0
$\begingroup$

Suppose that I have a sequence of $n$ numbers $x_1, \ldots, x_n$ all taken from the real interval $[0,1]$.

I am interested in minimizing the infinity norm of the vector $$ v = \left( \frac{x_{1}}{x_2}, \ldots, \frac{x_{i-1}+x_{i+1}}{x_i}, \dots, \frac{x_{n-1}}{x_n}\right).$$

A few remarks I have observed so far:

  • The vector is homogeneous in the $x_i$, so that we can always fix on variable or suppose that the product $x_1,\ldots, x_n$ is fixed to an arbitrary constant (this we can multiply everything by $x_1,\ldots, x_n$ and get coefficients which are polynomials of degree $n-1$).
  • The vector $v$ is invariant by the transformation $x_i \mapsto x_{n-i+1}$. I do therefore think that the values of $x_i$ minimizing $\|v\|_\infty$ satisfies $x_{n-i+1}=x_i$.
  • The analysis for n=3,4 reveals that this is indeed the case and that the coefficients of $v$ are all equal (which seems reasonable)
  • Using $x_i = \frac{1}{i(n-i+1)}$ gives an upper bound in $2-\Theta(n^{-2})$

Can we derive better bounds on the value of the minimum ? In particular on its asymptotic? I think that asymptotically the minimum tends to $2$ (kind of a concentration of the norm, somehow), but is $n^{-2}$ the correct speed of convergence?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ You wrote: "The vector $v$ is invariant by the transformation $x_i \mapsto x_{n-i+1}$." This is not true. Also, the vector $v$ would be "more palindromic" if $x_1/x_2$ were replaced by $x_2/x_1$ -- which way is it really? Also, for $x_i = \frac1{i(n-i+1)}$, we have $\|v\|_\infty>2$.Clearly, the optimal $n$-tuple $x$ must be strictly concave. $\endgroup$
    – Iosif Pinelis
    Commented Dec 8, 2020 at 14:25
  • 1
    $\begingroup$ Also, where does this problem arise? $\endgroup$
    – Iosif Pinelis
    Commented Dec 8, 2020 at 14:32

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .