03/11/2011, 12:35 AM
Surely, though, {0.25} should be weighted 3/4s towards the arithmetic mean, and 1/4 towards the geometric mean. Ah but is the weighting carried out arithmetically or geometrically? Apply a 3/4 arithmetic : 1/4 geometric weighting there too! And take the limiting case again.
In other words, for a {0.25} b, with a<b,
m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)
Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)
This seems like a good generalisation to real values between 0 and 1!
In other words, for a {0.25} b, with a<b,
m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)
Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)
This seems like a good generalisation to real values between 0 and 1!
(03/10/2011, 11:42 PM)JmsNxn Wrote: What would happen if we created this:
x {0} y = x + y
x {0.5} y = x @ y
x {1} y = x * y
x {2} y = x ^ y
And then {0.25} will be the same arithmetic-geometric algorithm of {0} and {0.5}; {0.75} will be the arith-geo-algo of {1} and {0.5}, so on and so forth.
We could then solve for x {1.5} n, n E N, since:
x {1.5} 2 = x {0.5} x
Perhaps Taylor series will be derivable giving us complex arguments.
It'd also be very interesting to see what happens with logs, i.e:
log(x {1.5} 2) = ? since normal operators undergo a transformation I wonder if something happens for these.
