泰勒級數，以及光滑 ≠ 解析

係數被迫等於導數

設 f(x) = sum c_n (x - a)^n 在 a 周圍某區間上成立。上一篇我們證明了 f 在那裡無窮可微且可逐項微分。反覆求導後令 x = a，會消去除一項外的所有項，從而釘死每個係數——這就是係數公式。

f(x) = c_0 + c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 + ...

Set x = a:                     f(a)   = c_0.
Differentiate once, set x = a: f'(a)  = c_1.
Differentiate twice:           f''(a) = 2 * c_2          -> c_2 = f''(a)/2!
Differentiate k times:         f^(k)(a) = k! * c_k       -> c_k = f^(k)(a)/k!

Hence the ONLY possible coefficients are  c_n = f^(n)(a) / n!.

Consequence (uniqueness): if two power series centered at a agree as
functions near a, they have identical coefficients term by term.

冪級數的係數就是其按比例縮放的導數——別無可能。

有泰勒級數 ≠ 等於它

任何光滑函數都有泰勒級數——只需作 sum f^(n)(a)/n! (x - a)^n。深層問題是該級數是否收斂回 f。在每點的某鄰域上等於自身泰勒級數的函數稱為實解析的。兩者之間的橋樑是泰勒定理：f 在 x 處等於其泰勒級數，當且僅當餘項 R_N(x) = f(x) -（第 N 個泰勒多項式）趨於 0。

這確實是不同的條件。令人震驚的事實是：存在光滑函數，其泰勒級數處處收斂，卻除中心外處處不等於該函數。這就是光滑但不解析的現象。

標準反例

Define  f(x) = exp(-1/x^2)  for x != 0,  and  f(0) = 0.

Claim 1: f is smooth on all of R, with f^(n)(0) = 0 for EVERY n.

  Near 0 each derivative has the form f^(n)(x) = P(1/x) * exp(-1/x^2),
  where P is a polynomial.  As x -> 0, the substitution t = 1/x^2 -> +inf
  makes  P(1/x) exp(-1/x^2)  behave like  (poly in sqrt(t)) * e^{-t} -> 0,
  because e^{-t} crushes any power of t.  An induction using the
  difference quotient at 0 then shows f^(n)(0) = 0 for all n.

Claim 2: the Maclaurin (Taylor at 0) series of f is
  sum_{n>=0} f^(n)(0)/n! x^n = sum 0 * x^n = 0  (the zero series).

The zero series converges everywhere, with sum 0.
But f(x) = exp(-1/x^2) > 0 for every x != 0.

So f equals its Taylor series ONLY at x = 0, and nowhere else.
f is smooth (C-infinity) but NOT real-analytic at 0.

光滑，泰勒級數處處收斂，卻沒捕捉到函數的任何資訊。