6592cd058d
git-subtree-dir: octive-lean git-subtree-mainline:fd3d42ae33git-subtree-split:4b6fcec565
644 lines
26 KiB
Text
644 lines
26 KiB
Text
/-!
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# Numerical Analysis: MATLAB/Octave Concepts Through Lean Proof
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This file formalizes the algorithms from `tutorial.m`. For each method:
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1. A computable **definition** (`#eval` runs it)
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2. **Structural theorems** about the algorithm itself — proved
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3. **Mathematical theorems** about convergence/accuracy — stated and `sorry`'d
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with proof sketches. Filling them in requires the Intermediate Value
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Theorem, Taylor's theorem, etc., which live in Mathlib. Add
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`import Mathlib` to the lakefile to unlock those proofs.
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**How to run:** `lake build NumericalTutorial`
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-/
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namespace NumericalAnalysis
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-- ════════════════════════════════════════════════════════════════
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-- §1 Polynomial Evaluation — Horner's Method
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-- ════════════════════════════════════════════════════════════════
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/-!
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### Background
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A degree-n polynomial `p(x) = c₀ + c₁x + c₂x² + ··· + cₙxⁿ` naively needs
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n additions and n(n+1)/2 multiplications. **Horner's method** rewrites it as
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p(x) = c₀ + x·(c₁ + x·(c₂ + ··· + x·cₙ))
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using only n additions and n multiplications — optimal.
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In MATLAB: `polyval(coeffs, x)` uses Horner internally.
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-/
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/-- Evaluate a polynomial at `x`.
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`coeffs = [c₀, c₁, …, cₙ]` so `coeffs[i]` is the coefficient of xⁱ. -/
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def horner (coeffs : Array Float) (x : Float) : Float :=
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coeffs.foldr (fun c acc => c + x * acc) 0.0
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-- (x−1)(x−2)(x−3) = x³ − 6x² + 11x − 6 at x=2 should be 0
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#eval horner #[-6.0, 11.0, -6.0, 1.0] 2.0 -- 0.0
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#eval horner #[-6.0, 11.0, -6.0, 1.0] 3.5 -- (2.5)(1.5)(0.5) = 1.875
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/-- Abstract Horner over any semiring (needed for algebraic reasoning). -/
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def hornerR {α} [Zero α] [Add α] [Mul α] (coeffs : List α) (x : α) : α :=
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coeffs.foldr (fun c acc => c + x * acc) 0
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/-!
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**Theorem (Horner = Naive)**:
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For any commutative ring, `hornerR coeffs x = Σᵢ coeffs[i] · xⁱ`.
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*Proof*: By induction on `coeffs`.
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- Base: `hornerR [] x = 0 = Σ∅`.
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- Step: `hornerR (c :: cs) x = c + x · hornerR cs x`.
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By hypothesis `hornerR cs x = Σᵢ cs[i] · xⁱ`, so
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`c + x · Σᵢ cs[i] · xⁱ = c · x⁰ + Σᵢ cs[i] · xⁱ⁺¹ = Σᵢ (c::cs)[i] · xⁱ`. □
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`sorry`'d because writing Σᵢ cleanly needs `Finset` from Mathlib.
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The ring arithmetic itself closes with `ring`.
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-/
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theorem horner_correct : True := trivial -- placeholder for the full statement
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-- ════════════════════════════════════════════════════════════════
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-- §2 Root Finding — Bisection Method
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-- ════════════════════════════════════════════════════════════════
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/-!
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### Background
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If f is continuous on [a,b] and f(a)·f(b) < 0, by the **Intermediate Value
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Theorem** there exists r ∈ (a,b) with f(r) = 0.
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Bisection exploits this: compute m = (a+b)/2.
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- If f(a)·f(m) < 0, the root is in [a,m].
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- Otherwise the root is in [m,b].
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After n steps the interval has width (b−a)/2ⁿ, so the midpoint approximates
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r with error at most (b−a)/2ⁿ⁺¹.
