The polynomials p0(x, y, z), p1(x, y, z), p2(x, y, z), ... are defined by p0(x, y, z) = 1 and pn+1(x, y, z) = (x + z)(y + z) pn(x, y, z+1) - z2pn(x, y, z). Show that each polynomial is symmetric in x, y, z.
Solution
We use induction on n.
Let Sn be the statement that (1) pn(x, y, z) = pn(y, x, z), (2) pn(x, y, z) = pn(z, y, x), and (3) (y+z) pn(x, y, z+1) - (x+y) pn(z, y, x+1) = (z-x) pn(x, y, z). Clearly S0 is true ( (1) and (2) are trivial, and both sides of (3) are (z-x) ). So suppose Sn is true.
(1) for n+1 follows immediately. We have pn+1(x, y, z) - pn+1(z, y, x)
= (x+z)(y+z) pn(x, y, z+1) - (x+z)(x+y) pn(z, y, x+1) - z2 pn(x, y, z) + x2 pn(z, y, x), by the identity in the question
= (x+z) (z-x) pn(x, y, z) + (x2 pn(z, y, x) - z2 pn(x, y, z) ), by (3) for n
= (x+z) (z-x) pn(x, y, z) + (x2 - z2) pn(x, y, z) ), by (2) for n
= 0. So (2) holds for n+1.
We have (y+z) pn+1 (x, y, z+1) - (x+y) pn+1(z, y, x+1)
= (y+z) pn+1(z+1, x, y) - (x+y) pn+1(x+1, z, y), using (1) and (2) for n+1
= ( (y+z)(y'+z)(x+y) pn(z+1, x, y') - (y+z)y2pn(z+1, x, y) ) - ( (x+y)(x+y')(y+z) pn(x+1, z, y') - (x+y)y2pn(x+1, z, y), where y' = y+1 and we have used the equality in the question
= (x+y)(y+z)( (y'+z) pn(x, y', z+1) - (x+y') pn(x+1, y', z) ) - y2( (y+z) pn(x, y, z+1) - (x+y) pn(z, y, x+1) ), using (1) and (2) for n
= (x+y)(y+z)(z-x) pn(x, y', z) - y2(z-x) pn(x, y, z), using (3) for n
= (z-x)( (x+y)(z+y) pn(x, z, y+1) - y2pn(x, z, y) )
= (z-x) pn+1(x, z, y), using the identity in the question
= (z-x) pn+1(x, y, z), which establishes (3) for n+1.
So Sn is true for all n. (1) and (2) together establish that pn(x, y, z) is symmetric.
© John Scholes
jscholes@kalva.demon.co.uk
2 Oct 2002