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Power series

  1. Apr 10, 2005 #1
    Two parts to this problem. On the first part I need someone to check my work, and I need help on solving the second part.

    (a) Find a power series representation for f(x) = ln(1+x).
    [tex] \frac{df}{dx} = \frac{1}{1+x} = \frac{1}{1-(-x)} = 1-x+x^2-x^3+x^4-...[/tex]
    [tex] \int_{n} \frac{1}{1+x} = \int_{n} (1-x+x^2-x^3+...)\dx = x-\frac{x^2}{2}+\frac{x^3}{3}-...+C = \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1} +C [/tex]
    sub x=0 in original equation: ln(1+0) = ln(1) = 0 = C.
    [tex]\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1}[/tex]
    with radius of convergence = 1.

    (b) Find a power series representation for f(x) = x*ln(1+x).
    If this function is differentiated, you get
    [tex] \ln(1+x) + \frac{x}{x+1}[/tex]
    which is the same as a sum of power series
    [tex] \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1} + \sum^{\infty}_{n=0} (-x)^{n+1} [/tex]
    have I gone too far? or where do I go from here?
    I know I will eventually have to integrate back to get the series for the original function.
     
  2. jcsd
  3. Apr 10, 2005 #2

    StatusX

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    Homework Helper

    The first series looks right except for a sign, but why don't you just replace n+1 by n and run the index from 1 to infinity? For the second one, you can just multiply each term from the expansion for ln(1+x) by x.
     
    Last edited: Apr 10, 2005
  4. Apr 10, 2005 #3

    Hurkyl

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    (a) looks fine (except for the sign), though it's usually good form to reindex the power series, and rearrange the terms, so that it looks like

    [tex]
    \sum_{n = ?}^{\infty} (\mathrm{something}) x^n
    [/tex]

    (b) You don't have to do any differentiation at all... :smile: but you could do it that way if you really want to, by combining the two sums.
     
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