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Macclaurin Series

  1. Jun 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the Maclaurin series for f(x)=(1-x)^-2 using the definition of a Macluarin series,[ Assume that f has a power series expansion. Also find

    2. Relevant equations

    Maclaurin Series


    3. The attempt at a solution

    Steps:

    1) I found the first 3 derivative of the function

    2) I plugged in 0 for x

    3) try to use the Macclaurin Series
    f(x)= f(0) + f'(0) x + f''(0) x^2 / 2! + f'''(0) x^3 / 3! +...


    From here, i'm am stuck.
     
  2. jcsd
  3. Jun 30, 2014 #2

    Zondrina

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    Let ##t = -x##, then ##f(x) = (1+t)^{-2}##

    You now have a case of the binomial theorem with ##\alpha = -2##.

    Do you know the binomial series?
     
  4. Jun 30, 2014 #3

    Dick

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    Ok, so what did you get for the first three derivatives evaluated at x=0? What do you get for the first three terms of the series? Don't you see a pattern to continue?
     
  5. Jun 30, 2014 #4
    Here is my work so far:

    f(0) = 1
    f'(x) = 2(1-x)^-3 ; f'(0) = 2
    f''(x) = 6(1-x)^-4 ; f''(0) = 6
    f'''(x) = 24(1-x)^-5 ; f'''(0) = 24


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

    (1-x)^-2 = 1+2x+6/2! (x^2)+24/3! (x^3)+...
     
  6. Jun 30, 2014 #5

    Dick

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    Good! Expand the factorials. So you've got 1+2x+3x^2+4x^3+... Do you see the pattern? Can you argue why it must continue?
     
  7. Jun 30, 2014 #6

    Zondrina

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    Notice every time you take a derivative, it has a "general form". Finding the ##n^{th}## derivative in a closed form, evaluated at ##a## will give you almost all of the work. Then every power series has the form:

    ##\sum_{n=0}^{∞} \frac{f^{(n)}(a)}{n!} (x-a)^n, \quad |x-a| < R##

    Alternatively, you could appeal to the binomial theorem:

    ##(1+x)^{\alpha} = \sum_{k=0}^{∞} \begin{pmatrix}
    \alpha \\
    k
    \end{pmatrix} x^k##

    Where ##\begin{pmatrix}
    \alpha \\
    k
    \end{pmatrix} = \frac{\alpha(\alpha-1)(\alpha-2) \cdots (\alpha - k + 1)}{k!}##

    and

    ##\begin{pmatrix}
    \alpha \\
    0
    \end{pmatrix} = 1##
     
  8. Jun 30, 2014 #7
    Thanks! This really made more sense than what i wrote on my class notes
     
  9. Jun 30, 2014 #8

    Zondrina

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    No problem. If you want an easy way to get ##R##, let ##c_n = \frac{f^{(n)}(a)}{n!}##. Then every power series has the form:

    ##\sum_{n=0}^{∞} c_n (x-a)^n, \quad |x-a| < R##

    With: ##R = \displaystyle \lim_{n→∞} |\frac{c_n}{c_{n+1}}|##
     
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