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de's question at Yahoo! Answers (Power series representation)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.

If you have further questions you can post them in the Calculus section.