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\(\ell ^1_\nu \) is a commutative Banach algebra over \(\mathbb {R}\):
Complete normed space (Banach)
Commutative ring under \(\star \)
\(\mathbb {R}\)-algebra with compatible scalar multiplication
Submultiplicative: \(\| a \star b\| \leq \| a\| \cdot \| b\| \)
Scalar-norm compatibility: \(\| c \cdot a\| = |c| \cdot \| a\| \)
\(\| 1\| = 1\)
This is the full Mathlib NormedAlgebra + CompleteSpace structure.
Consider an open set \(U \subset \mathbb {R}^n\). Let \(f: U \rightarrow \mathbb {R}^n\) be a \(C^1\) function. Fix a point \(x_0 \in U\) and assume \(\overline{B}_\rho (x_0) \subset U\) for some \(\rho {\gt} 0\). Then, for all \(x, y \in \overline{B}_\rho (x_0)\),
where \(\| \text{Df}(\cdot )\| \) denotes the operator norm.
A metric space \((X, d)\) is complete if every Cauchy sequence converges in \(X\), i.e., if given any \(\varepsilon {\gt} 0\) there exists \(N(\varepsilon )\) such that \(n, m {\gt} N(\varepsilon )\) implies \(d(x_n, x_m) {\lt} \varepsilon \), then there is some \(y \in X\) with \(\lim _{n\rightarrow \infty } x_n = y\).
For \(\lambda _0 = 1\) and \(N = 2\), the approximate solution is:
These are the first three Taylor coefficients of \(\sqrt{1 + z}\) at \(z = 0\).
Let \((X, d)\) be a metric space. A function \(T: X \rightarrow X\) is a contraction if there is a number \(\kappa \in [0, 1)\), called a contraction constant, such that
for all \(x, y \in X\).
For a function \(f: E \rightarrow F\) and an approximate inverse \(A: F \rightarrow E\), the Newton-like map is
Note that \(T: E \rightarrow E\) even though \(f\) maps between different spaces.
Let \((X, \| \cdot \| _X)\) and \((Y, \| \cdot \| _Y)\) be normed linear spaces and let \(A: X \rightarrow Y\) be a continuous linear map. The operator norm on \(A\) is given by
For \(\lambda _0 \in \mathbb {R}\), the parameter sequence is
This encodes the polynomial \(\lambda _0 + z\) (i.e., \(\lambda \) when \(z = \lambda - \lambda _0\)).
The \(Y_0\) bound measures the initial defect:
where \(A\) is the approximate inverse. For diagonal \(A\) with entries \(A_{nn} = 1/(2\bar{a}_0)\):
where \(Y_0^{\text{tail}}\) accounts for the tail contribution.
The \(Z_0\) bound measures deviation from identity:
Since \(D_{\bar{a}}F(h) = 2(\bar{a} \star h)\) and \(A\) is diagonal with \(A_{nn} = 1/(2\bar{a}_0)\):
where \(e_n\) is the \(n\)-th standard basis sequence.
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.
Given \(T: E \rightarrow E\) differentiable with: (a) \(\| T(\bar{x}) - \bar{x}\| \leq Y_0\); (b) \(\| \text{DT}(c)\| \leq Z(r_0)\) for all \(c \in \overline{B}_{r_0}(\bar{x})\); (c) \(Z(r_0) \geq 0\); (d) \(p(r_0) = (Z(r_0) - 1)r_0 + Y_0 {\lt} 0\). Then \(T: \overline{B}_{r_0}(\bar{x}) \rightarrow \overline{B}_{r_0}(\bar{x})\).
Suppose for all \(c \in \overline{B}_r(\bar{x})\):
Then \(\| \text{DT}(c)\| \leq Z_0 + Z_1 + Z_2(r) \cdot r = Z(r)\).
When \(A^\dagger = \text{Df}(\bar{x})\) (so \(Z_1 = 0\)), for all \(c \in \overline{B}_r(\bar{x})\): if \(\| I_E - A \circ \text{Df}(\bar{x})\| \leq Z_0\) and \(\| A \circ (\text{Df}(c) - \text{Df}(\bar{x}))\| \leq Z_2(r) \cdot r\), then \(\| \text{DT}(c)\| \leq Z_0 + Z_2(r) \cdot r\).
For \(c \in \mathbb {R}\) and \(a \in \ell ^1_\nu \):
Let \(E, F\) be vector spaces, \(f: E \rightarrow F\), and \(A: F \rightarrow E\) an injective linear map. Let \(T(x) = x - A(f(x))\) be the Newton-like operator. Then:
Let \(f: E \rightarrow F\) and \(A: F \rightarrow E\) be injective. Then for the Newton-like operator \(T(x) = x - A(f(x))\):
This holds even when \(E \neq F\); injectivity of \(A\) is sufficient.
