Explained: "Find the Maclaurin series for f(x)=x21−8x8"

untchick04tm

Answered question

2021-12-23

Find the Maclaurin series for

f (x) = \frac{x^{2}}{1 - 8 x^{8}}

Answer & Explanation

lalilulelo2k3eq

Beginner2021-12-24Added 38 answers

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

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}^{\infty} {(8 x^{8})}^{n} = \sum_{n = 0}^{\infty} 8^{n} x^{8 n + 2}

which is valid if

| 8 x^{8} | < z

, hence,

| x | < \frac{1}{8^{\frac{1}{8}}}

user_27qwe

Skilled2021-12-30Added 230 answers

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

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