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Taylor series for ln

  1. Jan 26, 2006 #1
    may i know from ln(1-x) how to become - [infinity (sum) k=1] x^k / k ???
    pls help
     
  2. jcsd
  3. Jan 26, 2006 #2

    HallsofIvy

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    You posted this in the "precalculus section" so I assume you wouldn't be able to use "Taylor Series". That's the only way I know to show it.
     
  4. Jan 26, 2006 #3

    VietDao29

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    Uhmm, I think it can also be done with geometric series:
    [tex]\sum_{i = 0} ^ \infty (a r ^ i) = \frac{a}{1 - r}, \quad |r| < 1[/tex].
    So if your final sum is:
    [tex]\frac{1}{1 - k}[/tex], what should your geometric series look like? Then integrating that should give you the result you want (remember to choose the appropriate C, i.e the constant of integration).
     
  5. Jan 26, 2006 #4

    HallsofIvy

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    I thought about that but then integrating is not "pre-calculus" either!
     
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