# How do you find the Taylor polynomial of degree 10 of the function arctan(x^3) at a = 0?

May 28, 2018

$\arctan {x}^{3} = {x}^{3} - {x}^{9} / 3 + {x}^{15} / 5 + o \left({x}^{15}\right)$

#### Explanation:

Start from the sum of the geometric series:

${\sum}_{n = 0}^{\infty} {q}^{n} = \frac{1}{1 - q}$

for $\left\mid q \right\mid < 1$.

Let $q = - {x}^{2}$:

$\frac{1}{1 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- {x}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n}$

converging for $\left\mid x \right\mid < 1$.

In the interval of convergence we can integrate term by term:

${\int}_{0}^{t} \frac{\mathrm{dx}}{1 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\int}_{0}^{t} {x}^{2 n} \mathrm{dx}$

$\arctan t = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{2 n + 1} / \left(2 n + 1\right)$

Let now $t = {x}^{3}$:

$\arctan {x}^{3} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left({x}^{3}\right)}^{2 n + 1} / \left(2 n + 1\right)$

$\arctan {x}^{3} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{6 n + 3} / \left(2 n + 1\right)$

and truncating at the third term:

$\arctan {x}^{3} = {x}^{3} - {x}^{9} / 3 + {x}^{15} / 5 + o \left({x}^{15}\right)$