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Experts,

I am a biologist and thus my natural strength is not math, yet I´m quite okay with statistics. Now I am facing the problem that I have to find an unusual (?) mathematical solution for a function with certain properties.

In biology, there is something called the 'Species-Area Relationship' (SAR) which describes the increase of species numbers with an increase in area investigated. Usually, a power functions is best describing this. Now there is the special case of very small areas where vales are close to zero for a number of area sizes.

One can use piecewise regressions such as

y = f1 (x) = c + (x ≤ T) z1 x + (x > T) [(z1 – z2) T + z2 x],

  y = number of species
  x = Area
  T = Breakpoint
  z1, z2 = slopes of breakpoint functions on LHS and RHS of T

Sadly, the breakpoint function has an unrealistic "break" which is quite rough. Hence, I am trying to find a smooth version of this breakpoint function that connects the two breakpoint functions on the left and right handside with a smooth transition function.

Trying a logistic function it becomes already quite close. However, it is unrealistically bound between 0 and 1.

Now I am looking for an integral of this logistic function

z = f4’ (x) = z1 + (z2 - z1) (1 /(1 +exp (–k x)))

Then I only had to add a constanct 'c' to lift it beyond 1...and here is where I have to stop since my brain starts hurting.

Could anyone help me out and try to find the integral of the second function? Or give a hint how to achieve a smooth transition in the BP-function?

many thanks already for all grey matter turned into words,

best,

Jens

PS: I am not sure if this post would better fit to 'crossvalidated' list on stackexchange?

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1 Answer 1

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I'm not quite sure that I understand the question, but if you're simply looking for $f_4(x)$ with $f_4'(x)$ given by your expression, you can find it by using $$ \int \frac{1}{1+e^{-kx}} dx = \int \frac{e^{kx}}{e^{kx}+1}dx = \frac{1}{k} \int \frac{(e^{kx}+1)'}{e^{kx}+1} dx = \frac{1}{k} \ln(e^{kx}+1) + C. $$

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Dear Hans, thank you so much for your answer. I will check it out! –  Jens Jan 25 '12 at 21:09
    
Dear Hans, this is precisely what I meant! Fantastic! do you think you could help me to get this one here integreated as well? f'(x) = z1 + (z2 - z1) / {(1 + exp [- k * (z2 - z 1) * x] * [(z2 - z1) / y0 - 1] I am sorry, my colleague for whom I ask here did not say that the f4(x) was only an example. You would really greatly help us out! –  Jens Jan 26 '12 at 10:00
    
@Jens: wolframalpha.com/input/… –  Hans Lundmark Jan 26 '12 at 11:03
    
Hans, many thanks! This link is extremely helpful for all biologists I guess. I definitely need to understand better the integration and I promise that I am working on it. –  Jens Jan 27 '12 at 9:28
    
@Jens: You're welcome. Note that you can click on the "Show steps" button in Wolfram Alpha to see how it was "thinking". (It's not always helpful, but it's good to know that the option exists.) –  Hans Lundmark Jan 27 '12 at 10:30
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