Mathematics / Bisection Method & Convergence
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LABORATORY 01 · MATHEMATICS

Root-Finding and Linear Error Convergence via the Bisection Method

Determine the real root of the nonlinear algebraic equation f(x) = x³ − x − 2 = 0 on the initial bracketing interval [1.0, 2.0] using the bisection method, evaluate interval halving dynamics across consecutive iterations, and extract the asymptotic error constant C = 0.500 and linear convergence rate via ordinary least-squares regression. Explore interactive numerical methods, evaluate error attenuation and verify convergence orders.

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INTERVAL BRACKET WIDTH ΔX_K—
COMPUTATION STATUS

Ready to execute k = 1.0.

1.0 steps
Test Function: Continuous cubic polynomial f(x) = x³ − x − 2.

Adjust k, then execute computation.

Mathematical operations are computed deterministically using 64-bit IEEE 754 precision. Residuals reflect ordinary least-squares regression across successive observations.

Numerical assumptions and limits

The bisection method cannot locate roots of even multiplicity where f(x) does not cross the X-axis (e.g. f(x) = (x−r)²), because no sign-reversing bracket can be established. The numerical simulation model evaluates the real root of the benchmark cubic polynomial f(x) = x³ − x − 2 = 0 on the initial bracketing interval [a₀, b₀] = [1.000, 2.000]. By Cardano's formula, the unique real root is x* = 1.521379706804568... Initial boundary function values: f(1.0) = −2.000 and f(2.0) = +4.000. Across the six standardized iteration steps k = [1, 2, 3, 4, 5, 6], the exact interval widths are Δx_k = (b₀ − a₀) / 2^k = [0.5000, 0.2500, 0.1250, 0.0625, 0.03125, 0.015625]. The base-2 logarithmic transform yields x = k = [1, 2, 3, 4, 5, 6] and y = log2(Δx_k) = [−1.000, −2.000, −3.000, −4.000, −5.000, −6.000]. Ordinary least-squares linear regression on these coordinates produces an exact empirical slope m = −1.00000 and Y-intercept b = 0.00000 with Pearson correlation coefficient R² = 1.000000000. The asymptotic error constant is evaluated as C = 2^m = 2^(−1.000) = 0.50000, demonstrating zero deviation (0.0000%) from the theoretical asymptotic error ratio C = 0.500.

Computational tools & numerical modules.

High-Precision IEEE 754 Double-Precision Arithmetic Unit

Executes 64-bit floating-point interval arithmetic with 53 bits of mantissa precision (machine epsilon ε_mach ≈ 2.22 × 10⁻¹⁶) to prevent round-off accumulation during repeated interval division.

Dynamic Continuous Function Evaluator and Sign Detector

Computes polynomial values f(c) = c³ − c − 2 and evaluates the Bolzano condition sign(f(a)) · sign(f(c)) < 0 to direct interval updates.

Bisection Interval Halving Coordinate Register

Tracks the active left bound a_k, right bound b_k, midpoint c_k = (a_k + b_k)/2, and interval span Δx_k = b_k − a_k across iteration sequences.

Asymptotic Convergence Rate and Error Log Analyzer

Computes consecutive error ratios e_{k+1}/e_k and linearizes logarithmic interval widths log2(Δx_k) to extract empirical convergence constants.

Interactive Function Plotter and Root Bracketing Canvas

Renders high-resolution SVG curves of f(x) alongside animated vertical bracketing bounds [a_k, b_k] and midpoint markers c_k.

Verification Matrix and Cardano Analytical Reference Benchmark

Supplies the closed-form algebraic root x* = (1 + √19/27)^(1/3) + (1 − √19/27)^(1/3) ≈ 1.5213797 to evaluate absolute true error.

FOLLOW THE METHOD

A reason behind every step.

Determine the real root of the nonlinear algebraic equation f(x) = x³ − x − 2 = 0 on the initial bracketing interval [1.0, 2.0] using the bisection method, evaluate interval halving dynamics across consecutive iterations, and extract the asymptotic error constant C = 0.500 and linear convergence rate via ordinary least-squares regression. Standardized input settings: 1.0, 2.0, 3.0, 4.0, 5.0, 6.0 steps.