Let \((X, \| \cdot \| )\) be a normed linear space and let \(A, B: X \rightarrow X\) be linear maps. Then:
(1) \(\| Ax\| \leq \| A\| \| x\| \) for all \(x \in X\)
(2) \(\| AB\| \leq \| A\| \| B\| \) (submultiplicativity)
For \(a, b \in \ell ^1_\nu \):
This is essentially definitional: multiplication in \(\texttt{PowerSeries}\ \mathbb {R}\) is the Cauchy product.
Let \((X, d)\) be a complete metric space. If \(T: X \rightarrow X\) is a contraction with contraction constant \(\kappa \), then there exists a unique fixed point \(\tilde{x} \in X\) of \(T\). Furthermore, \(\tilde{x}\) is globally attracting, and for any \(x \in X\),
For \(\lambda _0 = 1/3\), \(\nu = 1/4\), and \(N = 2\):
(Existence) There exists a unique \(\tilde{a} \in \ell ^1_\nu \) with \(\| \tilde{a} - \bar{a}\| _{\ell ^1_\nu } \leq r_0\) satisfying \(F(\tilde{a}) = \tilde{a} \star \tilde{a} - c = 0\).
(Analytic interpretation) The power series \(\tilde{x}(z) = \sum _{n=0}^\infty \tilde{a}_n z^n\) converges for \(|z| \leq \nu \) and satisfies \(\tilde{x}(\lambda - \lambda _0) = \sqrt{\lambda }\) for \(\lambda \in [\lambda _0 - \nu , \lambda _0 + \nu ]\) with \(\lambda {\gt} 0\).
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 \(T(\tilde{x}) = \tilde{x}\).
Let \(T: E \rightarrow E\) be Fréchet differentiable and \(\bar{x} \in E\). Suppose:
Define \(p(r) := (Z(r) - 1)r + Y_0\).
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 \(T(\tilde{x}) = \tilde{x}\).
Let \(E\) and \(F\) be Banach spaces and \(f: E \rightarrow F\) be Fréchet differentiable. Suppose \(\bar{x} \in E\), \(A^\dagger : E \rightarrow F\), and \(A: F \rightarrow E\) with \(A\) injective. Assume:
Define \(p(r) := Z_2(r)r^2 - (1 - Z_0 - Z_1)r + Y_0\).
If there exists \(r_0 {\gt} 0\) such that \(p(r_0) {\lt} 0\), then there exists a unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(f(\tilde{x}) = 0\).
Suppose that \(U \subset \mathbb {R}^n\) is open and that \(f: U \rightarrow \mathbb {R}^n\) is \(C^1\). Consider \(x, y \in U\) such that the line segment \((1-t)x + ty\), \(t \in [0,1]\), is contained in \(U\). Then,
where \(\text{Df}\) denotes the Jacobian of \(f\).
Given \(f: E \rightarrow F\) Fréchet differentiable and injective \(A: F \rightarrow E\) satisfying \(\| A(f(\bar{x}))\| \leq Y_0\), \(\| I_E - A \circ \text{Df}(\bar{x})\| \leq Z_0\), and \(\| A \circ [\text{Df}(c) - \text{Df}(\bar{x})]\| \leq Z_2(r) \cdot r\) for \(c \in \overline{B}_r(\bar{x})\). If \(p(r_0) = Z_2(r_0)r_0^2 - (1-Z_0)r_0 + Y_0 {\lt} 0\), then there exists unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(f(\tilde{x}) = 0\).
Consider \(f: E \rightarrow E\) Fréchet differentiable, \(\bar{x} \in E\), and \(A: E \rightarrow E\) injective. Assume \(\| A(f(\bar{x}))\| \leq Y_0\), \(\| I_E - A \circ \text{Df}(\bar{x})\| \leq Z_0\), and \(\| A \circ [\text{Df}(c) - \text{Df}(\bar{x})]\| \leq Z_2(r) \cdot r\) for \(c \in \overline{B}_r(\bar{x})\). Define \(p(r) := Z_2(r)r^2 - (1-Z_0)r + Y_0\).
If \(p(r_0) {\lt} 0\), then there exists unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(f(\tilde{x}) = 0\) and \(\text{Df}(\tilde{x})\) invertible.
Let \(\texttt{toPowerSeries}(a) = \sum _n a_n X^n\). Then:
This bridge lets us transport all ring axioms from PowerSeries R.