help on critical points

• To: mathgroup at smc.vnet.net
• Subject: [mg4037] help on critical points
• From: Jose Luis Lugo Goytia <lugo at hplara.iquimica.unam.mx>
• Date: Tue, 28 May 1996 01:46:01 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Could you please give me some advise?

I'm trying to find the critical points of a function
f:R^n -> R, I do the following:

In[2]:= ceros[f_,vars_]:=  vars /. Solve[Numerator[Together[f]] ==
Table[0,{i,1,Length[vars]}],vars];

and for example

In[3]:= ceros[gradiente[x^2 y - y^2 -2 x y,{x,y}][[1]],{x,y}] //N

Out[3]= {{1., -0.5}, {0, 0}, {2., 0}}

but if I define the function

In[4]:= PuntosCriticos[f_,vars_]:=

and apply it to the before example then obtain

In[4]:= PuntosCriticos[x^2 y - y^2 -2 x y,{x,y}]

2      2 (0,1)
Solve::dinv: The expression (-2 x y + x  y - y )     [x, y]
involves unknowns in more than one argument, so inverse functions can't
be used.

ReplaceAll::reps:
{<<1>>} is neither a list of replacement rules nor a valid dispatch table,
and so cannot be used for replacing.

2         2 (0,1)
Solve::dinv: The expression (-2. x y + x  y - 1. y )     [x, y]
involves unknowns in more than one argument, so inverse functions can't
be used.

ReplaceAll::reps:
{<<1>>} is neither a list of replacement rules nor a valid dispatch table,
and so cannot be used for replacing.

Out[13]= {x, y} /.

2         2 (1,0)
>    Solve[{(-2. x y + x  y - 1. y )     [x, y],

2         2 (0,1)
>       (-2. x y + x  y - 1. y )     [x, y]} == {0, 0}, {x, y}]

Could someone tell me where I have gone wrong? How can I do that?

Thanks

_______________________________________________________
Jose Luis Lugo Goytia
Instituto de Quimica,
lugo at hplara.iquimica.unam.mx
lugo at barajas.fciencias.unam.mx
_______________________________________________________

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