Certified Evaluations of Hölder Continuous Functions at Roots of Polynomials

Abstract

Various methods can obtain certified estimates for roots of polynomials. Many applications in science and engineering additionally utilize the value of functions evaluated at roots. For example, critical values are obtained by evaluating an objective function at critical points. For analytic evaluation functions, Newton’s method naturally applies to yield certified estimates. These estimates no longer apply, however, for Hölder continuous functions, which are a generalization of Lipschitz continuous functions where continuous derivatives need not exist. This work develops and analyzes an alternative approach for certified estimates of evaluating locally Hölder continuous functions at roots of polynomials. An implementation of the method in Maple demonstrates efficacy and efficiency.

Appendix

Proof of Theorem 1. Suppose that $C \ne 0$ such that $q(x) = C\cdot \prod _{i=1}^d (x - \alpha _i)$. Thus, we know $q'(x) = C\cdot \sum _{i=1}^d \prod _{j \ne i} (x - \alpha _j)$ and $q'(\alpha _i) = C\cdot \prod _{j \ne i} (\alpha _i - \alpha _j) \ne 0$ for all i. Let $p_i(x) = q(x)/(x - \alpha _i) = C\cdot \prod _{j \ne i} (x - \alpha _j)$. Hence, $p_i(\alpha _i) = q'(\alpha _i)$ and $p_i(\alpha _j) = 0$ if $j \ne i$. The polynomials $p_1,\dots ,p_d$ are linearly independent since, if $\sum _{i=1}^d a_i p_i(x) = 0$, then evaluating at $x = \alpha _j$ yields $a_j \cdot q'(\alpha _j) = 0$ which implies $a_j = 0$. Thus, they must form a basis for the d-dimensional vector space of polynomials of degree at most $d-1$.

Since p(x) has degree at most $d-1$, there are unique constants $a_i$ so that $\sum _{i=1}^d a_i p_i(x) = p(x)$. Evaluating at $x = \alpha _j$ yields $a_j q'(\alpha _j) = p(\alpha _j)$ so that $a_j = p(\alpha _j)/q'(\alpha _j)$. Therefore, for all $x\in \mathbb {C}\setminus \{\alpha _1,\dots ,\alpha _d\}$,

$$\begin{aligned} {\displaystyle \frac{p(x)}{q(x)} = \sum _{i=1}^d \frac{p(\alpha _i)}{q'(\alpha _i)} \frac{1}{x-\alpha _i} = \sum _{i=1}^d - \frac{p(\alpha _i)}{\alpha _i q'(\alpha _i)} \frac{1}{1 -x/ \alpha _i}.}\end{aligned}$$

(9)

The terms in (9) have a Taylor series expansion centered at the origin that converge for all x with $|x| < \min \{|\alpha _1|, \ldots , |\alpha _d|\}$ such that, as (8) claims,

$${\displaystyle \frac{p(x)}{q(x)} = \sum _{i=1}^d - \frac{p(\alpha _i)}{\alpha _i q'(\alpha _i)} \sum _{n=0}^\infty \alpha _i^{-n} x^n = \sum _{n=0}^\infty \left( -\sum _{i=1}^d \frac{p(\alpha _i)}{\alpha _i q'(\alpha _i)} \alpha _i^{-n} \right) x^n.} $$

Proof of Theorem 2. Clearly, one has $r_n = \frac{d^n}{dz^n}\left. \frac{p(z)}{q(z)}\right| _{z=0}$. Since p(x) and q(x) have real coefficients, $r_n$ is real for all $n \ge 0$. For $i\in \{1,\dots ,d\}$, let $t^i_n = C_i \alpha _i^{-n}$ so that (8) reduces to $r_n = \sum _{i=1}^d t^i_n$. Moreover, $\alpha _1\in \mathbb {R}\setminus \{0\}$ implies $C_1\in \mathbb {R}\setminus \{0\}$. Clearly, if $\alpha _1<0$, then $t^1_n$ is alternating in sign.

Consider the case when $\alpha _1>0$. First, note that $t^1_n$ and $C_1$ always have the same sign. The following derives a threshold N such that $|r_n - t^1_n| < |t^1_n|$ for all $n > N$. Given such an N, $r_n$ will have the same sign as $t^1_n$ and $C_1$ for $n > N$ and the theorem will be proved. To that end, since $(r_n - t^1_n)/t^1_n = \sum _{i=2}^d t^i_n/t^1_n$,

$${\displaystyle \frac{|r_n - t^1_n|}{|t^1_n|} \le \sum _{i=2}^d \frac{|C_i|}{|C_1|}\frac{|\alpha _1|^n}{|\alpha _i|^n} \le K \left( \frac{m}{M}\right) ^n}$$

for all n. Since, by assumption, $m/M < 1$, there is a threshold N so that $K (m/M)^n < 1$ and $|r_n - t^1_n| < |t^1_n|$ for all $n > N$. We may take N so that $K(m/M)^N=1$ or $N = \log (K)/\log (M/m)$ as claimed.