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 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!
Show symmetry of the critical points of this function
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multivariable-calculus
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0Isn't $(1, 1,...,1)$ a local (really global) minimum? – 2012-04-10
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0You 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. – 2012-04-10
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0Can we see the data for n=2,3,4 to get the idea of what additional symmetry might be there? – 2012-04-10
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0@Enrique: I am having trouble verifying your maximum for $n=2$. Does $f_2(x_1,x_2) = \frac{e^{-x_1^2-x_2^2} \left(x_1-x_2\right)^2}{\sqrt{1+4 x_1^2} \sqrt{1+4 x_2^2} \left(1+\left(x_1+x_2\right)^2\right)}$? – 2012-04-20
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0@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. – 2012-04-21
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0@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? – 2012-04-21
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0@oenamen: It is related to graph colorings. See http://www.sciencedirect.com/science/article/pii/037026939390075S – 2012-04-23
1 Answers
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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.
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0In 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. – 2012-04-10
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0I 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? – 2012-04-10
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0@Enrique: It doesn't, I misunderstood your question. Sorry. – 2012-04-10