Let $f(x)=x^5-x^4-x^3-x^2-x-c$, where $c>2$ is a real number. It is easy to prove that there exists a positive real root $\alpha>2$ of $f(x)$ and all the other roots are non real.

Also, by using Kakeya-Enestrom theorem, it is possible to prove that if $\beta$ is a non real root of $f(x)$, then $|\beta|\leq \alpha$.

However, I would like to prove that this inequality is strict, but I am not being able to prove this. Someone has some suggestion?

The same problem happens by defining $f_c(x)=x^k-x^{k-1}-\cdots - x- c$, where $c>2$. So, $f_c(x)$ has only one positive root, say $\alpha$, which must be dominant, i.e., if $\beta$ is another root, then $|\beta|<\alpha$.

Thanks in advance!

  • $\begingroup$ The question received two correct (and nice) answers -- any reason not to accept one of them? $\endgroup$ – Peter Mueller Jan 30 at 16:55

Write the equation in the form $1=x^{-1}+x^{-2}+x^{-3}+x^{-4}+cx^{-5}:=f(x) $. If $|\beta|\geqslant \alpha$, the RHS has absolute value at most $f(\alpha) =1$ with equality if and only if $|\beta|=\alpha$ and all five summands $\beta^{-1} $ etc are positive reals. That is, $\beta=\alpha$.


$f(x)$ is the characteristic polynomial of its companion matrix $$ A = \pmatrix{0 & 0 & 0 & 0 & c\cr 1 & 0 & 0 & 0 & 1\cr 0 & 1 & 0 & 0 & 1\cr 0 & 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 1 & 1\cr}$$ and $A^5$ has all entries $> 0$. Therefore by the Perron-Frobenius theorem there is a positive eigenvalue of multiplicity $1$ strictly greater in absolute value than all other eigenvalues.

This works for all $c > 0$.

  • $\begingroup$ Why all the entries of $A^5$ are positive? I didn't see this. $\endgroup$ – Jeremy Jan 27 at 22:30
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    $\begingroup$ It's easy to prove. Look at the directed hypergraph of $5$ vertices where arcs correspond to positive entries of $A$ (including a loop $5 \to 5$). Entry $(i,j)$ of $A^k$ is positive if you can get from $i$ to $j$ in $k$ steps. You can always do that here: the first step goes to $5$, then stay at $5$ for $j-1$ steps, then decrease by one $5-j$ times. $\endgroup$ – Robert Israel Jan 27 at 23:27
  • $\begingroup$ Where can I find this fact: 'Entry $(i,j)$ of $A^k$ is positive if you can get from $i$ to $j$ in $k$ steps?" By steps means a path with possibilty of cycles? Thanks! $\endgroup$ – Jeremy Jan 28 at 20:24
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    $\begingroup$ @Jeremy: That follows directly from the definition of matrix multiplication and the assumption that all entries of $A$ are non-negative. $\endgroup$ – Peter Mueller Jan 29 at 8:45

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