You're right about the integral of (i. 0). I was confused.
The underlying point is still valid, though. (i. 0) is a different
thing from (0) and there is no strong argument for allowing it as a
polynomial. A 'polynomial' of (i. 0) is a bug and should be caught as
soon as possible.
Henry Rich
On 1/3/2023 10:43 PM, Raul Miller wrote:
On Tue, Jan 3, 2023 at 10:40 PM Henry Rich <[email protected]> wrote:
What is the integral of (i. 0)?
When integrating we add an arbitrary constant to represent the unknown
constant which would have been lost during differentiation.
By convention, J tends to use 0 to represent arbitrary constants when
there's no overriding issue.
So here, the integral would either be ,0 or i.0 though of course, any
single element numeric list would technically be valid.
If (i. 0) as a polynomial is the same as 0, the answer has to be 0 1,
which is inconsistent with simple definitions.
I think you should double check your math here.
The integral of ,1 would be 0 1.
Thanks,
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