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Taylor Series question

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the Taylor Series for f(x) = ln(1-3x) about x = 0

    2. Relevant equations

    ln(1+x) = [tex]\sum\fract(-1)^n^+^1 x^n /{n}[/tex]

    3. The attempt at a solution

    ln(1-3x) = ln(1+(-3x))

    ln(1+(-3x)) = [tex]\sum\fract(-1)^n^+^2 x^3^n /{n}[/tex]

    Is that right?
  2. jcsd
  3. Mar 29, 2009 #2
    The -1 is in the right place, but I'm not sure why the 3 migrated to the exponent.
  4. Mar 29, 2009 #3
    So is it:
    \sum\fract(-1)^n^+^2 3x^n /{n}
  5. Mar 29, 2009 #4
    You check it yourself by computing the first couple of terms in the Taylor series.
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