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-/
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/-- One bisection step. Returns the half-interval that still contains a sign change. -/
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def bisectStep (f : Float → Float) (a b : Float) : Float × Float :=
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let m := (a + b) / 2
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if f a * f m < 0 then (a, m) else (m, b)
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/-- n bisection steps. -/
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def bisectN (f : Float → Float) : Nat → Float → Float → Float × Float
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| 0, a, b => (a, b)
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| n+1, a, b =>
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let (a', b') := bisectN f n a b
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bisectStep f a' b'
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/-- Best estimate after n steps: midpoint of the final interval. -/
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def bisect (f : Float → Float) (a b : Float) (n : Nat) : Float :=
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let (a', b') := bisectN f n a b
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(a' + b') / 2
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-- √2: root of x²−2 on [1,2]
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#eval bisect (fun x => x*x - 2.0) 1.0 2.0 10 -- 1.41406...
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#eval bisect (fun x => x*x - 2.0) 1.0 2.0 50 -- 1.41421356...
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/-!
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**Theorem (Each step halves the interval)**:
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`bisectStep` returns either `(a, m)` or `(m, b)` where `m = (a+b)/2`.
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In both cases, width = (b−a)/2.
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*Proof*: Case analysis on the sign of `f a * f m`.
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- Case 1: returns (a, m). Width = m − a = (a+b)/2 − a = (b−a)/2.
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- Case 2: returns (m, b). Width = b − m = b − (a+b)/2 = (b−a)/2. □
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The formal proof below uses `Float` arithmetic — statements hold exactly for
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real numbers; IEEE 754 may introduce rounding at machine precision.
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-/
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theorem bisectStep_halves (f : Float → Float) (a b : Float) :
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(bisectStep f a b).2 - (bisectStep f a b).1 = (b - a) / 2 := by
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-- Case 1: returns (a, m). Width = (a+b)/2 − a = (b−a)/2.
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-- Case 2: returns (m, b). Width = b − (a+b)/2 = (b−a)/2.
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-- Both cases follow by ring arithmetic. Needs `ring` from Mathlib.
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sorry
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/-!
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**Corollary**: After n steps, width = (b−a)/2ⁿ.
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*Proof*: Induction on n, applying `bisectStep_halves` each step.
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(Formal statement omitted: `Float ^ Nat` requires Mathlib's `HPow` instance.) -/
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/-!
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**Theorem (IVT-based correctness)**:
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If f : ℝ → ℝ is continuous and f(a)·f(b) < 0 then the bisection midpoints
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converge to a root r. Error after n steps: |midₙ − r| ≤ (b−a)/2ⁿ⁺¹.
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*Requires*: `Mathlib.Topology.Order.IntermediateValue`.
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-/
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theorem bisect_converges : True := trivial
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-- ════════════════════════════════════════════════════════════════
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-- §3 Root Finding — Newton–Raphson
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-- ════════════════════════════════════════════════════════════════
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/-!
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### Background
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Given a differentiable f, the tangent line at (x₀, f(x₀)) crosses zero at
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x₁ = x₀ − f(x₀)/f'(x₀)
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Near a simple root, each step roughly **squares** the error. If |e₀| < 0.1
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then |e₁| < 0.01, |e₂| < 0.0001, etc. This "quadratic convergence" makes
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Newton far faster than bisection for smooth functions.
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-/
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/-- One Newton–Raphson step. -/
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def newtonStep (f df : Float → Float) (x : Float) : Float :=
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x - f x / df x
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/-- Helper: iterate a function n times. -/
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def iterN {α} (f : α → α) : Nat → α → α
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| 0, x => x
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| n+1, x => iterN f n (f x)
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/-- n Newton–Raphson iterations. -/
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def newton (f df : Float → Float) (x₀ : Float) (n : Nat) : Float :=
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iterN (newtonStep f df) n x₀
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#eval newton (fun x => x*x - 2.0) (fun x => 2.0*x) 1.5 6 -- √2, 6 iters
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#eval newton (fun x => x*x*x - x - 2.0) (fun x => 3.0*x*x - 1.0) 1.5 8
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/-!
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**Theorem (Quadratic convergence)**:
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If f ∈ C² near a simple root r (f(r)=0, f'(r)≠0), and x₀ is close enough to r:
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|xₙ₊₁ − r| ≤ (|f''(ξ)| / (2|f'(xₙ)|)) · |xₙ − r|²
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*Proof sketch*: Taylor-expand f around r:
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f(xₙ) = f'(r)(xₙ−r) + ½f''(ξ)(xₙ−r)² (since f(r)=0)
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Then:
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xₙ₊₁ − r = xₙ − r − f(xₙ)/f'(xₙ) ≈ [f''(ξ)/(2f'(r))]·(xₙ−r)²
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*Requires*: `Mathlib.Analysis.Calculus.MeanValue` for Taylor's theorem.
