How do you find f^6(0) where f(x)=arctanx/x?

Mar 2, 2017

${f}^{\left(6\right)} \left(0\right) = - \frac{720}{7}$

Explanation:

We start by constructing the MacLaurin series for $\arctan x$.
Consider the function:

$f \left(x\right) = \frac{1}{1 + {x}^{2}}$

This is the sum of a geometric series of ratio: $- {x}^{2}$, so that we have:

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

with radius of convergence $R = 1$.

Within the interval $x \in \left(- 1 , 1\right)$ we can therefore integrate term by term:

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

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

and dividing by $x$ both sides:

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

Now consider the standard expression of the MacLaurin series of $f \left(x\right)$

f(x) = sum_(n=0)^oo f^((n))(0)/(n!)x^n

The two series are equal only if the coefficients of the same degree in $x$ are equal, so that for ${x}^{6}$ we have:

f^((6))(0)/(6!) = -1/7

and:

${f}^{\left(6\right)} \left(0\right) = - \frac{720}{7}$