Is $|x|+e^{-|x|}$ known?

Question

Consider the function $$ f(x)=|x|+e^{-|x|}. $$ Interestingly, $f\in C^2(\mathbb{R})$, and its plot is

This function emerges as a particular case of the discussion here, for example. Does this function appear elsewhere in mathematics? Are these types of curves studied in a specific field?

For instance, when restricted to $x\in (-1,1)$, the plot below relates it to the "nearest" parabola and catenary passing through the points

where the parabola $p$ and catenary $c$ functions are given by $$ \begin{align} p(x)&=1+e^{-1}x^2\\ c(x)&=\cosh(x\cosh^{-1}(1+e^{-1})) \end{align} $$ Does $f$ appear in some applications?

This question seems too vague. And sure $f$ can be thought as a $C^2$ approximation of absolute value — Liding Yao, Mar 16 '25 at 22:45
@LidingYao This is essentially a reference request. I'm wondering if this type of curve/approximation appears elsewhere in mathematics, under a different motivation/context than the linked question. — sam wolfe, Mar 17 '25 at 14:18
You may add it as a new answer to this question Approximate $|x|$ with a smooth function — Amir, Mar 18 '25 at 19:03
The function $y=|x|+e^{-|x|}$ is related to the LambertW function: $$|x|=y+W(-e^{-y})$$ by inversion on real branches. — JJacquelin, Mar 19 '25 at 11:03
Very nice and interesting approximations ! You made my day. (+1) and cheers :-) — Claude Leibovici, Mar 20 '25 at 14:28
An interesting (to me at least) thing I find is that $|x|+e^{-|x|}=|x| \star e^{-|x|}$ (convolution). I am omitting some likely scaling factor. — Srini, Mar 22 '25 at 20:57
@Srini Are there other examples of functions such as $$f(x)+g(x)=f(x)\circledast g(x)$$ ??? — Joako, Mar 24 '25 at 03:11

Claude Leibovici · Answer 1 · 2025-03-22T09:06:04.090

Too long for a comment.

If we consider the range $(0\leq x\leq 1)$, these functions could be used for an approximation of the solution of $$f(x)=k\qquad \text{for} \qquad 1 \leq k \leq 1+\frac 1e$$ without using Lambert function which has been already mentioned in comments by @JJacquelin.

In terms of infinite norm, they are equivalent $$\Phi_p=\int_0^1 \Big(f(x)-p(x)\Big)^2\, dx=1.22\times 10^{-4}$$ $$\Phi_c=\int_0^1 \Big(f(x)-c(x)\Big)^2\, dx=2.06\times 10^{-4}$$

In the same spirit, we could use an $[n,n]$ Padé approximant built around $x=0$ and get $$f(x)\sim \frac {1+\frac{1}{3}x+\frac{19}{36} x^2 } {1+\frac{1}{3}x+\frac{1}{36} x^2}=g(x)$$ whose error is $\frac{x^5}{1080}$ $$\Phi_g=\int_0^1 \Big(f(x)-g(x)\Big)^2\, dx=2.83\times 10^{-8}$$

which would give $$k+W\left(-e^{-k}\right)\sim \frac{6 \left(1-k-3 \sqrt{2(k-1)}\right)}{k-19}$$ giving $$\text{lhs - rhs}=-\frac{1}{270} (k-1)^2+O\left((k-1)^{5/2}\right)$$

Edit

Amazing would be to use the next Padé approximant $$f(x)\sim \frac {1+\frac{1}{4}x+\frac{1}{2}x^2-\frac{11}{240}x^3 } {1+\frac{1}{4}x-\frac{1}{240}x^3 }=h(x)$$ whose error is $\frac{x^7}{40320}$ $$\Phi_h=\int_0^1 \Big(f(x)-h(x)\Big)^2\, dx=1.31\times 10^{-11}$$ which would give $$k+W\left(-e^{-k}\right)\sim \frac{4 \sqrt{5} \sqrt{k^2-12 k+91} \sin(t)-40}{k-11}$$ where $$t=\frac 13 \sin^{-1}\Bigg(\frac{-3 k^3+99 k^2-789 k+2293 } {\sqrt 5 \left(k^2-12 k+91\right)^{3/2} } \Bigg)$$ giving $$\text{lhs - rhs}=\frac{(k-1)^3}{5040}+O\left((k-1)^{7/2}\right)$$

Is $|x|+e^{-|x|}$ known?

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