From: "Eldon Eller" <[EMAIL PROTECTED]>
> The arc tangent of _1r1 is different from the arc tangent of 1r_1.
From: "Randy MacDonald" <[EMAIL PROTECTED]>
> (_3 o. 1r_1) = _3 o. _1r1
>1
I believe Eldon means it in a mathematical sense. Probably in the same way
111r333 can differ from 1r3 (which neither J nor I can fathom). To J, 1r_1
and _1r1 are the same quantity:
_1r1
_1
1r_1
_1
_1r1 -: 1r_1
1
_1r1 -:&:(3!:3) 1r_1 NB. Identical to the last bit, even in
internal rep
1
So of course any operation on these quantities is going to produce the same
output: they're the same input. In fact, I shouldn't say "these quantities", I
should say "this quantity".
None the less, there are ways of looking at the two quantities where they'll
differ. For example, when I was in college, I always saw ^._1 defined to be
0 j. 1p1 . But that bugged me, because:
_1 -: % _1 NB. (A) The important part.
_1 -: ^ 0 j. 1p1 NB. (B) Euler
_1 -: % ^ 0 j. 1p1 NB. (C) Substitute (B) into (A)
_1 -: ^ 0 j. - 1p1 NB. (D) (%x^y) <==> (x^-y)
With (D), it becomes clear that ^._1 is also equivalent to ^.(^ 0 j.-1p1) .
And ^. and ^ are an inverse pair -- they cancel out. So that leaves us
with (^._1) -: 0 j.-1p1 . So it matters whether you think of _1 as _1r1
or as 1r_1 (i.e. %_1 ). See postscript.
Again, keep in mind, J cannot see the "differences" in these quantities. In
J's internals, they're represented identically. It is impossible for an
operation to reveal the difference, because no difference exists. So, despite
my little "proof":
(^._1) -: 0 j.-1p1
0
-Dan
PS:
I'm not a mathematician, but it seems to me that the right solution is to
express ^._1 as 0 j. (+,-) 1p1 . But no one does. At least not in front
of me.
Not that such would present a solution for J. If there are multiple solutions
to a mathematical operation, J picks one. Think of 0%0 or %:y :
4 -: *: _2 NB. Yep
1
_2 -: %: 4 NB. Nope
0
I wonder if Perl6 will leverage its fancy new junctions to produce answers like
(+,-)2
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