Little o notation in Taylor's formula : o(1)

Question

I see multiple versions of Taylor's formula (some call it Taylor's series ( I dont really know whats the difference ) but one I know pretty well is :

$f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+f^{3}(a)\frac{(x-a)^3}{3!}+...+f^{n}(a)\frac{(x-a)^n}{n!}+(x-a)^n o(1)$

I get the idea of approximating a function using its derivatives , but I cant really understand what is the $(x-a)^n$ o(1) part doing . I know it means the left is negligble to 1 somewhere , but is it a function or other notation ? Why choosing $(x-a)^n$ on the left ? And what does it really mean here? Thank you

Edit: Converted formulas to Latex formatting - Felix

Answer 1

In this context, $o(1)$ is an unknown function of $x$ that converges to $0$ when $x\to a$. This takes a while to get used to, but it is an incredibly useful notation.

The complete term $(x-a)^no(1)$ is a $(x-a)^n$ times a function that converges to $0$. In other words it is a function that converges to $0$ after being divided by $(x-a)^n$. This can be rewritten as $o((x-a)^n)$, which can be interpreted as "goes to $0$ faster than $(x-a)^n$.

Remark that you never know what function the $o(1)$ exactly is. Only that it converges to $0$. So for instance an equation $o(1) + o(1) = o(1)$ is valid.It means that the sum of two function that goes to $0$ still goes to $0$. Something important to know is that $o(1)-o(1) \neq 0$, because the two small o's are not necessarily the same functions. The only thing you can say is $o(1)-o(1) = o(1)$. Another weird thing is that you can replace a $o((x-a)^5)$ by a $o(1)$, but not the other way around (the first one is a stronger condition).

Let me give you a list of examples of manipulations of small o's, when $x\to a$ like in your example. $$ o(1) + o(x-a) = o(1)+o(1) = o(1).$$ $$ o(1)/(1+o(1)) = o(1)/1 = o(1).$$ $$ (x-a)^3o(x-a) = (x-a)^3(x-a)o(1) = o((x-a)^4).$$ If $a\neq 0$, $$ xo(1) = ao(1) = o(1).$$ If $a=0,$ $$ xo(1) = o(x).$$

Answer 2

If $f\in o(g(x)),\, x\rightarrow 0$ then $\lim_{x\rightarrow 0}\left|\frac{f(x)}{g(x)}\right|=0$ now $h(x)o(g(x))=o(h(x)g(x))$ because: $$ f \in o(g(x)) \Rightarrow 0=\lim\left|\frac{f(x)}{g(x)}\right|=\lim\left|\frac{h(x)f(x)}{h(x)g(x)}\right| $$ Thus for every $f\in o(g),\ h(x)f(x)\in o(h(x)g(x))$

So in your case $(x-a)^no(1)=o((x-a)^n), \, x\rightarrow a$ which is simply the peano residual. In words: when a tends to x the point your taylor approximation is approximating the error will tend to 0 with a speed faster than a polynomial with degree n.