If you were trying to explain, ((0&^)d.1) _1 __
then I am afraid your reasoning is also beyond my comprehension considering that, ((0&^)d.1) _1 is supposed to evaluate the derivative function of 0^x at the point _1. I introduced the derivative of 0^x ((0&^)d.1) within the discussion of the integral of 0^x ((0&^)d._1) to highlight the difficulties I find while (attempting to) apply calculus (at least the calculus I learned) to functions which evaluate to _ for all x in an interval of length greater than zero (such as, f(x) = 0^x for x < 0, according to J). On Mon, Oct 2, 2017 at 5:32 AM, Raul Miller <rauldmil...@gmail.com> wrote: > (Hopefully the reasoning is clear? Integral is basically a sum, and if > you have infinities in your sum, adding 0 (or 1) to it isn't going to > make a significant difference.) > > Thanks, > > -- > Raul > > > On Mon, Oct 2, 2017 at 5:28 AM, Raul Miller <rauldmil...@gmail.com> wrote: > > Oh, oops, I should have looked at that - definitely not dirac delta. > > > > Integral of 0^x should be _ > > > > Thanks, > > > > -- > > Raul > > > > > > On Mon, Oct 2, 2017 at 1:43 AM, Jose Mario Quintana > > <jose.mario.quint...@gmail.com> wrote: > >> " > >> My main point was that I'm not sure how the best way to handle integral > of > >> 0^x is. I guess it could be considered the > >> integral of 0 (i.e. 0^x = 0). So ... > >> " > >> It seems that J evaluates 0^x as follows: > >> > >> _ if x < 0 > >> 1 if x = 0 > >> 0 if x > 0 > >> > >> So, which is the function, apart from adding a constant, whose > derivative > >> is 0&^? If x > 0 I would say it is 0"0; otherwise, it might be wise to > >> regard it as undefined. > >> Consider the derivative of 0^x, according to J, > >> > >> (0&^)d.1 > >> __&*@(0&^) > >> > >> If x > 0 the evaluation makes sense to me; however, for instance, > >> > >> ((0&^)d.1) _1 > >> __ > >> > >> is beyond my comprehension. Can anybody explain this result? > >> > >> NB. Wolfram Alpha > >> NB. https://www.wolframalpha.com/input/?i=d%2Fdx(+0%5Ex) > >> NB. d/dx(0^x) > >> NB. 0 if x >0 > >> NB. indeterminate otherwise > >> > >> Incidentally, although Wolfram Alpha regards 0^0 as undefined, it is > fine > >> with integral x^0 dx and agrees with J... > >> > >> NB. Wolfram Alpha > >> NB. https://www.wolframalpha.com/input/?i=integral+x%5E0+dx > >> NB. integral x^0 dx > >> NB. x + constant > >> > >> (^&0)d._1 > >> [ > >> > >> NB. Wolfram Alpha > >> NB. https://www.wolframalpha.com/input/?i=d%2Fdx(+integral+x%5E0+dx) > >> NB. d/dx( integral x^0 dx) > >> NB. 1 > >> > >> (^&0)d._1 d.1 > >> 1 > >> > > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