01

Verification of Bolzano's Intermediate Value Theorem

Verify that f(x) = x³ − x − 2 is continuous on [a₀, b₀] = [1.000, 2.000]. Evaluate endpoints: f(1.0) = (1)³ − 1 − 2 = −2.000 < 0 and f(2.0) = (2)³ − 2 − 2 = +4.000 > 0. Because f(a₀) · f(b₀) = −8.000 < 0, Bolzano's Theorem guarantees at least one real root x* in (1.0, 2.0).

02

First Midpoint Evaluation and Sign Determination (k = 1)

Compute initial midpoint c₁ = (a₀ + b₀)/2 = (1.0 + 2.0)/2 = 1.5000. Evaluate f(c₁) = (1.5)³ − 1.5 − 2 = 3.375 − 3.500 = −0.1250 < 0. Because sign(f(c₁)) matches sign(f(a₀)), the root lies in [c₁, b₀]. Update new interval [a₁, b₁] = [1.5000, 2.0000] with span Δx₁ = 0.5000.

03

Iterative Bisection across Steps k = 2 to k = 6

Execute subsequent bisection cycles: at k = 2, c₂ = 1.7500, f(c₂) = +1.6094 > 0, yielding [a₂, b₂] = [1.5000, 1.7500], Δx₂ = 0.2500; at k = 3, c₃ = 1.6250, f(c₃) = +0.6660 > 0, yielding [1.5000, 1.6250], Δx₃ = 0.1250; at k = 4, c₄ = 1.5625, f(c₄) = +0.2522 > 0, yielding [1.5000, 1.5625], Δx₄ = 0.0625; at k = 5, c₅ = 1.53125, f(c₅) = +0.0591 > 0, yielding [1.5000, 1.53125], Δx₅ = 0.03125; at k = 6, c₆ = 1.515625, f(c₆) = −0.0340 < 0, yielding [1.515625, 1.53125], Δx₆ = 0.015625.

04

Tabulation of Bracketing Intervals and Error Bounds

Record the iteration step k, left bound a_k, right bound b_k, root estimate c_k, function value f(c_k), bracket width Δx_k = (b₀ − a₀)/2^k, and maximum possible absolute error ε_max = Δx_k / 2 in the observation register.

05

Logarithmic Linearization of Interval Widths

Compute the base-2 logarithm of the bracket width y = log2(Δx_k) for each step k = 1 to 6. For Δx = [0.5000, 0.2500, 0.1250, 0.0625, 0.03125, 0.015625], the coordinates are y = [−1.000, −2.000, −3.000, −4.000, −5.000, −6.000].

06

Ordinary Least-Squares Regression and Convergence Rate Extraction

Perform ordinary least-squares regression of y = log2(Δx_k) against x = k. Extract the empirical slope m and intercept b. Verify that m = −1.00000 identically, proving linear convergence with asymptotic error constant C = 2^m = 0.5000 (gain of 1 binary bit of precision per iteration).

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Points are your recorded observations. The solid regression line validates Bisection Method & Convergence mathematical scaling.

LINEAR ASYMPTOTIC FIT · y = mx + b
\log_2(\Delta x_k) = -1.000 \cdot k + 0.000 \implies C = 2^{-1} = 0.500
#Iteration Step k (steps)Interval Bracket Width Δx_k ()Iteration Step kBase-2 Logarithm of Interval Width log2(Δx_k)Remove

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CHECK YOUR UNDERSTANDING

Explain what the mathematical metrics mean.

Given an initial bracketing interval [a₀, b₀] = [0.0, 4.0], how many bisection iterations are required to guarantee that the root error is less than 0.001?

What is the order of convergence α and asymptotic error constant C of the standard bisection method?

Why will the bisection method fail to find the root of f(x) = (x − 3)² on the interval [1.0, 5.0]?

In floating-point arithmetic, why is the formula c = a + (b − a) / 2 preferred over c = (a + b) / 2 for computing the midpoint?