1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    fresh_42

    Staff: Mentor

    Reconsider your differentiation. Isn't ##\frac{d}{dx} x^{-2} = -2x^{-3}##?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Taylor series representation
  1. Series representation (Replies: 3)

  2. Taylor series (Replies: 15)

Loading...