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

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SUMMARY

The Maclaurin Series for the function g(x) = 4/(4 + 2x + x^2) can be derived using polynomial long division and completing the square. The first three terms of the series are 1 - (1/2)x - (1/4)x². To express g(x) in a form suitable for series expansion, it can be rewritten as (4/3)/((1/3)(x + 1)² + 1), which leads to a Taylor series expansion about x = -1. The unique representation of the series is confirmed through the method of power series.

PREREQUISITES
  • Understanding of Maclaurin Series and Taylor Series
  • Knowledge of polynomial long division
  • Ability to complete the square in algebraic expressions
  • Familiarity with power series and convergence concepts
NEXT STEPS
  • Study the derivation of Maclaurin Series for rational functions
  • Learn about polynomial long division techniques in calculus
  • Explore the process of completing the square for quadratic expressions
  • Investigate the convergence criteria for power series
USEFUL FOR

Students studying calculus, particularly those focusing on series expansions, algebraic manipulation, and convergence of series. This discussion is beneficial for anyone looking to deepen their understanding of Maclaurin and Taylor series.

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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!
 

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