The answer 3j2 looks sane.
3j2 * 0j1 0j_1
and the only one in the first quadrant.
Nice!
Am 16.05.20 um 12:32 schrieb Raul Miller:
Oh, I see -- or, I think I see what you were trying to illustrate now.
That said, this is kind of interesting:
+./(,-)2j_3
3j2
But I also noticed that +/(,-) seems to bring back associativity (we
already had commutativity with +.):
(i.6) A."0 1]4.57 4.34 4.44
4.57 4.34 4.44
4.57 4.44 4.34
4.34 4.57 4.44
4.34 4.44 4.57
4.44 4.57 4.34
4.44 4.34 4.57
+./@(,-)"1 (i.6) A."0 1]4.57 4.34 4.44
8.88178e_16 8.88178e_16 8.88178e_16 8.88178e_16 8.88178e_16 8.88178e_16
Thanks,
--
Raul
On Sat, May 16, 2020 at 5:51 AM Hauke Rehr <hauke.r...@uni-jena.de> wrote:
sorry, I was not sufficiently precise about my example
I meant to talk about atomic a only
if $ a is empty, then
+./ (, -) a
will work
Thanks for pointing out the lack of precision.
Am 16.05.20 um 11:46 schrieb Raul Miller:
I was talking about the implementation.
These are different results:
+./@(,-)4.57 4.34 4.44
8.88178e_16
+./@(,-)&.x:4.57 4.34 4.44
0.01
+./@(,-)&.:(*&1p1)4.57 4.34 4.44
5.65432e_16
The reason is that binary floating point cannot represent 5^_1 nor
5^_2 accurately.
Thanks,
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