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Thread: Analytic function - estimation

  1. #1
    Junior Member
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    Analytic function - estimation

    Hi,

    How can I show the following estimation:

    $\displaystyle |c_n| \leq (\log||\phi||+||a||r)\frac{2}{r^n}$,

    where

    $\displaystyle c_n=\frac{1}{\pi i} \int_{|\lambda|=r}\log|\psi (\lambda)|\frac{d\lambda}{\lambda^{n+1}}$;
    $\displaystyle \psi (\lambda)=\phi[exp(\lambda a)]$

    I can show that

    $\displaystyle \log|\psi (\lambda)| \leq \log||\phi||+|\lambda|\cdot||a||$

    So that I guess that I should show

    $\displaystyle \frac{1}{\pi i} \int_{|\lambda|=r}\frac{d\lambda}{\lambda^{n+1}} \leq \frac{2}{r^n}$

    Probably I have to use the following estimation

    $\displaystyle \|\int_c f(z)dz \| \leq ML$,

    where

    $\displaystyle |f(z)| \leq M$ and $\displaystyle L$ - lenght of the contour (in this problem $\displaystyle L=2\pi r$)

    It is everything what I know and I've still problem with this estimation. From where we have $\displaystyle \frac{2}{r^n}$?
    Last edited by Arczi1984; Mar 21st 2010 at 10:30 AM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Champaign, Illinois
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    Quote Originally Posted by Arczi1984 View Post
    Hi,

    How can I show the following estimation:

    $\displaystyle |c_n| \leq (\log||\phi||+||a||r)\frac{2}{r^n}$,

    where

    $\displaystyle c_n=\frac{1}{\pi i} \int_{|\lambda|=r}\log|\psi (\lambda)|\frac{d\lambda}{\lambda^{n+1}}$;
    $\displaystyle \psi (\lambda)=\phi[exp(\lambda a)]$

    I can show that

    $\displaystyle \log|\psi (\lambda)| \leq \log||\phi||+|\lambda|\cdot||a||$

    So that I guess that I should show

    $\displaystyle \frac{1}{\pi i} \int_{|\lambda|=r}\frac{d\lambda}{\lambda^{n+1}} \leq \frac{2}{r^n}$

    Probably I have to use the following estimation

    $\displaystyle \|\int_c f(z)dz \| \leq ML$,

    where

    $\displaystyle |f(z)| \leq M$ and $\displaystyle L$ - lenght of the contour (in this problem $\displaystyle L=2\pi r$)

    It is everything what I know and I've still problem with this estimation. From where we have $\displaystyle \frac{2}{r^n}$?
    Let $\displaystyle f(\lambda)=\frac{\log\left|\phi e^{\lambda a} \right|}{\lambda^{n+1}} $.

    $\displaystyle |f(\lambda)| = \frac{\log\left|\phi e^{\lambda a} \right|}{r^{n+1}} $ since $\displaystyle |\lambda| = r $.

    So $\displaystyle |f(\lambda)| \leq \frac{\log||\phi||+|\lambda|\cdot ||a||}{r^{n+1}} = \frac{\log||\phi||+||a||r}{r^{n+1}} = M $.

    Thus $\displaystyle |c_n| \leq \frac{1}{\pi}ML = 2rM = (\log||\phi||+||a||r)\frac{2}{r^n} $.
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