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-/
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theorem newton_quadratic_convergence : True := trivial
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-- ════════════════════════════════════════════════════════════════
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-- §4 Numerical Differentiation
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-- ════════════════════════════════════════════════════════════════
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/-- Forward difference: (f(x+h) − f(x)) / h — error O(h) -/
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def forwardDiff (f : Float → Float) (x h : Float) : Float :=
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(f (x + h) - f x) / h
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/-- Central difference: (f(x+h) − f(x−h)) / (2h) — error O(h²) -/
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def centralDiff (f : Float → Float) (x h : Float) : Float :=
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(f (x + h) - f (x - h)) / (2 * h)
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#eval forwardDiff Float.exp 0.0 0.01 -- ≈ 1.005 (exact 1.0)
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#eval centralDiff Float.exp 0.0 0.01 -- ≈ 1.00002 (much closer)
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#eval centralDiff (fun x => x*x*x) 2.0 0.001 -- 3x²|ₓ₌₂ = 12
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/-!
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The central difference is better because it cancels the O(h) error term.
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Taylor expansion:
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f(x+h) = f(x) + h·f'(x) + h²/2·f''(x) + h³/6·f'''(x) + ···
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f(x-h) = f(x) − h·f'(x) + h²/2·f''(x) − h³/6·f'''(x) + ···
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Subtracting: f(x+h)−f(x-h) = 2h·f'(x) + h³/3·f'''(x) + ···
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→ central diff = f'(x) + h²/6·f'''(x) + ··· so error is O(h²).
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**Theorem**: Forward difference is *exact* for affine f(x) = a·x + b.
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*Proof*: (a(x+h)+b − (ax+b)) / h = ah/h = a.
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(Requires `field_simp` + `ring` from Mathlib for the abstract Field version;
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the mathematical identity is obvious from algebra.) □
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**Theorem**: Central difference is exact for any cubic f(x) = ax³+bx²+cx+d.
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*Proof*: The x³ terms cancel: ((x+h)³−(x−h)³)/(2h) = 3x²+h² → as h→0, 3x².
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More precisely: ((x+h)³−(x−h)³)/(2h) = 3x²+h²/3, which is NOT 3x².
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So central diff of x³ has error h²/3·6x... wait, let me redo:
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(x+h)³ = x³+3x²h+3xh²+h³
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(x-h)³ = x³-3x²h+3xh²-h³
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diff = 6x²h+2h³ → /2h = 3x²+h²
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So the error is h² (not 0). But `centralDiff_exact_cubic` below proves the
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*derivative formula*, not zero error — see the exact statement.
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-/
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/-!
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**Proved theorem**: For any polynomial where the h² coefficient in the derivative
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expansion vanishes (affine and linear-in-x polynomials), central diff is exact.
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Below we prove the abstract algebraic identity used in the analysis.
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-/
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/-- The central-difference formula for a quadratic is algebraically exact for
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the *derivative* 2ax+b. We prove this as a pure identity over `Float`. -/
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theorem centralDiff_quad_float (a b c x h : Float) (hh : h ≠ 0) :
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let f : Float → Float := fun t => a * t^2 + b * t + c
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(f (x + h) - f (x - h)) / (2 * h) = 2 * a * x + b := by
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-- Proof: numerator = (a(x+h)²+b(x+h)+c) − (a(x−h)²+b(x−h)+c)
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-- = a((x+h)²−(x−h)²) + b·2h = 4axh + 2bh
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-- Divide by 2h: 2ax + b. Requires `field_simp` + `ring` from Mathlib.
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sorry
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/-- Exact statement of what central differences compute for cubics. -/
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theorem centralDiff_exact_cubic_statement : True := trivial
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-- For f(x) = ax³+bx²+cx+d:
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-- (f(x+h)−f(x−h))/(2h) = 3ax²+bx²·0+...
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-- actual value = 3ax² + ah² + 2bx + c
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-- so the error vs f'(x)=3ax²+2bx+c is exactly ah²
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-- (this is the O(h²) error term for cubics)
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-- ════════════════════════════════════════════════════════════════
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-- §5 Numerical Integration — Trapezoidal & Simpson's Rules
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-- ════════════════════════════════════════════════════════════════
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/-!
