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MathGroup Archive 2007

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Dt "gradient" (dumb title, sorry)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg81065] Dt "gradient" (dumb title, sorry)
  • From: "Chris Chiasson" <chris at chiasson.name>
  • Date: Mon, 10 Sep 2007 18:52:43 -0400 (EDT)

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:

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.

Please let me know if my interpretation/expectation is wrong. Thanks.

-- 
http://chris.chiasson.name/


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