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MathGroup Archive 2004

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Re: Problem with the Derivative of a Arg-function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48628] Re: Problem with the Derivative of a Arg-function
  • From: klishko at mail.ru (Alex Klishko)
  • Date: Tue, 8 Jun 2004 00:48:23 -0400 (EDT)
  • References: <c9k37o$fbo$1@smc.vnet.net> <c9pfq9$t79$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

ab_def at prontomail.com (Maxim) wrote:

> The short answer is that the derivative of Arg is undefined: Arg is
> not an analytical function. Just consider what happens when you move
> along different directions, for example, from 1 to 1+eps and from 1 to
> 1+I*eps, with real eps.

I think you mean that derivative is Limit[(f[z+dz]-f[z])/dz,dz->0]
and for f=Arg it's depends on Re[dz] and Im[dz].
The fisrt case (from 1 to 1+eps) corresponds to Im[dz]=0, the second
case(from 1 to 1+I*eps) Re[dz]=0.
I think the problem in another field, namely, in that fact that Arg is
discontinuous function.
A complex number z=Re+I*Im, we may represend also as z=Abs*E^(I*Arg).
Where Abs=Squre root[Re^2+Im^2], Arg=ArcTan[Im/Re].

Hense as my function is f=E^(I*z)=E^(-Im)*E^(I*Re)
so Arg[f]=Re[z] by definition.
Therefore Arg'[E^(I*z)]=Re'[z].
So in that case the derivative is define only for real z, but
Mathematica does not show any error message for not real z and gives
definite number.
I think it's a bug.

Examples:
f[x_] := Arg[E^(-I*x)] ; N[D[f[x], x]] /. x -> 5 +3*I
Clear[f]
Out: -0.919536 + 0.272011*I

f[x_] := Arg[E^(-I*x)] ; N[D[f[x], x]] /. x -> 5 
Clear[f]
Out: -0.919536 + 0.272011*I

f[x_] := ComplexExpand@Arg[E^(-I* x)] ; N[D[f[x], x]] /. x -> 5 
Clear[f]
Out: -1.



By the way
---------------------
dArg=(1/(1+(Im/Re)^2))*d(Im/Re)=(-Im*dRe+Re*dIm)/Abs^2

> 
> In[1]:=
> ComplexExpand[(Arg[z + h] - Arg[z])/h, z,
>     TargetFunctions -> Abs] //
>   Limit[#, h -> 0] & // FullSimplify
> 
> Out[1]=
> -Im[z]/Abs[z]^2

In that case Mathematica suppose that dRe=dz, and dIm=0.
-----------------------

Best wishes
Alex Klishko


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