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Re: PolynomialQ (wrong) behavior ?


In article <7otmvl$4g9 at smc.vnet.net>, "Tomas Garza"
<tgarza at mail.internet.com.mx> wrote:

:Drago Ganic [drago.ganic at in2.hr] wrote:
:
:> f = 2x +1
:>
:> PolynomialQ [ f, x ]
:> True
:>
:> That's OK. But why the following
:>
:> PolynomialQ [ f, y ]
:> True
:>
:> PolynomialQ [ f, Sin[x] ]
:> True
:
:I guess you're trying to substitute Sin[x} in your f so as to obtain 2
:Sin[x] + 1 which is certainly not a polynomial in x. But what you are asking
:is whether it is a polynomial in Sin[x], so that the test gives True, as it
:should. I think you can avoid the problem by being more precise, for
:example:

etc.

The Help browser gives the example

  PolynomialQ[x + Sin[x],x]

as returning False, which of course is correct.  If f is free of y then
PolynomialQ[f, y] SHOULD return True, since it's a constant polynomial in
y.

But there are some anomolous situations.  For example, PolynomialQ[Cos[x],
Sin[x]] returns True.  Obviously PolynomialQ[f, y] doesn't look at
dependencies between f and y (that's probably undecidable), but only looks
to see if there are expressions A, B, etc. such that f = A + B y + etc.

--Ron Bruck


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