|
5
2
|
given a homogeneous polynomial p of dgree n on $R^d$, there is a unique symmetric n-linear functional $B$ on $(R^d)^n$ such that $p(x)=B(x,..,x)$. The question is: Can we get $B$ by means of a polarization formula as in the case $n=2$ for quadratic forms ? Thanks in advance. |
||
|
|
|
4
|
I'm assuming that $R$ are the reals. In any case, you need to be able to divide by $n!$. Given that, the answer is yes. I can't locate a reference, but here's the formula for $n=3$, say: $$ 6 B(x,y,z) = p(x+y+z) - p(x+y) - p(y+z) - p(z+x) + p(x) + p(y) + p(z) $$ which should give you a hint as to the general case. Edit by Denis Serre. This suggests the general formula $$n!B(x_1,\ldots,x_n)=\sum_I(-1)^{n-|I|}p(x_I),\qquad x_I:=\sum_{i\in I}x_i.$$ Further edit by JMF. The formula is proved in this preprint by Erik G.F. Thomas A polarization identity for multilinear maps |
|||||
|
