Flat function ($f^{(n)}(a)=0(n\ge 0)$)

A smooth function is flat at $a$ when every derivative, including order zero, vanishes at $a$. The function $e^{-1/x^2}$ off zero, extended by zero at zero, is flat there but not identically zero near zero. Thus its Taylor series does not determine it. A real analytic function that is flat at a point vanishes on a neighborhood of that point.

In any finite-dimensional Euclidean space, this extension is a smooth function and all its partial derivatives at the origin vanish. Away from zero, repeated differentiation gives finite sums of coordinate polynomials divided by powers of $|x|$, multiplied by $e^{-1/|x{|}^2}$. Every such term is bounded in absolute value by $C|x{|}^{-N}{e}^{-1/|x{|}^2}$ near zero. Since Gaussian decay dominates every power, this bound tends to zero faster than any prescribed positive power of $|x|$. Induction using difference quotients then proves smoothness and zero derivatives of every order at zero.

The origin is nevertheless a strict global minimum, because $F(x)>0$ for $x\ne 0$. All its Taylor series coefficients vanish, so the function is not real analytic at zero. Adding a constant preserves the stationary minimum but changes the zeroth-order value. An expression written only for $x\ne 0$ has no stationary point at the excluded origin until an extension is specified.

Extend this expression by zero at zero. Gaussian decay dominates every power, so the resulting function and its derivative difference quotients vanish to every order at zero. Away from zero, differentiation replaces $p(t)$ by $2t^{3}p(t)-t^2{p}^{\prime }(t)$, another polynomial. Induction therefore gives a smooth function flat at zero. If $p$ is nonzero, its Taylor series is zero while the function is nonzero at every sufficiently small nonzero argument, so it is not real analytic there.