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

AI Thread Summary
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 $\displaystyle\sum_{n=0}^{\infty}x^{4n+1}.$
 
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