(07/18/2019, 05:13 PM)Syzithryx Wrote: As for intermediate / fractional ops, I have thought of various approaches. My initial concept relied on using averages. For instance, with 1.5ation, the arithmetic average of (a+b) and (ab) obviously in some sense "favors" addition, because it's summing them before dividing by two; but the geometric average "favors" multiplication because it multiplies them before taking the square root; so properly we want to go right between those two averages.That kind of mean is the arithmetic–geometric mean.
So, define two functions: f(a,b) = (a+b)/2, and g(a,b) = ²√(ab). Then make an iterative sequence: c₀ = a+b, d₀ = ab; c_i = f(c_(i-1), d_(i-1)); d_i = g(c_(i-1), d_(i-1)). Then c_i and d_i should *** closer together with each iteration and converge on a single value, then defined to be the 1.5ation of a and b; but unfortunately I don't know enough math to actually *prove* that or find any kind of closed expression for what that actually is.
There was a thread about using it to do 1.5ation here.
Quote:Another way of doing 1.5ation, which probably wouldn't end up with the same values (but of course, who knows), is something I came up with yesterday looking at graphs of y=x+a versus y=xa - they are both lines, which meet at some point except when a=1; so, one could define a new line which bisects the angle they form, or which is exactly intermediate between them in the a=1 case, and that would be 1.5ation. I haven't really had a chance to analyze the properties of this variant yet though - and I really don't know how to extend this method to other pairs of ops.Then that would equal b*tan(π/8+arctan(a)/2)+a+a/(a-1)-a/(a-1)*tan(π/8+arctan(a)/2), using radians.
That definition of 1.5ation is not the same as the other one.
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ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\

