Re: Problem with the Derivative of a Arg-function

*To*: mathgroup at smc.vnet.net*Subject*: [mg48535] Re: Problem with the Derivative of a Arg-function*From*: Bill Rowe <readnewsciv at earthlink.net>*Date*: Fri, 4 Jun 2004 04:50:18 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

On 6/2/04 at 4:21 AM, klishko at mail.ru (Alex Klishko) wrote: >Let us define a simple complex function Exp[-j*x], j is a square >root of -1. Rather than using j to be square root of -1 it would be better to use the built in symbol I. Of course, if you want you could add an additional statement, j = I, which now defines j to be what you want. >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 If you set j = I or use I instead of j in the above you will get what you expect with this syntax. But, it would probably be better to get in the habit of using DelayedSet instead of Set when defining functions. That is, using f[x_]:= Arg[E^(-I x)] The difference is Set sets the value of the function to the RHS at the time the function definition is entered. DelayedSet doesn't evaluate the function until it is used later. In general, this is what you want functions to do. -- To reply via email subtract one hundred and four