泰勒定理与导数的局限

Taylor's theorem (Lagrange remainder). Suppose f has n+1 derivatives on an
open interval containing a and x. Then there is a point c strictly between a and x
with
    f(x) = P_n(x) + R_n(x),
where the degree-n Taylor polynomial at a is
    P_n(x) = f(a) + f'(a)(x-a) + f''(a)/2! (x-a)^2 + ... + f^(n)(a)/n! (x-a)^n,
and the remainder is
    R_n(x) = f^(n+1)(c) / (n+1)! · (x - a)^(n+1).

Worked estimate: f(x) = e^x at a = 0, n = 2, on [0, 1].
    P_2(x) = 1 + x + x^2/2,   R_2(x) = e^c/6 · x^3  for some c in (0, x).
For x in [0,1], c < 1 so e^c < e < 3, giving the honest bound
    |e^x - (1 + x + x^2/2)| = |R_2(x)| <= 3/6 · x^3 = x^3 / 2.
At x = 0.1:  error <= (0.1)^3 / 2 = 0.0005.  (True error ≈ 0.00017 — within bound.)

The classic smooth-but-not-analytic example:
    f(x) = exp(-1/x^2)  for x ≠ 0,    f(0) = 0.

One can show (by induction, using exp decay beating every polynomial) that
    f^(n)(0) = 0   for EVERY n = 0, 1, 2, ...
So every Taylor coefficient at 0 is 0, and the Taylor polynomial P_n is the
zero polynomial for all n. Its Taylor 'series' is therefore 0.

But f(x) > 0 for every x ≠ 0. So the Taylor series (= 0) does NOT equal f
near 0. Here the remainder R_n(x) = f(x) - 0 = f(x) does NOT go to 0.

Moral: infinitely differentiable (smooth) is strictly weaker than real-analytic.