RE: PolynomialQ ?

*To*: mathgroup at smc.vnet.net*Subject*: [mg34903] RE: [mg34866] PolynomialQ ?*From*: "David Park" <djmp at earthlink.net>*Date*: Wed, 12 Jun 2002 02:15:36 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Juan, I guess it is the way that mathematicians think. If p = c is a constant, then p is a polynomial in z because it can be written as c + 0*z + 0*z^2 +... p = x^3 - 2*x^2 + x - 1; so PolynomialQ[p, z] True by the above reasoning because the x terms are all treated as constants. So you want to do an extra test to see if p contains z. MemberQ[p, z, Infinity] && PolynomialQ[p, z] False MemberQ[p, x, Infinity] && PolynomialQ[p, x] True David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ > From: Juan [mailto:erfa11 at hotmail.com] To: mathgroup at smc.vnet.net > > > Hi,I tried to check is a polynomial have a variable, ussing the function > PolynomialQ. > > In[1]:=p = x^3 - 2*x^2 + x - 1; > In[2]:=PolynomialQ[p, x] > Out[2]=True > In[3]:=PolynomialQ[p, y] > Out[3]=True > In[4]:=PolynomialQ[p, z^2] > Out[4]=True > In[5]:=PolynomialQ[p, {u, v}] > Out[5]=True > > What is the thing I am doing wrong? > > Regards.Juan > > _________________________________________________________________ > Descargue GRATUITAMENTE MSN Explorer en > http://explorer.yupimsn.com/intl.asp. > >