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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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