Re: Problem with the Derivative of a Arg-function
- To: mathgroup at smc.vnet.net
- Subject: [mg48546] Re: Problem with the Derivative of a Arg-function
- From: ab_def at prontomail.com (Maxim)
- Date: Fri, 4 Jun 2004 04:51:07 -0400 (EDT)
- References: <c9k37o$fbo$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
klishko at mail.ru (Alex Klishko) wrote in message news:<c9k37o$fbo$1 at smc.vnet.net>...
> Hello group,
>
> Let us define a simple complex function Exp[-j*x], j is a square root of -1.
> Fase of this function is computed by Arg -function.
> If x is a real number the fase is equal -x and it's derivative is -1.
> If x is a complex number the fase is equal Re[-x].
> But if we would write:
>
> f[x_] = Arg[E^(-j*x\)]; N[D[f[x], x]]\ /. \ x -> 5+j*3
>
> we would get a complex number not equal -1.
>
> In case of a real x, this problem may be solved by ComplexExpand-function:
>
> f[x_] = ComplexExpand[Arg[E^(-j*x\)]]; N[D[f[x], x]]\ /. \ x -> 5
>
> we would get -1.
>
> In case of a complex x, Matematica gives a wrong solution.
>
> Would you correct me if I make a mistake.
>
> Wishes
> Alex Klishko
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.
There's the question of whether the value of, say, N[Arg'[1+I]] has
any meaning (Mathematica 5.0 gives -0.5). It turns out that it does:
it gives the directional derivative along the real axis. Here's the
explicit expression for this directional derivative:
In[1]:=
ComplexExpand[(Arg[z + h] - Arg[z])/h, z,
TargetFunctions -> Abs] //
Limit[#, h -> 0] & // FullSimplify
Out[1]=
-Im[z]/Abs[z]^2
For z=1+I we get -1/2 as expected. Note that by default ComplexExpand
assumes that all variables are real.
I believe that earlier versions of Mathematica gave similar results
for numerical values of Im', Re', Abs', but those just remain
unevaluated now; N[Arg'] seems to be a relict. This leads to some
contradictions: for example,
In[2]:=
Limit[(Arg[1 + I + I*h] - Arg[1 + I])/(I*h), h -> 0]
Out[2]=
Arg'[1 + I]
while in fact here we're taking the directional derivative along a
different direction, which means that -1/2 is an incorrect value for
this limit.
Maxim Rytin
m.r at inbox.ru