Results 1 to 2 of 2

Thread: Multiplicative comutativity of power series

  1. #1
    Super Member
    Joined
    Feb 2008
    Posts
    535

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Sep 29th 2010, 06:11 AM
  2. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  3. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 19th 2009, 09:01 AM
  4. Multiplying power series - Maclaurin series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 7th 2009, 11:24 PM
  5. Replies: 10
    Last Post: Apr 18th 2008, 10:35 PM

Search Tags


/mathhelpforum @mathhelpforum