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Math Help - Multiplicative comutativity of power series

  1. #1
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    Multiplicative comutativity of power series

    A(x) = sum(n=0 -> infinity) of a(subn)x^n
    B(x) = sum(n=0 -> infinity) of b(subn)x^n

    A(x) * B(x) = sum(n=0 -> infinity) of a(subn)x^n * sum(n=0 -> infinity) of b(subn)x^n

    = sum(k=0 -> infinity) of (a(subk)*b(sub(n-k))x^n (by definition of mulitiplication)

    = ?

    = sum(k=0 -> infinity) of (b(subk)*a(sub(n-k))x^n

    = sum(n=0 -> infinity) of b(subn)x^n * sum(n=0 -> infinity) of a(subn)x^n

    = B(x) * A(x)

    I can't figure out the missing step... Any advice? Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If R and R' are respectively the radius of convergence of the series



    then, the radius of convergence R'' of the product series



    satisfies



    Now, use that both series are absolutely convergent in | x | < R'' and apply a well known theorem.
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