OK, this looks better!
(1) -> D(abs(x),x)
_
x + x
(1) -------
2abs(x)
Type:
Expression(Integer)
(2) -> D(conjugate(x),y)
(2) 0
Type:
Expression(Integer)
(3) -> D(conjugate(x),x)
(3) 1
Type:
Expression(Integer)
(4) -> f:=operator 'f
(4) f
Type:
BasicOperator
(5) -> D(abs(f(x)),x)
, _ _ ,
f(x)f (x) + f(x)f (x)
(5) ---------------------
2abs(f(x))
Type:
Expression(Integer)
(6) -> D(abs(log(x)),x)
_ _
xlog(x) + x log(x)
(6) ------------------
_
2xxabs(log(x))
Type:
Expression(Integer)
On 19 November 2014 10:19, Bill Page <bill.p...@newsynthesis.org> wrote:
>
> On 2014-11-19 9:36 AM, "Bill Page" <bill.p...@newsynthesis.org> wrote:
> > ...
> > Then I noticed that if we have f=z we get
> >
> > conjugate(z).diff(z)
> >
> > which is 0. So the 2nd term is 0 and the result is just the first
> Wirtinger derivative.
> >
> > Perhaps I am misinterpreting something?
> >
>
> Oops, my fault. According to your definition
>
> conjugate(z).diff(z) = 1
>
> Bill.
>
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