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Taylor polynomial question

  1. Aug 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Write down the Taylor Polynomial of degree n of the function f(x) at x=0

    2. Relevant equations

    f(x) = ln(1-x)

    3. The attempt at a solution

    f(x) = ln(1-x)

    f'(x) = (-1)((1-x)^(-1))

    f``(x) = (-1)((1-x)^(-2))

    f```(x) = (-2)((1-x)^(-3))

    f````(x) = (-6)((1-x)^(-4))

    f`````(x) = (-24)((1-x)^(-5))

    Pn(x) = -1x - ((x^2)/(2!)) - ((2(x^3))/(3!)) - ((6(x^4))/(4!)) - ((24(x^5))/(5!))

    now, I rechecked all my derivatives, but I still can't find a pattern to make an nth term with.

    any help would be appreciated.

  2. jcsd
  3. Aug 21, 2007 #2


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    1, 2, 6, 24 etc is 1!, 2!, 3!, 4! etc. So you have (n-1)!/n!. What's that?
  4. Aug 21, 2007 #3
    I think I got it

    I got the numbers in the factorials to cancel out, and came up with

  5. Aug 21, 2007 #4
    Thanks a lot, Dick, I appreciate it.
  6. Aug 21, 2007 #5
    Thanks a lot.
  7. Aug 22, 2007 #6
    Another way to do it:
    You know that

    [tex]\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}[/tex]

    So by integration you get that

    [tex]ln(1-x)=-\int{\sum^{\infty}_{n=0}x^n dx}[/tex]
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