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Coefficients in the power series

  1. Jan 15, 2010 #1
    Hi,

    I am trying to prove something, but I need some kind of a result on the coefficients of a power series.

    Suppose f(z) has a power series expansion about zero (converges). What can I say about the sum of the absolute values of the coefficients? Ideally I would like to show this sum is less than 1.

    thank you
     
  2. jcsd
  3. Jan 15, 2010 #2
    [tex]
    \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n for \ all \ | x | < 1
    [/tex]

    If we denote a_n to be the coefficient of the nth term, then then a_n = 1 for all n, in which case,

    [tex]
    \sum_{n = 1}^{\infty} 1 \rightarrow \infty
    [/tex]

    Or in the case when it does converge, consider:

    [tex]
    e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}
    [/tex]

    Letting x = 1, we'd be summing the coefficients, i.e.

    [tex]
    \sum_{n=0}^{\infty} \frac{1}{n!} = e
    [/tex]

    It seems your hypothesis fails.
     
  4. Jan 15, 2010 #3
    Thank you L'Hospital, that is very clear
     
  5. Jan 15, 2010 #4
    Re: Scharz lemma/coefficients in the power series/ order of zero

    In light of L'Hospital's comment I need change my plan of of attack on the question. I think I need to state the question I was trying to prove:

    If f(z) : D--->D where D is the open unit disk, and

    the first (k-1) derivatives at zero vanish,

    I would like to show that

    If(z)I <= IzI^k Tha is abs{f(z)} \leq abs{z}^k

    I believe one can (an the question is possibly intended to be solved this way) approach it using Schwarz, I was trying to do it by brute force whereby expanding the power series and saying
    f(x)=z^k g(z) and somehow show that I g(z) I <= 1. I am failing as of now.
    Any help? I thank you for your time
     
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