Re: PolynomialQ (wrong) behavior ?

• To: mathgroup at smc.vnet.net
• Subject: [mg19228] Re: [mg19199] PolynomialQ (wrong) behavior ?
• From: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>
• Date: Wed, 11 Aug 1999 02:06:45 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```The function PolynomialQ is pretty simple minded. The expression 2x+1 is
considered as a constant polynomial in x. A little more controversially, it
is also considered as a constant polynomial in Sin[x], since it does not
explicitely involve Sin[x] as variable. This is quite consistent with the
behaviour of the derivative:

In[1]:=
D[2x + 1, y]
Out[1]=
0

In[2]:=
D[2x + 1, Sin[x]]
Out[2]=
0

--
Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp
http://eri2.tuins.ac.jp

----------
>From: "Drago Ganic" <drago.ganic at in2.hr>
To: mathgroup at smc.vnet.net
>To: mathgroup at smc.vnet.net
>Subject: [mg19228] [mg19199] PolynomialQ  (wrong) behavior ?
>Date: Tue, Aug 10, 1999, 8:52 AM
>

> Hi,
>
> f = 2x +1
>
> PolynomialQ [ f, x ]
> True
>
> That's OK. But why the following
>
> PolynomialQ [ f, y ]
> True
>
> PolynomialQ [ f, Sin[x] ]
> True
>
> Greetings,
> Drago Ganic
>
>
>
>
>
>
>

```

• Prev by Date: Re: Mathematica Link for Excel and Excel 2000
• Next by Date: RE: Automatic Display in MatrixForm
• Previous by thread: PolynomialQ (wrong) behavior ?
• Next by thread: Re: PolynomialQ (wrong) behavior ?