No, I was trying to explain
((0&^)d._1) _1
I suspect the derivative should give a NaN error.
Thanks,
--
Raul
On Mon, Oct 2, 2017 at 7:03 PM, Jose Mario Quintana
<jose.mario.quint...@gmail.com> wrote:
> 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
>> >>
>>
>>
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