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The Taylor formula with integral remainder isFor $n=0$, this is the fundamental theorem of calculus. Integration by parts givesRearranging and applying mathematical induction proves the displayed expansion. The oriented integral also covers $x<a$.

For $x\ne 0$, the chain rule and product rule giveThusBecause Gaussian decay dominates every power, $p(1/x)e^{-1/x^2}/x\to 0$ as $x\to 0$. Hence $f^{\prime }(0)=0$, and the derivative formula extends continuously to zero. Repeating the argument replaces $p$ by another polynomial at each step: $p_{j+1}(t)=2t^3{p}_j(t)-t^2{p}_j^{\prime }(t)$. All derivative difference quotients at zero tend to zero, so the polynomial multiples of a Gaussian flat function are smooth functions withThe Taylor polynomial about zero is therefore identically zero for every degree, andA nonzero polynomial has only finitely many roots of a polynomial, so $p(1/x)\ne 0$ 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.

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.