# Find a power series for the function, centered at c, and determine the interval

Find a power series for the function, centered at c, and determine the interval of convergence.
$$\displaystyle{g{{\left({x}\right)}}}=\frac{{3}}{{2}}{x}-{1},{c}={2}$$

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Geometric Power Series Centered at c: The geometric power series centered at c is a series of the form
$$\displaystyle{\frac{{{a}}}{{{1}-{\left({r}-{c}\right)}}}}={\sum_{{{n}={0}}}^{\infty}}{a}{\left({r}-{c}\right)}^{{n}},{\left|{r}-{c}\right|}{<}{1}$$
where a is the first term and r-c is the common ratio.
The function is given by
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{3}}}{{{2}{x}-{1}}}},{c}={2}$$
Writing f(x) in the form $$\displaystyle{\frac{{{a}}}{{{1}-{r}}}}$$ produces
$$\displaystyle{\frac{{{3}}}{{{2}{x}-{1}}}}={\frac{{{3}}}{{{2}{\left({x}--{2}\right)}+{3}}}}$$
$$\displaystyle={\frac{{{3}}}{{{3}{\left({1}+{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}\right)}}}}$$
$$\displaystyle={\frac{{{1}}}{{{\left({1}+{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}\right)}}}}={\frac{{{a}}}{{{1}-{r}}}}$$
which implies that a=1 and $$\displaystyle{r}=-{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}$$.
So, the power series for f(x) is
$$\displaystyle{\frac{{{3}}}{{{2}{x}-{1}}}}={\sum_{{{n}={0}}}^{\infty}}{a}{r}^{{n}}$$
$$\displaystyle={\sum_{{{n}={0}}}^{\infty}}{\left[-{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}\right]}^{{n}}$$
$$\displaystyle={\sum_{{{n}={0}}}^{\infty}}{\left(-{\frac{{{2}}}{{{3}}}}\right)}^{{n}}{\left({x}-{2}\right)}^{{n}}$$
$$\displaystyle={1}-{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}+{\frac{{{4}}}{{{9}}}}{\left({x}-{2}\right)}^{{2}}-{\frac{{{8}}}{{{27}}}}{\left({x}-{2}\right)}^{{3}}+\ldots$$
This power series converges when
$$\displaystyle{\frac{{{2}}}{{{3}}}}{\left|{x}-{2}\right|}{<}{1}\Rightarrow{\left|{x}-{2}\right|}{<}{\frac{{{3}}}{{{2}}}}$$
$$\displaystyle\Rightarrow-{\frac{{{3}}}{{{2}}}}{<}{x}-{2}{<}{\frac{{{3}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{\frac{{{1}}}{{{2}}}}{<}{x}{<}{3}{\frac{{{1}}}{{{2}}}}$$
which implies that the interval of convergence is $$\displaystyle{\left({\frac{{{1}}}{{{2}}}},{3}{\frac{{{1}}}{{{2}}}}\right)}$$
The power series converges when
$$\displaystyle{\frac{{{2}}}{{{3}}}}{\left|{x}-{2}\right|}{<}{1}\Rightarrow{\left|{x}-{2}\right|}{<}{\frac{{{3}}}{{{2}}}}$$
$$\displaystyle\Rightarrow-{\frac{{{3}}}{{{2}}}}{<}{x}-{2}{<}{\frac{{{3}}}{{{2}}}}$$
$$\displaystyle{2}-{\frac{{{3}}}{{{2}}}}{<}{x}{<}{2}+{\frac{{{3}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{\frac{{{1}}}{{{2}}}}{<}{x}{<}{3}{\frac{{{1}}}{{{2}}}}$$
which implies that the interval of converges is $$\displaystyle{\left({\frac{{{1}}}{{{2}}}},{3}{\frac{{{1}}}{{{2}}}}\right)}$$
Therefore, the power series for the function $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{3}}}{{{2}{x}-{1}}}},{c}={2}$$ centered at c is
$$\displaystyle{\frac{{{3}}}{{{2}{x}-{1}}}}={1}-{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}+{\frac{{{4}}}{{{9}}}}{\left({x}-{2}\right)}^{{2}}-{\frac{{{8}}}{{{27}}}}{\left({x}-{2}\right)}^{{3}}+\ldots$$
Also, the interval of convergence is (1,3).
Final answer: $$\displaystyle{\frac{{{3}}}{{{2}{x}-{1}}}}={1}-{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}+{\frac{{{4}}}{{{9}}}}{\left({x}-{2}\right)}^{{2}}-{\frac{{{8}}}{{{27}}}}{\left({x}-{2}\right)}^{{3}}+\ldots$$