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Math Help - Asmptotic analysis problem

  1. #1
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    Asmptotic analysis problem

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

    show by induction \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}}
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  2. #2
    MHF Contributor chisigma's Avatar
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    A complex variable function \varphi(s) analytic in s=0 can be written as...

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

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

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

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

    \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...

    \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 s=\frac{1}{z} so that is...

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

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

    The conclusion is that if...

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

    ... then it is also...

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

    Kind regards

    \chi \sigma
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  3. #3
    Junior Member
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    idont

    i dont understand y u prove (7)
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