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Suppose that I have a function $f(x_1,x_2,\ldots,x_n)$ that is a multivariate polynomial linear in each argument, i.e. each additive term contains $x_i$ raised to the power 1 or zero, but no more than one. Formally, for the set $\mathcal{B}_n$ of all $n$-bit binary vectors, and vector $\beta\in \mathcal{B}_n$, denote $\beta_i$ as the $i$-th element of $\beta$ (either 0 or 1). Then, $f(x_1,x_2,\ldots,x_n)=\sum_{\beta\in \mathcal{b}_n}c_\beta\prod_{i=1}^nx_i^{\beta_i}$, where $c_\beta$ is a coefficient associated with each $\beta$.

Now suppose also that $f(\cdot)$ is shift-symmetric, that is: $f(x_1,x_2,\ldots,x_n)=f(x_2,x_3,\ldots,x_n,x_1)=\ldots=f(x_n,x_1,\ldots,x_2)$

I am interested in properties of the stationary points of such $f(\cdot)$ for non-negative arguments $x_i\geq 0$. It seems to me (from experimentation) that, if they exist (see comment from @copper.hat) at least one is achieved with equal $x_i$'s: $x_1=x_2=\ldots=x_n$ (at least in the cases I tested). I am wondering if this is true for all functions $f(\cdot)$ defined above, and if so, how to prove that. Also, when it exists, does each $f(\cdot)$ have a unique non-negative stationary point? ($f(\cdot)$ that has two stationary points, one being negation of the other, is pointed out by @copper.hat in the comments) I am also interested in the conditions for $f(\cdot)$ for existence of at least one stationary point (I think that has to do with the signs of the coefficients).

This problem arises in the study of the trace of the inverse of the sum of circulant and diagonal matrices... Reason I ask this is because I can't find any literature on the properties of the shift-symmetric multivariate polynomials. Maybe I am not looking in the right places.

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You could characterize the polynomial as "linear in each argument". –  joriki Apr 14 '12 at 5:03
    
This is probably not relevant to the question, but "for a set $\mathcal B_n$ of all $n$-bit binary vectors" makes no sense. There's only one set of all $n$-bit binary vectors, so either the article should be definite or the "all" should be deleted. –  joriki Apr 14 '12 at 5:05
    
@joriki Thanks for the wording suggestions, I've changed the question appropriately (and corrected the article in front of "set $\mathcal{B}_n$ as well). –  M.B.M. Apr 14 '12 at 5:16
    
$f$ may not have any stationary points; $f(x_1,x_2) = x_1+x_2$. –  copper.hat Apr 14 '12 at 5:46
    
Sometimes people use the notation $x^{\beta} = x_1^{\beta_1}...x_n^{\beta_n}$. –  copper.hat Apr 14 '12 at 6:00
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