I am struggling to understand how my math book explains how to establish the Maclaurin polynomial of the fourth order to the function: $f(x) = \ln(1+x) \arctan2x$
We have:
$\ln(1+x) = x -\frac{1}{2}x^2+\frac{1}{3}x^3+x^4B_1(x)$
$\arctan2x = 2x -\frac{8}{3}x^3+32x^5B_2(x)$
by multiplying $\ln(1+x)$ with $\arctan2x$ we get:
$f(x) = x*2x - x * \frac{8}{3}x^3 + x*32x^5B_2(x) +$
$\frac{1}{2}x^2 * 2x - \frac{1}{2}x^2* \frac{8}{3}x^3 + \frac{1}{2}x^2*32x^5B_2(x) +$
$\frac{1}{3}x^3 * 2x - +\frac{1}{3}x^3* \frac{8}{3}x^3 + +\frac{1}{3}x^3*32x^5B_2(x) +$
$x^4B_1(x) * 2x - +x^4B_1(x)* \frac{8}{3}x^3 + +x^4B_1(x)*32x^5B_2(x)$
According to my math book his can be simplified to:
$f(x) =x*2x - x * \frac{8}{3}x^3 - \frac{1}{2}x^2 * 2x + \frac{1}{3}x^3 * 2x + x^5B(x) = 2x^2 - x^3 -2x^4 + x^5B(x)$
They motivate this by saying that all terms that looks like $x^5 * (bounded function)$ can be combined to $x^5B(x)$. Can someone please explain to me the reasoning behind this? Why does this also hold true for a term like this one: $\frac{1}{3}x^3* \frac{8}{3}x^3$?
Thank you very much!