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Analytic functions proofs.

  1. Apr 17, 2007 #1
    i need to prove that if f and and g are analytic functions in (-a,a) then so is fg.

    well basically i need to find the radius of convergence of fg, which its coefficient is: [tex]c_n=\sum_{i=0}^{n} b_i*a_{n-i}[/tex], by using cauchy hadamard theorom for finding the radius of convergence, and to show that it's not greater than a.
    well limsup |c_n|^1/n, then [tex]c_n=a_0b_n+...+a_nb_0[/tex]
    now i think that [tex](c_n)^{\frac{1}{n}}<= ((n+1)(\max_{n \in N}(|a_n|,||b_n|))^2)^{1/n}[/tex]
    i think this inequality also applies to alternating sequences.
    anyway i dont know how bound it below.
     
    Last edited: Apr 17, 2007
  2. jcsd
  3. Apr 17, 2007 #2
    wait a minute i think that's enough iv'e shown that the radius cannot be greater than a.
    stupid me. (-;
     
  4. Apr 17, 2007 #3

    matt grime

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    But this isn't true. If f is analytic in some region, then gf is analytic in that region if f maps into some region in which g is analytic. And the converse is certainly false.
     
  5. Apr 17, 2007 #4

    AKG

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    He's talking about the product of f and g, not the composition.
     
  6. Apr 17, 2007 #5

    matt grime

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    Ah! Now why didn't I think of that? (Suggestions not necessary.)
     
  7. Apr 18, 2007 #6
    yes im talking of the prodcut.
    anyway, i think this appraoch of mine isn't correct.
    i think that bassically in order to prove it you need to use here cauchy's product.
    i.e if two partial sums f_n and g_n converges absolutely then also their cauchy's product converges absolutely.
    bassically here we assume that both g and f converges in 0<r<a, in [-r,r] they converge absolutely so also their cauchy product converges absolutely at [-r,r].
    is this better than my other obvoiusly faulty answer?
     
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