LemmaStatement: AI-generatedProof: AI-generated

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Finite exponential and smooth quotient remainder control

Statement

Let $d$ be a nonnegative integer and let $A$ be a complex $d \times d$ matrix acting on $C^{d}$ with its Euclidean norm $∥ x ∥_{2} = (\sum_{j < d} ∣ x_{j} ∣^{2})^{1 / 2}$ , let $t \in R$ , and let $n$ be a nonnegative integer. With $∥ A ∥ = \sup_{∥ x ∥_{2} \leq 1} ∥ A x ∥_{2}$ and $e^{t A} = \sum_{k = 0}^{\infty} t^{k} A^{k} / k!$ , one has $∥ e^{t A} - \sum_{k = 0}^{n} t^{k} A^{k} / k! ∥ \leq e^{∣ t ∣ ∥ A ∥} (∣ t ∣ ∥ A ∥)^{n + 1} / (n + 1)!$ .

For $d \geq 1$ the same operator norm is the supremum over unit vectors. For $d = 0$ , the space has only the zero vector, $A = I = 0$ as operators on this space, and the unit-ball supremum and the operator remainder are both zero.

For scalar functions near $t = 0$ , write $h (t) = O (t^{n + 1})$ when there are constants $C, δ > 0$ such that $∣ h (t) ∣ \leq C ∣ t ∣^{n + 1}$ for $∣ t ∣ < δ$ . If $f (t) = \sum_{k = 0}^{n} a_{k} t^{k} + O (t^{n + 1})$ and $g (t) = \sum_{k = 0}^{n} b_{k} t^{k} + O (t^{n + 1})$ , their product has the expansion through order $n$ obtained by truncating the polynomial product. If $b_{0} \neq 0$ , their quotient has an expansion through order $n$ with coefficients $c_{0} = a_{0} / b_{0}$ and $c_{j} = (a_{j} - \sum_{k = 1}^{j} b_{k} c_{j - k}) / b_{0}$ for $1 \leq j \leq n$ . These assertions apply at fixed parameter values. Uniform remainders on a parameter set require uniform bounds on the input coefficients and remainders and, for quotients, a uniform positive lower bound on $∣ b_{0} ∣$ . A denominator vanishing at zero requires separate cancellation or extension before this quotient assertion is applied.

Facts & Assumptions

[F1]

On a complex Banach space the bounded-operator exponential is the absolutely convergent operator-norm series given in the statement. The exponential series of a bounded operator

[F2]

A bounded linear operator has finite operator norm defined by the unit-ball supremum; this equals the unit-sphere supremum on a nonzero domain and is zero on the zero space. Composition satisfies $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ . A bounded linear operator between normed spaces The operator norm as the least bound and as the unit-sphere or unit-ball supremum Composition satisfies ‖ST‖≤‖S‖ ‖T‖

[F3]

For $u \geq 0$ , the real exponential is $e^{u} = \sum_{j = 0}^{\infty} u^{j} / j!$ . The real exponential function and the number e by a power series

[F4]

The coordinate pairing $⟨ x, y ⟩ = \sum_{j < d} x_{j} \overline{y_{j}}$ is an inner product, including when $d = 0$ , and its induced length is a complex norm. The standard formulas ⟨ x,y⟩=∑_k<nxₖ yₖ on ℝⁿ and ∑_k<nxₖyₖ on ℂⁿ are inner products The induced length is a norm

[F5]

The standard coordinate vectors form an ordered basis of $C^{d}$ , with the empty basis at $d = 0$ . Every real or complex normed space with an ordered basis of finite length is Banach. The standard list e : n → Fⁿ with eᵢ(i) = 1_F and eᵢ(j) = 0_F for j ≠ i is an ordered basis of Fⁿ; hence dim_F Fⁿ = n, and F⁰ is the zero space with basis ∅ and dimension 0 Every finite-dimensional normed space is Banach

[F6]

A linear map from a normed space with an ordered basis of finite length into a normed space over the same scalar field is bounded. A linear map from a finite-dimensional normed space is bounded

Proof

Given: The matrix, integer, and scalar expansions in the statement.

Goal: Establish the operator tail estimate and the scalar expansion rules with their stated uniformity conditions.

1.1givenF1F2F4F5F6algebra

By [F4], the displayed Euclidean length makes $C^{d}$ a complex normed space, and its finite ordered basis in [F5] makes it Banach. Matrix multiplication is complex linear: for each $i < d$ , $(A (α x + β y))_{i} = \sum_{j < d} A_{i j} (α x_{j} + β y_{j}) = α (A x)_{i} + β (A y)_{i}$ . Thus [F6] makes $A$ bounded and [F1] applies. If $d = 0$ , there is only one vector and only one operator, so $A = I = 0$ and the asserted norm and remainder are zero by [F2]. If $d \geq 1$ , the unit-ball norm equals the unit-sphere supremum by [F2].

1.2givenalgebra

Write $f = P + r$ and $g = Q + s$ , where $P, Q$ are the displayed polynomials and $r, s = O (t^{n + 1})$ . For $∣ t ∣$ sufficiently small, $P$ and $Q$ are bounded, so $P s + Q r + r s = O (t^{n + 1})$ . Terms of $P Q$ above degree $n$ are also $O (t^{n + 1})$ ; thus the stated product expansion follows.

2.1F1F2F3step 1.1algebra

Put $u = ∣ t ∣ ∥ A ∥$ . By absolute convergence and the triangle inequality, the operator tail is at most $\sum_{k = n + 1}^{\infty} u^{k} / k!$ . Since $(n + 1 + j)! \geq (n + 1)! j!$ for every $j \geq 0$ , this sum is at most $u^{n + 1} e^{u} / (n + 1)!$ . This also covers $u = 0$ .

2.2givenstep 1.2algebra

If $b_{0} \neq 0$ , define $C (t) = \sum_{j = 0}^{n} c_{j} t^{j}$ by the stated recurrence. Its coefficients make every coefficient through degree $n$ of $P - Q C$ zero, so $P - Q C = O (t^{n + 1})$ . Also $g (t) = b_{0} + O (t)$ when $n \geq 1$ , while for $n = 0$ the given remainder gives the same conclusion; hence $∣ g (t) ∣ \geq ∣ b_{0} ∣ / 2$ for sufficiently small $∣ t ∣$ . Consequently $f / g - C = (f - g C) / g = O (t^{n + 1})$ .

3.1step 1.2step 2.2algebra∎

If coefficient and remainder bounds are uniform, the bounds used for the product are uniform after restricting to a common sufficiently small neighborhood. If additionally $∣ b_{0} ∣$ has a uniform positive lower bound, the same is true of $∣ g (t) ∣$ , and the recurrence and quotient estimate are uniform. Without that lower bound the argument gives only fixed-parameter control. Cancellation or extension at a zero denominator must be established separately before invoking the nonzero-denominator result.

Depends on

Used by · 12

Dependency tree · 2 levels

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