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Taylor Series, and Smooth ≠ Analytic

Run the previous guide backward: if a function IS a power series, its coefficients must be its derivatives at the center. That forces the Taylor series. But having a Taylor series is not the same as equaling it — the famous counterexample exp(-1/x^2) is smooth yet not analytic.

Coefficients are forced to be derivatives

Suppose f(x) = sum c_n (x - a)^n on some interval around a. We proved last time that f is infinitely differentiable there and that we may differentiate term by term. Repeatedly differentiating and then setting x = a kills every term but one and pins down each coefficient — this is the coefficient formula.

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.
The coefficients of a power series are its scaled derivatives — nothing else works.

Having a Taylor series ≠ equaling it

Any smooth function has a Taylor series — just form sum f^(n)(a)/n! (x - a)^n. The deep question is whether that series converges back to f. A function that equals its own Taylor series on a neighborhood of every point is called real-analytic. The bridge between the two is Taylor's theorem: f equals its Taylor series at x precisely when the remainder R_N(x) = f(x) - (N-th Taylor polynomial) tends to 0.

These are genuinely different conditions. The shocking fact is that there exist smooth functions whose Taylor series converges everywhere yet equals the function nowhere except at the center. This is the phenomenon smooth but not analytic.

The canonical counterexample

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.
Smooth, with an everywhere-convergent Taylor series that captures none of the function.