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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.

For $x\ne 0$, induction giveswhere each $P_n$ is a polynomial. This recurrence follows from the product rule and $d(1/x)/dx=-1/x^2$. Since Gaussian decay dominates every power,These limits also describe the derivatives of the extension at zero. If $f^{(n)}(0)=0$, its next difference quotient is $P_n(1/h)e^{-1/h^2}/h$, which again tends to zero. Induction proves existence and continuity of every derivative, all zero there. This is a flat function, covered more generally by polynomial multiples of a Gaussian flat function.