I would like to re-express the following bivariate (symmetric) function (defined over the unit square)

f[x, y] = Piecewise[{{p1, 0 < y < x < 1}, {p2, y == x}, {p3, 0 < x < y < 1}}]


p1 = -(x - 1)^3 (x (x + 3) - 5 y^2 + 1); p3 = -(y - 1)^3 (-5 x^2 + y (y + 3) + 1);
p2 = p3 /. y -> x;

as a (bivariate) polynomial in the two symmetric polynomials, s1 = x + y and s2 = x y.

Can the SymmetricReduction command be applied to such a problem (involving a function defined in Piecewise form)?

  • $\begingroup$ Why not apply SymmetricReduction[] to p1 and p3 before feeding 'em to Piecewise[]? $\endgroup$ – J. M.'s technical difficulties Feb 29 '16 at 5:38
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  • $\begingroup$ Thanks! I did try what J. M. suggested. No apparent simplification, however. $\endgroup$ – Paul B. Slater Feb 29 '16 at 14:39
  • $\begingroup$ Thanks! I did try what J.M. suggested. No apparent simplification, however. I simply want to obtain a bivariate polynomial in s1 and s2 in "standard" form. That is, without the use of the Piecewise command. Something like "s1^3 +s1 s2 +s2^2.....", by way of illustration. $\endgroup$ – Paul B. Slater Feb 29 '16 at 14:42

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