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/