The equation x³ − x − 1 = 0 has no tidy algebraic solution, yet an engineer needs the root to four decimal places. How can an approximate answer be PROVED correct to a stated accuracy?
Start with a prediction. State what a sign change shows, and what could make a decimal answer trustworthy.
Build the model
A continuous function with opposite signs at two endpoints has a root between them. Bisection 二分法 halves the interval, keeping the half whose endpoints still differ in sign.
Iteration 迭代 x next = g(x) refines an estimate from a rearrangement; a limit must satisfy L = g(L) and be checked in the original equation. Newton–Raphson 牛顿迭代 uses x next = x − f(x)/f′(x) with a nonzero derivative. The trapezium rule 梯形法则 estimates a definite integral: (h/2)(y₀ + 2y₁ + … + 2yₙ₋₁ + yₙ).
bisection: Halving a root bracket, keeping the half with a sign change; iteration: Repeating an update x next = g(x) from the previous estimate; Newton–Raphson: The root update x next = x − f(x)/f′(x); trapezium rule: Estimating an integral by adding trapezium strips.
Choose evidence that can test it
A sign change proves a root only where the function is CONTINUOUS: 1/x changes sign across x = 0 with no root there. Report a rounded root with its bracket, such as [1.3245, 1.3255] for four decimal places, so the accuracy is provable.
Iteration can diverge or cycle: check the iterates, the domain and the original equation. The trapezium rule over- or under-estimates depending on curvature, not on whether the function merely increases.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: f(x) = x² − 2 and x₀ = 1.5 for Newton–Raphson. f′(x) = 2x, so x₁ = x₀ − f(x₀)/f′(x₀) = 1.5 − (2.25 − 2)/3 = 1.5 − 0.0833… = 1.4166667. A second step: f(1.4166667) ≈ 0.0069444, f′ ≈ 2.8333333, so x₂ ≈ 1.4142157, closing on √2.
Example:
Estimate ∫₀² x² dx with two equal trapezium strips and compare with the exact value, using the same sequence. (Answer: h = 1, heights 0, 1, 4, so A = (1/2)(0 + 2 × 1 + 4) = 3. The exact value is 8/3 ≈ 2.667 — the convex curve makes the estimate an overestimate.)
Check the conclusion and its limits
Each method has conditions: bisection needs a continuous sign change, Newton needs a nonzero derivative and can fail from a poor start, and a rearranged iteration converges only near a fixed point with |g′| < 1. More strips improve a trapezium estimate but never make it exact by assertion.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
Every sign change proves a root, including one across a discontinuity. This claim is false: the intermediate-value reasoning needs continuity on the whole interval — across an asymptote the function can jump from negative to positive with no root at all.
Key:
Numerical methods: bracket a root with a continuous sign change and halve with bisection; iterate with checked convergence; Newton–Raphson uses the tangent; the trapezium rule estimates integrals with curvature deciding the error direction.
