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From: "John D'Errico" <woodchips@rochester.rr.com>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Is there any polynomial which satisfies some properties
Date: Mon, 29 Dec 2008 14:02:03 +0000 (UTC)
Organization: John D'Errico (1-3LEW5R)
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"zedong 
" <zdongwu@gmail.com> wrote in message <gjah3m$fin$1@fred.mathworks.com>...
> "John D'Errico" <woodchips@rochester.rr.com> wrote in message <gjag02$54m$1@fred.mathworks.com>...
> > "zedong 
> > " <zdongwu@gmail.com> wrote in message <gjaej2$9o5$1@fred.mathworks.com>...
> > > I have known a function g(x,y)=(x^m)*(y^n)
> > > Could I find a polynomial f which is order m+n+1
> > > and 
> > > diff(f,'x')+diff(f,'y')=g
> > > And more simple more good(The less term more good)
> > > Thank you all
> > > 
> > 
> > I don't know if this is just an example that you made
> > up, not thinking if it is possible, and that you really
> > have some other problem in mind.
> > 
> > But the specific problem as posed has no solution.
> > 
> > John
> 
> Thank you for your attention.In fact I am working on an integration of a polynomial over a polygon(I am writing a finite element code with matlab.).so if the above integration is possible.then  I can do that.and still I Think it's possible.because  If not I can make the order of the polynomial  more higher.I think.But actually.I don't know how to solve it.It must be solved by a matrix equation.I guess

NO. You are not thinking clearly here.

Given a polynomial of the form

g(x,y)=(x^m)*(y^n)

If some polynomial f(x,y) exists such that

diff(f,'x') + diff(f,'y') = g

then we must have f(x,y) be order m+1 in x
and order n in y, but at the same time, f(x,y)
must be of order m in x and order n+1 in y.

We cannot have both of these things true at
the same time. A "matrix equation" will not
help. There is no magic here, unless you have
again failed to be accurate in your question.

John