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Math Help - series representation

  1. #1
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    series representation

    Represent the function f(x) = x/(1+4x)^2 as a power series and determine the radius of convergence.

    so the correct answer is (-1)^n * (n+1) *4^n*x^(n+1) but this is what I got:

    x/(1+4x)^2

    x/[1-(-4x)^2

    x series sign (-4x)^n

    series sign (-1)^n * 4^n x^(n+1)

    Where did the (n+1) in front of the 4^n come from?
    Last edited by smray7; April 12th 2011 at 09:49 AM.
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  2. #2
    Senior Member Sambit's Avatar
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    Sorry, but I can't understand what you mean. What is the question exactly?
    so the correct is what I have multiplied by (n+1)
    And what does this mean? What have you multiplied by (n+1) and where?
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  3. #3
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    Are you starting with the power series

    \dfrac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots

    and if so, what are you doing to this series to result in your series?
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  4. #4
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    OK so I fixed the question. What I get from this section is that you want to make your function look like 1/1-x and go from there according to the definition.
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  5. #5
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    Right, so differentiate my series

    \dfrac{1}{1+x} = 1 - x + x^2 - x^3 \cdots + (-1)^{n}x^{n}

    so

    \dfrac{-1}{(1+x)^2} = -1 + 2x - 3x^2 + 4x^3  - \cdots +  (-1)^{(n+1)} (n+1)x^n

    or

    \dfrac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots -  (-1)^{(n+1)} (n+1)x^n

    or

    \dfrac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots +  (-1)^n (n+1)x^n


    Now replace x with 4x and multiply the whole thing by x.
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  6. #6
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    So once you have the power series representation of a function, you the take the derivative?
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  7. #7
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    Quote Originally Posted by smray7 View Post
    So once you have the power series representation of a function, you the take the derivative?
    You showed in the OP that you already know the series for 1/(1 - x). The derivative of that series is the series of that derivative, -1/(1 - x)^2. This more closely resembles the original problem which you are working towards.

    This is just one tool you can use to find a series for a function you didn't know.
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  8. #8
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    You could show that, if \displaystyle f(x) = \frac{x}{(1+4x)^2}, then:

    \displaystyle f^{(k)}(x) = \frac{4^{k-1}(-1)^k(k)!}{(1+4x)^{k+1}}-\frac{4^{k-1}(-1)^k(k+1)!}{(1+4x)^{k+2}}

    and use the general formula for Taylor/Maclaurin series.
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