Thread Subject:

Is there any polynomial which satisfies some properties

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

"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

"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

"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

"John D'Errico" <woodchips@rochester.rr.com> wrote in message <gjal8r$r3a$1@fred.mathworks.com>...
> "zedong
> " <zdongwu@gmail.com> wrote in message <gjah3m$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



Thank you for your attention.Yes ,you are right.But I can make the order of the polynomial more higher.(I have pointed out,But the page divided it into two)
 Thank you for your help.
And thank you all.
I have getten the solution.as a matter of fact.I only need x^my^n,in which m<=3;n<=3;
I have done it with the unknown coefficient method.
This is the result:
(1)g=x f=x^2
(2)g=xy f=x^2y/2-x^3/6
(3)g=x^2y f=x^3y/3-x^4/12
(4)g=x^2y^2 f=x^3y^2/3-x^4y/6+x^5/30
(5) g=x^3 f=x^4/4
(6)g=x^3y f=x^2y^3/2-x^3y^2/2+x^4y/4-x^5/20
Then change the position of x and y get others.(If m>3 and n>3 we can still find out,I think)
Thank you for your attention and kind help for another time

"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
>

  Yes, Zedong, there is always a solution but it is not unique. We get m+n+1 equations and m+n+2 unknown coefficients. Were you looking for a unique answer?

Roger Stafford

"zedong

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

Let us define u:= x+ y
You can write the above as df/du = g

This is the first order PDE (convection). If no boundary is specified, infinity solution exist. Solution is unique if and only if boundary condition is specified on half part of the boundary (where the set of lines u=x+y=cte toucjh it). And so on...

Bruno

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)?

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