Re: Dt "gradient" (dumb title, sorry)
- To: mathgroup at smc.vnet.net
- Subject: [mg81097] Re: Dt "gradient" (dumb title, sorry)
- From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
- Date: Wed, 12 Sep 2007 03:47:58 -0400 (EDT)
- References: <fc4ht0$e8q$1@smc.vnet.net> <46E64FE6.8030903@gmail.com>
Chris Chiasson wrote:
> I agree with your assessment. I also think the present behavior should
> be changed. Do you?
Yes, I think so.
> On 9/11/07, Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote:
> > Chris Chiasson wrote:
> >
> > > I wanted a list of different total derivatives, so I fed in the
> > > familiar {{vars},n} syntax to Dt. Dt took it without complaint, but I
> > > don't trust the results:
> >
> > The "familiar {{vars, n} syntax" is explicitly documented for the
> > partial derivative operator D only. I have found no reference or mention
> > of this syntax for the total derivative operator in the documentation
> > center, the 5.2 online help, the Mathematica book, and Michael Trott's
> > work. (Of course this does not mean that such references do not exist,
> > but they seem hard to find.)
> >
> > > In[1]:= Dt[z[x,y],{x,1}]
> > > Dt[z[x,y],{y,1}]
> > > Dt[z[x,y],{{x,y},1}]
> > >
> > > (*this paste of "copy as plain text" shows derivatives as exponents;
> > > it's not my doing and isn't the subject of this post*)
> > > Out[1]= Dt[y,x] (z^(0,1))[x,y]+(z^(1,0))[x,y]
> > > Out[2]= (z^(0,1))[x,y]+Dt[x,y] (z^(1,0))[x,y]
> > > Out[3]= {(z^(1,0))[x,y],(z^(0,1))[x,y]}
> > >
> > > I think out 3 should be a list containing the expression for out 1 and
> > > the expression for out 2. It seems like Dt decided to ignore the
> > > dependence of y on x in the first entry, and the dependence of x on y
> > > in the second entry.
> >
> > Another explanation (or interpretation of this behavior) is that
> > whenever Dt is fed with an argument of the form {{vars}, n} Dt calls D
> > instead and we get a list of partial derivatives. That would explain why
> > Dt does not complain and produces a result identical to D's.
> >
> > In[1]:= Dt[z[x, y], {{x, y}, 1}]
> >
> > Out[1]=
> > (1,0) (0,1)
> > {z [x, y], z [x, y]}
> >
> > In[2]:= D[z[x, y], {{x, y}, 1}]
> >
> > Out[2]=
> > (1,0) (0,1)
> > {z [x, y], z [x, y]}
> >
> >
> > > Please let me know if my interpretation/expectation is wrong. Thanks.
> >
> > You can mimic the behavior of D and get the expected result with Dt
> > using Thread and a one dimensional list of variables as in
> >
> > In[3]:= Thread[Dt[z[x, y], #] &[{x, y}]]
> >
> > Out[3]=
> > (0,1) (1,0)
> > {Dt[y, x] z [x, y] + z [x, y],
> >
> > (0,1) (1,0)
> > z [x, y] + Dt[x, y] z [x, y]}
> >
> >
> > Best regards,
> > --
> > Jean-Marc
> >
>
>
> --
> http://chris.chiasson.name/
>
--
Jean-Marc