Root Finding — Open Methods

Convergence Comparison

Bisection and the secant method are compared on f(x)=x²-2 over [1,2] with starting points x0=1, x1=2. After 3 steps bisection reaches |f|=0.1094 while the secant method reaches |f|=0.0012 — demonstrating the secant method's faster superlinear convergence versus bisection's linear convergence.

Example

Compare residual tables and convergence rates. Bisection is guaranteed to converge for a valid bracket, but the secant method can diverge from poor starting points.

highlighted = computed this step

Step 1 — Set up

Compare residuals for bisection and secant.

f(x)=x2−2f(x)= x^{2}-2

Step 2 — Bisection residuals

Bisection has linear convergence.

nxabs f11.50000.250021.25000.437531.37500.1094\begin{array}{ccc}\text{n} & \text{x} & \text{abs f} \\ \hlmath{1} & \hlmath{1.5000} & \hlmath{0.2500} \\ \hlmath{2} & \hlmath{1.2500} & \hlmath{0.4375} \\ \hlmath{3} & \hlmath{1.3750} & \hlmath{0.1094}\end{array}

Step 3 — Secant residuals

Secant has superlinear convergence.

nxabs f11.33330.222221.40000.040031.41460.0012\begin{array}{ccc}\text{n} & \text{x} & \text{abs f} \\ \hlmath{1} & \hlmath{1.3333} & \hlmath{0.2222} \\ \hlmath{2} & \hlmath{1.4000} & \hlmath{0.0400} \\ \hlmath{3} & \hlmath{1.4146} & \hlmath{0.0012}\end{array}

Step 4 — Compare rates

Secant is superlinear while bisection is linear.

secant superlinearbisection linear\hl{secant superlinear} \quad \hl{bisection linear}
convergence-comparison Bisection halves the error each step (linear convergence, rate 1/2). The secant method has superlinear convergence (order ≈1.618). For the same number of function evaluations the secant method typically achieves much lower residuals, at the cost of needing two starting points rather than a bracket.