From here is the following excerpt

Suppose one is given a homogeneous symmetric polynomial $P$ and asked to prove that $P(x_1, \ldots , x_n) ≥ 0$ How should one proceed? Our first step is purely formal, but may be psychologically helpful. We introduce the following notation: $$\sum_\mathrm{sym}Q(x_i)=\sum_\sigma Q(x_{\sigma(i)})$$ where $\sigma$ runs over all permutations of $1, \ldots , n$ (for a total of $n!$ terms). For example, if $n = 3$, and we write $x, y, z$ for $x_1, x_2, x_3$, $$\sum_\mathrm{sym}x^3=2x^3+2y^3+2z^3$$ $$\sum_\mathrm{sym}x^2y=x^2y+y^2z+z^2x+x^2z+y^2x+z^2y$$ $$\sum_\mathrm{sym}xyz=6xyz$$

I cannot understand it properly and my doubt is regarding this:

Permuting third example, let$f(x_1,y_1,z_1)=x_1y_1z_1$ Then permutations are$$f(x,y,z),f(x,z,y),f(y,x,z),f(y,z,x),f(z,y,x),f(z,x,y).\text{ Same as example }$$

Permuting second example, let $f(x_1,y_1)=x_1^2y_1$ Then permutations are$$f(x,y),f(x,z),f(y,x),f(y,z),f(z,y),f(z,x).\text{ Same as example }$$

Permuting second example, let$f(x_1)=x_1^3$ Then permutations are $$f(x),f(y),f(z).\text{ Not same as example }$$ As then $$\sum_\mathrm{sym}x^3=x^3+y^3+z^3$$


Let's try $Q(x,y,z)=f(x)$. So you have permutations:

$$Q(x,y,z),Q(x,z,y), Q(y,x,z),Q(y,z,x),Q(z,x,y),Q(z,y,x).$$

They give $f(x),f(x),f(y),f(y),f(z),f(z)$ correspondingly.

  • $\begingroup$ sorry but why do you consider duplicates? $\endgroup$ – RE60K Jul 31 '14 at 16:52
  • $\begingroup$ aha i understood $\endgroup$ – RE60K Jul 31 '14 at 17:28

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