Bounds for the Roots of Some Polynomials
Abstract Generated abstract
This paper derives location and modulus bounds for the zeros of two families of denominator polynomials arising from convergents to a continued fraction expansion of the confluent hypergeometric series. Using positivity properties of explicit formulas for the squared modulus in a half-plane, the author shows that the roots lie in a prescribed open half-plane. Further coefficient ratio estimates for associated real polynomials evaluated on the imaginary axis yield lower bounds for the absolute values of the roots, including separate radius formulas for the odd and even denominator cases under the parameter condition \(0 \leq \alpha < 1\).
Full Text
Bounds for the Roots of Some Polynomials
V. E. Kornilov
(Presented by the Department of Higher Mathematics)
The article studies bounds for the roots of the polynomials (1) and (10), which are the denominators of the following convergent fraction
\[ \Phi(1;\,1+\alpha;\,cx)=\frac{p_m(cx)}{q_m(cx)}+r_m(cx),\quad m=1,\,2,\ldots, \]
where
\[ \Phi(1;\,1+\alpha;\,cx)=\sum_{n=0}^{\infty}\frac{(cx)^n}{(1+\alpha)_n},\quad \alpha\ne -1,-2,\ldots \]
1. Theorem. All roots of the polynomial \((c=a+bi)\)
\[ q_{2k+1}(cx)=\sum_{n=0}^{k} C_k^n\,\frac{(cx)^n}{(-\alpha-2k)_n},\quad k=0,\,1,\,2,\ldots \tag{1} \]
are located in the region \(\operatorname{Re}z>0,\ z=ax+bxi\), and outside the circle centered at the origin with the following radius:
\[ R=2\sqrt{2k+\alpha-1},\quad k=1,\,2,\ldots;\ 0\leq \alpha<1, \tag{2} \]
Proof. In the region \(\operatorname{Re}z=ax\leq 0\), the square of the modulus [1]
\[ |q_{2k+1}(cx)|^2= \sum_{n=0}^{k} \frac{C_k^n(-\alpha-k)_n|cx|^{2n}} {(-\alpha-2k)_n(-\alpha-2k)_{2n}} \times \]
\[ {}\times \sum_{m=0}^{k-n} C_{k-n}^m\,\frac{(2ax)^m}{(2n-\alpha-2k)_m} \tag{3} \]
has all terms greater than zero; therefore all roots of the polynomial (1) are located in the region \(\operatorname{Re}z>0\).
Let \(\alpha_1,\ldots,\alpha_k\) be the roots of the polynomial (1); then
\[ q_{2k+1}(cx)=\prod_{l=1}^{k}\left(1-\frac{cx}{\alpha_l}\right). \tag{4} \]
If \(cx=ix\), then on the basis of equality (3)
\[ |q_{2k+1}(ix)|^2=\prod_{l=1}^{k}\left(1+\frac{x^2}{\alpha_l^2}\right), \tag{5} \]
and also (here \(x^2=u^{-1}>0\))
\[ u^k\left|q_{2k+1}(i\sqrt{u})\right|^2 = \sum_{n=0}^{k} \frac{C_k^n(-a-k)_n u^{k-n}} {(-a-2k)_n(-a-2k)_{2n}} = \]
\[ = \sum_{n=0}^{k} a_n u^{k-n},\quad k=1,2,\ldots \tag{6} \]
Let us find the ratio of the preceding coefficient to the following one for the polynomial (6):
\[ \frac{a_{n-1}}{a_n} = \frac{n(2k+a+1-n)(2k+a+1-2n)_2} {(k-n+1)(k-n+1+a)}; \quad n=1,\ldots,k. \tag{7} \]
The minimum of the right-hand side of equality (7) for all \(n\) simultaneously is established by the following easily verified inequalities:
\[ \frac{a_{n-1}}{a_n}> \frac{4n(2k+a+1-n)(2k+a+1-2n)} {2k+a+2-2n} > \]
\[ >4(2k+a-1); \quad n=1,\ldots,k;\quad 0\le a<1. \tag{8} \]
By virtue of (5), (8), and ([2], ed. 4, pp. 295–296),
\[ \min|\alpha_l|\ge 2\sqrt{2k+a-1}=R,\quad k=1,2,\ldots, \tag{9} \]
as was required to be proved.
2. Theorem. All roots of the polynomial
\[ q_{2k}(cx)= \sum_{n=0}^{k} C_k^n \frac{(cx)^n}{(1-a-2k)_n}, \quad k=1,2,\ldots \tag{10} \]
are located in the half-plane \(\operatorname{Re} z>0,\ z=ax+bxi\), and in the exterior of the circle with center at the origin and the following radius:
\[ R_1= \begin{cases} \min\{\sqrt{k(k+a)(a+1)},\,2\sqrt{2k+a-2}\},\\ \qquad k=1,\ldots,6;\\ 2\sqrt{2k+a-2},\quad k=7,8,\ldots;\quad 0\le a<1. \end{cases} \tag{11} \]
Proof. In the domain \(\operatorname{Re} z=ax\le 0\), the square of the modulus (3) (\(a\) must be replaced by \(a-1\)) has positive summands; therefore all roots of the polynomial (10) are located in the domain \(\operatorname{Re} z<0\).
By virtue of (1), (3)–(6), we have
\[ u^k\left|q_{2k}(i\sqrt{u})\right|^2 = \sum_{n=0}^{k} \frac{C_k^n(1-a-k)_n u^{k-n}} {(1-a-2k)_n(1-a-2k)_{2n}} = \]
\[ = \sum_{n=0}^{k} b_n u^{k-n},\quad k=1,2,\ldots \tag{12} \]
Next we find:
\[ \frac{b_{n-1}}{b_n} = \frac{n(2k+a-n)(2k+a-2n)_2} {(k-n+1)(k-n+a)}; \quad n=1,\ldots,k. \tag{13} \]
It is not difficult to verify the following inequalities:
\[ \frac{b_{n-1}}{b_n} > 8n\left(k+\frac{a}{2}-n\right) \left(k+\frac{a}{2}-\frac{n}{2}\right) : \left(k+\frac{a}{2}+\frac{1}{2}-n\right) > \]
\[ >8n\left(k+\frac{a}{2}-n\right), \quad n=1,\ldots,k;\quad 0\le a<1. \tag{14} \]
According to the right-hand side of inequality (14), the minima of \(b_{n-1}:b_n\) will occur at \(n=1\) and \(n=\kappa\), as follows:
\[ \min \frac{b_{n-1}}{b_n}=\frac{b_0}{b_1}>4(2\kappa+\alpha-2),\quad \kappa=2,3,\ldots, \tag{15} \]
\[ \min \frac{b_{n-1}}{b_n}=\frac{b_{\kappa-1}}{b_\kappa}=\kappa(\kappa+\alpha)(\alpha+1),\quad \kappa=1,\ldots,6. \tag{16} \]
On the basis of inequality (15) and equalities (12) and (16), we obtain equality (11); thus the theorem is proved.
References
- V. E. Kornilov. Application of the theory of continued fractions to the computation of certain types of integrals. Izv. TPI, vol. 205, 1972.
- A. G. Kurosh. Course of Higher Algebra. 6th ed., Fizmatgiz, 1959.