# Maclaurin polynomial question.

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!

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Why the "Theoretical" in the title? – Nameless Dec 9 '12 at 16:52
I thought that it would be appropriate. But I can see how it might not be, I will remove that. – Lukas Arvidsson Dec 9 '12 at 16:53

By definition $f(x)=O(g(x))$ means that $\left| \frac{f(x)}{ g(x)} \right|\leq C$ as $x\to x_0$, $g(x)\neq 0$.

In the above notations $x^5 B_1(x)= O(x^5)$ as $x \to 0.$

If $|\psi(x)| \leq M$, then $|x^5 \psi(x)| \leq M |x^5|$ which implies that $x^5 \psi(x)= O(x^5)$ as $x \to 0$.

In other words, $x^5 \psi(x)$ can be combined with $x^5B_1(x)$.

Similarly, $$\frac{x^3}{3} \frac{8 x^3}{3}= \frac{8x^6}{9}= O(x^5), \; x \to 0.$$

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Thank you for the answer! It makes sense now! – Lukas Arvidsson Dec 16 '12 at 19:08
You're welcome. – Ragnar Dec 16 '12 at 20:15