are there any polynomial-exponential, bell-shaped functions?

Question

I am looking for a polynomial-exponential, bell-shaped function under the restrictions below.

Definition: By polynomial-exponential function I mean something of the sort $g(x)^{h(x)}$ where $g(\cdot)$ and $h(\cdot)$ are polynomial functions with integer coefficients.

Essentially, in terms of shape, something like a Radial Basis Function, e.g. the Gaussian $\exp(-x^2)$ or the inverse quadratic $(1+x^2)^{-1}$ RBFs, going to 0 at $-\infty$ and $+\infty$, and peaking at x=0.

RESTRICTIONS:

• no divisions

• no non-polynomial-exponential functions (exp, log, absolute value, things requiring evaluation of logical statements,...)

• only integer coefficients (if any)

• decays to 0: $\lim_{x\to-\infty} f(x) = \lim_{x\to\infty} f(x) = 0$

• peaks at $x=0$ (the peak value $f(0)$ is not important)

• is symmetric around $x=0$

RELAXATION: (last-resort solution)

• In a last-resort case, it may be possible to work with a restricted domain, say $x \in (-10,10)$, as long as the tails are reasonably close to zero at the extremes of the domain. This is not ideal, but I may just have to fall back to it if nothing more general is available.

It does seem like a lot of restrictions but I was wondering if maybe there's something out there that I've overlooked.

Many thanks!

Answer 1

From wiki spline, the 'Irwin-Hall distribution'

$y=\begin{cases} \begin{align} &\frac{1}{4}(x+2)^3 \quad&-2\leq x\leq-1\\ &\frac{1}{4}(3\lvert x\rvert^3-6x^2+4) \quad&-1\leq x\leq1\\ &\frac{1}{4}(2-x)^3 \quad&1\leq x\leq2\\ &\,0 \quad & \text{otherwise} \end{align} \end{cases}$

This seems to check all your boxes. Polynomials, it's zero outside $(-2,2)$, peaks at $x=0$, and has value $1$ there, and is symmetric about $x=0$. Thanks @N74.

Although if you count $-2\leq x\leq-1$ as evaluating a logical statement, and you don't want that, then I stand my original comment that $f\equiv0$ is the only polynomial whose limit at $\pm\infty$ is $0$.

Answer 2

@N74 @snulty Unfortunately, can't use inequality checking, nor absolute values. Although this comes very close and will definitely be useful in a different scenario. Thank you!

What we can do is to use polynomials with degree dependent on $x$, as @CountIblis suggested. So, something like $$f(x) = (11+x)^{27-x}$$ seems to have roughly the right shape (slightly asymmetric, graph below) and the constants can be tweaked I guess.

Q: Is this a known class of functions? Perhaps they're used in some scientific field...?

It can also be written as $$f(x) = e^{(27-x)\ln(11+x)}$$ which reveals the exponential shape. Importantly, however, the first one is computable in the case of interest (it's enough that it works for $x \in \mathbb{Z}$ and $11+x>0$) while the second is not.