The logistic function has the differential equation:

dy / dt = ky(1 - (y / L)) Solving this differential equation by separation of variables and integration yields the Logistic Equation:

f(t) = L / [1 + b * e^(-k * t)]

in which b = ± e^(-C), where "C" is a constant of integration.

The maximum height of the "S" shaped curve is "L", "k" controls the steepness and "b" moves the curve horizontally. The "S" shaped curve models some biological phenomena well such as the spread of disease or the maximum carrying capacity of population. It rises to a maximum height "L" and stays there.

However when populations die off quickly the logistic curve does not model that very well because it stays at a constant maximum height "L" forever.

I have noticed that the Witch of Agnesi curve does a good job of illustrating decay from a maximum, which is governed by the function:

g(t) = (8 * a^3) / (t^2 + 4a^2), where "a" is the radius of the circle that forms this curve which depends upon the size of the circle which is variable.

if a = 0.5, and I combine these curves:

h(t) = f(t) * g(t) I get a curve preserves some of the steepness of the Logistic function and drops off like the Witch of Agnesi, but it is not as steep as the logistic function and its maximum is smaller than the logistic function, although it looks a "S" shaped dropping off from a maximum

Combined Graph

Red Curve: Logistic Function: f(t)

Black Curve: Witch of Agnesi: g(t)

Blue Curve: h(t) = f(t) * g(t)

Desired curve: Logistic curve (red) dropping off in the "green" region like the Witch of Agnesi. (Sorry if the green curve is not drawn to scale. I meant to make it look like the decay of the black curve.) The logistic curve models the steepness "k" better than the Witch of Agnesi curve, but the logistic curve does not drop off.

What kind of mathematical skills are required to form such a function? Do there exist any functions like this?

Thanks.

  • 1
    Your mention of the Witch of Maria Agnesi compelled me to upvote, I could not help myself. – James Phillips Oct 22 at 14:22
  • I would propose, in all seriousness, that the very first skill should be to learn how to state your objectives quantitatively. That is because the qualitative phrasing of this question makes it vague and subject to many different, but equally valid, answers. – whuber Oct 23 at 18:38
up vote 2 down vote accepted

What do you think of this function, more general than the logistic function : $$f(t)=\frac{L}{1+b \:e^{-k\,t}+c\: e^{h\,t}}$$ The new parameters allow to fit to what you are expecting. A few examples below :

enter image description here

  • 1
    I love it! Thanks! – xyz123 Oct 22 at 13:54
  • 2
    I am interested in adding this equation to the hundreds of known, named equations on my online curve and surface fitting web site zunzun.com. Do you know of a formal name used for this equation? If not, I can refer to it on the site as "JJacquelin's Generalized Logistic" or something similar. – James Phillips Oct 22 at 14:37
  • @James Phillips : "Generalized Logistic" is sufficient. I am not calling for royalties. Cheers ! – JJacquelin Oct 22 at 14:58
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    There is a method of regression without initial guess of parameters and without iterative calculus to compute approximates of $L$, $b$ , $k$ , $c$ , and $h$. The general principle is introduced in the paper : fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . The direct calculus is rather simple, but this function is not yet treated in the referenced paper. A new updated edition should be necessary. – JJacquelin Oct 22 at 15:15
  • I cannot read French, but you use the form of so many statistical words that I can almost read it. It seems like you write rather clearly, and I think if I thought about this for a bit, it would be understandable. Thanks. – xyz123 Oct 22 at 15:47

This is not another answer but a comment too long to be edited in the comments section.

Since the proposed function is accepted, a straightforward calculus of regression can be proposed according to the method introduced in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales $$f(t)=\frac{L}{1+b e^{-kt}+c e^{hx}}$$

$$y(t)=\frac{1}{f(t)}=\frac{1}{L}+\frac{b}{L}e^{-kt}+\frac{c}{L}e^{ht}$$ $$y(x)=A+Be^{px}+Ce^{qx}$$ with $\quad x=t\quad;\quad A=\frac{1}{L}\quad;\quad B=\frac{b}{L}\quad;\quad C=\frac{c}{L}\quad;\quad p=-k\quad;\quad q=h$

Double exponential regression : See pages 71-72 of the paper referenced above.

Note that there is a constant parameter in addition of the two exponentials. As a consequence the 4X4 matrix becomes a 5X5 matrix as shown below.

Data : $(t_1\:,\:f_1)\:,\:(t_2\:,\:f_2)\:,\:...\:,\:(t_k\:,\:f_k)\:,\:...\:,\:(t_n\:,\:f_n)$

enter image description here

NUMERICAL EXAMPLE :

enter image description here

enter image description here

Remember : This method provides an approximative result without specified criteria of fitting. If a particular criteria is specified and imperative, then iterative calculus must be used, thanks to other methods and of course will leads to a better fitting depending on the criteria considered.

  • 1
    I understand - when I restrict h and k to be positive only, I have similar results to yours. L = 1.0020686869786672E+00, b = 2.6949842058468221E+00, k = 2.1917600868860445E+00, c = 5.0689130096851585E-03, h = 7.9965918633555377E-01 – James Phillips Oct 24 at 20:18
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    @James Phillips. Nice, I feel reassured. On the drawing the two curves (from your result and from mine) are undistinguishable. From numerical comparison, your fitting is slightly better than mine. This is what was expected since iterative calculus logically tends to a better limit than a direct calculus without iteration. I delete .the wrong curve. – JJacquelin Oct 24 at 20:45

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