# What is the right way to approximate $e^{-1/x^2}$ by polynomials?

It's well known that $f(x) = e^{-1/x^2}$ (with zero added) is a smooth function that's not analytic at $x=0$, because every derivative at zero is zero, and so all of its Taylor polynomials are zero. For the sake of simplicity fix the center at zero for the rest of the question.

This function isn't all that pathological and it seems like there should still be a principled way to approximate it by polynomials by using some other natural data about $f$.

More concretely, what is a method for approximating a smooth function $f(x)$ by polynomials that has the following properties:

• Optimal by some natural criterion (analogous to how the degree-$k$ Taylor polynomial is optimal among degree $k$ polynomials on a sufficiently small interval)
• Graded, i.e. there is a parameter $n$ so that larger values of $n$ use higher-degree polynomials and improve the approximation.
• Efficiently computable, i.e., there is a $\text{poly}(n)$-time algorithm which constructs the polynomial representation from the input parameter of $n$
• Nontriviality for $e^{-1/x^2}$
• Approximate over which interval? Weierstrass' approximation theorem grants that any continuous function, over a compact interval, can be approximated with a sequence of polynomials with a uniform error converging to zero. Only continuity is requested, not analyticity. – Jack D'Aurizio Oct 16 '16 at 21:46
• Additionally, the shifted Legendre polynomials provide a orthogonal base of $L^2(0,1)$ made by polynomials, so a sequence of polynomials approximating a $C^\infty$ function $f(x)$ over $(0,1)$ is also provided by $$\sum_{n=0}^{N}c_n P_n(2x-1)$$ with $c_n = (2n+1)\int_{0}^{1}f(x)P_n(2x-1)\,dx.$ – Jack D'Aurizio Oct 16 '16 at 21:49
• What about an expansion at $x \to \infty$? That would be well-defined and give the result you want. – valerio Oct 16 '16 at 21:53
• An interesting (but probably easy) question follows from here. Consider taking approximations of $e^{-1/x^2}$ on $[-\delta,\delta]$ for $\delta>0$. One can do this in various ways, e.g. using Bernstein polynomials or projection onto a subspace of $L^2$ consisting of polynomials. For each $\delta>0$ these are not trivial. (For example, the orthogonal projection onto the constants is $f(x)=\frac{1}{2\delta} \int_{-\delta}^\delta e^{-1/y^2} dy$.) But do they vanish as $\delta \to 0^+$? – Ian Oct 16 '16 at 21:54
• (Cont.) I think they do, indeed I think if you do this to any $C^k$ function you will recover the degree $k$ Maclaurin polynomial. – Ian Oct 16 '16 at 21:58

$$f(x) = e^{-\frac1{x^2}}$$ is in fact quite pathological at $x=0$: If you look at the values of $f(it)$ (for real $t$) near $it=0$ you see not only an approach to infinity but an essential singularity. By restricting your vision to the real line you can put on blinders to this singularity, but it is still there and it still affects any attempt to expand in series that demand analyticity.

If you insist on a series of approximations that are good near $x=0$, then consider $$f_n(x) = \sum_{k=0}^n\frac1{k!}\left.\frac{d^k\left(e^{-\frac1{y^2}}\right)}{dx}\right|_{y=x}\left(x-\frac1n\right)^k$$ This meets all your criteria.

• Does that approximation have a name? It's like the $n$-th Taylor polynomial around $1/n$, but with the derivatives evaluated in $x$ instead of $1/n$... Are you sure you want to evaluate them in $x$? – punctured dusk Oct 16 '16 at 22:15

Notice that as $x$ approaches $0$ along the imaginary axis, this function approaches $\infty$, so its bad behavior is close to home even if not on the real axis.

For every polynomial function $p(x)$ that is not identically $0$, there exists $\varepsilon>0$ such that for every $x$ within the neighborhood $(-\varepsilon,+\varepsilon)$ with the possible exception of $x=0$, the value of $p(x)$ is farther from $f(x)$ than $0$ is from $f(x)$.

Thus for any polynomial approximation to be better than the identically $0$ function, you'd have to have some other criterion of what is "better".

I think that the polynomial of best approximation fits your requirements.

If you fix an interval and a degree $n$, the polynomial you get is the polynomial of degree $n$ that minimizes $$\{\|f-p\|_\infty:\ p \ \text{ has degree } n \} .$$

• What do you get when you approximate this function by Chebyshev polynomials? – Mark Fischler Oct 16 '16 at 21:48
• Chebyshev polynomial interpolants are not minimax polynomial interpolants. They are roughly speaking the closest thing to the minimax interpolant that is easy to calculate. – Ian Oct 16 '16 at 21:54
• No, but they are very useful and computationally convenient. – marty cohen Oct 16 '16 at 21:56
• What do you mean by $\|f-p\|$? In some contexts that would mean $$\sup\{|f(x)-p(x)| : x \in \text{some specified set}\}$$ (but which specified set??) and in some contexts it means $$\int\limits_{\text{some set}} |f(x)-p(x)|\,dx$$ (but again, which set do you integrate over?), and in some contexts it means $$\int\limits_{\text{some set}} |f(x)-p(x)|^2\,dx,$$ etc. $\qquad$ – Michael Hardy Oct 16 '16 at 21:58
• It's the uniform norm. The set, in any case, is obvious from context since one is prescribing an interval. – Martin Argerami Oct 16 '16 at 22:01