So, I'm reading Pugh's Real Mathematical Analysis, and he has a theorem about Taylor polynomials that I can't find in many other places. It is:
Assume that f : (a, b) → R is rth order differentiable at x. Then: (a) The rth order Taylor Polynomial P approximates f to order r at x in the sense that the Taylor remainder $R(h)=f(x+h)−P(h)$ is rth order flat at h = 0; i.e., $\frac{R(h)}{h^r}$ → 0 as h → 0.
(b) The Taylor polynomial is the only polynomial of degree ≤ r with this approximation property
To prove this, Pugh says:
(a) The first r derivatives of R(h) exist and equal 0 at h = 0. If h > 0 then repeated applications of the Mean Value Theorem give $R(h)=R'(θ_1)h =R''(θ_2)θ_1h=R^{(r−1)}(θ_{r−1})(θ_{r−2})...θ_1h$, where $0<θ_{r−1}<...< θ_1< h$. Thus $|\frac{R(h)}{h^r}|=|\frac{R^{(r−1)}(θ_{r−1})(θ_{r−2})...θ_1h}{h^r}|\leq |\frac{R^{(r−1)}(θ_{r−1})−0}{θ_{r−1}}|$, which goes to 0 as h goes to 0. If $h<0$ the same is true with $h<θ_1<...<θ_{r−1}<0$.
(b) If $Q(h)$ is a polynomial of degree $\leq r$, $Q \neq P$, then $Q−P$ isn't rth-order flat at h = 0, so $f(x+h)−Q(h)$ cannot be rth-order flat either.
Here are my questions:
I understand the upper bound on $|\frac{R(h)}{h^r}|$, but why does that go to zero as h goes to zero? The best I can do is see that $R^{(r-1)}$ is differentiable and thus continuous, so L'Hopital's rule applies, but where do I go from here?
I understand why $Q-P$ isn't rth-order flat, but how does that have any bearing on $f(x+h)-Q(h)$ being rth-order flat?
Thank you in advance.
