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

Maiclubk 2021-11-03 Answered
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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Expert Answer

unessodopunsep
Answered 2021-11-04 Author has 3871 answers

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: \(\displaystyle{\frac{{{3}}}{{{2}{x}-{1}}}}={1}-{\frac{{{2}}}{{{3}}}}{\left({x}-{2}\right)}+{\frac{{{4}}}{{{9}}}}{\left({x}-{2}\right)}^{{2}}-{\frac{{{8}}}{{{27}}}}{\left({x}-{2}\right)}^{{3}}+\ldots\)

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