1. ## Asmptotic analysis problem

$\displaystyle f(z) \approx \sum_{n=0}^{\infty}{a_{n}z^{-n}} \ \ as \ \ z\rightarrow\infty$

show by induction $\displaystyle \frac{1}{f(z)} \approx \frac{1}{a_{0}}\sum_{n=0}^{\infty}{d_{n}z^{-n}} \ \ where \ \ \sum_{k=0}^{n}{d_{n-k}a_{k}}$

2. A complex variable function $\displaystyle \varphi(s)$ analytic in $\displaystyle s=0$ can be written as...

$\displaystyle \varphi(s)=\sum_{n=0}^{\infty} a_{n}\cdot s^{n}$ (1)

If in (1) is $\displaystyle a_{0} \ne 0$ then $\displaystyle \frac{1}{\varphi(s)}$ is also analytic in $\displaystyle s=0$ and is...

$\displaystyle \frac{1}{\varphi(s)}=\sum_{n=0}^{\infty} d_{n}\cdot s^{n}$ (2)

A direct way to compute the $\displaystyle d_{n}$ from the $\displaystyle a_{n}$ is given by the identity we obtain combining (1) and (2)...

$\displaystyle \varphi(s) \cdot \frac{1}{\varphi(s)} = \sum_{n=0}^{\infty} a_{n}\cdot s^{n} \cdot \sum_{n=0}^{\infty} d_{n}\cdot s^{n} = \sum_{n=0}^{\infty}s^{n}\cdot \sum_{k=0}^{n}a_{k}\cdot d_{n-k} = 1$ (3)

... so that is...

$\displaystyle \sum_{k=0}^{n} a_{k}\cdot d_{n-k}= \left\{\begin{array}{cc}1, &\mbox {if } n=0\\0, & \mbox {if } n>0\end{array}\right.$ (4)

If we perform the substitution of variable $\displaystyle s=\frac{1}{z}$ so that is...

$\displaystyle \lim_{z \rightarrow \infty} \varphi (z)= \lim_{s \rightarrow 0} \varphi (s)$

$\displaystyle \lim_{z \rightarrow \infty} \frac{1}{\varphi (z)}= \lim_{s \rightarrow 0} \frac{1}{\varphi (s)}$ (5)

The conclusion is that if...

$\displaystyle \lim_{s \rightarrow 0} \frac{\varphi(s)} {f(s)}=1$ (6)

... then it is also...

$\displaystyle \lim_{s \rightarrow 0} \frac{f(s)}{\varphi(s)} =1$ (7)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## idont

i dont understand y u prove (7)