Hi Raul,
I am not sure, I can help specifically, but it still could
clarify things, if you said what specific form of result you
are looking for: is it an operator that rotates any function
or transforms polynomials? Your original question was a little
confusing, at least to me:
> My question is: how do I find the rank 0 functions which
> correspond to general cases of that plot?
What are the inputs, parameters, form of result, how it is applied,
examples, etc?
> From: Raul Miller <[EMAIL PROTECTED]>
>
> 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.
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