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