On Fri, Dec 5, 2014 at 1:20 PM, Ondřej Čertík wrote:
> Hi Bill,
>
> I thought about this a lot (essentially I studied complex analysis
> from several books as well as consulted with many colleagues) and I
> figured out some answers to my questions.
>
> In the approach (A), you have:
>
> log(a*b) =
Hi Bill,
I thought about this a lot (essentially I studied complex analysis
from several books as well as consulted with many colleagues) and I
figured out some answers to my questions.
In the approach (A), you have:
log(a*b) = log(a) + log(b)
What that means is that log() is multivalued, so yo
On 26 November 2014 at 12:58, Ondřej Čertík wrote:
> On Wed, Nov 26, 2014 at 10:17 AM, Bill Page
> wrote:
>>
>> Does it help if a say the operations are defined "symbolically"?
>
> All I want is if you can give me an algorithm of your approach
> in sufficient detail, so that it can be implemente
On Wed, Nov 26, 2014 at 10:17 AM, Bill Page wrote:
> On 25 November 2014 at 14:51, Ondřej Čertík wrote:
>> On Tue, Nov 25, 2014 at 11:30 AM, Bill Page
>> wrote:
> ...
> Try it this way:
>
> a*b = exp(?1)
> a = exp(?2)
> b = exp(?3)
>
> I think 'normalize' is sa
On 25 November 2014 at 15:14, Erik Massop wrote:
> On Tue, 25 Nov 2014 13:30:33 -0500
> Bill Page wrote:
>
>> On 25 November 2014 at 01:11, Ondřej Čertík wrote:
>> > On Mon, Nov 24, 2014 at 10:23 PM, Bill Page
>> > wrote:
> ...
>> >> But I don't want to be forced to make a choice of branch unt
On 25 November 2014 at 14:51, Ondřej Čertík wrote:
> On Tue, Nov 25, 2014 at 11:30 AM, Bill Page
> wrote:
...
Try it this way:
a*b = exp(?1)
a = exp(?2)
b = exp(?3)
I think 'normalize' is saying that there is a solution that makes
?1 - ?2 - ?3
On Tue, 25 Nov 2014 13:30:33 -0500
Bill Page wrote:
> On 25 November 2014 at 01:11, Ondřej Čertík wrote:
> > On Mon, Nov 24, 2014 at 10:23 PM, Bill Page
> > wrote:
...
> >> But I don't want to be forced to make a choice of branch until
> >> I actually need to evaluate an expression numerically
On Tue, Nov 25, 2014 at 11:30 AM, Bill Page wrote:
> On 25 November 2014 at 01:11, Ondřej Čertík wrote:
>> On Mon, Nov 24, 2014 at 10:23 PM, Bill Page
>> wrote:
>>> ...
>>> I am not very interested in real numbers. I am interested in the
>>> algebra. Would you say that
>>>
>>> sqrt(x^2).dif
>
> >
> > This discussion is about how a CAS should handle (complex)
> > differentiation. Since it started here, I would finish it here, so
> > that the whole thread is in one mailinglist for future reference.
> >
>
> OK. It would be nice to know if other sage-devel subscribers actually
> r
On 25 November 2014 at 01:11, Ondřej Čertík wrote:
> On Mon, Nov 24, 2014 at 10:23 PM, Bill Page
> wrote:
>> ...
>> I am not very interested in real numbers. I am interested in the
>> algebra. Would you say that
>>
>> sqrt(x^2).diff(x) = sqrt(x^2)/x
>>
>> is OK?
>
> I think so, using the fol
On Mon, Nov 24, 2014 at 10:23 PM, Bill Page wrote:
> On 24 November 2014 at 17:43, Ondřej Čertík wrote:
>> On Mon, Nov 24, 2014 at 1:57 PM, Bill Page
>> wrote:
>> ...
>>>
>>> In FriCAS 'abs' is already a kernel function and it implemented the
>>> derivative of 'abs' even before my proposed patc
On 24 November 2014 at 17:43, Ondřej Čertík wrote:
> On Mon, Nov 24, 2014 at 1:57 PM, Bill Page wrote:
> ...
>>
>> In FriCAS 'abs' is already a kernel function and it implemented the
>> derivative of 'abs' even before my proposed patch but I think the
>> current definition is wrong:
>>
>> (14) ->
On Mon, Nov 24, 2014 at 1:57 PM, Bill Page wrote:
> On 22 November 2014 at 12:34, Ondřej Čertík wrote:
>> On Sat, Nov 22, 2014 at 7:23 AM, Bill Page
>> wrote:
>>> ...
