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Power Series

  1. Mar 7, 2007 #1
    If f(x) has a power series: a_n(x-a)^n (centered at a)

    what does the power series for f(2x) look like?
     
  2. jcsd
  3. Mar 7, 2007 #2

    HallsofIvy

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    You have a problem with this? If "f(x) has power series a_n(x-a)^n" (by which I assume you mean
    [tex]\Sigma a_n (x-a)^n[/tex]
    Doesn't it follow that:
    [tex]f(2x)= \Sigma a_n (2x-a)^n[/tex]?
     
  4. Mar 8, 2007 #3

    arildno

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    Hmm..
    I would think he is after something like this:
    [tex]f(2x)=\sum_{n=0}^{\infty}a_{n}(2x-a)^{n}=\sum_{n=0}^{\infty}{a_{n}}{2^{n}}\sum_{l=0}^{n}\binom{n}{l}(\frac{a}{2})^{n-l}(x-a)^{l}[/tex]
    and then reorganize this double sum into some expression:
    [tex]f(x)=\sum_{n=0}^{\infty}b_{n}(x-a)^{n}[/tex]
    with a known sequence b_n
     
  5. Mar 8, 2007 #4

    D H

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    I would think he is after something like this:

    [tex]f(2x) = \sum_{n=0}^{\infty}a_n(2x-a)^n = \sum_{n=0}^{\infty}a_n2^n\left(x-\frac a 2\right)^n =\sum_{n=0}^{\infty}b_n\left(x-\frac a 2\right)^n[/tex]

    where [itex]b_n=a_n2^n[/itex]
     
  6. Mar 8, 2007 #5

    JasonRox

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    You did all the work!

    Oh well, I hope he sees how it came around to this.
     
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