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
--------------------------------------------------------------------------------
(Too long for a comment.)
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 $\,K_{n-1} \le \dfrac{n-1}{2n} K_n\,$.