MHB De's question at Yahoo Answers (Power series representation)

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To find the first five non-zero terms of the power series representation for the function f(x) = 3x^3/(x-3)^2 centered at x=0, the discussion introduces the function g(x) = 1/(x-3) and its derivative g'(x). By applying the uniform convergence of the power series and the geometric series sum, g(x) is expressed as a series. The transformation leads to the expression for f(x) in terms of a power series, allowing the extraction of the first five non-zero terms. The response encourages further questions to be posted on a dedicated math help forum.
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Hello de,

Denote $g(x)=\dfrac{1}{x-3}$, then $g'(x)=-\dfrac{1}{(x-3)^2}$. Using the uniform convergence of the power series and the sum of the geometric series: $$g(x)=-\frac{1}{3}\frac{1}{1-x/3}=-\frac{1}{3}\sum_{n=0}^{+\infty}\frac{x^n}{3^n} \Rightarrow g'(x)=-\frac{1}{3}\sum_{n=1}^{+\infty}\frac{nx^{n-1}}{3^n}\quad (|x|<3)$$ Then, $$f(x)=\frac{3x^3}{(x-3)^2}=(3x^3)\frac{1}{3}\sum_{n=1}^{+\infty}\frac{nx^{n-1}}{3^n}=\sum_{n=1}^{+\infty}\frac{nx^{n+2}}{3^n} \quad (|x|<3)$$ and now, you'll easily find the first five non-zero terms.

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