Find Maclaurin Series for g(x): Interval of Conv.

wilcofan3
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Homework Statement



Find the Maclaurin Series for g(x)= (4)/(4+2x+x^2) and its interval of convergence.

Homework Equations



I know the Maclaurin Series usually involves taking derivatives but every other problem I've done so far has had a degree that I've solved to. So, other than the general equation for the Maclaurin Series, I'm not sure what other equation is relevant.

The Attempt at a Solution



At first I thought I might need to use partial fraction decomposition but then I realized the bottom did not factor easily. Do I just start taking derivatives of g(x)? And if so, where do I go from there? I would greatly appreciate a step-by-step breakdown, I think I will understand if someone breaks down the problem. Thanks!
 
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Try dividing 4 by 4 + 2x + x^2, using ordinary polynomial long division. With a Maclaurin series, however you get the series doesn't much matter, since the series representation is unique.

Doing what I suggested above, I get the first three terms as 1 - 1/2 * x - 1/4 * x^2.
 
Another thing you can do is this: complete the square in the denominator to write the fraction as 4/((x+1)^2+ 3). Divide both numerator and denominator by 3 to get (4/3)/((1/3)(x+1)^2+ 1) or (4/3)/(1- (-1/3)(x+1)^2) and think of that as the sum of a power series with first term 4/3 and common ration (-1/3)(x+1)^2. That gives the general term as (4/3)(-1/3)^n(x+1)^}{2n}. Unfortunately, that is now a Taylor's series, about x=-1, rather than a MacLaurin series!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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