# Can we prove that all roots of those polynomials are real negative?

Suppose I have a strictly decreasing sequence of positive constants:

$$K = \{K_1, \dots, K_n \, | K_i \in (0;1) \wedge K_i > K_{i+1} \}$$

And I create polynomials as follow (I read they are monic polynomials):

\begin{align} p_1(x) &= x + K_1 \\ p_2(x) &= x^2 + K_2 x + K_2K_1 \\ p_3(x) &= x^3 + K_3 x^2 + K_3K_2 x + K_3K_2K_1 \\ p_4(x) &= x^4 + K_4 x^3 + K_4K_3 x^2 + K_4K_3K_2x + K_4K_3K_2K_1 \end{align}

Because all coefficients are positive, using Descartes' Rule I can say that there is no positive roots. Applying Descartes' Rule on $p_k(-x)$ we have an alternated serie, thus the sign will change $k$ times, then we have at maximum $k$ negative real roots, or maybe some are pair of complex conjugates.

If we choose some typical values of $K_i$:

$$K = \{ 10^{-5}, 10^{-7}, 10^{-9}, 10^{-10}, 10^{-11} \}$$

Polynomials look like: Taking the inverse of polynomials shows there must have some vertical asymptotes: Looking for roots numerically, we have: Here are my observations:

• Functions seems to have exactly $k$ negative roots, but Descartes' Rule is not enough to ensure it;
• Functions seems to pass by $(1,1)$ and $(-1,\pm 1)$ but it is not the case, they pass near by;
• Roots are close to original value of constants $K_i$ but not equals;

My questions are:

• Can it be proven that for this class of polynomials all roots have strictly negative real part (degree can exceed 5)?
• Why roots are located near by the value of constants?
• What makes those polynomials converging near by $(1,1)$?

Update

As pointed out in comment:

\begin{align} p_k(x=1) & = \sum\limits_{i=0}^k a_i \simeq 1 \\ p_k(x=1) & > 1 \end{align}

Because $K_i$ are small before 1, where $a_i$ are polynomial coefficients.

Polynomials info

Bellow a log of polynomial coefficients, roots and values:

--------------------------------------------------------------------------------
coeffs = [1.e+00 1.e-05]
r = [-1.e-05]
p(r) = [0.]
p(-1) = -0.99999
p(1) = 1.00001
--------------------------------------------------------------------------------
coeffs = [1.e+00 1.e-05 1.e-12]
r = [-9.89897949e-06 -1.01020514e-07]
p(r) = [1.41363874e-27 0.00000000e+00]
p(-1) = 0.999990000001
p(1) = 1.0000100000010002
--------------------------------------------------------------------------------
coeffs = [1.e+00 1.e-05 1.e-12 1.e-21]
r = [-9.89898980e-06 -1.00000000e-07 -1.01020409e-09]
p(r) = [-1.03330881e-30 -9.57322599e-35  0.00000000e+00]
p(-1) = -0.999990000001
p(1) = 1.0000100000010002
--------------------------------------------------------------------------------
coeffs = [1.e+00 1.e-05 1.e-12 1.e-21 1.e-31]
r = [-9.89898980e-06 -1.00001031e-07 -8.96490095e-10 -1.12683195e-10]
p(r) = [ 1.09000040e-35 -2.62828854e-42 -2.18952885e-47 -2.18952885e-47]
p(-1) = 0.999990000001
p(1) = 1.0000100000010002
--------------------------------------------------------------------------------
coeffs = [1.e+00 1.e-05 1.e-12 1.e-21 1.e-31 1.e-42]
r = [-9.89898980e-06 -1.00001031e-07 -8.97935890e-10 -9.99857205e-11
-1.12517827e-11]
p(r) = [-1.06899029e-40  3.30811304e-49 -7.69622702e-55 -5.25720331e-57
7.96545956e-58]
p(-1) = -0.999990000001
p(1) = 1.0000100000010002
--------------------------------------------------------------------------------

• The fact that the polynomials seem to converge to $(1,1)$ simply is due to the fact that you chose your constants $K_i$ to be so small. Take $p_k$, change $x$ by $1$, and it is rather clear that $1$ is the dominant term numerically. – Suzet Aug 3 '18 at 15:10
• Since these polynomials satisfy an identity $p_n(x) = x^n + K_n p_{n-1}(x)$, I wonder if there is a simple Sturm sequence you can construct for $p_n$, or alternatively if the underlying ideas in the proof of Sturm's theorem can be applied directly. – Hurkyl Aug 3 '18 at 15:26
• For all $K_i$ close to $1$ the roots are close to $(n+1)$-th roots of unity. – WimC Aug 3 '18 at 16:54
• @Hurkyl, Thank you for pointing this out. I will check next week about Sturm Sequence. – jlandercy Aug 3 '18 at 20:11
• @WimC, I am not sure I have understood your comment. All $K_i$ are really small, they are more like zero than unity. – jlandercy Aug 3 '18 at 20:13

## 1 Answer

(Too long for a comment.)

$\quad K = \{K_1, \dots, K_n \, | K_i \in (0;1) \}$

This condition is not enough to ensure that all roots are real.

For example, the quadratic $\,p_2(x) = x^2 + K_2 x + K_2K_1\,$ has no real roots if $\,K_1 \gt \dfrac{K_2}{4}\,$.

A necessary (but not sufficient) condition for all roots of $\,p_n(x)\,$ to be real is that the quadratic $\,p_n^{(n-2)}(x)\,$ must have both roots real, which translates into $\,K_{n-1} \le \dfrac{n-1}{2n} K_n\,$.

• Thanks for answering and pointing this out. Effectively, when my constant are too close from each other, I get complex roots. What do you mean by $p_n^{(n-2)}$: a power of polynomial or its second derivative? How can $p_{5}^{3}$ be a quadratic, it has a a term in $x^{15}$? Could you elaborate a bit more: how did you get the last inequality? – jlandercy Aug 4 '18 at 6:09
• @jlandercy What do you mean by I mean $\,p_n^{(n-2)}\,$ to be the $\,(n-2)^{th}\,$ derivative of $\,p_n\,$. It has degree $\,n - (n-2)=2\,$, so it is a quadratic indeed. – dxiv Aug 4 '18 at 6:14
• This observation can be extended for each adjacent pair. Note that if $p$ has only real roots then so has $p’$. A quick check shows that if $K_1, \ldots K_n$ is “admissible” then so is the shorter sequence $$\tfrac12K_2, \tfrac23K_3, \ldots, \tfrac{n-1}nK_n.$$ Repeating this we get that each sequence $$\tfrac1mK_m, \tfrac2{m+1}K_{m+1}, \ldots, \tfrac{n-m+1}nK_n$$ is admissible. Therefore $$K_{m+1}\geq 2\left(1+\frac1m\right)K_m$$ for $m\in\{1,\ldots,n-1\}$. – WimC Aug 4 '18 at 8:26
• @WimC Right. Also, using the same argument for $\,x^np_n(1/x)\,$ gives more necessary conditions. – dxiv Aug 4 '18 at 17:14