Re: Assumption -> quadratic multivariate function
- To: mathgroup at smc.vnet.net
- Subject: [mg47188] Re: Assumption -> quadratic multivariate function
- From: "Peter Pein" <no at spam.no>
- Date: Mon, 29 Mar 2004 04:22:34 -0500 (EST)
- References: <c40spe$mqu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Damir Herman" <damir at 2d.com> schrieb im Newsbeitrag news:c40spe$mqu$1 at smc.vnet.net... > Hi, > > Is there a way to tell Mathematica to assume that a function of more > than one variable is at most quadratic in its arguments? I do not want > to specify the variables, so in that regard I think I have two > options: to use Dt on f or partial derivatives acting on Hold[f]. > > For instance, approach that works for a function of one variable > (1) f/: Dt[f, {x, 3}] = 0; > does not work for a function of two variables > (2) f/:Dt[f, {x, 3}] = 0; > (3) f/:Dt[f, {y,3}] = 0; > I don't understand that, because these are not conditions for a > particular x or y. > > For example, > > Clear[f]; > SetAttributes[p, {Constant}]; > Dt[f^p, {x, 4}, {y,4}] > > seems to work for x only and does not care about y. > > Note that Mathematica returns > > In[79]:= Dt[Dt[f, {x, 3}], y] > Out[79]:= 0 > > however > > In[80]:= Dt[f, {x,3}, y] > Out[80]:= Dt[f, {x,3}, y] > > How do I make mathematica use the comutativity of these two > operations? > > Thanks, > > Damir > Damir, try f /: Dt[f, ___, {x | y, 3}, ___] = 0; instead of f /: Dt[f, {x, 3}] = 0; f /: Dt[f, {y, 3}] = 0; I suppose you would like to add x/: Dt[x, ___, y | {y, _}, ___] = 0; y/: Dt[y, ___, x | {x, _}, ___] = 0; too ;-) Peter -- Peter Pein, Berlin StringReplace["petsie at arcand.de", Rule@@(ToLowerCase@ToString@Head@#&)/@{William&&S,2b||!2b}]