>>> FriCAS currently does not implement a symbolic 'conjugate' operator.
>>> The issue concerns whether adding 'conjugate' is a
On 22 November 2014 at 12:34, Ondřej Čertík wrote:
> On Sat, Nov 22, 2014 at 7:23 AM, Bill Page wrote:
>> ...
>> FriCAS currently does not implement a symbolic 'conjugate' operator.
>> The issue concerns whether adding 'conjugate' is a good idea and only
>> secondly how to differentiate it.
>
> A
On Sat, Nov 22, 2014 at 7:23 AM, Bill Page wrote:
> On 21 November 2014 at 20:18, Ondřej Čertík wrote:
>>
>> I am still confused about one thing: is this issue is already
>> present in FriCAS before your changes? Because you can
>> already use conjugate, sin, +, *, ..., even without defining the
On 21 November 2014 at 20:18, Ondřej Čertík wrote:
> On Fri, Nov 21, 2014 at 9:37 AM, Bill Page wrote:
>>
>> You are right about the derivative. But my limited understanding
>> is that the strategy is not to avoid 'abs(x)' but rather to avoid 'sin'.
>> We cannot similarly avoid 'conjugate' and i
On 21 November 2014 at 20:18, Ondřej Čertík wrote:
>
> I am still confused about one thing: is this issue is already
> present in FriCAS before your changes? Because you can
> already use conjugate, sin, +, *, ..., even without defining the
> derivative for abs(x). I fail to see how defining the a
On Fri, Nov 21, 2014 at 9:37 AM, Bill Page wrote:
> On 20 November 2014 22:08, Ondřej Čertík wrote:
>> On Thu, Nov 20, 2014 at 7:53 PM, Bill Page
>> wrote:
>> ...
>>> This problem can be reduced to finding an algorithm to determine
>>> if f(x) is everywhere non-negative. Richardson proves that
On 20 November 2014 22:08, Ondřej Čertík wrote:
> On Thu, Nov 20, 2014 at 7:53 PM, Bill Page wrote:
> ...
>> This problem can be reduced to finding an algorithm to determine
>> if f(x) is everywhere non-negative. Richardson proves that no such
>> algorithm exists.
>
> I see. But what does this ha
I've written up all the equations from this thread together with
detailed step by step derivation:
http://www.theoretical-physics.net/dev/math/complex.html
e.g. the derivatives are here:
http://www.theoretical-physics.net/dev/math/complex.html#complex-derivatives
Most of the examples from this
On Thu, Nov 20, 2014 at 7:53 PM, Bill Page wrote:
> On 20 November 2014 12:56, Ondřej Čertík wrote:
>> ...
>> Can you give an example of an expression that cannot be decided by
>> the Richardson's theorem?
>
> Well, no not exactly. Richardson's theorem is not about individual
> expressions, it i
On 20 November 2014 12:56, Ondřej Čertík wrote:
> ...
> Can you give an example of an expression that cannot be decided by
> the Richardson's theorem?
Well, no not exactly. Richardson's theorem is not about individual
expressions, it is about decidability, i.e. computability, in general.
Conside
On Thu, Nov 20, 2014 at 9:59 AM, Bill Page wrote:
> Perhaps this is more or less where Richardson's theorem enters.
>
> http://en.wikipedia.org/wiki/Richardson%27s_theorem
>
> We badly want a reliable way to determine when an expression is
> identically zero. In general this is not possible, but i
Perhaps this is more or less where Richardson's theorem enters.
http://en.wikipedia.org/wiki/Richardson%27s_theorem
We badly want a reliable way to determine when an expression is
identically zero. In general this is not possible, but if we restrict
our selves to a subset of "elementary" function
On Thu, Nov 20, 2014 at 9:16 AM, Ondřej Čertík wrote:
> On Thu, Nov 20, 2014 at 7:52 AM, Bill Page wrote:
>> So here (20) is a simpler expression for derivative of arg:
>>
>> (16) -> abs(x)==sqrt(x*conjugate(x))
>>Compiled code for abs has been cleared.
>>Compiled code for arg has been cl
On Thu, Nov 20, 2014 at 7:52 AM, Bill Page wrote:
> So here (20) is a simpler expression for derivative of arg:
>
> (16) -> abs(x)==sqrt(x*conjugate(x))
>Compiled code for abs has been cleared.
>Compiled code for arg has been cleared.
