Logan Wyatt

2022-07-10

How to divide the fraction $1/1+1$
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
$\frac{1}{something}+\frac{1}{e}$

Allison Pena

The problem is that there isn't really a good way to do that. Things that do work with fractions are the following:
1.$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$
2.$\frac{a\cdot b}{c\cdot d}=\frac{a}{c}\cdot \frac{b}{d}$
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
$\frac{1}{1+e}=1-e+{e}^{2}-{e}^{3}+{e}^{4}+\cdots =\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}{e}^{n}$
but I'm not so certain this is what you're looking for.

Patatiniuh

I suppose that the question is less elementary than only finding $\frac{1}{1+e}=\frac{1}{e}-\frac{1}{e\left(1+e\right)}$
May be you want to express $\frac{1}{1+e}$ in terms of $\frac{1}{e}$?
If so, use a geometric series :
$\frac{1}{1+e}=\frac{\frac{1}{e}}{1+\frac{1}{e}}=\frac{1}{e}-{\left(\frac{1}{e}\right)}^{2}+{\left(\frac{1}{e}\right)}^{3}-{\left(\frac{1}{e}\right)}^{4}+...$
$\frac{1}{1+e}=-\sum _{n=1}^{\mathrm{\infty }}{\left(-\frac{1}{e}\right)}^{n}$

Do you have a similar question?