= Solution
The <Taylor formula with integral remainder> is
$$
\boxed{f(x)=\sum_{j=0}^n\frac{f^{(j)}(a)}{j!}(x-a)^j+R_n(f,a,x),\qquad
R_n(f,a,x)=\frac1{n!}\int_a^x(x-t)^n f^{(n+1)}(t)\,dt.}
$$
For $n=0$, this is the <fundamental theorem of calculus>. <Integration by parts> gives
$$
R_n(f,a,x)=-\frac{f^{(n)}(a)}{n!}(x-a)^n+R_{n-1}(f,a,x).
$$
Rearranging and applying <mathematical induction> proves the displayed expansion. The oriented integral also covers $x<a$.
For $x\ne0$, the <chain rule> and <product rule> give
$$
f'(x)=\left[2x^{-3}p(1/x)-x^{-2}p'(1/x)\right]e^{-1/x^2}.
$$
Thus
$$
\boxed{q(t)=2t^3p(t)-t^2p'(t).}
$$
Because <Gaussian decay dominates every power>, $p(1/x)e^{-1/x^2}/x\to0$ as $x\to0$. Hence $f'(0)=0$, and the derivative formula extends continuously to zero. Repeating the argument replaces $p$ by another polynomial at each step: $p_{j+1}(t)=2t^3p_j(t)-t^2p_j'(t)$. All derivative difference quotients at zero tend to zero, so the <polynomial multiples of a Gaussian flat function> are <smooth functions> with
$$
f^{(j)}(0)=0\qquad(j\ge0).
$$
The <Taylor polynomial> about zero is therefore identically zero for every degree, and
$$
\boxed{R_n(f,0,x)=f(x)\quad\text{for every }n.}
$$
A nonzero polynomial has only finitely many <roots of a polynomial>, so $p(1/x)\ne0$ for every sufficiently small nonzero $x$. At each such fixed $x$, this remainder never tends to zero. The <flat function> is therefore smooth but not <real analytic> at zero.
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