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From: Walter Roberson <roberson@hushmail.com>
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Subject: Re: Is there any polynomial which satisfies some properties
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Date: Fri, 02 Jan 2009 02:59:46 -0600
Xref: news.mathworks.com comp.soft-sys.matlab:509478


zedong wrote:
> 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)

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

collect(diff(f,x) + diff(f,y),[x,y]) = 
(2*y*a2_2+a2_1)*x^2 + (2*y^2*a2_2+(2*a1_2+2*a2_1)*y + a1_1+2*a2_0)*x +
y^2*a1_2 + (2*a0_2+a1_1)*y + a0_1+a1_0

focus first on the expression that would have to yield the x*y part
and do a comparison by parts:

(2*y^2*a2_2+(2*a1_2+2*a2_1)*y + a1_1+2*a2_0)*x = x*y
so a2_2 = 0 to zero the x^1*y^2,
and (a1_1 + 2*a2_0) = 0 to zero x^1 * y^0,
and 2*(a1_2 + a2_1) = 1 to match x^1 * y^1 
By this we see that at least one of a1_2 or a2_1 is non-zero.

Now examine the complete summation of the two diff() expressions
and look for a1_2 and a2_1 . We see a2_1 in the first expression,
(2*y*a2_2+a2_1)*x^2 . Matching by parts we know this overall expression
is 0. Now, a2_2 is not allowed to be dependent upon y (because otherwise
you would not be dealing with a polynomial) so to zero the x^2*y part, we
see that a2_2 must be 0, leaving a2_1*x^2 which again must be 0, and
since a2_1 cannot depend upon x (or else we are not dealing with a polynomial)
we see that a2_1 must be 0. We determined that at least one of a1_2 and a2_1
must be non-zero to get proper matching on x^1*y^1, so we have found by
elimination that a1_2 must be non-zero. But when we look back at the
summation, we see that a1_2 appears as the coefficient in x^0*y^2*a1_2
and again with comparison by parts we know that must be 0, showing that
a1_2 must be 0 after-all. With a1_2 and a2_1 both being 0, the
(2*a1_2+2*a2_1)*y portion of the by-parts matching for x^1*y^1 must be 0.

We are thus forced to the conclusion that if we are dealing with strict
polynomials (coefficients independent of x and y), there is no
polynomial f with terms {c, x, y, x*y, x^2*y, x*y^2} such that
diff(f,x) + diff(f,y) = x*y .

I have little doubt that with some effort, this counter-example could be
expanded into a general proof, probably based upon examining the coefficients
a{m+1,n} and a{m,n+1} for all positive m, n.

-- 
.signature note: I am now avoiding replying to unclear or ambiguous postings.
Please review questions before posting them. Be specific. Use examples of what you mean,
of what you don't mean. Specify boundary conditions, and data classes and value
relationships -- what if we scrambled your data or used -Inf, NaN, or complex(rand,rand)?