This is from NuVoc (for 806)
1. For finite x, x | y is extended to negative numbers by the definition
x|y ==> y-x*<. y % x+0=x and to complex
<http://code.jsoftware.com/wiki/Vocabulary/Glossary#Complex> numbers by
the definition x|y ==> y-x* <.@(0.5&+)&.+. y % x+0=x .
which agrees with your algebra.
The Vocabulary uses
y-x*<. y % x+0=x extends the residue to a zero left argument, and to
negative and fractional finite arguments.
but says that the dyad | extends to complex numbers
In either case the result appears to be different from the Complex Residue
In mathematics, more specifically *complex* analysis, the *residue* is a
*complex number* proportional to the contour integral of a meromorphic
function along a path enclosing one of its singularities. Don Kelly
Residue (complex analysis) - Wikipedia
<https://www.google.ca/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjwqYKwxNLUAhUJ8WMKHVhgCQsQFggoMAE&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FResidue_%28complex_analysis%29&usg=AFQjCNGJ2qCb-juMCMekL7Vd4znD0H9xnA>
https://en.wikipedia.org/wiki/Residue_(complex_analysis)
*SO why do we **attempt to use rhe "common residue" for a complex
number- does it have meaning?
*
Don Kelly
On 2017-06-22 2:53 PM, Henry Rich wrote:
What it actually is (when x and y are not infinite and x is not 0) is
d =. |x
ab =. (+x) * y
q =. <.@(0.5&+)&.+. ab%d
y - x * q
If my algebra is right, this is equivalent to
(x|y)=y-x* <.@(0.5&+)&.+.y%x+0=x
In other words, the Dictionary description incorrectly says that the
complex quotient will be rounded using complex floor, while in reality
each component of the quotient is (tolerantly) rounded independently.
Henry Rich
On 6/22/2017 4:53 PM, Louis de Forcrand wrote:
Hi,
There's been recent discussion on the GNU APL forums about how to
define the residue function on complex arguments. I wondered how it
was specified in J, but the dictionnary page seems kind of vague.
Nuvoc says that
(x|y)=y-x*<.y%x+0=x
for all scalars x and y (including complex numbers).
However, this is untrue for
'x y'=: 1j1 3j4
(tested on the old J for iOS)
While it would seem like a logical extension, the dictionnary only
says that this is true for rational numbers.
So how is the residue of two complex numbers defined in J?
Thanks,
Louis
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