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Taylor series representation

  1. Aug 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Find a power series that represents $$ \frac{x}{(1+4x)^2}$$

    2. Relevant equations
    $$ \sum c_n (x-a)^n $$

    3. The attempt at a solution
    $$ \frac{x}{(1+4x)^2} = x* \frac{1}{(1+4x)^2} $$
    since [tex] \frac{1}{1+4x}=\frac{d}{dx}\frac{1}{(1+4x)^2} [/tex]
    $$ x*\frac{d}{dx}\frac{1}{(1+4x)^2} =x\frac{d}{dx}\sum_{n=0}^\infty(-4)^nx^n=x\sum_{n=0}^\infty(-4)^nnx^{n-1}=\sum_{n=0}^\infty(-4)^nnx^{n}$$

    The solution suggests $$\sum_{n=0}^\infty(-4)^n(n+1)x^{n+1}$$

    Am i doing something incorrect?
  2. jcsd
  3. Aug 5, 2016 #2


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    2017 Award

    Staff: Mentor

    Reconsider your differentiation. Isn't ##\frac{d}{dx} x^{-2} = -2x^{-3}##?
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