>1 old definition(s) deleted for function or rule a
On Thu, Nov 20, 2014 at 7:41 AM, Bill Page wrote:
> On 20 November 2014 01:54, Ondřej Čertík wrote:
>>
>> What you posted looks good. But we need to test it for arg(z), re(z),
>> im(z) and any other non-analytic function that we can find.
>>
>
> (1) -> re(x)==(conjugate(x)+x)/2
>
So here (20) is a simpler expression for derivative of arg:
(16) -> abs(x)==sqrt(x*conjugate(x))
Compiled code for abs has been cleared.
Compiled code for arg has been cleared.
1 old definition(s) deleted for function or rule abs
On 20 November 2014 01:54, Ondřej Čertík wrote:
>
> What you posted looks good. But we need to test it for arg(z), re(z),
> im(z) and any other non-analytic function that we can find.
>
(1) -> re(x)==(conjugate(x)+x)/2
Type:
Void
On Wed, Nov 19, 2014 at 7:36 PM, Bill Page wrote:
> On 19 November 2014 21:23, kcrisman wrote:
>>
>>
>>> Since this mostly concerns FriCAS I am cross posting to that group. I will
>>> also post the patch there. For FriCAS list reference the original email
>>> thread is here:
>>>
>>
>> But if
On 19 November 2014 21:23, kcrisman wrote:
>
>
>> Since this mostly concerns FriCAS I am cross posting to that group. I will
>> also post the patch there. For FriCAS list reference the original email
>> thread is here:
>>
>
> But if you come up with a solution Sage (or Ginac, or whatever) can
> Since this mostly concerns FriCAS I am cross posting to that group. I
> will also post the patch there. For FriCAS list reference the original
> email thread is here:
>
>
But if you come up with a solution Sage (or Ginac, or whatever) can
implement too, please let us know!
--
You receive
Since this mostly concerns FriCAS I am cross posting to that group. I will
also post the patch there. For FriCAS list reference the original email
thread is here:
https://groups.google.com/forum/#!topic/sage-devel/6j-LcC6tpkE
Here is the result of compiling the patch against the current SourceF
On Wed, Nov 19, 2014 at 9:42 AM, Ondřej Čertík wrote:
> On Wed, Nov 19, 2014 at 9:32 AM, Bill Page wrote:
>> OK, this looks better!
>>
>> (1) -> D(abs(x),x)
>>
>> _
>> x + x
>>(1) ---
>> 2abs(x)
>> Type:
>> Exp
On Wed, Nov 19, 2014 at 9:32 AM, Bill Page wrote:
> OK, this looks better!
>
> (1) -> D(abs(x),x)
>
> _
> x + x
>(1) ---
> 2abs(x)
> Type:
> Expression(Integer)
> (2) -> D(conjugate(x),y)
>
>(2) 0
>
On Wed, Nov 19, 2014 at 8:19 AM, Bill Page wrote:
>
> On 2014-11-19 9:36 AM, "Bill Page" 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 mis
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) -
On 2014-11-19 9:36 AM, "Bill Page" 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 defi
On 18 November 2014 21:22, Ondřej Čertík wrote:
> On Tue, Nov 18, 2014 at 6:51 PM, Bill Page
wrote:
>> On 18 November 2014 17:40, Ondřej Čertík wrote:
>>>
>>> In my notation, the Wirtinger derivative is d f(z) / d z and d f(z) /
>>> d conjugate(z). The Df(z) / Dz is the complex derivative taking
On Tue, Nov 18, 2014 at 6:51 PM, Bill Page wrote:
> On 18 November 2014 17:40, Ondřej Čertík wrote:
>>
>> In my notation, the Wirtinger derivative is d f(z) / d z and d f(z) /
>> d conjugate(z). The Df(z) / Dz is the complex derivative taking in
>> direction theta (where it could be theta=0). Giv
On 18 November 2014 17:40, Ondřej Čertík wrote:
>
> In my notation, the Wirtinger derivative is d f(z) / d z and d f(z) /
> d conjugate(z). The Df(z) / Dz is the complex derivative taking in
> direction theta (where it could be theta=0). Given the chain rule, as
> I derived above using chain rules
Hi,
With Sage 6.3, I am getting:
sage: abs(x).diff(x)
x/abs(x)
sage: abs(I*x).diff(x)
-x/abs(I*x)
But abs(I*x) == abs(x). So also abs(x).diff(x) and abs(I*x).diff(x)
must be the same. But in the first case we get x/abs(x), and in the
second we got -x/abs(x).
