1.4 Radii Polynomials in Finite Dimensions
This section develops the radii polynomial method for proving existence of zeros in finite dimensions. The method provides explicit domains of existence and uniqueness.
Consider a map \(T \in C^1(\mathbb {R}^n, \mathbb {R}^n)\) and let \(\bar{x} \in \mathbb {R}^n\). Let \(Y_0 \geq 0\) and \(Z: (0,\infty ) \rightarrow [0,\infty )\) be a non-negative function satisfying
Define
If there exists \(r_0 {\gt} 0\) such that \(p(r_0) {\lt} 0\), then there exists a unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) such that \(T(\tilde{x}) = \tilde{x}\).
The assumption \(p(r_0) {\lt} 0\) implies \(Z(r_0)r_0 + Y_0 {\lt} r_0\) and hence \(Z(r_0) {\lt} 1\).
Step 1: \(T\) maps \(\overline{B}_{r_0}(\bar{x})\) into itself. Let \(x \in \overline{B}_{r_0}(\bar{x})\). By the Mean Value Inequality:
Step 2: \(T\) is a contraction on \(\overline{B}_{r_0}(\bar{x})\). For \(a, b \in \overline{B}_{r_0}(\bar{x})\), by the Mean Value Inequality: \(\| T(a) - T(b)\| \leq Z(r_0)\| a - b\| \). Since \(Z(r_0) {\lt} 1\), \(T\) is a contraction.
Step 3: Apply Contraction Mapping Theorem. By Theorem 1.1.3, there exists a unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(T(\tilde{x}) = \tilde{x}\).
Given constants \(Y_0, Z_0 \geq 0\) and a function \(Z_2: (0,\infty ) \rightarrow [0,\infty )\), the radii polynomial is defined by
The combined bound is defined as
The radii polynomial can be rewritten as
Direct calculation: \((Z_0 + Z_2(r)r - 1)r + Y_0 = Z_2(r)r^2 + Z_0 r - r + Y_0 = Z_2(r)r^2 - (1-Z_0)r + Y_0\).
If \(Y_0 \geq 0\), \(r_0 {\gt} 0\), and \(p(r_0) {\lt} 0\), then \(Z(r_0) {\lt} 1\).
By Lemma 1.4.4, \(p(r_0) = (Z(r_0) - 1)r_0 + Y_0 {\lt} 0\). Since \(Y_0 \geq 0\), we have \((Z(r_0) - 1)r_0 {\lt} 0\). Since \(r_0 {\gt} 0\), dividing gives \(Z(r_0) - 1 {\lt} 0\), i.e., \(Z(r_0) {\lt} 1\).
The main radii polynomial theorem (Theorem 2.4.2) is stated and proved in full generality for Banach spaces in Section 2.2 as Theorem 2.2.23. Here we apply it to the finite-dimensional case \(E = \mathbb {R}^n\).
1.4.1 Example 2.4.5: Finding \(\sqrt{2}\)
We demonstrate the radii polynomial method on the simplest nonlinear function \(f(x) = x^2 - 2\), verifying the existence of a unique zero near \(\bar{x} = 1.3\).
Consider \(f(x) = x^2 - 2\). Choose initial guess \(\bar{x} = \frac{13}{10} = 1.3\), approximate inverse \(A = \frac{19}{50} = 0.38 \approx (f'(\bar{x}))^{-1} = (2\bar{x})^{-1}\), bounds \(Y_0 = \frac{3}{25} = 0.12\), \(Z_0 = \frac{3}{250} = 0.012\), \(Z_2 = \frac{19}{25} = 0.76\), and radius \(r_0 = \frac{3}{20} = 0.15\).
The radii polynomial \(p(r) = 0.76r^2 - 0.988r + 0.12\) satisfies \(p(0.15) {\lt} 0\).
Therefore, there exists a unique \(\tilde{x} \in \overline{B}_{0.15}(1.3) = [1.15, 1.45]\) with \(\tilde{x}^2 = 2\) and \(f'(\tilde{x}) = 2\tilde{x}\) invertible.
Step 1: Compute bounds. For the \(Y_0\) bound: \(|Af(\bar{x})| = |0.38(1.3^2 - 2)| = |0.38 \cdot (-0.31)| = 0.1178 \leq 0.12 = Y_0\).
For the \(Z_0\) bound: \(|1 - A \cdot 2\bar{x}| = |1 - 0.38 \cdot 2.6| = |1 - 0.988| = 0.012 = Z_0\).
For the \(Z_2\) bound: If \(c \in \overline{B}_r(\bar{x})\), then \(|A[f'(c) - f'(\bar{x})]| = |A(2c - 2\bar{x})| = 2|A||c - \bar{x}| \leq 2 \cdot 0.38 \cdot r = 0.76r = Z_2 r\).
Step 2: Verify polynomial negativity. \(p(0.15) = 0.76 \cdot 0.15^2 - 0.988 \cdot 0.15 + 0.12 = 0.0171 - 0.1482 + 0.12 = -0.0111 {\lt} 0\).
Step 3: Apply Theorem 2.2.23. The theorem guarantees a unique zero \(\tilde{x} \in [1.15, 1.45]\) with invertible derivative. Since \(\sqrt{2} \approx 1.414 \in [1.15, 1.45]\), this zero is \(\sqrt{2}\).
There exists a unique \(\tilde{x} \in \overline{B}_{3/20}(13/10)\) with \(\tilde{x}^2 = 2\).
Optimal choice: If \(\bar{x} = \sqrt{2}\) (exact) and \(A = (2\bar{x})^{-1}\) (exact inverse), the radii polynomial becomes \(p(r) = \frac{\sqrt{2}}{2}r^2 - r\), giving \(\text{EI}(p) = (0, \sqrt{2})\).
Approximate choice: With \(\bar{x} = 1.3\) and \(A = 0.38\), the existence interval is approximately \(\text{EI}(p) \approx (0.136, 1.164)\), still sufficient to verify the zero.