Use Geometric Series to find a power series for $f$ based at $x = 0$ and $x=1$, and also their interval of convergence for $f(x)= 1/(x+3)$.
I don't understand--what's the "based at $x=0$ and $x=1$" mean? Do I just plug in?
Thanks
Use Geometric Series to find a power series for $f$ based at $x = 0$ and $x=1$ also their interval of convergence
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real-analysis
1 Answers
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A power series for an analytic function $f(x)$ based at $x = x_0$ is an infinite series $\sum_{n=0}^\infty a_n(x-x_0)^n$ such that for some radius $r > 0$, $f(x) = \sum_{n=0}^\infty a_n(x-x_0)^n$ whenever $|x-x_0| The classic example is given by the geometric series:
$$
\frac{1}{1-x} = \sum_{n=0}^\infty x^n, \quad |x| < 1.
$$
This is a power series for $f(x) = \frac{1}{1-x}$ based at $x=0$. So to find a power series for $f(x) = \frac{1}{x+3}$ based at $x=0$, observe that by the geometric series equation above,
$$
\frac{1}{x+3} = \frac{1}{3-(-x)} = \frac{1}{3} \cdot \frac{1}{1-(-\frac{x}{3})} =
\frac{1}{3} \cdot \sum_{n=0}^\infty \left( -\frac{x}{3} \right)^n
$$
whenever $|\frac{x}{3}| < 1$. Hence for all $|x| < 3$,
$$
f(x) = \sum_{n=0}^\infty (-1)^n\left(\frac{1}{3}\right)^{n+1}x^n.
$$
Similarly, for all $|x-1| < 4$,
$$
f(x) =
\frac{1}{x+3} =
\frac{1}{4} \cdot \frac{1}{1-\frac{1-x}{4}} =
\frac{1}{4} \cdot \sum_{n=0}^\infty \left( \frac{1-x}{4} \right)^n =
\sum_{n=0}^\infty (-1)^n \left(\frac{1}{4}\right)^{n+1}(x-1)^n.
$$