-1
$\begingroup$

Let $t\in (0,1)$ and

  • ${a_n}{x^n} + .... + {a_1}{x^1} + f(t) = 0$
  • $f(t) $ is continuous decreasing function of $t$.
  • $a_i\ge0$ for all $i$.
  • $y(t)$ is positive real zero of the first equition.

Can we say that $y(t)$ is continuous decreasing function of $t$?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Which zero? In any case, the answer is no. $\endgroup$
    – Alex Degtyarev
    Commented Feb 8, 2016 at 14:57
  • $\begingroup$ @AlexDegtyarev - I edited this post. $\endgroup$
    – Roger
    Commented Feb 8, 2016 at 15:01
  • 1
    $\begingroup$ You may well consider the dependence on $s$ of the solutions to $\sum a_kx^k+s=0$. Assuming you have an isolated positive root, its location will be a differentiable function of $s$ in a small neighborhood. You will be able to see the sign of its derivative from the usual chain rule. Then it's just a matter of precomposing with $f(t)$. $\endgroup$
    – Lev Borisov
    Commented Feb 8, 2016 at 15:40

1 Answer 1

Reset to default
1
$\begingroup$

Counterexample: Let $n=1, a_1 = 1, f(t) = -t$.

Then $y(t) = t$ which is a continuous increasing function of $t$.

Did you mean to say increasing? Because in that case, the conjecture is true in intervals where $y(t)$ is strictly greater than zero.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ The conjecture is never true. Take your favorite polynomial p(x) with at least one extremum and let $f(t)=t$ or $-t$, whichever fits the conjecture. As is obvious from the graph, there would be both increasing and decreasing branches. One might be able to say something about the minimal/maximal root (and then, obviously, it suffices to consider the case $f(t)=\pm t$). $\endgroup$
    – Alex Degtyarev
    Commented Feb 8, 2016 at 15:42
  • $\begingroup$ Ah, but the problem states that all $a_i > 0$. So there will not be an extremum at any $x>0$. $\endgroup$
    – Mark Fischler
    Commented Feb 8, 2016 at 19:10
  • $\begingroup$ Oops! Sorry. Then one of them is certainly true (probably, yours), and it seems sufficient to consider $f(t)=t$. $\endgroup$
    – Alex Degtyarev
    Commented Feb 8, 2016 at 20:08

Not the answer you're looking for? Browse other questions tagged .