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I have an idea about producing a genetic algorithm. But it involves using a relationship which believe could be defined as a horizontal Asymptote. I have found an image that depicts the type of relationship I will attempt to apply. However finding out what it is called and what the standard equation is that defines it is escaping me.

Decay relationship of interest

a not so busy cat

I am interested in what the standard format for the green dashed graph would be. It was a long time since I took any calculus. So any help would be appreciated. AEA

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    $\begingroup$ Is this what you are looking for? en.wikipedia.org/wiki/… $\endgroup$ – sashkello Nov 27 '13 at 0:21
  • $\begingroup$ The red decay is an exponential function with negative exponent. The green is the complement of that (they add to 1). That is, the green curve will be of the form $f(t) = 1-\exp(-\alpha t)$ as a function of $t$, for some constant $\alpha>0$, which is related to the half-life (I think $\alpha = \ln 2/t_{0.5}$). There are many other functions with horizontal asymptotes! $\endgroup$ – Glen_b Nov 27 '13 at 0:24
  • $\begingroup$ Thanks Glen, I thought of it being an exponential type relationship but standard exponential relationships start low and increase exponentially. Perfect answer. If you want to post it as an answer it will get the required tick. :) $\endgroup$ – AEA Nov 27 '13 at 0:34
  • $\begingroup$ @Glen_b is of course correct. What seems less conventional but is also valid is to think of the curve as $1 - 2^{-t/\tau}$, i.e. to write the equation in terms of a power of 2 with the nice consequence that it can parameterised directly with the half-life $\tau$ ($= t_{0.5}$ in Glen_b's notation). $\endgroup$ – Nick Cox Nov 27 '13 at 0:54
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The red decay is an exponential function with negative exponent (as you might discern from sashkello's link, which I highly recommend).

The green is the complement of that (they add to 1).

That is, the green curve will be of the form $f(t)=1−\exp(−\alpha t)$ as a function of $t$, for some constant $α>0$, which is related to the half-life ($α=\ln(2)/t_{1/2}$).

As Nick Cox rightly points out, if you're actually interested in that as a decay function (or anything else where the actual half-life may be of interest), it can be useful to work in terms of powers of 2 rather than powers of $e$, since the half-life is then a natural parameter of the function.

There are, of course, many other monotonic increasing functions with horizontal asymptotes. Indeed continuous probability distributions on $(0,\infty)$ are a great source of such functions, if you need to consider other such functions.

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  • $\begingroup$ Ahh i cant upvote this answer 15 rep required. Once i get 15 i will revist this answer and upvote :) $\endgroup$ – AEA Nov 27 '13 at 1:31
  • $\begingroup$ I'm sure the rep you need will come along soon. Thanks for accepting the answer. $\endgroup$ – Glen_b Nov 27 '13 at 2:36

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