Fig 1. Logistic Function
Fig 2. Sigmoid Function
is it more like generalized kind of sigmoid function where you could have a higher maximum value?
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$\begingroup$ $S$ is an increasing function in $t$, $f$ is increasing for values of $x \leq x_0$, decreasing for values of $x \geq x_0$ $\endgroup$– micMar 30, 2016 at 7:25
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2 Answers
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Yes, the sigmoid function is a special case of the Logistic function when $L=1$, $k=1$, $x_0 =0$.
If you play around with the parameters (Wolfram Alpha), you will see that
$L$ is the maximum value the function can take. $e^{-k(x-x_0)}$ is always greater or equal than 0, so the maximum point is achieved when it it 0, and is at $L/1$.
$x_0$ controls where on the $x$ axis the growth should the, because if you put $x_0$ in the function, $x_0 - x_0$ cancel out and $e^0 = 1$, so you end up with $f(x_0) = L/2$, the midpoint of the growth.
the parameter $k$ controls how steep the change from the minimum to the maximum value is.
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The logistic function is: $$ f(x) = \frac{K}{1+Ce^{-rx}} $$ where $C$ is the constant from integration, $r$ is the proportionality constant, and $K$ is the threshold limit.
Assuming the limits are between $0$ and $1$, we get $\frac{1}{1+e^{-x}}$ which is the sigmoid function.