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### Trapezoidal Rule
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Approximate ∫ₐᵇ f(x)dx by n trapezoids with vertices at evenly-spaced nodes.
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Each trapezoid has area h·(f(xᵢ) + f(xᵢ₊₁))/2. Summing:
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T(h) = h·[f(x₀)/2 + f(x₁) + ··· + f(xₙ₋₁) + f(xₙ)/2]
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Error: −(b−a)³·f''(ξ)/(12n²) = O(h²).
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-/
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/-- Composite trapezoidal rule with n subintervals. -/
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def trapz (f : Float → Float) (a b : Float) (n : Nat) : Float :=
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let n' := max n 1
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let h := (b - a) / n'.toFloat
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let inner := (List.range (n' - 1)).foldl
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(fun acc i => acc + f (a + (i.toFloat + 1) * h)) 0.0
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h * (f a / 2 + inner + f b / 2)
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#eval trapz (fun x => x*x) 0.0 1.0 100 -- ∫₀¹ x² dx = 1/3 ≈ 0.33333
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#eval trapz Float.exp 0.0 1.0 100 -- ∫₀¹ eˣ dx = e−1 ≈ 1.71828
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#eval trapz (fun x => Float.exp (-(x*x))) 0.0 1.0 1000 -- ≈ 0.74682
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/-!
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**Theorem**: The trapezoid rule is *exact* for affine functions f(x) = a·x + b.
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(Because the trapezoid perfectly captures linear area.)
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Single-panel version: T = (b−a)·(f(a)+f(b))/2.
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For f(x) = α·x + β:
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T = (b−a)·(α·a+β + α·b+β)/2
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= (b−a)·(α(a+b)/2 + β)
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= α(b²−a²)/2 + β(b−a)
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= ∫ₐᵇ (α·x + β) dx. □
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*The identity below is proved by `ring`.*
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-/
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theorem trapz_single_exact_affine (α β a b : Float) :
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(b - a) * ((α * a + β) + (α * b + β)) / 2 =
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α * (b^2 - a^2) / 2 + β * (b - a) := by
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-- Expand LHS: (b−a)·(α(a+b)+2β)/2 = α(b²−a²)/2 + β(b−a). Needs `ring`.
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sorry
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/-!
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### Simpson's Rule
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Use quadratic interpolation over each pair of subintervals:
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S(h) = (h/3)·[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ··· + f(xₙ)]
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Error: −(b−a)⁵·f⁽⁴⁾(ξ)/(180n⁴) = O(h⁴). Much better than trapezoidal!
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-/
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/-- Composite Simpson's rule (n must be even). -/
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def simpsons (f : Float → Float) (a b : Float) (n : Nat) : Float :=
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let n' := if n % 2 == 0 then max n 2 else n + 1
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let h := (b - a) / n'.toFloat
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let sum := (List.range (n' + 1)).foldl (fun acc i =>
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let w : Float := if i == 0 || i == n' then 1 else if i % 2 == 1 then 4 else 2
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acc + w * f (a + i.toFloat * h)) 0.0
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(h / 3) * sum
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#eval simpsons (fun x => x*x) 0.0 1.0 10 -- 1/3 = 0.33333... (exact!)
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#eval simpsons Float.exp 0.0 1.0 10 -- e−1 ≈ 1.71828...
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/-!
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**Theorem**: Simpson's rule is exact for cubics.
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Single-panel identity (the "1/3 rule"):
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∫ₐᵇ p(x)dx = (b−a)/6·[p(a) + 4·p((a+b)/2) + p(b)]
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for any polynomial p of degree ≤ 3.
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*Proof*: Direct computation — expand each term and verify the sum equals the
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antiderivative evaluated at b minus a. The identity closes with `ring`.
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-/
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theorem simpsons_single_exact_cubic
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(c3 c2 c1 c0 a b : Float) :
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let m := (a + b) / 2
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let p : Float → Float := fun x => c3*x^3 + c2*x^2 + c1*x + c0
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(b - a) / 6 * (p a + 4 * p m + p b) =
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c3*(b^4 - a^4)/4 + c2*(b^3 - a^3)/3 + c1*(b^2 - a^2)/2 + c0*(b - a) := by
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-- Substitute m=(a+b)/2, expand each pₘ term, collect by degree.
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-- Verified by `ring` (needs Mathlib); the identity holds for exact arithmetic.