In SymPy, the answer is:
-
On Tue, Nov 18, 2014 at 2:50 PM, Bill Page wrote:
> On 18 November 2014 15:19, Ondřej Čertík wrote:
>> On Tue, Nov 18, 2014 at 12:14 PM, Bill Page
>> wrote:
>>>
>>> abs(x).diff(x)
>>>
>>> would return the symbolic expression
>>>
>>> conjugate(x)/(2*abs(x)) + conjugate(x)/(2*abs(x))* e^{-2*i
On 18 November 2014 15:19, Ondřej Čertík wrote:
> On Tue, Nov 18, 2014 at 12:14 PM, Bill Page
> wrote:
>>
>> abs(x).diff(x)
>>
>> would return the symbolic expression
>>
>> conjugate(x)/(2*abs(x)) + conjugate(x)/(2*abs(x))* e^{-2*i*theta}
>
> I think you made a mistake, the correct expressio
On Tue, Nov 18, 2014 at 1:19 PM, Ondřej Čertík wrote:
> On Tue, Nov 18, 2014 at 12:14 PM, Bill Page
> wrote:
>> On 18 November 2014 13:41, Ondřej Čertík wrote:
>>> On Tue, Nov 18, 2014 at 11:08 AM, Bill Page
>>> wrote:
...
Have you had a chance to consider the issue of the chain-rul
On Tue, Nov 18, 2014 at 12:14 PM, Bill Page wrote:
> On 18 November 2014 13:41, Ondřej Čertík wrote:
>> On Tue, Nov 18, 2014 at 11:08 AM, Bill Page
>> wrote:
>>> ...
>>> Have you had a chance to consider the issue of the chain-rule yet?
>>
>> Yes. Very straightforward, as I suggested in my last
On 18 November 2014 14:14, Bill Page wrote:
> On 18 November 2014 13:41, Ondřej Čertík wrote:
>> On Tue, Nov 18, 2014 at 11:08 AM, Bill Page
>> wrote:
>>> ...
>>> Have you had a chance to consider the issue of the chain-rule yet?
>>
>> Yes. Very straightforward, as I suggested in my last email.
On 18 November 2014 13:41, Ondřej Čertík wrote:
> On Tue, Nov 18, 2014 at 11:08 AM, Bill Page
> wrote:
>> ...
>> Have you had a chance to consider the issue of the chain-rule yet?
>
> Yes. Very straightforward, as I suggested in my last email. Just start with:
>
> D f / D z = df/dz + df/d conjug
On Tue, Nov 18, 2014 at 11:08 AM, Bill Page wrote:
> On 18 November 2014 12:29, Ondřej Čertík wrote:
>> On Tue, Nov 18, 2014 at 9:28 AM, David Roe wrote:
>>> ...
>>> Because derivative is not just used in the context of functions of a
>>> complex variable (whether they are analytic or not). Pro
On 18 November 2014 12:29, Ondřej Čertík wrote:
> On Tue, Nov 18, 2014 at 9:28 AM, David Roe wrote:
>> ...
>> Because derivative is not just used in the context of functions of a
>> complex variable (whether they are analytic or not). Probably more
>> than 90% of Sage users don't know any comple
On Tue, Nov 18, 2014 at 9:28 AM, David Roe wrote:
> On Tue, Nov 18, 2014 at 8:05 AM, Bill Page wrote:
>> On 18 November 2014 09:02, David Roe wrote:
>>> On Tue, Nov 18, 2014 at 5:57 AM, Bill Page
>>> wrote:
> I think you are overly focused on trying to define a derivative that
>
On Tue, Nov 18, 2014 at 8:05 AM, Bill Page wrote:
> On 18 November 2014 09:02, David Roe wrote:
>> On Tue, Nov 18, 2014 at 5:57 AM, Bill Page
>> wrote:
>>>
>>> > I think you are overly focused on trying to define a derivative that
>>> > reduces to the conventional derivative of non-analytic fun
On 18 November 2014 09:02, David Roe wrote:
> On Tue, Nov 18, 2014 at 5:57 AM, Bill Page wrote:
>>
>> > I think you are overly focused on trying to define a derivative that
>> > reduces to the conventional derivative of non-analytic functions
>> > over the reals.
>>
>> I've just been casually fol
>
> > I think you are overly focused on trying to define a derivative that
> > reduces to the conventional derivative of non-analytic functions over
> > the reals.
