Re: Revised version of Level 1 text (draft)
Nate
Checking your quadratic convolution kernel example as best I can, I agree that with xx=[1/5,1] you get h(xx) = ([-1/4,7/8],D3). The given h(x) has continuous first derivative on the whole line, so D3 is telling the truth.
But suppose thanks to a typo, my h(x) is discontinuous: say, I mistakenly write 9/4 for 9/8 in the formula. I STILL get decoration D3 on the answer don't I? What is your D3 telling us? Or is the onus entirely on the programmer to get the code right?
Another Q: does, say, [1,2]*Empty, when done with decorated intervals, get decoration D3 or D4?
John
On 10 May 2011, at 06:58, Nate Hayes wrote:
> In signal processing, a quadratic convolution kernel has the piecewise
> definition:
>
> h(x) = 3/4-x^2 if |x| <= 1/2
> 1/2*x^2-3/2*|x|+9/8 if 1/2 < |x| < 3/2
> 0 otherwise
>
> A valid interval extension h(xx) of h(x) is:
>
> h(xx) = u(xx) union v(xx) union w(xx)
>
> where
>
> u(xx) = 3/4-(|xx| intersect [0,1/2])^2
> v(xx) = 1/2*(|xx| intersect [1/2,3/2])^2-3/2*(|xx| intersect
> [1/2,3/2])+9/8
> w(xx) = 0*(|xx| intersect [3/2,Inf])
>
> By the new definitions and appropriate application of "Theorem 1" and
> "Theorem 8" in my Nov. 14 paper, if xx=[1/5,1] then
>
> u(xx) = ([1/2,3/4],D3)
> v(xx) = ([-1/4,7/8],D3)
> w(xx) = (Empty,D3)
>
> and then likewise by "Theorem 8"
>
> h(xx) = ([-1/4,7/8],D3).