Talk:Polarization of an algebraic form

The polar form of f is a polynomial

F(u⁽¹⁾, u⁽²⁾, ..., u^(d))

which is linear separately in each u⁽ⁱ⁾ (i.e., F is multilinear) and such that

F(u,u, ..., u)=f(u).

Is this supposed to be a definition of the polar form? In this case, I tend to believe that we need a condition more - namely, F should be symmetric in its d (vectorial) variables. Is this correct? - darij

I've added |_{\lambda=0} to the end of the construction of the polar form of f, as per p. 41 of Procesi's "Lie Groups". It might be confusing since there are multiple \lambda, but it's cleaner to write the one to indicate that they are all zero.209.189.246.35 (talk) 16:23, 3 June 2011 (UTC) — Preceding unsigned comment added by 209.189.246.35 (talk) 16:21, 3 June 2011 (UTC)[reply]

Mathematica code for the given two examples in the article

Here, I would like to show the mathematica code of the example given in the article.

The first example is

$f(x,y)=x^{2}+3xy+2y^{2}$ and the polarization of f is $x1x2+{\frac {3x2y1}{2}}+{\frac {3x1y2}{2}}+2y1y2$ which can be obtianed by the following mathematica code

  f[x_, y_] := x^2 + 3 x y + 2 y^2
  Coefficient[
  f[λ1 x1 +λ2 x2, λ1 y1 + λ2 y2], λ1 λ2 
  ]/Exponent[f[x, y], x]! // ExpandAll

Similarly, the second example can be obtianed by

  g[x_, y_] := x^3 + 2 x y^2
  Coefficient[
  g[λ1 x1 + λ2 x2 +λ3 x3,  λ1 y1 + λ2 y2 +λ3 y3], λ1 λ2  λ3
  ]/Exponent[g[x, y], x]! // ExpandAll

The output will be

$x1x2x3+{\frac {2x3y1y2}{3}}+{\frac {2x2y1y3}{3}}+{\frac {2x1y2y3}{3}}$

I hope this will explain the words in article" F is a constant multiple of the coefficient of λ₁ λ₂...λ_d in the expansion of f(λ₁u⁽¹⁾ + ... + λ_du^(d))."

Vanabel (talk) 02:39, 27 November 2012 (UTC)[reply]