I might have caused inadvertently some confusion by inserting the derivative of 0^x ((0&^)d.1) into the discussion of the integral of 0^x (see my reply to Raul).
" ... Which is fine... as long as the log of a is defined. But at a = 0, log(a) = __. Which blows up the whole thing. J's d. operator has a few issues, handling this is just one of them. " Right, that is why I suggested to look directly at the evaluation, according to J, of the function 0^x (0&^): "J evaluates 0^x as follows: _ if x < 0 1 if x = 0 0 if x > 0 " and proceed from there. The integral of 0^x for the interval x > 0 presents no problem; however, the situation for its complement, the interval x <= 0, is quite another matter. (My suggestion is to evaluate ((0&^)d._1) as _. for any value in that troublemaker interval). Wolfram Alpha's response, see the comment (the NB's) I inserted afterward, makes more sense to me. " ((_9&^)d.1) NB. integral of (-9)^x gives _0.244136j_0.349606 That is clearly wrong. My fix should clear up a few of these issues. " The sentence ((_9&^)d.1) is supposed to produce the derivative, as opposed to the integral, of (-9)^x, and I get, ((_9&^)d.1) 2.1972245773362196j3.1415926535897931&*@(_9&^) (I do not know what to make of that answer anyway.) On Mon, Oct 2, 2017 at 8:27 AM, 'Jon Hough' via Source <sou...@jsoftware.com > wrote: > > however, for instance, > > > ((0&^)d.1) _1 > > __ > > > is beyond my comprehension. Can anybody > > explain this result? > > Standard (or naive) way to evaluate integral of some number to the power x. > integral of a^x dx > can be rewritten as > integral exp(log(a)*x) dx > which can be easily evaluated to > exp(log(a)*x) / log(a) > and simplified to > a^x/log(a) > > Which is fine... as long as the log of a is defined. But at a = 0, log(a) > = __. Which blows up the whole thing. J's d. operator has a few issues, > handling this is just one of them. > Another issue: > > ((_9&^)d.1) NB. integral of (-9)^x > gives > _0.244136j_0.349606 > That is clearly wrong. My fix should clear up a few of these issues. > -------------------------------------------- > On Mon, 10/2/17, Jose Mario Quintana <jose.mario.quint...@gmail.com> > wrote: > > Subject: Re: [Jsource] d. fix > To: sou...@jsoftware.com > Date: Monday, October 2, 2017, 2:43 PM > > " > 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 > > > On Sun, Oct 1, 2017 at 9:18 PM, 'Jon > Hough' via Source <sou...@jsoftware.com > > wrote: > > > Yes, the sentence I wrote is essentially > gibberish. I'll put it down to > > > tiredness. (J only defines 0^0 as 1). > > > 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 > > > > > (0&^) d. _1 > > could either be 0 (+ > some constant) > > or > > > undefined. > > > > Wolfram > Alpha decides to go with undefined. Seems reasonable, since > if you > > are trying to integrate 0^x > anyway, you probably have other issues with > > your code. > > > -------------------------------------------- > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
