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$$ \begin{equation} f(x)=\frac{L}{1+e^{-k(x-x_0)}} \end{equation} $$
Fig 1. (img) Logistic Function
$$ \begin{equation} S(x)= \frac{1}{1+e^{-t}} \end{equation} $$
Fig 2. (img) Sigmoid Function
What are the differences between Logistic Function and Sigmoid Function? Is the logistic function a generalized kind of sigmoid function where you could have a higher maximum value?
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
$L$ controls 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 $e^{-k(x-x_0)}$ is 0 so the logistic function has value $L/1$.
$x_0$ controls where on the $x$ axis the growth is centered, 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.
$k$ controls how steep the growth from the minimum to the maximum value is.
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.
I would like to say in opposite way to the answer "the sigmoid function is a special case of the Logistic function" into "The Logisitic function is a special case of the sigmoid function". All S shape curved monotonically increasing fuction being confined a and b are sigmoid functions.
