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Pade(2,1)

Have you taken a look at the Pade(2,1) approximant? I compute it as $\frac{1}{4} \frac{(r-1)(r+5)}{2r+1}$. Calculations show it beats the quadratic approximant on the interval $[1,4]$ at the endpoints of which the two approximants agree. (The two are nearly identical on that interval, with a max difference of $\approx .006$.) The approximants are fairly simple (generally simpler than the Pade(1,2) approximants). E.g. for r = 2/1 we have the approximant 7/5 (ignoring the constant 1/4). A first computation shows that this approximation spots 34EDO by comparing the approximants for 5/4 and 3/2: 25/56 and 13/16 resp. We don't get equality, but minimizing the difference m*(13/16)-n*(25/56) gives m=11, n=20 with 11*(13/16) - 20*(25/56) = 1/112, a unit fraction with denominator the LCM of the approximant denominators.

- mattyhawthorn April 04, 2015, 05:08:22 PM UTC-0700

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