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.

If you have further questions you can post them in the http://www.mathhelpboards.com/f10/ section.http://www.mathhelpboards.com/f10/
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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