Secant Method - Numerical Methods Step by Step

The secant method approximates the Newton step by replacing the derivative with a finite-difference slope through two previous iterates. Applied to f(x)=x²-2 from x0=1, x1=2: after 3 secant steps the root approximation is 1.4146, with residual |f|≈0.0012 < 0.01.

secant-method Given two iterates x_{n-1} and x_n, the next secant iterate is x_{n+1} = x_n - f(x_n)·(x_n-x_{n-1})/(f(x_n)-f(x_{n-1})). No derivative is required. Convergence is superlinear (order ≈1.618) — slower than Newton but faster than bisection.

Example

Use two previous points to approximate the next root estimate.

highlighted = computed this step

Step 1 — Set up

Start secant method from x 0 equals 1 and x 1 equals 2.

f(x)= x^{2}-2 \quad x_{ 0 }= 1 \quad x_{ 1 }= 2

Step 2 — Secant iteration 1

Add secant iteration 1 to the table.

\begin{array}{ccc}\text{n} & \text{x} & \text{f} \\ \hlmath{1} & \hlmath{1.3333} & \hlmath{-0.2222}\end{array}

Step 3 — Secant iteration 2

Add secant iteration 2 to the table.

\begin{array}{ccc}\text{n} & \text{x} & \text{f} \\ 1 & 1.3333 & -0.2222 \\ \hlmath{2} & \hlmath{1.4000} & \hlmath{-0.0400}\end{array}

Step 4 — Secant iteration 3

Add secant iteration 3 to the table.

\begin{array}{ccc}\text{n} & \text{x} & \text{f} \\ 1 & 1.3333 & -0.2222 \\ 2 & 1.4000 & -0.0400 \\ \hlmath{3} & \hlmath{1.4146} & \hlmath{+0.0012}\end{array}

Step 5 — Root estimate

After 3 iterations the root is about 1.4146.

x\approx \hlmath{1.4146} \quad \text{iters}= 3