系数被迫等于导数
设 f(x) = sum c_n (x - a)^n 在 a 周围某区间上成立。上一篇我们证明了 f 在那里无穷可微且可逐项微分。反复求导后令 x = a，会消去除一项外的所有项，从而钉死每个系数——这就是系数公式。
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.
有泰勒级数 ≠ 等于它
任何光滑函数都有泰勒级数——只需作 sum f^(n)(a)/n! (x - a)^n。深层问题是该级数是否收敛回 f。在每点的某邻域上等于自身泰勒级数的函数称为实解析的。两者之间的桥梁是泰勒定理：f 在 x 处等于其泰勒级数，当且仅当余项 R_N(x) = f(x) -（第 N 个泰勒多项式）趋于 0。
这确实是不同的条件。令人震惊的事实是：存在光滑函数，其泰勒级数处处收敛，却除中心外处处不等于该函数。这就是光滑但不解析的现象。
标准反例
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.