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I am trying to prove that the global maxima of the following function $$f_n(x_1,\ldots,x_n):=\exp(-\sum_{i=1}^n x_i ^2)\prod_{1\leq i<j\leq n}(x_i-x_j)^2\prod_{1\leq i,j\leq n}\frac{1}{\sqrt{1+(x_i+x_j)^2}}$$ (note the second product does not have the restriction $i\lt j$) satisfy the following symmetry property: If $x^*=(x_1^*,\ldots,x_n^*)$ is a global maximum of $f_n$, then the set $\{x_1^*,\ldots,x_n^*\}$ is symmetric around the origin in the sense that if $a\in \{x_1^*,\ldots,x_n^*\}$, then $-a\in \{x_1^*,\ldots,x_n^*\}$ too.

I expect this to be true, but I have no proof. I have checked it numerically for $n=2,3,4$, and I am trying to come up with a nice proof of this fact without much success.

Any suggestions would be appreciated!

Enrique Acosta
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  • Isn't $(1, 1,...,1)$ a local (really global) minimum? – Jose27 Apr 10 '12 at 04:35
  • You are right. I meant the global maxima and not all the global minima along all sorts of hyperplanes $x_i=x_j$. I edited the question. – Enrique Acosta Apr 10 '12 at 05:50
  • Can we see the data for n=2,3,4 to get the idea of what additional symmetry might be there? – zyx Apr 10 '12 at 18:44
  • @Enrique: I am having trouble verifying your maximum for $n=2$. Does $f_2(x_1,x_2) = \frac{e^{-x_1^2-x_2^2} \left(x_1-x_2\right)^2}{\sqrt{1+4 x_1^2} \sqrt{1+4 x_2^2} \left(1+\left(x_1+x_2\right)^2\right)}$? – user26872 Apr 20 '12 at 06:14
  • @oenamen: Yes that is the f. You are right, there is something wrong with my numbers. I'll remove them for the time being while I check them. I am getting maxima for $f_2$ at (-0.5,0.5) and (0.5,-0.5) which have the expected property. – Enrique Acosta Apr 21 '12 at 16:18
  • @Enrique: I agree. The maxima for $f_2$ are at $(-0.5,0.5)$ and $(0.5, -0.5)$. The value of the function there is about $0.303$. In what context did this function arise? – user26872 Apr 21 '12 at 17:02
  • @oenamen: It is related to graph colorings. See http://www.sciencedirect.com/science/article/pii/037026939390075S – Enrique Acosta Apr 23 '12 at 17:59

1 Answers1

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Since $f_n(x_1,\ldots,x_n) = f_n(-x_1,\ldots,-x_n)$, if $x^*$ is a critical point, then so is $-x^*$. The symmetry property follows from this.

copper.hat
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  • In fact, it seems there is much more symmetry, if $\sigma$ is a permutation of ${1,...n}$, then $f_n(x_{\sigma(1)},\ldots,x_{\sigma(n)}) = f_n(x_1,\ldots,x_n)$, and so the function is independent of permutations of its arguments. Hence the critical points have the same symmetry. – copper.hat Apr 10 '12 at 07:36
  • I had noticed that, but how does the symmetry property follow from what you say? Why are the sets ${x_1^,\ldots,x_n^}$ and ${-x_1^,\ldots,-x_n^}$ equal? – Enrique Acosta Apr 10 '12 at 14:53
  • @Enrique: It doesn't, I misunderstood your question. Sorry. – copper.hat Apr 10 '12 at 16:31