I've just been trying to understand taylor polynomials more intuitively from a visual perspective. But as soon as the terms start using second derivatives, it becomes unclear.
1, Some thought's I've had:
Consider the expression
$\frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2}$
Using L'Hopital's rule, we know this goes to $\frac{1}{2}f''(a)$ as $x$ goes to $a$.
We can simply rewrite the expression:
$\frac{\frac{f(x)-f(a)}{(x-a)} - f'(a)}{(x-a)}$
We know from mean value theorem that $\frac{f(x)-f(a)}{(x-a)} = f'(\xi)$ for some $\xi$ in $(a,x)$. But if the limit of the expression is meant to be $\frac{1}{2} f''(a)$, then I would expect that $\xi$ gets close to $\frac{(x+a)}{2}$ in some sense - but that doesn't seem true for a function like $x^3$ - there, when considering $a=0$, we get that $f'(\frac{x+a}{2}) = \frac{3}{4}x^2$, while $\frac{f(x)-f(a)}{x-a}=x^2$ for all $x$.
Either way, I'd like to know if there's some nice explanation for why the limit of that expression is $\frac{1}{2} f''(a)$, without using L'Hopital's rule.
2, Is there some sort nice visual explanation of what exactly happens when using taylor polynomials of 2nd degree, and why it works out so neatly? I understand it in a handwavy-way of "well the first derivative gives you a linear aproximation, and the second gives you a linear approximation for the derivative, and if you want the derivative to change with that speed, you put $x^2$ in there, so that it's the first derivative that changes".
But that doesn't really make things perfectly clear. I'd like to understand why $\frac{f(x)-T_2(x)}{(x-a)^2} $ goes to zero, why the 3rd term of the taylor polynomial turns out to be exactly $\frac{1}{2} f''(a)(x-a)^2$ - but keeping the explanation visual, just like it can be done when approximating with an only linear function, ideally without resorting to something like l'Hopital's rule or using algebraic manipulations that don't get visually explained.