untchick04tm

2021-12-23

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

lalilulelo2k3eq

Beginner2021-12-24Added 38 answers

Notice

and this is valid for

Jordan Mitchell

Beginner2021-12-25Added 31 answers

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$\left|8{x}^{8}\right|<z$ , hence, $\left|x\right|<\frac{1}{{8}^{\frac{1}{8}}}$

which is valid if

user_27qwe

Skilled2021-12-30Added 375 answers

You can just write

The interval of convergence will be for all x such that

which is equivalent to

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