MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Problem with the Derivative of a Arg-function

  • To: mathgroup at
  • Subject: [mg48546] Re: Problem with the Derivative of a Arg-function
  • From: ab_def at (Maxim)
  • Date: Fri, 4 Jun 2004 04:51:07 -0400 (EDT)
  • References: <c9k37o$fbo$>
  • Sender: owner-wri-mathgroup at

klishko at (Alex Klishko) wrote in message news:<c9k37o$fbo$1 at>...
> 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:

ComplexExpand[(Arg[z + h] - Arg[z])/h, z,
    TargetFunctions -> Abs] //
  Limit[#, h -> 0] & // FullSimplify


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,

Limit[(Arg[1 + I + I*h] - Arg[1 + I])/(I*h), h -> 0]

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

  • Prev by Date: Re: Help Browser
  • Next by Date: Re: Extract substrings using metacharacters
  • Previous by thread: Re: Problem with the Derivative of a Arg-function
  • Next by thread: Re: Problem with the Derivative of a Arg-function