Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Is there any polynomial which satisfies some properties
Date: Fri, 2 Jan 2009 16:47:01 +0000 (UTC)
Organization: The MathWorks, Inc.
Lines: 26
Message-ID: <gjlge5$r7g$1@fred.mathworks.com>
References: <gjaej2$9o5$1@fred.mathworks.com> <GIk7l.8487$u17.2700@newsfe20.iad>
Reply-To: <HIDDEN>
NNTP-Posting-Host: webapp-02-blr.mathworks.com
Content-Type: text/plain; charset="ISO-8859-1"
Content-Transfer-Encoding: 8bit
X-Trace: fred.mathworks.com 1230914821 27888 172.30.248.37 (2 Jan 2009 16:47:01 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Fri, 2 Jan 2009 16:47:01 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 1187260
Xref: news.mathworks.com comp.soft-sys.matlab:509526


Walter Roberson <roberson@hushmail.com> wrote in message <GIk7l.8487$u17.2700@newsfe20.iad>...
> .......
> No, you cannot.
> .......
> Consider even the very simple case g(x,y) = x*y.
> The general polynomial f that has to be considered is
> f = x^2*y^2*a2_2 + x^2*y*a2_1 + x*y^2*a1_2 + x^2*a2_0 + x*y*a1_1 +
> y^2*a0_2 + x*a1_0 + y*a0_1 + a0_0
> .......

  Walter, yes he can!  There seems to be some misunderstanding here about the word 'order'.  It should be clear that when Zedong says for g(x,y) = x^m*y*n he is seeking a polynomial f of order m+n+1, he means polynomials in terms x^p*y^q in which the sums of their exponents, p+q, are not in excess of m+n+1.  (In his particular case of course p+q needs to be exactly equal to m+n+1.  Nothing else would make sense.)

  The definition given in Wikipedia agrees with that at

 http://en.wikipedia.org/wiki/Degree_of_a_polynomial ,

in which they state clearly that "The words degree and order are used interchangeably" and "The degree of a term is the sum of the powers of each variable in the term."  That is also my own understanding from my training in mathematics.  What it does NOT mean is that only terms x^p*y^q are allowed in which p and q are each individually limited to m+1 and n+1, respectively.  x^(m+1)*y^(n+1) is not allowed but x^(m+n+1)*y^0 is.

  Your "counterexample" did not allow for the x^3 or y^3 terms which are of the appropriate order, namely 3, while improperly allowing x^2*y^2 which is of order 4.  The example Zedong has already given (yesterday) of

 f(x,y) = -1/6*x^3+1/2*x^2*y

is a valid solution to his problem in accordance with this definition of 'order'.

Roger Stafford