Find a power series representation for the function and determine the interval o

ossidianaZ 2021-10-25 Answered
Find a power series representation for the function and determine the interval of convergence.
\(\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{3}}-{x}\)

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Expert Answer

SkladanH
Answered 2021-10-26 Author has 26876 answers

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{2}}}{{{3}-{x}}}}\)
Divide Numerator and denominator by 3. To get
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{\frac{{{2}}}{{{3}}}}}}{{{1}-{\frac{{{x}}}{{{3}}}}}}}\)
Note that the sum of the geometric series with Initial term a and common ratio r is
\(\displaystyle{S}={\sum_{{{n}={0}}}^{\infty}}{a}{r}^{{n}}={\frac{{{a}}}{{{1}-{r}}}}\)
The given function can be interpred as
\(\displaystyle{\frac{{{\frac{{{2}}}{{{3}}}}}}{{{1}-{\frac{{{x}}}{{{3}}}}}}}={\frac{{{a}}}{{{1}-{r}}}}\)
Therefore, we can say that f(x) is a sum of a geometric series with initial term \(\displaystyle{a}={\frac{{{2}}}{{{3}}}}\) and common ratio \(\displaystyle{r}={\frac{{{x}}}{{{3}}}}\)
Therefore,
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{a}{r}^{{n}}={\sum_{{{n}={0}}}^{\infty}}{\left({\frac{{{2}}}{{{3}}}}\right)}{\left({\frac{{{x}}}{{{3}}}}\right)}^{{n}}={\sum_{{{n}={0}}}^{\infty}}{\left({\frac{{{2}}}{{{3}^{{{n}+{1}}}}}}\right)}\cdot{x}^{{n}}\)
This is the power series representation of f(x)
We know that the geometric series converges when \(\displaystyle{\left|{r}\right|}={\left|{\frac{{{x}}}{{{3}}}}\right|}{<}{1}\)
\(\displaystyle{\left|{x}\right|}{<}{3}\)
Intterval of converges is \(\displaystyle{\left(-{3},{3}\right)}\)
Radius of converges is 3

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