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Re: Dt@x@1

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  • Subject: [mg119449] Re: Dt@x@1
  • From: Chris Chiasson <chris.chiasson at>
  • Date: Sat, 4 Jun 2011 06:20:58 -0400 (EDT)
  • References: <is4tgc$b3t$> <is95cc$61r$>

On Jun 2, 6:12 pm, Roland Franzius <roland.franz... at> wrote:
> Am 01.06.2011 10:33, schrieb Chris Chiasson:
> > Why does Dt@x@1 return zero? I would expect it to return unevaluated.
> The chain rule for Dt acting on a chain of functions of a single
> argument says
> Dt@x@1 = x'[1] Dt[1]  ~  Dt[1]=0
> Compare
> Trace[Dt[x[y[w[u]]], Constants -> {u, v}]] // TreeForm
> Trace[Dt[x[y[w[z]]], Constants -> {u, v}]] // TreeForm
> to see that Dt[mostinnnerargument]->0 is used as a rule without
> calulating superflous inner derivatives x', y', w' first.
> Its of course the simplifying use of those general cancelling rules,
> easy to recognize and to apply, that makes the CAS working at all (in a
> limited collection of cases in finite time ;-( ).
> --
> Roland Franzius

Thank you for the explanation Roland. I appreciate it. In another
reply, I posted a method to get x[1] to behave as a symbol (i.e. as if
the computer understood x to have SubValues).

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