Interesting Paper available on Moog VCF

[Eli Brandt]

> And for some strange reason, the rest of them can be computed from
> the relation C_n(x) = 2x*C_{n-1}(x) - C_{n-2}(x).

Oh, this is pretty cute.  Not on-topic, necessarily, but cute:

C_n(x) = cos (n arccos x)
C_n(a) = cos na,    a = arccos x

C_n+1(a) = cos (n+1)a = cos na cos a - sin na sin a
C_n-1(a) = cos (n-1)a = cos na cos a + sin na sin a

C_n+1(a) + C_n-1(a) = 2 cos a cos na = 2 cos(arccos x) C_n(a) = 2x C_n(a)

C_n+1 = 2x C_n - C_n-1

--
   Eli Brandt
   eli+ at cs.cmu.edu