prove a polynomial identity..

Question

The equation is that

$h_m(x_1, \cdots, x_n, a)-h_m(x_1, \cdots, x_n, b)=(a-b)h_{m-1}(x_1, \cdots, x_n, a, b)$ where $h_m$ is a complete homogeneous symmetric polynomial.

See and find several examples then yes, it looks obvious, though I don't know how to prove the identity.

Thanks.

Answer 1

Notice that we have $$ h_m(x_1, \dotsc, x_n, y_1, \dotsc, y_k) = \sum_{i=0}^m h_{m-i}(x_1, \dotsc, x_n) h_i(y_1, \dotsc, y_k). $$

We thus have $$ h(x_1, \dotsc, x_n,a) = \sum_{i=0}^m h_{m-i}(x_1, \dotsc, x_n) a^i. $$ Therefore \begin{align*} &\, h(x_1, \dotsc, x_n,a) - h(x_1, \dotsc, x_n,b) \\ =&\, \sum_{i=1}^m h_{m-i}(x_1, \dotsc, x_n) (a^i-b^i) \\ =&\, (a-b) \sum_{i=1}^m h_{m-i}(x_1, \dotsc, x_n) \sum_{j=0}^{i-1} a^{i-1-j} b^j. \end{align*} Now we finish with \begin{align*} &\, \sum_{i=1}^m h_{m-i}(x_1, \dotsc, x_n) \sum_{j=0}^{i-1} a^{i-1-j} b^j = \sum_{i=1}^m h_{m-i}(x_1, \dotsc, x_n) h_{i-1}(a,b) \\ &= \sum_{i=0}^{m-1} h_{m-1-i}(x_1, \dotsc, x_n) h_i(a,b) = h(x_1, \dotsc, x_n, a, b). \end{align*}