Free42 and cube root of complex number

02232021, 11:05 PM
Post: #1




Free42 and cube root of complex number
Why the result is different?
Code:
Code:


02242021, 02:02 AM
Post: #2




RE: Free42 and cube root of complex number
In the first case, the imaginary part is 0, while in the second case, it is 0.
0, also known as signed zero, is a feature of IEEE754 floating point. The difference between 0 and 0 shouldn't matter in Free42, but apparently this is a case where it does and where there is no specialcase code for zero to prevent this behavior. What happens internally is that the number is first converted to polar using the hypot() and atan2() functions, and atan2(0, 8) => pi, while atan2(0, 8) => pi. After multiplying that angle by 1/3, you end up on opposite sides of the real axis. I'll fix this in the next release. (You won't see this behavior in >POL, because it does contain specialcase code for handling zero in either coordinate.) 

02242021, 02:52 AM
Post: #3




RE: Free42 and cube root of complex number
From a numerical viewpoint, the current behaviour is correct.


02242021, 08:28 AM
(This post was last modified: 02242021 12:33 PM by JF Garnier.)
Post: #4




RE: Free42 and cube root of complex number
The 1984 HP71 was supporting the signed zero, but it was sometimes a source of confusion for users and the signed zero was not kept in later machines (starting with the 28C).
So yes, the HP71, with the Math ROM, gives the same (correct) results: > (8,0) ^ (1/3) (1,1.73205080757) > ((8,0)) ^ (1/3) (1,1.73205080757) JF 

02242021, 08:59 AM
Post: #5




RE: Free42 and cube root of complex number
Yes, the sign of zero is important!
Read Kahan's 'How Java’s FloatingPoint Hurts Everyone Everywhere'. Please, leave it as it is  Free42 would be one of the only ones to get it right ;) Werner 

02242021, 10:26 AM
Post: #6




RE: Free42 and cube root of complex number
(02242021 02:02 AM)Thomas Okken Wrote: atan2(0, 8) => pi, while atan2(0, 8) => pi pi < arg(z) ≤ pi Since polar form and rectangular form should match, I think 8 ± 0i should be adjusted to 8 + 0i = 8 exp(i*pi) For the same reason, sqrt(4  0i) = sqrt(4 + 0i) = 2i 4 [ENTER] 0 [COMPLEX] [+/] [SQRT] → 2i 

02242021, 01:35 PM
Post: #7




RE: Free42 and cube root of complex number
(02242021 08:59 AM)Werner Wrote: Please, leave it as it is  Free42 would be one of the only ones to get it right ;) There is that! But the HP42S doesn't support 0, and when I did the review of complex transcendentals a few years ago, I added special cases for purereal and pureimaginary numbers that made the not42Scompatible behavior with 0 disappear. That review only covered the transcendentals, though. I missed SQRT, for which I ended up creating a special case for pureimaginary a bit later. And this issue with Y^X follows the same pattern. 

02242021, 02:47 PM
Post: #8




RE: Free42 and cube root of complex number
(02242021 01:35 PM)Thomas Okken Wrote:(02242021 08:59 AM)Werner Wrote: Please, leave it as it is  Free42 would be one of the only ones to get it right ;) Hi, Thomas Okken Can you elaborate what is decided ? If Free42 do support 0, then sqrt(40i) = sqrt(4*exp(pi*i)) = 2*exp(pi/2*i) = 2i But, user need to know "0" is actually inputed. 0 should displayed as "0", not "0" 40*i should displayed as "4 i0", not "4 i0" If not, (80i)^(1/3) = (8+0i)^(1/3) = 1 + sqrt(3)*i 

02242021, 02:53 PM
Post: #9




RE: Free42 and cube root of complex number
(02242021 02:47 PM)Albert Chan Wrote: Can you elaborate what is decided ? The behavior reported by Ajaja is a bug. Free42 does not support 0, and so the two calculations he showed should return the same result. The only case where the sign of zero actually matters in Free42 is when it gets passed as the first parameter to atan2(). That scenario is supposed to be prevented by specialcase logic, but that check is missing in Y^X. I'll fix that in the next release. 

02252021, 06:08 AM
Post: #10




RE: Free42 and cube root of complex number
.. and, not unimportant, that way it is consistent with the 42S.
Werner 

02252021, 04:40 PM
Post: #11




RE: Free42 and cube root of complex number
FYI, we can remove 0 parts of x, by doing xx+x = 0+x
(assumed 0+x does not get optimized away, to plain x) Redo OP example: 8 [Enter] 0 [Complex] [+/] → x = 8  0i [Enter]  [LastX] + → xx+x = 8 + 0i 3 [1/x] [y^x] → ≈ 1 + 1.73205080757*i 

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