I thought I figured out what I was looking for.
The points of discontinuity occur where the slope of the rotated curve
is vertical, which corresponds to the zeros of the derivative if I rotate
my curve an additional 90 degrees.
discon=: (#~ 0 = {:"1@:+.)@(1 {:: p.)@((}.@:* [EMAIL PROTECTED])@] -/@,:
0,3&[EMAIL PROTECTED]@[)
Unfortunately, this is not very efficient, nor does it produce a viable
result.
p=: ($&0 1 0 _1 % [EMAIL PROTECTED]) 30
0.375 discon p NB. very slow
19.2422 _19.2422
But from inspection, I have three discontinuities between _9 and 9.
And I expect them to correspond to my control parameter being between...
dot=: +/ .*
rot=: (1 _1,:1 1) * (2 1,:1 2)&[EMAIL PROTECTED]
({~ (0 1 +/~0 1 [EMAIL PROTECTED] (] = >./\)@{.)@((rot 0.375)&dot)@(,:
p&p.)) 0.1*i:90
_3.8 _3.7
2.4 2.5
8.7 8.8
So I still have not figured this out.
--
Raul
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