Inflection points of bell shaped curves

The standard normal probability density function {f(x) = \exp(-x^2/2)/\sqrt{2\pi}} has inflection points at {x = \pm 1} where {y=\exp(-1/2) /\sqrt{2\pi}\approx 0.24} which is about 60.5% of the maximum of this function. For this, as for other bell-shaped curves, the inflection points are also the points of steepest incline.

Standard Gaussian with inflection points marked in red

This is good to know for drawing an accurate sketch of this function, but in general, the Gaussian curve may be scaled differently, like {A\exp(-B x^2)}, and then the inflection points will be elsewhere. However, their relative height is invariant under scaling: it is always 60.5% of the maximum height of the curve. Since it is the height that we focus on, let us normalize various bell-shaped curves to have maximum {1}:

Gaussian rescaled to height 1

So, the Gaussian curve is inflected at the relative height of {\exp(-1/2)\approx 0.605}. For the Cauchy density {1/(x^2+1)} the inflection is noticeably higher, at {3/4 = 0.75} of the maximum:

Cauchy distribution

Another popular bell shape, hyperbolic secant or simply {2/(e^x+e^{-x})}, is in between with inflection height {1/\sqrt{2}\approx 0.707}. It is slightly unexpected to see an algebraic number arising from this transcendental function.

sech is such a cool name for a function

Can we get inflection height below {1/2}? One candidate is {1/(x^n+1)} with large even {n}, but this does not work: the relative height of inflection is {\displaystyle \frac{n+1}{2n} > \frac{1}{2}}. Shown here for {n=10}:

A smooth trapezoid

However, increasing the power of {x} in the Gaussian curve works: for example, {\exp(-x^4)} has inflection at relative height {\exp(-3/4) \approx 0.47}:

Does this distribution have a name?

More generally, the relative height of inflection for {\exp(-x^n)} is {\exp(1/n - 1)} for even {n}. As {n\to \infty}, this approaches {1/e\approx 0.37}. Can we go lower?

Well, there are compactly supported bump functions which look bell-shaped, for example {\exp(1/(x^2-1))} for {|x|<1}. Normalizing the height to {1} makes it {\exp(x^2/(x^2-1))}. For this function inflection occurs at relative height about {0.255}.

This one actually looks like a bell

Once again, we can replace {2} by an arbitrary positive even integer {n} and get relative inflection height down to {\displaystyle\exp\left(-\left(1+\sqrt{5-4/n}\right)/2\right)}. As {n} increases, this height decreases to {\displaystyle e^{-\varphi}} where {\varphi} is the golden ratio. This is less than {0.2} which is low enough for me today. The smallest {n} for which the height is less than {0.2} is {n=54}: it achieves inflection at {y\approx 0.1999}.

A tough one for derivative-based numerical solvers

In the opposite direction, it is easy to produce bell-shaped curve with high inflection points: either {1/(|x|^p + 1)} or {\exp(-|x|^p)} will do, where {p} is slightly larger than {1}. But these examples are only once differentiable, unlike the infinitely smooth examples above. Aside: as {p\to 1}, the latter function converges to the (rescaled) density of the Laplace distribution and the former to a non-integrable function.

As for the middle between two extremes… I did not find a reasonable bell-shaped curve that inflects at exactly half of its maximal height. An artificial example is {\exp(-|x|^p)} with {p=1/(1-\log 2) \approx 3.26} but this is ugly and only {C^3} smooth.

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