6
$\begingroup$

I am interested in the following expression: $$ F_{k_1,\ldots,k_n}(t):=\sum_{\sigma\in S_n}\cos(\sigma(1)k_1t)\cos(\sigma(2)k_2t)\cdots\cos(\sigma(n)k_nt) $$ where $k_1, \ldots, k_n$ are natural numbers (can be assumed all different) and $\sigma$ runs over the full permutation group on $\{1,...,n\}$.

Does this resemble anything simpler? I want any kind of additional information. To ask something slightly more concrete: expanding this into powers of $t$, coefficients are obviously symmetric functions of $k_1,\ldots,k_n$; they look quite impenetrable. For example, $$ F_{k_1,k_2,k_3,k_4}(t) = 24-90s_1t^2 + \left(\frac{177}2s_1^2+96s_2\right)t^4 - \left(\frac{163}4s_1^3 + \frac{233}2s_1s_2 + 21s_3 \right)t^6 + \text{(more and more horrible fractions)}, $$ where $s_1, s_2, \ldots$ are the elementary symmetric functions of the $k_i$'s, $s_1=k_1+k_2+k_3+k_4$, $s_2=k_1k_2+k_1k_3+...+k_3k_4$, etc. How to treat this?

What would be ideal for me is a decomposition into a product of some nice terms (preferably into sums of sines or cosines, something like that).

$\endgroup$
  • $\begingroup$ Where you wrote $F_{k_1,k_2,k_3,k_4}(t) = \cdots$, did you have in mind some specific values of $k_1,k_2,k_3,k_4$? $\qquad$ $\endgroup$ – Michael Hardy Feb 29 '16 at 17:02
  • $\begingroup$ Fixed but arbitrary. I need natural numbers but they might be just some variables too. Actually after all I have some (formal) infinite sum over all finite tuples $k_1,k_2,...,k_n$ for all $n$. $\endgroup$ – მამუკა ჯიბლაძე Feb 29 '16 at 17:02
  • $\begingroup$ Then why don't we see $k_1,k_2,k_3,k_4$ on the right side? And where do the numbers on the right side come from? $\qquad$ $\endgroup$ – Michael Hardy Feb 29 '16 at 17:03
  • $\begingroup$ On the right, elementary symmetric functions in these are present - e. g. $s_1=k_1+k_2+k_3+k_4$, etc. $\endgroup$ – მამუკა ჯიბლაძე Feb 29 '16 at 17:04
  • 1
    $\begingroup$ It's the permanent of the matrix $A=\bigl[A_{jl}\bigr]$ with $A_{jl}=\cos(jk_l t)$. Permanents are notoriously difficult to compute. $\endgroup$ – Christian Blatter Mar 1 '16 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.