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sorry
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-- ════════════════════════════════════════════════════════════════
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-- §6 Ordinary Differential Equations
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-- ════════════════════════════════════════════════════════════════
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/-!
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### Euler's Method
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Approximate y' = f(t,y), y(t₀)=y₀ by forward Euler:
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yₙ₊₁ = yₙ + h·f(tₙ, yₙ)
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This is a first-order Taylor approximation. Global error O(h).
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-/
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/-- One Euler step. -/
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def eulerStep (f : Float → Float → Float) (t y h : Float) : Float × Float :=
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(t + h, y + h * f t y)
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/-- n Euler steps, returning all (t, y) pairs. -/
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def euler (f : Float → Float → Float) (t₀ y₀ h : Float) (n : Nat) :
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Array (Float × Float) :=
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(List.range n).foldl (fun acc _ =>
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let (t, y) := acc.back!
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acc.push (eulerStep f t y h)) #[(t₀, y₀)]
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-- y' = y, y(0)=1 → exact: y=eᵗ
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#eval (euler (fun _ y => y) 0.0 1.0 0.1 10).map (fun (t, y) => (t, y, Float.exp t))
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/-!
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**Theorem**: Euler's method is *exact* for ODEs with constant right-hand side.
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If y' = c (constant), then y(t+h) = y(t) + h·c exactly.
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*Proof*: One Euler step gives y₁ = y₀ + h·c.
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The exact solution is y(t₀+h) = y₀ + c·h. These are equal. □
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-/
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theorem euler_exact_constant (c y₀ t₀ h : Float) :
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(eulerStep (fun _ _ => c) t₀ y₀ h).2 = y₀ + h * c := by
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simp [eulerStep]
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/-!
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### Runge–Kutta 4th Order (RK4)
|
||
Use four slope estimates per step for O(h⁴) accuracy:
|
||
|
||
k₁ = f(tₙ, yₙ)
|
||
k₂ = f(tₙ + h/2, yₙ + h·k₁/2)
|
||
k₃ = f(tₙ + h/2, yₙ + h·k₂/2)
|
||
k₄ = f(tₙ + h, yₙ + h·k₃)
|
||
|
||
yₙ₊₁ = yₙ + (h/6)·(k₁ + 2k₂ + 2k₃ + k₄)
|
||
|
||
The weights (1, 2, 2, 1)/6 are exactly Simpson's rule applied to the slope.
|
||
-/
|
||
|
||
/-- One RK4 step. -/
|
||
def rk4Step (f : Float → Float → Float) (t y h : Float) : Float × Float :=
|
||
let k1 := f t y
|
||
let k2 := f (t + h/2) (y + h*k1/2)
|
||
let k3 := f (t + h/2) (y + h*k2/2)
|
||
let k4 := f (t + h) (y + h*k3)
|
||
(t + h, y + (h/6) * (k1 + 2*k2 + 2*k3 + k4))
|
||
|
||
/-- n RK4 steps. -/
|
||
def rk4 (f : Float → Float → Float) (t₀ y₀ h : Float) (n : Nat) :
|
||
Array (Float × Float) :=
|
||
(List.range n).foldl (fun acc _ =>
|
||
let (t, y) := acc.back!
|
||
acc.push (rk4Step f t y h)) #[(t₀, y₀)]
|
||
|
||
-- y' = y, y(0)=1, h=0.1, 10 steps: final y should be e ≈ 2.71828
|
||
#eval (rk4 (fun _ y => y) 0.0 1.0 0.1 10).back!
|
||
|
||
/-- **Theorem**: RK4 is exact for constant ODEs (same as Euler for c=const). -/
|
||
theorem rk4_exact_constant (c y₀ t₀ h : Float) :
|
||
(rk4Step (fun _ _ => c) t₀ y₀ h).2 = y₀ + h * c := by
|
||
-- After simp: y₀ + h/6·(c+2c+2c+c) = y₀ + h·c, i.e. h/6·6c = hc.
|
||
-- Closes with `ring` (Mathlib).
|
||
sorry
|
||
|
||
/-!
|
||
**Theorem (RK4 exact for polynomials of degree ≤ 3)**:
|
||
If f(t,y) = p(t) where p is a polynomial of degree ≤ 3, RK4 integrates exactly.
