# How do you find the taylor polynomial of degree 4 of the function f(x)= arctan(x^1) at a=0?

Mar 9, 2017

$\arctan x \cong x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + {x}^{9} / 9$

#### Explanation:

Start from the integral:

$\arctan x = {\int}_{0}^{x} \frac{\mathrm{dt}}{1 + {t}^{2}}$

Now note that the integrand function is in the form of the sum of a geometric series of ratio $- {t}^{2}$:

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

which has radius of convergence $R = 1$. Thus in the interval $\left(- 1 , 1\right)$ we can integrate term by term and have:

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

For $n = 4$ we have:

$\arctan x = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + {x}^{9} / 9 + {R}_{4} \left(x\right)$