Yulia

2021-02-12

Use the Geometric Series Test to help you find a power series representation of $f\left(x\right)=\frac{x}{\left(2+{x}^{3}\right)}$ centered at 0. Find the interval and radius of convergence.

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Given a function
$f\left(x\right)=\frac{x}{\left(2+{x}^{3}\right)}$
To Find: power of series
Note: The geometric power series is a series of the form
$\frac{a}{1-r}=\sum _{n=0}^{\mathrm{\infty }}a{r}^{n},|r|<1$
Where a is the first term and r is the common ratio.
This implies $f\left(x\right)=\frac{x}{2+{x}^{3}}$
$=\frac{x}{2}\cdot \left(\frac{1}{1+\frac{{x}^{3}}{2}}\right)$
Now, the power series for f(x) is
$\frac{1}{1+\frac{{x}^{3}}{2}}=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}$
$\sum _{n=0}^{\mathrm{\infty }}\left(-\left(\frac{x}{{2}^{\frac{1}{3}}}{\right)}^{3}{\right)}^{n}$
$\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{3n}}{{2}^{n}}$
Converging for
$|\frac{{x}^{3}}{2}<1|,$ i.e. $-{2}^{\frac{1}{3}}
Now,
$\frac{x}{2+{x}^{3}}=\frac{x}{2}\left(\frac{1}{1+\frac{{x}^{3}}{2}}\right)=\frac{x}{2}\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{3n}}{{2}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{3n+1}}{{2}^{n+1}}$
With radius of radius of convergence $R={2}^{\frac{1}{3}}$
Therefore, the power series of the function is
$\frac{x}{2+{x}^{3}}=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{3n+1}}{{2}^{n+1}}$

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