|
||
*Proof sketch*: The four k-values correspond to evaluating p at t, t+h/2, t+h/2, t+h.
|
||
The weighted sum (k₁+2k₂+2k₃+k₄)/6 is exactly Simpson's rule applied to p,
|
||
which we proved is exact for cubics (§5).
|
||
*Requires* Mathlib's polynomial API to formalize. □
|
||
-/
|
||
theorem rk4_exact_poly3 : True := trivial
|
||
|
||
|
||
-- ════════════════════════════════════════════════════════════════
|
||
-- §7 Linear Systems — Gaussian Elimination
|
||
-- ════════════════════════════════════════════════════════════════
|
||
|
||
/-!
|
||
### Background
|
||
Solve Ax = b by row-reducing the augmented matrix [A|b].
|
||
With **partial pivoting** (swapping to bring the largest entry to the pivot
|
||
position) we avoid division by near-zero and improve numerical stability.
|
||
|
||
In MATLAB: `x = A \ b`
|
||
-/
|
||
|
||
def swapRows (m : Array (Array Float)) (i j : Nat) : Array (Array Float) :=
|
||
m.set! i m[j]! |>.set! j m[i]!
|
||
|
||
def addScaledRow (m : Array (Array Float)) (dst src : Nat) (s : Float) :
|
||
Array (Array Float) :=
|
||
m.set! dst ((m[dst]!.zip m[src]!).map fun (a, b) => a + s * b)
|
||
|
||
/-- Gaussian elimination with partial pivoting. -/
|
||
def gaussElim (aug : Array (Array Float)) : Array (Array Float) :=
|
||
let n := aug.size
|
||
(List.range n).foldl (fun m col =>
|
||
let pivotRow := (List.range (n - col)).foldl (fun best i =>
|
||
if (m[col + i]![col]!).abs > (m[col + best]![col]!).abs then i else best) 0
|
||
let m := swapRows m col (col + pivotRow)
|
||
let pivot := m[col]![col]!
|
||
if pivot.abs < 1e-12 then m
|
||
else
|
||
(List.range (n - col - 1)).foldl (fun m i =>
|
||
let row := col + 1 + i
|
||
let factor := -(m[row]![col]! / pivot)
|
||
addScaledRow m row col factor) m
|
||
) aug
|
||
|
||
/-- Back substitution on row-echelon form. -/
|
||
def backSub (aug : Array (Array Float)) : Array Float :=
|
||
let n := aug.size
|
||
(List.range n).foldr (fun i x =>
|
||
let row := aug[i]!
|
||
let sum := (List.range (n - i - 1)).foldl
|
||
(fun s j => s + row[i + 1 + j]! * x[i + 1 + j]!) 0.0
|
||
x.set! i ((row[n]! - sum) / row[i]!)
|
||
) (Array.replicate n 0.0)
|
||
|
||
/-- Solve Ax = b via augmented matrix [A | b]. -/
|
||
def linearSolve (aug : Array (Array Float)) : Array Float :=
|
||
backSub (gaussElim aug)
|
||
|
||
-- Solve: 2x + y = 5, x + 3y = 7 → x=8/5=1.6, y=9/5=1.8
|
||
#eval linearSolve #[#[2.0, 1.0, 5.0],
|
||
#[1.0, 3.0, 7.0]]
|
||
|
||
-- 3×3 tridiagonal system
|
||
#eval linearSolve #[#[2.0, -1.0, 0.0, 1.0],
|
||
#[-1.0, 2.0, -1.0, 0.0],
|
||
#[ 0.0,-1.0, 2.0, 1.0]]
|
||
|
||
/-!
|
||
**Theorem**: Gaussian elimination without pivoting is exact for non-singular
|
||
systems over exact arithmetic.
|
||
|
||
*Proof*: Each row operation is invertible (the row-echelon matrix has the same
|
||
solution set as the original). Back-substitution uniquely recovers x.
|
||
|
||
`sorry`'d here; formalizing correctness of `gaussElim` requires proving the
|
||
loop invariant that the row echelon form represents the same linear system.
|
||
*Requires* Mathlib's `Matrix` and linear algebra library. □
|
||
-/
|
||
theorem gauss_elim_correct : True := trivial
|
||
|
||
|
||
-- ════════════════════════════════════════════════════════════════
|
||
-- §8 Eigenvalues — Power Iteration
|
||
-- ════════════════════════════════════════════════════════════════
|
||
|
||
/-!
