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Re: Derivative of a Conjugate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg87244] Re: Derivative of a Conjugate
*From*: dh <dh at metrohm.ch>
*Date*: Sat, 5 Apr 2008 04:23:25 -0500 (EST)
*References*: <ft4n6o$3vk$1@smc.vnet.net>
Hi David,
Conjugate is not an analytic function. Consider:
r=x+I y; z[r]= x+I y; Conjugate[z[r]]= x-I y;
dz= dx + I dy; dConjugate[z]= dx -I dy
now take the differential quotient and you see that the derivative
depends on the direction. That is why Mathematica does not evaluate
Conjugate'.
If dr is real you may e.g. use a rule to get what you want:
expression /. Conjugate'[y_[x_]]:= Conjugate[y'[x]]
hope this helps, Daniel
David Forehand wrote:
> Hi All,
>
> My first posting here, so please forgive me if I am being a bit stupid.
>
> I'm entering the following input:
>
> D[f[t0, t1], t0, t1] /. {f -> ((#1^2)*Conjugate[a[#2]] &)}
>
> and Mathematica gives the following output:
>
> 2 t0 a'[t1] Conjugate'[a[t1]]
>
> I would have expected:
>
> 2 t0 Conjugate[a'[t1]]
>
> i.e. the derivative of a conjugate is the conjugate of the derivative.
>
> Any idea how a force Mathematica to give the result I am expecting? In the above, I am assuming the variables "t0" and "t1" are real and the variable "a" is complex, although I have not explicitly told Mathematica this.
>
> Thanks Very Much in advance,
>
> David
>
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