Other functions for this curve?

by Ξ

I got a call this afternoon from Marc, a friend of mine from college in Minnesota who now lives back East and runs his own business. He was trying to find a model for a collection of functions, but couldn’t figure out what kind of function it would be.

The basic scenario was that the function should start at (0,0), increase rapidly to a point [say, (10,10)], and then slowly decrease. The x-axis could be a practical asymptote, although it didn’t really matter since this would only be looked at in finite time.

My first thought was surge function (something of the form $Axe^{-Bx}$ ), and sure enough that works. But I was on Homework Patrol, so I handed it off to TwoPi, and he came up with $xe^{1-0.1x}$ and $\frac{80x}{(x+6)(0.01x^2+4)}$ . Walphaing these shows that both work well:

This seemed to help. So here’s what I’m wondering — are there any other simple functions that fit the bill? Neither function is too complicated, but it would be fun to be able to share other examples of functions that rise quickly, but then that taper off after a while. [I think the tapering should be more gradual compared to the climb, though I believe that can be controlled with more constants.

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This entry was posted on December 13, 2009 at 8:12 pm and is filed under Problems. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

8 Responses to “Other functions for this curve?”

L Zoel Says:
December 13, 2009 at 9:49 pm | Reply
My initial thought was something along the lines of 150*x/(x+4)^2, but since you already basically have that, how about:

10*sin((x/50-3.14^.5)^2)
jd2718 Says:
December 13, 2009 at 9:53 pm | Reply
I’d like to just add two parts – one to dominate near the origin (linear, positive slope), one in the tail (reciprocal, most likely)

But all my tricks make the tail explode, and make the origin explode.

Could we take the reciprocal of what I am thinking of? Hmm.

The reciprocal of your graph would have a vertical asymptote at x=0, come down to a minimum at your peak, and monotonically increase after that. So say f(x) = kx + b/x

1/f = x/(kx^2 + b)

And I am on the road to two pi, who has already done it better.

Jonathan
Jaime Says:
December 14, 2009 at 4:54 am | Reply
There are several classical statistical distributions that have such a behavior:

http://en.wikipedia.org/wiki/Rayleigh_distribution
http://en.wikipedia.org/wiki/Maxwell_distribution
http://en.wikipedia.org/wiki/Chi_distribution
toss255 Says:
December 14, 2009 at 7:26 am | Reply
Let me add the Weibull distribution which lets you play with two parameters.
Dan M Says:
December 14, 2009 at 2:03 pm | Reply
Is this the first real recorded use of “Walphaing”? Be sure to submit to the OED.
Ξ Says:
December 14, 2009 at 3:51 pm | Reply
Thanks for all the curve suggestions! I’ll be passing them along.

Dan, sadly I’m nowhere near first to use Walphaing — the term dates back nearly to the launch of Wolfram Alpha (I myself first heard of it from Mike Croucher, though as damon365 points out, if there are any updates it would need to be changing to wbetaing and that just doesn’t have quite the same right to it.)
ric Says:
December 14, 2009 at 6:51 pm | Reply
I vote for the beta distribution :
http://www.aiaccess.net/French/Glossaires/GlosMod/f_gm_beta_distri.htm

(sorry, it’s in french, but the applet might help.)
Milo Schield Says:
December 27, 2009 at 12:55 pm | Reply
I think that the log-normal would fit your constraints. It certainly fits a number of social distributions having those constraints.

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