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Thread: power series representation of (2 + x)^(-2)

  1. #1

    power series representation of (2 + x)^(-2)

    Hello all,

    My class went over power series representation a few days ago and I am having some trouble. Here is the problem I'm working on, and how I have tried to solve it:


    "use differentiation to find a power series representation of f(x) = (2+x)^(-2)"

    (2+x)^(-2) = d/dx (-1)/(2+x) = d/dx (-1/2)/(1-(-x/2))

    d/dx (summation from n = 0 to infinity) (-1/2)(-x/2)^n

    (summation from n = 1 to infinity) (n)(-1/2)(-x/2)^(n-1)

    ANSWER: (summation from n = 0 to infinity) (n + 1)(-1/2)(-x/2)^n


    Can someone please explain what it is that I am doing wrong?

  2. #2
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    Quote Originally Posted by nsievers View Post
    Hello all,

    My class went over power series representation a few days ago and I am having some trouble. Here is the problem I'm working on, and how I have tried to solve it:


    "use differentiation to find a power series representation of f(x) = (2+x)^(-2)"

    (2+x)^(-2) = d/dx (-1)/(2+x) = d/dx (-1/2)/(1-(-x/2))

    d/dx (summation from n = 0 to infinity) (-1/2)(-x/2)^n

    (summation from n = 1 to infinity) (n)(-1/2)(-x/2)^(n-1)

    ANSWER: (summation from n = 0 to infinity) (n + 1)(-1/2)(-x/2)^n


    Can someone please explain what it is that I am doing wrong?

    You didn't take the derivative of (-x/2)^n correctly.

  3. #3
    if d/dx (x/2)^n != (n)(x/2)^(n-1), then I'm in more trouble than I thought

    The course I am taking is a single-variable calculus course. How would I solve this using only knowlege you would find in a Calculus I textbook?
    Last edited by nsievers; 06-09-2012 at 06:53 AM.

  4. #4
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    Quote Originally Posted by nsievers View Post
    if d/dx (x/2)^n != (n)(x/2)^(n-1), then I'm in more trouble than I thought

    The course I am taking is a single-variable calculus course. How would I solve this using only knowlege you would find in a Calculus I textbook?
    Since your 'inner' function is (x + 2), whose d/dx = 1, stop worrying so much.

    BUT:

    If f(x) = (x + 2)^-2 [ = zeroth derivative D0]

    D1 = -2(x + 2)^-3

    D2 = +6(x + 2)^-4

    D3 = -24(x + 2)^-5 = (-1)^3 * 4! (x + 2)^-5

    Looks as if

    Dn = (-1)^n * (n + 1)! (x + 2)^(-(n+2))

    Now do your power series stuff, using x = 0. Just plug in:

    Dn(0) = (-1)^n * (n + 1)! (x + 2)^(-(n+2))

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