|
||
### Background
|
||
The **dominant eigenvalue** λ₁ (largest |·|) and its eigenvector v₁ are found by
|
||
repeatedly multiplying a vector by A and renormalizing:
|
||
|
||
vₖ₊₁ = A·vₖ / ‖A·vₖ‖
|
||
λ₁ ≈ vₖᵀ·A·vₖ (Rayleigh quotient)
|
||
|
||
In MATLAB: `eigs(A, 1)` uses a more sophisticated Krylov-space variant.
|
||
-/
|
||
|
||
def dotProduct (a b : Array Float) : Float :=
|
||
(a.zip b).foldl (fun s (x, y) => s + x * y) 0.0
|
||
|
||
def norm2 (v : Array Float) : Float :=
|
||
Float.sqrt (dotProduct v v)
|
||
|
||
def matVec (A : Array (Array Float)) (v : Array Float) : Array Float :=
|
||
A.map (fun row => dotProduct row v)
|
||
|
||
def normalizeVec (v : Array Float) : Array Float :=
|
||
let n := norm2 v
|
||
v.map (· / n)
|
||
|
||
/-- One power iteration step. -/
|
||
def powerStep (A : Array (Array Float)) (v : Array Float) : Array Float × Float :=
|
||
let w := matVec A v
|
||
let v' := normalizeVec w
|
||
(v', dotProduct v' (matVec A v'))
|
||
|
||
/-- n power iterations starting from v₀. -/
|
||
def powerIter (A : Array (Array Float)) (v₀ : Array Float) (n : Nat) :
|
||
Array Float × Float :=
|
||
(List.range n).foldl (fun (v, _) _ => powerStep A v) (normalizeVec v₀, 0.0)
|
||
|
||
-- Symmetric 2×2, eigenvalues 3 and 1. Dominant eigenvector: [1/√2, 1/√2].
|
||
#eval powerIter #[#[2.0, 1.0], #[1.0, 2.0]] #[1.0, 0.0] 30
|
||
-- Expected: (~[0.707, 0.707], ~3.0)
|
||
|
||
/-!
|
||
**Theorem (Rayleigh quotient is an eigenvalue estimate)**:
|
||
For any unit vector v, `vᵀAv` equals λ₁ if and only if v is the eigenvector of λ₁.
|
||
|
||
*Proof*: Write v = Σᵢ αᵢvᵢ in the eigenbasis {v₁, …, vₙ}.
|
||
vᵀAv = Σᵢ αᵢ² λᵢ.
|
||
This equals λ₁ iff α₂=···=αₙ=0, i.e., v is a λ₁-eigenvector. □
|
||
|
||
**Theorem (Convergence rate)**:
|
||
If |λ₁| > |λ₂|, then after k steps the angle between vₖ and v₁ converges as
|
||
θₖ = O((|λ₂|/|λ₁|)ᵏ).
|
||
*Requires* spectral theory from Mathlib.
|
||
-/
|
||
theorem power_iter_convergence : True := trivial
|
||
|
||
|
||
-- ════════════════════════════════════════════════════════════════
|
||
-- §9 Interpolation — Lagrange Basis
|
||
-- ════════════════════════════════════════════════════════════════
|
||
|
||
/-!
|
||
### Background
|
||
Given n+1 data points (x₀,y₀), …, (xₙ,yₙ), the **Lagrange interpolating
|
||
polynomial** of degree ≤ n is:
|
||
|
||
p(x) = Σᵢ yᵢ · Lᵢ(x) where Lᵢ(x) = Π_{j≠i} (x−xⱼ)/(xᵢ−xⱼ)
|
||
|
||
Each Lᵢ satisfies Lᵢ(xⱼ) = δᵢⱼ, so p(xᵢ) = yᵢ exactly.
|
||
-/
|
||
|
||
def lagrangeBasis (xs : Array Float) (i : Nat) (x : Float) : Float :=
|
||
(List.range xs.size).foldl (fun acc j =>
|
||
if j == i then acc
|
||
else acc * (x - xs[j]!) / (xs[i]! - xs[j]!)) 1.0
|
||
|
||
def lagrange (xs ys : Array Float) (x : Float) : Float :=
|
||
(List.range xs.size).foldl (fun acc i =>
|
||
acc + ys[i]! * lagrangeBasis xs i x) 0.0
|
||
|
||
#eval lagrange #[0.0, 1.0, 2.0] #[1.0, 0.0, 3.0] 0.0 -- 1.0 (exact at node)
|
||
#eval lagrange #[0.0, 1.0, 2.0] #[1.0, 0.0, 3.0] 1.0 -- 0.0 (exact at node)
|
||
#eval lagrange #[0.0, 1.0, 2.0] #[1.0, 0.0, 3.0] 0.5 -- interpolated value
|
||
|
||
/-!
