ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The pendulum linearisation bound at $g = l = 1$ : exact error bound for the data $δ = 1 / 100$ , $T = 1$

Example

Take $g = l = 1$ in The pendulum field near the downward equilibrium: bounded force derivative, cubic remainder and the finite-time small-angle error bound, so that $ω = 1$ , $F (θ, v) = (v, - \sin θ)$ , $A = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ , $L = 1$ and $∥ A ∥ = 1$ . Choose the ball radius $r = 1 / 2$ , the datum $z_{0} = (δ, 0)$ with $δ = 1 / 100$ , and the horizon $T = 1$ . Then $δ e^{L T} = e / 100 < 3 / 100 < 1 / 2 = r$ because $2 < e < 3$ (The elementary numerical bound $2 < e < 3$ ), so the proposition applies and gives

$∥ z (t) - y (t) ∥_{2} \leq \frac{1}{12 \cdot 10^{4}} (e^{2 t} - e^{t}) (0 \leq t \leq 1),$

where $y (t) = δ (\cos t, - \sin t)$ solves $y^{'} = A y$ , $y (0) = z_{0}$ ; at $t = 1$ the bound reads $(e^{2} - e) / (12 \cdot 10^{4})$ and is at most $e^{3} / (12 \cdot 10^{4})$ . The remainder checks at $∥ z ∥_{2} \leq r$ are: $∣ \sin θ - θ ∣ \leq ∣ θ ∣^{3} / 6 \leq 1 / 48$ , and $∣ \sin θ ∣ \leq ∣ θ ∣ \leq r = 1 / 2$ . All constants are exact; the single inequality $e < 3$ is the only numerical input.

Facts & Assumptions

Given: The parameters $g = l = 1$ , the field $F (θ, v) = (v, - \sin θ)$ , the matrix $A$ , the constants $r = 1 / 2$ , $δ = 1 / 100$ , $T = 1$ , and the solution $y (t) = δ (\cos t, - \sin t)$ of the linear problem.

[L1]

The proposition (The pendulum field near the downward equilibrium: bounded force derivative, cubic remainder and the finite-time small-angle error bound): for this field, $∥ F (z) - A z ∥_{2} = ∣ \sin θ - θ ∣ \leq ∣ θ ∣^{3} / 6 \leq r ∥ z ∥_{2}^{2} / 6$ and $∥ F (z) ∥_{2} \leq L ∥ z ∥_{2}$ with $L = 1$ and $∥ A ∥ = 1$ on $∥ z ∥_{2} \leq r$ ; for $δ e^{L T} < r$ the estimate $∥ z (t) - y (t) ∥_{2} \leq (ω^{2} r / 6) δ^{2} \int_{0}^{t} e^{∥ A ∥ (t - s)} e^{2 L s} d s$ holds on $[0, T]$ (Finite-time linearization error bound: quadratic remainder, exponential growth bound and the variation-of-constants estimate).

[L2]

Exponential and trigonometric calculus (The exponential function is smooth and $(\exp)^{'} = \exp$ , The real exponential function and the number $e$ by a power series, The elementary numerical bound $2 < e < 3$ , The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, Signs, monotonicity intervals, and ranges of sine and cosine, A uniform derivative bound gives a uniform Taylor remainder bound, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ): $2 < e < 3$ ; $e^{t}$ is its own derivative, and $e^{u + v} = e^{u} e^{v}$ for real $u, v$ . For fixed $t > 0$ , $G (s) = e^{t + s}$ has derivative $G^{'} (s) = e^{t + s}$ on $[0, t]$ ; this derivative is continuous, hence integrable, so the fundamental theorem gives $\int_{0}^{t} e^{t + s} d s = e^{2 t} - e^{t}$ . At $t = 0$ the integral and expression both vanish. $\sin$ and $\cos$ have range $[- 1, 1]$ and $\sin 0 = 0$ ; and the Taylor bound gives $∣ \sin θ - θ ∣ \leq ∣ θ ∣^{3} / 6$ .

[L3]

Norm data (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Open ball, closed ball and sphere in a metric space): use the one-based component labels $w_{j} : = w (j - 1)$ for $w \in R^{2}$ , $j \in {1, 2}$ . Then $∥ (w_{1}, w_{2}) ∥_{2}^{2} = w_{1}^{2} + w_{2}^{2}$ ; $y_{1} (t) = δ \cos t$ and $y_{2} (t) = - δ \sin t$ satisfy $y_{1}^{'} = y_{2}$ and $y_{2}^{'} = - y_{1}$ .

Verification

Proof technique: direct; instantiate the proposition at the stated constants, verify the smallness condition with $e < 3$ , evaluate the integral with the fundamental theorem, and check the pointwise remainder bounds.

1.1givenL1L2algebra

Smallness condition: $L = 1$ and $δ = 1 / 100$ give $δ e^{L T} = e / 100$ , and $e < 3$ gives $e / 100 < 3 / 100 < 1 / 2 = r$ , so the datum lies in the range of the proposition with $T = 1$ .

1.2givenL1L2algebra

Pointwise bounds at this data: on $∥ z ∥_{2} \leq 1 / 2$ one has $∣ θ ∣^{3} / 6 \leq (1 / 2) ∥ z ∥_{2}^{2} / 6 = ∥ z ∥_{2}^{2} / 12$ and $∣ \sin θ ∣ \leq ∣ θ ∣ \leq 1 / 2$ , so the two hypotheses of [L1] hold with $C = 1 / 12$ and $L = 1$ ; at $z = (1 / 2, 0)$ this reads $∣ \sin (1 / 2) - 1 / 2 ∣ \leq 1 / 48$ .

2.1step 1.1step 1.2L1L2algebra

The error bound: by [L1] with $ω = 1$ , $r = 1 / 2$ , $δ = 1 / 100$ , $L = ∥ A ∥ = 1$ , $∥ z (t) - y (t) ∥_{2} \leq \frac{1}{12} \cdot \frac{1}{10^{4}} \int_{0}^{t} e^{t - s} e^{2 s} d s = \frac{e^{2 t} - e^{t}}{12 \cdot 10^{4}}$ for $0 \leq t \leq 1$ . By [L2], $e^{t - s} e^{2 s} = e^{t + s}$ and $\int_{0}^{t} e^{t + s} d s = e^{2 t} - e^{t}$ , including $t = 0$ . This evaluates the upper bound, not the actual nonlinear error.

3.1step 2.1L1L3algebra∎

Reading at $t = 1$ : the bound becomes $(e^{2} - e) / (12 \cdot 10^{4}) < e^{3} / (12 \cdot 10^{4})$ , a finite quantity of order $δ^{2} = 10^{- 4}$ , and $y_{1} (t) = δ \cos t$ satisfies $∣ z_{1} (t) - y_{1} (t) ∣ \leq ∥ z (t) - y (t) ∥_{2}$ by [L3]; the estimate is restricted to $[0, 1]$ and asserts nothing about the nonlinear pendulum beyond it.

The pendulum linearisation bound at $g = l = 1$ : exact error bound for the data $δ = 1 / 100$ , $T = 1$

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

The pendulum linearisation bound at g=l=1g=l=1: exact error bound for the data δ=1/100δ=1/100, T=1T=1

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

The pendulum linearisation bound at $g = l = 1$ : exact error bound for the data $δ = 1 / 100$ , $T = 1$