c_0=\lim{z\to a} f(z),\quad c_{N+1}=\lim_{z\to a} \frac{1}{(z-a)^{N+1}}\left(f(z)-\sum_{n=0}^{N} c_n (z-a)^n\right)$$
+$$c_0=\lim_{z\to a} f(z),\quad c_{N+1}=\lim_{z\to a} \frac{1}{(z-a)^{N+1}}\left(f(z)-\sum_{n=0}^{N} c_n (z-a)^n\right)$$
 However there may be different functions with the same asymptotic powerseries: for example $e^{1/x}$ and $f(x)=0$ both have the asymptotic powerseries $c_n=0$ at 0.

@@ Line 1: / Line 1: @@
 Let $f$ be holomorphic on $D$ and let $a$ be a point on the boundary $\partial D$.
-We say that $f$ has an asymptotic powerseries (development) at $a$, iff there exist $(c_n)_{n\in\N_0}$ such that
+We say that $f$ has the asymptotic powerseries (development) $\sum_{n=0}^\infty c_n (z-a)^n$ at $a$, iff
 $$\lim_{z\to a} \frac{1}{(z-a)^N}\left(f(z)-\sum_{n=0}^N c_n (z-a)^n\right) = 0$$
 for all $N\in\N_0$.
 Every function can have at most one asymptotic powerseries at $a$ because