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I am using a logistic function to model observed response values (growth) as function of an input value:

$$\text{growth} = \displaystyle{\frac{A}{1+e^{\frac{k-\text{input}}{s}}}}$$

This functional form has lower asymptote of growth=$0$, but at input=$0$ it can assume a positive growth value.

I would like to use a similar sigmoid function with the constraint of growth=$0$ at input=$0$.

How can I modify the function?

I found this but it did not solve the problem: Logistic function passing through two points AND the origin?

mechanodroid
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Simone Bianchi
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  • This is not possible. To have $\mathrm{growth} = 0$, you need the denominator to be $+\infty$ (or $A = 0$ but in that case, $\mathrm{growth}$ is identically zero). This happens when $\mathrm{input} \to -\infty$ (if $s > 0$). – pitchounet Sep 19 '17 at 08:43
  • thanks. is there any different sigmoid function that can be constrained to pass through the origin? – Simone Bianchi Sep 19 '17 at 08:57
  • If the sigmoid function is given by:

    $$ s(x) = \frac{1}{1+\exp(-x)} $$

    then:

    $$ \tilde{s}(x) = s(x) - \frac{1}{2} $$

    goes through the origin at $x=0$. $\tilde{s}$ is a translated sigmoid function.

     – pitchounet Sep 19 '17 at 09:15

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