MHB Nick's question at Yahoo Answers (Maclaurin series)

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The discussion focuses on finding the Maclaurin series for the function f(x) = x/(1-x^4). It establishes that for |t|<1, the series expansion of 1/(1-t) can be utilized. By applying this to f(x), the series is expressed as f(x) = x * (1/(1-x^4)), leading to the summation form. The resulting Maclaurin series is identified as the summation from n=0 to infinity of x^(4n+1), valid for |x|<1. This provides a clear solution to the original question posed by Nick.
Fernando Revilla
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Here is the question:

for f(x)= (x)/(1-x^4)

= summation from 0 to infinity:

Here is a link to the question:

Find the Maclaurin series? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello nick,

If $|t|<1$ we know that $\dfrac{1}{1-t}=\displaystyle\sum_{n=0}^{\infty}t^n$. Then, using the Algebra of series: $$f(x)=x\cdot\dfrac{1}{1-x^4}=x\sum_{n=0}^{\infty}(x^4)^n=\sum_{n=0}^{ \infty}x^{4n+1}\quad (|x|<1)$$ So, necessarily the Maclaurin series for $f(x)$ is $\displaystyle\sum_{n=0}^{\infty}x^{4n+1}.$
 
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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