Consider $f(x)=x^n-x^s-1$ and $g(x)=x^i-x^j-1$ , I want to find $Resultant(f,g)$. It is well known that it is determinant of a Sylvester matrix but, I am finding it to obscure to evaluate in that way.

Is there some known result for such special cases ?

It is known that the Discriminant of a trinomial has a closed form expression. (LINK)

  • 1
    $\begingroup$ Posted some months ago to m.se, math.stackexchange.com/questions/1943631/… but with no answers. $\endgroup$ May 16, 2017 at 6:48
  • 7
    $\begingroup$ But why the votes to close? Wouldn't it be nice to have a closed-form formula for the resultant as a function of $n,s,i,j$ – or to have an acknowledgement that no such formula is known? $\endgroup$ May 16, 2017 at 6:56
  • 1
    $\begingroup$ @GerryMyerson I would like to add that, if we can find a formula for Resultant of general trinomials, then the resilt will subsume this paper. ijpam.eu/contents/2012-74-1/5/5.pdf $\endgroup$
    – xyz
    May 16, 2017 at 10:03
  • 1
    $\begingroup$ I made some calculations for the easiest non-trivial case $s=j=1$. Let's call the resultant $A(n,i)$. Then $A(n,2)=- floor(\varphi^{n-2})+((n-1)\mod 2)$ (oeis.org/A001350) and $A(n,3)$ is oeis.org/A001945 (with an offset). $A(n,4)$ is not in OEIS. $\endgroup$ Jan 12, 2018 at 23:15
  • 1
    $\begingroup$ And another observation: $|(A(n,n+k))_k|$ seems to be periodic with period $n-1$. $\endgroup$ Jan 12, 2018 at 23:31


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.