Little-$O$ and Taylor Polynomial

Question

Why do I need to put $o(x^n)$ at the end of a Taylor polynomial? For instance: $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)$$

I'm having troubles to understand the utility of the little-$o$ notation.

Answer 1

If you just wrote $$ e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6} $$ then it would be a false equality. The goal of Taylor series is to approximate complicated functions with simple functions, namely polynomials, around a given point. Suppose you're working around $0$, you would have $f(x)=a_0+a_1x+\cdots+a_nx^n+\varepsilon_n(x)$ for some $n\geqslant 0$ and $\varepsilon_n$ a function that is negligeable around $0$ so that as $n$ is chosen to be large, the approximation gets better. What does it mean for $\varepsilon_n$ to be negligeable ? It should mean that $\varepsilon_n$ is way smaller than the approximation, that is $a_0+a_1x+\cdots+a_nx^n$, around $0$. For Taylor series, it is asked that $$ \lim\limits_{x\rightarrow 0}\frac{\varepsilon_n(x)}{a_0+a_1x+\cdots+a_nx^n}=0 $$ which is equivalent to $\lim\limits_{x\rightarrow 0}\frac{\varepsilon_n(x)}{x^n}=0$ i.e $\varepsilon_n(x)=o(x^n)$.