I'm so confused about the Taylor's Polynomial and Series (including approximation of a function using Taylor's Series) and derivative tests and come up with a few questions.
So, we approximate a function using Taylor's polynomials. And all of the documents and books that I've read are saying 'higher order polynomials mean better approximation i.e. more accurate'. However, I don't really see why. I know that as the order gets higher, $$\ f(a) = P_n(a), f'(a) = P_n'(a), f''(a) = P_n^{''}(a), ... , f^{(n)}(a) = P_n^{(n)}(a)$$ But this doesn't seem like a proof for me. It seems like just a consequence of higher order Taylor's polynomial rather than a proof. Why is it getting more accurate? Is there any mathematical proof of this? Is the absolute value of the error getting smaller as the order gets higher?
Another question is about derivative test (not only the second derivative test but all higher order derivative test). When we test the nature of stationary points, we check $f''(x)$ at the point whether it's positive or negative (or sometimes zero). I've been taught this is because of Taylor series expansion at the point. $$\ f(x) = f(c) + f'(c)(x-c) + \frac {f''(c)}{2!}(x-c)^2 + ... $$ and since $f'(c)=0$ $$ f(x) = f(c) + \frac{f''(c)}{2!}(x-c)^2+...$$ $$=> f(x) - f(c) = \frac {f''(c)}{2!}(x-c)^2+...$$ and then we approximate $$f(x) - f(c) \approx \frac {f''(c)}{2!}(x-c)^2 $$ because the higher order doesn't really matter. I really don't understand how do we approximate like this because for me it seems like higher order might matter. I mean what if $f''(c) > 0$ but $f^{3}(c), f^{4}(c), ..., f^{n}(c), ... $ are all negative so that it actually makes $f(x) - f(c) < 0 $ ? I guess this is not the case since the method works, but I don't really know how to explain this (not only the second derivative test but all higher order derivative tests) in terms of Taylor series. PLEASE HELP ME! I'M SO CONFUSED AND FRUSTRATED! THANK YOU!