On 12 August 2014 15:53, Waldek Hebisch <[email protected]> wrote:
> Bill Page wrote:
>>
>> On 6 August 2014 21:13, Waldek Hebisch <[email protected]> wrote:
>> > Bill Page wrote:
>> >> I think that
>> >>
>> >>   abs(x)/x ~= x/abs(x)
>> >>
>> >> If  D(abs(x),x) = x/abs(x) then D(abs(x),[x,x]) = (abs(x)^2 - x^2 ) / 
>> >> abs(x)^3.
>> >> This is zero only for x is real.
>> >
>> > But what D(abs(x),x) mean for non-real x?  'abs' considered
>> > as function of complex variable is not differentiable.  AFAICS
>> > one reasonable interpretation of D(abs(x),x) is that it
>> > automatically forces x to be real.
>>
>> Yes, D(abs(x),x) is not complex-differentiable. But D(abs(x),x) =
>> x/abs(x) is nonetheless a derivative in the direction of the real
>> axis.
>
> No, derivative in the direction of the real axis is real(x)/abs(x).

OK.

> z/abs(z) = \partial_x abs(z) + i*\partial_y abs(z) where
> z = x + iy.

I believe that this is called a Wirtinger derivative (one of two
related derivatives).

Ref.

http://en.wikipedia.org/wiki/Wirtinger_derivatives#Functions_of_one_complex_variable

Strictly speaking I should write

D(abs(z),x) = conjugate(z)/abs(z)/2

This remains suitable for my purpose.

> Note that neither definition fits FriCAS.  In
> fact, once you try to use FriCAS to compute derivatives
> of functions containig abs of complex argument you will
> get nonsense regardless of what you take as D(abs(x), x).
> More precisely, for a given function f you may be able to
> tweak D(abs(x), x) so that D(f, x) will be sensible.
> But for any value of D(abs(x), x) you will find some f
> so that D(f, x) would be a nonsense.  The reason is
> that FriCAS assumes chain rule and Leibintz rule
> and they are valid only for differentiable functions.

I do not see any problem with Leibniz or chain rule provided we use
both Wirtinger derivatves appropriately.

http://en.wikipedia.org/wiki/Wirtinger_derivatives#Product_rule

>
>>  My thinking is this:  If we can "define" signum(x) as x/abs(x)
>> then perhaps we can also "define" diracDelta(x) as (abs(x)^2 - x^2 ) /
>> abs(x)^3 / 2 ?  To do this the derivative D(abs(x),[x,x]) must not be
>> zero. Of course these expressions are not actual distributions but
>> perhaps we can also define integration of expressions involving abs so
>> that we may use them to represent distributions in a consistent way.
>
> You probably can use (abs(x)^2 - x^2 ) / abs(x)^3 / 2 to
> represent delta distribution, but there are plenty of alternative
> representations, so I do not see why you want this one.
> Also, connection of this with D(abs(x),[x,x]) is still
> unclear.

I would like derivative of expression to be compatible with derivative
of distribution. In particular I noted the claim: "they can be readily
extended to every space of generalized functions".

> Concerning integration: if you want to integrate
> something on _real_ interval, then integrator should
> simplify your expression to zero since x is real on
> te interval.  ATM integrator is not doing this, but
> sensible handling of integrals containing abs will
> do this.   So your representation is really unsuitable
> for purpose of integrating delta distribution.
>

That is not clear to me.

> Note that your expression probably would represent delta
> via intagration along some complex contours.

As contour I would choose the real line.

> But we
> do not want to use such contours in normal integration,
> because results are discontinous and we would risk
> wrong integrals for real function.  Also, if delta
> is represented explicitely, then integrating it
> is quite easy.  I do not remember formula by heart,
> but it is relatively simple.
>

I would like to refer to

http://axiom-wiki.newsynthesis.org/SandBoxDiracDelta

Cheers,
Bill Page.

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