polarization formula for homogeneous polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:51:58Z http://mathoverflow.net/feeds/question/61884 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61884/polarization-formula-for-homogeneous-polynomials polarization formula for homogeneous polynomials mostafa 2011-04-16T00:21:21Z 2011-04-16T08:56:30Z <p>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 ?</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/61884/polarization-formula-for-homogeneous-polynomials/61891#61891 Answer by José Figueroa-O'Farrill for polarization formula for homogeneous polynomials José Figueroa-O'Farrill 2011-04-16T04:18:28Z 2011-04-16T08:56:30Z <p>I'm assuming that $R$ are the reals. In any case, you need to be able to divide by $n!$.</p> <p>Given that, the answer is <strong>yes</strong>. 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.</p> <p><em>Edit by Denis Serre</em>. This suggests the general formula $$n!B(x_1,\ldots,x_n)=\sum_I(-1)^{n-|I|}p(x_I),\qquad x_I:=\sum_{i\in I}x_i.$$</p> <p><em>Further edit by JMF</em>. The formula is proved in this preprint by Erik G.F. Thomas <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.3481" rel="nofollow"><em>A polarization identity for multilinear maps</a></em></p>