This has to do with re-calculating the sigmoid function in ai. It isn't really important, but the simplest way to put it is I need a math guru to help my monkey brain do this:
$$\frac{1}{1+e}$$
to like
$$\frac{1}{something} + \frac{1}{e}$$
Please help me remember my math from high-school if this was ever taught to us.
On
solve $\frac{1}{1+e}=\frac{1}{x}+\frac{1}{e}$
so $\frac{1}{x}=\frac{e-1-e}{(1+e)e}=-\frac{1}{(1+e)e}$ so $x=-e(1+e)$
On
I suppose that the question is less elementary than only finding $\frac{1}{1+e}=\frac{1}{e}-\frac{1}{e(1+e)}$
May be you want to express $\frac{1}{1+e}$ in terms of $\frac{1}{e}$ ?
If so, use a geometric series : $$\frac{1}{1+e}=\frac{\frac{1}{e} }{1+\frac{1}{e}} = \frac{1}{e}-\left(\frac{1}{e}\right)^2+\left(\frac{1}{e}\right)^3-\left(\frac{1}{e}\right)^4+...$$ $$\frac{1}{1+e}=-\sum_{n=1}^\infty \left(-\frac{1}{e}\right)^n$$
The problem is that there isn't really a good way to do that. Things that do work with fractions are the following:
but there isn't a way to separate when there is a sum in the denominator.
I suppose perhaps one thing you could do, although this isn't likely what you have in mind, is the following: if $e$ is small in your description (that is, if $|e| < 1$) then there is a geometric series expansion $$ \frac{1}{1 + e} = 1 - e + e^2 - e^3 + e^4 + \cdots = \sum_{n=0}^\infty (-1)^n e^n $$ but I'm not so certain this is what you're looking for.