Maclaurin Series with arctan

Quote:

Originally Posted by Grimey

Find a Maclaurin series for f(x) = (arctan(x^3))/(x^3)

(Angry)

any help appreciated (Bow)

Build this up in stages, starting with

$\frac1{1+x^2} = 1-x^2+x^4-x^6+\ldots$ (binomial series).

Integrate both sides to get

$\arctan x = x - \frac{x^3}3 + \frac{x^5}5 - \frac{x^7}7 + \ldots.$

Divide both sides by x:

$\frac{\arctan x}x = 1 - \frac{x^2}3 + \frac{x^4}5 - \frac{x^6}7 + \ldots.$

Finally, replace x by $x^3$ :

$\frac{\arctan x^3}{x^3} = 1 - \frac{x^6}3 + \frac{x^{12}}5 - \frac{x^{18}}7 + \ldots.$