MHB Power Series for f(x) and Radius of Convergence

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The power series representation of f(x) = 4x/(x-3)^2 centered at x = 0 yields the first five non-zero terms as 4x/9, 8x^2/27, 12x^3/81, 16x^4/243, and 20x^5/729. The series is derived from the geometric series expansion of 1/(x-3) and its derivative. The radius of convergence for this series is determined to be 3, as it is the distance from the center of the series to the nearest singularity at x = 3. The discussion emphasizes the importance of correctly deriving the series and identifying the radius of convergence. Understanding these concepts is crucial for analyzing the behavior of the function near its center.
joshuapeterson
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f(x) = 4x/(x-3)^2
Find the first five non-zero terms of power series representation centered at x = 0.
Also find the radius of convergence.
 
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let $y' = \dfrac{1}{(x-3)^2} \implies y = -\dfrac{1}{x-3} = \dfrac{\frac{1}{3}}{1-\dfrac{x}{3}} = \dfrac{1}{3}\bigg[1+\dfrac{x}{3} + \dfrac{x^2}{3^2} + \dfrac{x^3}{3^3} + \, ... \bigg]$

$y'= \dfrac{1}{3}\bigg[\dfrac{1}{3} + \dfrac{2x}{3^2} + \dfrac{3x^2}{3^3} + \dfrac{4x^3}{3^4}+ \, ... \bigg]$

$f(x) = \dfrac{4x}{(x-3)^2} = \dfrac{4x}{3}\bigg[\dfrac{1}{3} + \dfrac{2x}{3^2} + \dfrac{3x^2}{3^3} + \dfrac{4x^3}{3^4}+ \, ... \bigg] = \dfrac{4x}{3^2} + \dfrac{8x^2}{3^3} + \dfrac{12x^3}{3^4} + \dfrac{16x^4}{3^5}+ \, ... $

I'll leave the 5th non-zero term for you to figure out ...
 

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