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I have a question related to working with symmetric polynomials in some variables. Let us say, I have an expression
expr = x^2 + y^2 + 2*x^3 + 2*y^3 + 3 x^2 y + 3 y^2 x
this is symmetric as x <-> y
I want to express this as expr1 + expr2 where
expr1 = x^2 + 2 x^3 + 3 x^2 y
and
expr2 = expr1/.{x -> y, y-> x}
How to use Mathematica to achieve this?
The original question is not clear. There is a wide range of functionality to work with polynomials symbolically e.g. PolynomialReduce, SymmetricReduction, etc. however it is not clear why one just assumes that expr1 and expr2 are as given above and if they are unique representants of possible reductions in terms of third order polynomials of two variables, most likely they are not.
So I find that this question is asking: how can I find appropriate coefficients {a, b, c} such that expr == p[x,y] + p[y,x] where:
p[x_, y_] := a x^3 + b x^2 + c x y^2
There is SolveAlways which helps in such cases:
{p[x, y], p[y, x]} /. SolveAlways[ expr == p[x, y] + p[y, x], {x, y}]
{{x^2 + 2 x^3 + 3 x y^2, 3 x^2 y + y^2 + 2 y^3}}
PolynomialReduce demonstrates that expr1 and expr2 can express the original polynomial expr
PolynomialReduce[ expr, {expr1, expr2}, {x, y}]
{{1, 1}, 0}
SymmetricReduction serves a different purpose, see e.g.
SymmetricReduction[expr, {x, y}, {expr1, expr2}]
The usual way is to make a Groebner basis from the polynomials that define the relationships you intend to apply ("reduce with", if you will), and then do that with PolynomialReduce.
poly = x^2 + 2 x^3 + 3 x^2 y;
polys = {poly - e1, (poly /. {x -> y, y -> x}) - e2};
gb = GroebnerBasis[polys, {x, y}];
Now for the example in question.
expr = x^2 + y^2 + 2*x^3 + 2*y^3 + 3 x^2 y + 3 y^2 x;
PolynomialReduce[expr, gb, {x, y}][[2]]
(* e1 + e2 *)
Using knowledge from CoefficientList docs (Properties and Relations section) of how to recreate multivariate polynomial from coefficient list:
coe = CoefficientList[expr, {x, y}][[ All , ;; 2]];
expr1 = Fold[FromDigits[Reverse[#1], #2] &, coe, {x, y}] // Expand
expr2 = expr1 /. {x -> y, y -> x}
x^2 + 2 x^3 + 3 x^2 y y^2 + 3 x y^2 + 2 y^3
One way to do that is :
expr = x^2 + y^2 + 2*x^3 + 2*y^3 + 3 x^2 y + 3 y^2 x;
expr1 = x^2 + 2 x^3 + 3 x^2 y;
expr2 = expr1 /. {x -> y, y -> x}
expr1 + expr2 == expr
Another way is to define functions :
expr1[x_,y_]:=x^2 + 2 x^3 + 3 x^2 y;
expr2[x_,y_]:=expr1[y,x];
Together[expr]gives what you want. $\endgroup$expr1=expr2=expr/2? The two add up to your original expression and can be obtained from each other by doingx$\leftrightarrow$y. The class of functions that does that is rather broad! How do you pick your target from within that? $\endgroup$