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An expansion/interval of convergence problem

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data


    The function f(x) = |x| is not differentiable at x = 0, so computing a Taylor Expansion of this function just isn't possible by taking derivatives of |x|. Through the use of clever substitution, you can still obtain the polynomial expansion of this function to be:

    f(x) = |x| = 1 - (1/2)(1-x^2) - (1/24)(1-x^2)^2 - (13/246)(1-x^2)^3

    for a certain interval of convergence. Establish this formula and the interval of convergence.

    2. Relevant equations



    3. The attempt at a solution

    I went about solving this by expanding sqrt(1-x) and plugging in (-x^2 + 1) for x. This gave me the expansion:

    1 - (1/2(1-x^2) - (1/8)(1 - x^2)^2 - (1/16)(1-x^2)^3

    This is not the given expansion but I graph it and it is a better approximation than the given expansion.

    In order to find the interval of convergence I differentiated sqrt(1-x) and notice that its derivatives all have singularities at x = 1. Does this mean that the interval of convergence for the absolute value function I found is (-1, 1).? Also, I like my expansion better, but does anyone know how to find the given expansion?

    Any help would be appreciated.
     
  2. jcsd
  3. Jun 22, 2010 #2
    Bob Busby Hi !

    I don't nearly understand where you got this:

    f(x) = |x| = 1 - (1/2)(1-x^2) - (1/24)(1-x^2)^2 - (13/246)(1-x^2)^3

    it doesn't seem strange to you? :)

    1 - (1/2)(1-x^2) - (1/24)(1-x^2)^2 - (13/246)(1-x^2)^3 <-- this does not equals |x|, try to plot it on a graph, and in no way behaves like the function "absolute value of x" (|x|) they are both even functions that's it.

    May be you can simulate |x| by sqrt(x^2) but i don't think you can write it in a closed form as a Taylor expansion.
     
  4. Jun 22, 2010 #3

    Gib Z

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    Homework Helper

    Try expanding sqrt(x^2) about x=1.
     
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