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Math Help - complex analysis question

  1. #1
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    complex analysis question

    p is the radius of convergence of Σa_n*z^n (n≥0) and let r be a real number less than p. prove that the series converges normally on the disc abs (z)<r and diverges for abs (z)>p. *converging normally means that the sup-norm converges.
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    What would this mean? \sum_{n=0}^{\infty} || a_n z^n || converges? If so then how do we interpert ||a_n z^n||?

    I can show \sum_{n=0}^{\infty} |a_nz^n| would converge, for |z|<r. But I am not sure if this is what you want.
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    i'm sorry for not being clear, but this would mean that llull=sup abs (u(z)) where z is in E(subset of complex C) converges. here llull would be called the sup-norm of u.
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    Quote Originally Posted by squarerootof2 View Post
    i'm sorry for not being clear, but this would mean that llull=sup abs (u(z)) where z is in E(subset of complex C) converges. here llull would be called the sup-norm of u.
    The sup-norm, ||a_nz^n||, over E = \{ z\in \mathbb{C}: |z| < r\} would be \sup\{ |a_n z^n| : z\in E \} = \sup \{ |a_n||z^n| : z\in E\} = |a_n|r^n.
    Thus, we need to show \sum_{n=0}^{\infty} |a_n| r^n converges.

    It does converge by using the theorem that if \sum_{n=0}^{\infty} a_n z^n is convergent on |z|<\rho then it is absolutely convergent on |z| \leq r < \rho.
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    is this theorem we're supposed to use the abel's lemma?
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    Quote Originally Posted by squarerootof2 View Post
    is this theorem we're supposed to use the abel's lemma?
    I am not sure what Abel's lemma you are using here.
    If you never seen that before, how about you try proving this.
    It happens to be not so difficult to show.
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  7. #7
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    oh and how would we deal with divergence of the series? i was actually getting more stuck on that part... thanks.
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    Quote Originally Posted by squarerootof2 View Post
    oh and how would we deal with divergence of the series? i was actually getting more stuck on that part... thanks.
    Same idea. If E = \{ z\in \mathbb{C} : |z| < r\} then ||a_n z^n||_E = \sup \{ |a_n z^n|: z\in E \} = |a_n|r^n.

    Suppose that \sum_{n=0}^{\infty} |a_n|r^n would converge then by the comparision test it would mean \sum_{n=0}^{\infty}a_n r^n would converge. But that is clearly impossible since r>\rho, so it diverges.
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