untchick04tm

2021-12-23

Find the Maclaurin series for $f\left(x\right)=\frac{{x}^{2}}{1-8{x}^{8}}$

### Answer & Explanation

lalilulelo2k3eq

Notice $\frac{1}{1-x}=\sum {x}^{n}$. Hence
$\frac{{x}^{2}}{1-8{x}^{8}}={x}^{2}\sum {\left(8{x}^{8}\right)}^{n}=\sum {8}^{n}{x}^{8n+2}$
and this is valid for $|8{x}^{8}|<1$

Jordan Mitchell

What you have got is almost the right answer.
$\frac{{x}^{2}}{1-8{x}^{8}}={x}^{2}\sum _{n=0}^{\mathrm{\infty }}{\left(8{x}^{8}\right)}^{n}=\sum _{n=0}^{\mathrm{\infty }}{8}^{n}{x}^{8n+2}$
which is valid if $|8{x}^{8}|, hence, $|x|<\frac{1}{{8}^{\frac{1}{8}}}$

user_27qwe

You can just write
$\frac{{x}^{2}}{1-8{x}^{8}}={x}^{2}\left(1+8{x}^{8}+\left(8{x}^{8}{\right)}^{2}+...\right)={x}^{2}+8{x}^{10}+64{x}^{18}+...$
The interval of convergence will be for all x such that
$|8{x}^{8}|<1$
which is equivalent to
$|x|<\left(\frac{1}{8}{\right)}^{\frac{1}{8}}$

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