>
> I've just been casually following this conversation, but I think it's
> important that the derivative of abs(x) be sign(x) not
On Tue, Nov 18, 2014 at 5:57 AM, Bill Page wrote:
> On 17 November 2014 23:16, Ondřej Čertík wrote:
>> Hi Bill,
>>
>> Thanks for the clarification. So your point is that 2) is not
>> sufficient, that we really need two Wirtinger derivatives --- it's
>> just that one can be expressed using the oth
On 17 November 2014 23:16, Ondřej Čertík wrote:
> Hi Bill,
>
> Thanks for the clarification. So your point is that 2) is not
> sufficient, that we really need two Wirtinger derivatives --- it's
> just that one can be expressed using the other and a conjugate,
> so perhaps CAS can only return one,
Hi Bill,
Thanks for the clarification. So your point is that 2) is not
sufficient, that we really need two Wirtinger derivatives --- it's
just that one can be expressed using the other and a conjugate, so
perhaps CAS can only return one, but a chain rule needs modification
and probably some other
On 17 November 2014 15:17, Ondřej Čertík wrote:
> On Sat, Nov 15, 2014 at 9:18 AM, Bill Page wrote:
>>
>> I am sorry for the confusion. What I am proposing is that the
>> Wirtinger derivative(s) be considered the fundamental case (valid
>> for complex or even quaternion variables). As you noted
> I still don't understand exactly your proposal. We've played with a
> few ideas above, in particular we have considered at least (below d/dz
> is the Wirtinger derivative, d/dx and d/d(iy) are partial derivatives
> with respect to "x" or "iy" in z=x+i*y) :
>
> 1) d/dz
> 2) d/dz + d/d conjugate(z)
Hi Bill,
On Sat, Nov 15, 2014 at 9:18 AM, Bill Page wrote:
> On 14 November 2014 14:29, Ondřej Čertík wrote:
>>
>> On Nov 14, 2014 11:30 AM, "Bill Page" wrote:
>>>
>>> What do you mean by "the real derivative"?
>>
>> The absolute value doesn't have a complex derivative, but it has a real
>> der
> Vladimir V. Kisil kisilv's patch
>
> http://www.ginac.de/pipermail/ginac-devel/2013-November/002053
>
> looks like a good start to me especially if one doesn't want to
> consider the issue of derivatives of non-analytic functions in
> general.
>
>
http://www.ginac.de/pipermail/ginac-devel/
Vladimir V. Kisil kisilv's patch
http://www.ginac.de/pipermail/ginac-devel/2013-November/002053
looks like a good start to me especially if one doesn't want to
consider the issue of derivatives of non-analytic functions in
general.
On 17 November 2014 10:14, kcrisman wrote:
> For reference (sin
For reference (since Sage uses Ginac for most derivatives)
see http://www.cebix.net/pipermail/ginac-devel/2014-April/002105.html
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To unsubscribe from this group and stop receiving emails from it, sen
On 14 November 2014 14:29, Ondřej Čertík wrote:
>
> On Nov 14, 2014 11:30 AM, "Bill Page" wrote:
>>
>> What do you mean by "the real derivative"?
>
> The absolute value doesn't have a complex derivative, but it has a real
> derivative, over the real axis.
>
It seems to me that the concept of "re
On Nov 14, 2014 11:30 AM, "Bill Page" wrote:
>
> On 14 November 2014 13:18, Ondřej Čertík wrote:
> >
> > On Nov 14, 2014 8:57 AM, "Bill Page" wrote:
> >>
> >> It seems to me that we should forget about x and y. All we really
need is
> >>
> >> |z|' = d |z| / d z = conjugate(z) / (2*|z|)
> >>
>
On 14 November 2014 13:18, Ondřej Čertík wrote:
>
> On Nov 14, 2014 8:57 AM, "Bill Page" wrote:
>>
>> It seems to me that we should forget about x and y. All we really need is
>>
>> |z|' = d |z| / d z = conjugate(z) / (2*|z|)
>>
>> and the appropriate algebraic properties of conjugate.
>
> Sur
On Nov 14, 2014 8:57 AM, "Bill Page" wrote:
>
> On 14 November 2014 02:19, Ondřej Čertík wrote:
> > On Fri, Nov 14, 2014 at 12:14 AM, Ondřej Čertík
wrote:
> >> ...
> >> Ok, thanks for the confirmation.
