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Feb 4, 2018 at 13:52 history edited Cedron Dawg CC BY-SA 3.0
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Feb 4, 2018 at 10:23 comment added Guest Or a more accurate 2-param version of that, $\frac{a_0^x-a_1^x}{a_0^x+a_1^x}$ looks pretty good with $a_0\approx\frac{1}{3}$ and $a_1\approx\frac{10}{9}$
Feb 4, 2018 at 8:42 comment added Guest ...or $f_0(x)$ can be pretty much any other odd-symmetrical function; sigmoids seem to work well, like $\frac{a^x-1}{a^x+1}$ (but then the right value for $a$ needs to be found, of course). Here's a plot... as Olli mentions, this probably isn't practical for on-the-fly computation, but I guess it could be useful for building a lookup table.
Feb 4, 2018 at 8:38 comment added Guest Nice update, this makes more sense to me now. The $x-\frac{x^e}{6}$ probably doesn't need to be tested, I just threw it out there because I was trying to figure out the significance of $e$ which seemed to keep popping up while I was playing with this. A better rational expression to test might be something like this: $f_0\left(x\right)=\left|x\right|^a\operatorname{sign}\left(x\right)$ ; $b=f_0'\left(1\right)$ ; $f_1\left(x\right)=f_0\left(x\right)-bx$ ; $c=\frac{1}{f_1\left(1\right)}$ ; $f_2\left(x\right)=f_1\left(x\right)c$ ... now $a$ should be set to about $2\frac{2}{3}$...
Feb 3, 2018 at 20:12 comment added Olli Niemitalo Good work! I fixed that bug ("176 of 120").
Feb 3, 2018 at 20:09 history edited Cedron Dawg CC BY-SA 3.0
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Jan 23, 2018 at 13:50 history edited Cedron Dawg CC BY-SA 3.0
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Jan 23, 2018 at 13:42 history answered Cedron Dawg CC BY-SA 3.0