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Math Help - Analytic function - estimation

  1. #1
    Junior Member
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    Analytic function - estimation

    Hi,

    How can I show the following estimation:

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

    where

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

    I can show that

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

    So that I guess that I should show

    \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

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

    where

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

    It is everything what I know and I've still problem with this estimation. From where we have \frac{2}{r^n}?
    Last edited by Arczi1984; March 21st 2010 at 10:30 AM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
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    Quote Originally Posted by Arczi1984 View Post
    Hi,

    How can I show the following estimation:

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

    where

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

    I can show that

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

    So that I guess that I should show

    \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

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

    where

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

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

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

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

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