On 10/11/20 9:45 AM, eki...@gmail.com wrote:
> On Sunday, October 11, 2020 at 12:29:47 PM UTC-4, eki...@gmail.com wrote:
>> On Saturday, October 10, 2020 at 6:24:33 PM UTC-4, Sam Sirlin wrote:
>>> On Friday, October 9, 2020 at 5:37:44 AM UTC-7, Andrew Sengul wrote:
>>>> In Dyalog:
>>>> ¯7○¯2
>>>> 0.5493061443J¯1.570796327
>>>>
>>>> In GNU APL:
>>>> ¯7○¯2
>>>> ¯0.5493061443J1.570796327
>>>>
>>>> In Common Lisp:
>>>> * (atanh -2)
>>>> #C(-0.54930615 1.5707964)
>>>>
>>>> In Wolfram Alpha:
>>>> arctanh(-2)
>>>>
>>>> Decimal approximation:
>>>>
>>>> -0.54930614433405484569762261846126285232374527891137472586734716... +
>>>> 1.5707963267948966192313216916397514420985846996875529104874722... i
>>>>
>>>> Why is Dyalog's result negated and conjugated?
>>>>
>>>> Andrew
>>> A Bug. Is the tanh of the result correct? In J
>>>
>>> _7 o. _2
>>> _0.549306j1.5708
>>> a=. _7 o. _2
>>> 7 o. a
>>> _2
>>> 7 o. -a
>>> 2
>>>
>>> Also in octave:
>>>
>>> octave:2> a = atanh(-2)
>>> a = -5.4931e-01 + 1.5708e+00i
>>> octave:3> tanh(a)
>>> ans = -2.0000e+00 + 1.8370e-16i
>>> octave:4> tanh(-a)
>>> ans = 2.0000e+00 - 1.8370e-16i
>>>
>>> Note there is ambiguity on the imaginary part:
>>>
>>> octave:12> A = log( 1/sqrt(3))
>>> A = -5.4931e-01
>>> octave:16> tanh( A + i*pi/2)
>>> ans = -2.0000e+00 + 1.8370e-16i
>>> octave:17> tanh( A - i*pi/2)
>>> ans = -2.0000e+00 - 1.8370e-16i
>> Circular and hyperbolic functions are all periodic, so for each such function f ,
>> there are infinitely many choice for f x for each x. The same is true (in the complex
>> plane) for the exponential function. It's not possible to define inverses for these
>> functions that are both single valued and continuous on the complex plane. It is
>> possible, though too specify where the discontinuities should be, and what values
>> should be taken at points of discontinuity, in a way that is consistent across all
>> the inverses.
>>
>> IIRC the APL Extended standard follows Paul Penfield's APL 81 paper "Principal
>> values and branch cuts in complex APL" in specifying how these choice are to be
>> made, where the discontinuities should be and values at the points of discontinuity.
>> There is a long discussion (which cites and is largely based on the Penfield paper),
>> with illustrations, in Guy Steele's "Common Lisp, The Language (2nd Ed.)",
>> at section 12.5.3.
>>
>> IMHO the J implementation is almost certainly correct, because ... Roger.
>
> Or maybe not, since J and CL differ.
>
J and CL are the same: negative real part, positive imaginary.
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