MHB Don's question at Yahoo Answers (Taylor series)

AI Thread Summary
The discussion focuses on finding the Taylor series for the function f(x) = 1/x centered at a = -3. A substitution is made by letting t = x + 3, which simplifies the function using the geometric series. The resulting series is expressed as f(x) = -Σ((x + 3)^n / 3^(n + 1)), valid for |x + 3| < 3, or x in the interval (-6, 0). A link to the original question on Yahoo Answers is provided for further reference. This method offers a concise way to derive the Taylor series expansion.
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Hello Don,

Denoting $t=x+3$ and using the geometric series:

$$\frac{1}{x}=\frac{1}{t-3}=-\frac{1}{3}\cdot\frac{1}{1-\frac{t}{3}}=-\frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{t}{3}\right)^n \qquad \left(\;\left|\frac{t}{3}\right|<1\;\right)$$ Hence, $f(x)=-\displaystyle\sum_{n=0}^{\infty}\frac{(x+3)^n}{3^{n+1}}$, valid expasion for $|x+3|<3$, or equivalently for $x\in (-6,0).$
 
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