Re: Dt@x@1

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

On Jun 2, 6:12 pm, Roland Franzius <roland.franz... at uos.de> 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).