- #1
Bob Busby
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Homework Statement
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.
Homework Equations
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.