> >>
> >> There is an issue though --- since |z| is not analytic, the
> >> derivatives depend
On 14 November 2014 02:19, Ondřej Čertík wrote:
> On Fri, Nov 14, 2014 at 12:14 AM, Ondřej Čertík
> wrote:
>> ...
>> Ok, thanks for the confirmation.
>>
>> There is an issue though --- since |z| is not analytic, the
>> derivatives depend on the direction. So along "x" you get
>
> |z|' = \partial
On Fri, Nov 14, 2014 at 12:14 AM, Ondřej Čertík wrote:
> On Thu, Nov 13, 2014 at 6:56 PM, Bill Page wrote:
>> Sorry, I hit send before I was quite ready. To continue ...
>>
>> On 13 November 2014 19:24, Ondřej Čertík wrote:
>>> On Thu, Nov 13, 2014 at 2:00 PM, Ondřej Čertík
>>> wrote:
>>> ...
On Thu, Nov 13, 2014 at 6:56 PM, Bill Page wrote:
> Sorry, I hit send before I was quite ready. To continue ...
>
> On 13 November 2014 19:24, Ondřej Čertík wrote:
>> On Thu, Nov 13, 2014 at 2:00 PM, Ondřej Čertík
>> wrote:
>> ...
>> For example, for |z| we get:
>>
>> |z|' = \partial |z| / \pa
Sorry, I hit send before I was quite ready. To continue ...
On 13 November 2014 19:24, Ondřej Čertík wrote:
> On Thu, Nov 13, 2014 at 2:00 PM, Ondřej Čertík
> wrote:
> ...
> For example, for |z| we get:
>
> |z|' = \partial |z| / \partial x = d |z| / d z + d |z| / d
> conjugate(z) = conjugate(z
On 13 November 2014 19:24, Ondřej Čertík wrote:
> On Thu, Nov 13, 2014 at 2:00 PM, Ondřej Čertík
> wrote:
>>
>> As you said, the function is analytic if it doesn't functionally
>> depend on conjugate(z), as can be shown easily. So |z| or
>> Re z are not analytic, while z^2 is. If the function is
On Thu, Nov 13, 2014 at 2:00 PM, Ondřej Čertík wrote:
> Hi Bill,
>
> On Thu, Nov 13, 2014 at 10:16 AM, Bill Page
> wrote:
>> It has always seemed very inconvenient to me that "computer algebra
>> programs such as Mathematica" choose to define derivative as
>> complex-derivative. I believe a rea
Hi Bill,
On Thu, Nov 13, 2014 at 10:16 AM, Bill Page wrote:
> It has always seemed very inconvenient to me that "computer algebra
> programs such as Mathematica" choose to define derivative as
> complex-derivative. I believe a reasonable alternative is what is
> known as a Wirtinger derivative.
On 13 November 2014 12:16, Bill Page wrote:
>
> The Wirtinger derivative of abs(x) is 1/2 x/abs(x). Its total
> Wirtinger derivative is x/abs(x).
>
Sorry, I should have written that the Wirtinger derivative of abs(x) is
1/2 conjugate(x)/abs(x)
Bill.
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You received this message because you
It has always seemed very inconvenient to me that "computer algebra
programs such as Mathematica" choose to define derivative as
complex-derivative. I believe a reasonable alternative is what is
known as a Wirtinger derivative. Wirtinger derivatives exist for all
continuous complex-valued functio
Yes. Note also here:
http://mathworld.wolfram.com/AbsoluteValue.html
which says that complex derivative of d|z|/dz does not exist, as
Cauchy-Riemann equations do not hold for Abs(z). And:
"As a result of the fact that computer algebra programs such as
Mathematica generically deal with complex va
possibly related to http://trac.sagemath.org/ticket/12588 ?
Regards, CH
Am 2014-11-13 um 06:19 schrieb Ondřej Čertík:
> Hi,
>
> With Sage 6.3, I am getting:
>
> sage: abs(x).diff(x)
> x/abs(x)
> sage: abs(I*x).diff(x)
> -x/abs(I*x)
>
> But abs(I*x) == abs(x). So also abs(x).diff(x) and abs(I*
Hi,
With Sage 6.3, I am getting:
sage: abs(x).diff(x)
x/abs(x)
sage: abs(I*x).diff(x)
-x/abs(I*x)
But abs(I*x) == abs(x). So also abs(x).diff(x) and abs(I*x).diff(x)
must be the same. But in the first case we get x/abs(x), and in the
second we got -x/abs(x).
In SymPy, the answer is:
In [1]: ab
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