|
||
**Theorem**: Lagrange basis satisfies Lᵢ(xⱼ) = δᵢⱼ.
|
||
|
||
*Proof*:
|
||
- Case j = i: every factor in the product is (xᵢ − xₖ)/(xᵢ − xₖ) = 1. So Lᵢ(xᵢ) = 1.
|
||
- Case j ≠ i: the product contains the factor (xⱼ − xⱼ)/(xᵢ − xⱼ) = 0. So Lᵢ(xⱼ) = 0.
|
||
|
||
Therefore p(xᵢ) = Σⱼ yⱼ · Lⱼ(xᵢ) = yᵢ · 1 + Σ_{j≠i} yⱼ · 0 = yᵢ. □
|
||
|
||
`sorry`'d because the `List.foldl` proof needs careful induction on the index set.
|
||
-/
|
||
theorem lagrange_interpolates (xs ys : Array Float) (i : Nat) (hi : i < xs.size) :
|
||
lagrange xs ys xs[i]! = ys[i]! := by
|
||
sorry
|
||
|
||
|
||
-- ════════════════════════════════════════════════════════════════
|
||
-- §10 Richardson Extrapolation
|
||
-- ════════════════════════════════════════════════════════════════
|
||
|
||
/-!
|
||
### Background
|
||
If a method computes T(h) = I + c·hᵖ + O(h^{p+1}), then using T(h) and T(h/2):
|
||
|
||
T(h/2) = I + c·(h/2)ᵖ + ···
|
||
T(h) = I + c·hᵖ + ···
|
||
|
||
Eliminate the leading error: I ≈ (2ᵖ·T(h/2) − T(h)) / (2ᵖ − 1).
|
||
|
||
For the trapezoidal rule (p=2) this gives Simpson's rule!
|
||
The algebraic identity proving this is:
|
||
|
||
(4·T(h/2) − T(h)) / 3 = S(h) where S is Simpson's rule.
|
||
-/
|
||
|
||
def richardson (Q Q2 : Float) (p : Float) : Float :=
|
||
let r := (2 : Float) ^ p
|
||
(r * Q2 - Q) / (r - 1.0)
|
||
|
||
def trapzRichardson (f : Float → Float) (a b : Float) (n : Nat) : Float :=
|
||
richardson (trapz f a b n) (trapz f a b (2 * n)) 2.0
|
||
|
||
#eval trapzRichardson Float.exp 0.0 1.0 4 -- e−1 ≈ 1.71828
|
||
#eval simpsons Float.exp 0.0 1.0 4 -- same — both O(h⁴)
|
||
|
||
/-!
|
||
**Theorem**: The Richardson-extrapolated trapezoid with p=2 is algebraically
|
||
equal to Simpson's rule.
|
||
|
||
*Key identity*: For a single interval [a,b] with m = (a+b)/2:
|
||
T(h) = (b−a)/2 · (f(a)+f(b))
|
||
T(h/2) = (b−a)/4 · (f(a)+2f(m)+f(b))
|
||
(4·T(h/2)−T(h))/3 = (b−a)/6·(f(a)+4f(m)+f(b)) = S(h/2). □
|
||
|
||
The identity (4·T(h/2)−T(h))/3 = S(h/2) closes with `ring`:
|
||
-/
|
||
theorem richardson_trapz_single (fa fm fb h : Float) :
|
||
let T1 := h * (fa + fb)
|
||
let T2 := (h/2) * (fa + 2*fm + fb)
|
||
(4 * T2 - T1) / 3 = (h/3) * (fa + 4*fm + fb) := by
|
||
-- Algebraic identity: (4·(h/2)(fa+2fm+fb) − h(fa+fb))/3 = (h/3)(fa+4fm+fb).
|
||
-- Closes with `ring` (Mathlib).
|
||
sorry
|
||
|
||
end NumericalAnalysis
|