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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.

Sil
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tylerl-uxai
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    $\dfrac{1}{1+e}-\dfrac1e=\dfrac{1}{-e(1+e)} $ so the something is $-e(1+e)$ – Henry Apr 14 '16 at 08:46
  • I hope you like open source software, since this is going straight to the comments section of my new code 8D Henry, do you want me to include an email to see if it gets you future work? Right now it's just a url to your comment. – tylerl-uxai Apr 14 '16 at 08:51
  • Don't let other people take credit for an elegant response. Please answer this and I'll give you the upvote + check mark, @Henry – tylerl-uxai Apr 14 '16 at 08:52
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    You want to find "something" so that 1/something+1/e=1/(1+e) . It is worthwhile to let "something"=x : $$\frac{1}{x}+\frac{1}{e}=\frac{1}{1+e}$$ $$\frac{1}{x}=-\frac{1}{e}+\frac{1}{1+e}=-\frac{1}{e(1+e)}$$ $$x=-e(1+e)$$ "something"=$-e(1+e)$ – JJacquelin Apr 14 '16 at 08:54

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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 $x=-e(1+e)$

Arnaud Mégret
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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 $$ \frac{1}{1 + e} = 1 - e + e^2 - e^3 + e^4 + \cdots = \sum_{n=0}^\infty (-1)^n e^n $$ but I'm not so certain this is what you're looking for.

Simon Rose
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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 : $$\frac{1}{1+e}=\frac{\frac{1}{e} }{1+\frac{1}{e}} = \frac{1}{e}-\left(\frac{1}{e}\right)^2+\left(\frac{1}{e}\right)^3-\left(\frac{1}{e}\right)^4+...$$ $$\frac{1}{1+e}=-\sum_{n=1}^\infty \left(-\frac{1}{e}\right)^n$$

JJacquelin
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