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MathGroup Archive 2004

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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}]



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