In Google Image, I've seen that Taylor "expansion" of a function $f$ around $0$ (i.e. $f(x) = \sum_{i=0}^{n-1} \frac{f^{i}(0)}{i!}x^i + \frac{f^{n}(t) }{n!}x^n$ for $t\in (0,x)$ or $t\in (x,0)$ depending on where $x$ stands) shows a polynomial of degree $n$, which is everywhere equal to $f$ between $0$ and $x$ (or in the interval $[x,0]$ if $x<0$) and that only after $x$ (or before if $x<0$) becomes different from $f$.
What I would have thought is that the polynomial of degree $n$, which is undoubtfully equal to $f$ at $x$, is only also equal to $f$ at $0$ (that would be because at $x=0$ the polynomial of the RHS of Taylor's formula is equal to $f(0)$) but not inbetween $0$ and $x$.
I don't know how can one see that inbetween $0$ and $x$ (or inbetween $x$ and $0$ if $x<0$) the polynomial of degree $n$ is equal to $f$.
I first thought about considering Taylor's formula at $\frac{x}{2}$ but this doesn't lead me to a convincing result.
What intrigues me is that the highest coefficient of the polynomial of the RHS of Taylor's formula is $x$ dependent, so this polynomial should be a different one at $\frac{x}{2}$.
