$L$ 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.
If you minimize the mean squared error, then it's different from logistic regression. Logistic regression is normally associated with the cross entropy loss, here is an introduction page from the scikit-learn library.
(I'll assume multilayer perceptrons are the same thing called neural networks.)
If you used the cross entropy loss (with regularization) for a single-layer neural network, then it's going to be the same model (log-linear model) as logistic regression.
If you use a multi-layer network instead, it can be thought of as logistic regression with parametric nonlinear basis functions.
However, in multilayer perceptrons, the sigmoid activation function is
used to return a probability, not an on off signal in contrast to
logistic regression and a single-layer perceptron.
The output of both logistic regression and neural networks with sigmoid activation function can be interpreted as probabilities. As the cross entropy loss is actually the negative log likelihood defined through the Bernoulli distribution.
Best Answer
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.