Raul, note that this works:
(+.&.(*&1p1))/2 3
1
even though the numbers
0j16":(*&1p1)2 3
6.2831853071795862 9.4247779607693793
do not satisfy your demands of being exactly represented.
Thanks!
Bo
Den lørdag den 16. maj 2020 13.03.21 CEST skrev Hauke Rehr
<[email protected]>:
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 <[email protected]> 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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