---
title: "Secant method: explain and derive"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:secant-method-explain-derive"
source: https://www.bctengineer.com/q/semester_4/numerical_method/secant-method-explain-derive
site: BCT Engineering
---

# Secant method: explain and derive

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2069 Bhadra, 4 marks; 2069 Poush, 4 marks; 2068 Bhadra, 4 marks (asked 3 times)
- Video: [Chapter 03.05: Lesson: Derivation of Secant Method: Approach 1](https://www.youtube.com/watch?v=jXIUi7zxWIU) — numericalmethodsguy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/secant-method-explain-derive
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Explain the mechanism (graphically) of finding a real root of a non-linear equation using the Secant method and hence obtain the iteration formula.

## Short answer

Starting with two initial points, the secant line through them intersects the x-axis to give the next approximation. The iterative formula is x(n+1) = x(n) - f(x(n)) * (x(n) - x(n-1)) / (f(x(n)) - f(x(n-1))).

## Answer

The Secant method finds a real root of f(x) = 0 by using a straight line (secant) through two recent points on the curve. It needs two starting values but no derivative, and it does not need the root to stay bracketed.

#### Graphical mechanism
```text
 y
 |           / f(x1)
 |          /*  (x1, f1)
 |        _/
 |  x2   /
 |---+--/-------- x
 |   | * (x0, f0)
 |   root
```

1. Take two initial approximations x0 and x1 and the points P0(x0, f(x0)) and P1(x1, f(x1)) on the curve.
2. Draw the straight line (secant) through P0 and P1. The point where it cuts the x-axis is the next approximation x2.
3. Now use x1 and x2 (discard x0), draw the secant again and find x3.
4. Repeat until two successive values agree to the required accuracy. The cut points move toward the root where the curve crosses the x-axis.

#### Derivation of the formula
The line through (x0, f0) and (x1, f1) is

```text
y − f1 = [(f1 − f0)/(x1 − x0)] (x − x1)
```

Where it cuts the x-axis, y = 0 and x = x2:

```text
−f1 = [(f1 − f0)/(x1 − x0)] (x2 − x1)
x2 = x1 − f1 (x1 − x0)/(f1 − f0)
```

Equivalent form: x2 = (x0 f1 − x1 f0) / (f1 − f0). In general:

```text
x(n+1) = x(n) − f(x(n)) · [x(n) − x(n−1)] / [f(x(n)) − f(x(n−1))]
```

This is Newton-Raphson with the derivative f′(x(n)) replaced by the slope of the secant. Its order of convergence is 1.618.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The secant method finds the next approximation by:
   A. Taking where the line through two recent points cuts the x-axis
   B. Drawing a tangent at one point
   C. Halving an interval
   D. Using g(x) = x
   Answer: A. A secant through P0 and P1 is extended to the x-axis. (Key point: Idea)

2. Secant method requires:
   A. One value and f'(x)
   B. Two starting values, no derivative
   C. A bracketing interval with sign change
   D. Three starting values
   Answer: B. It replaces the derivative by the secant slope. (Key point: Requirements)

3. The secant iteration is:
   A. x(n+1) = x(n) - f(x(n))/f'(x(n))
   B. x(n+1) = (x(n)+x(n-1))/2
   C. x(n+1) = x(n) - f(x(n))[x(n) - x(n-1)]/[f(x(n)) - f(x(n-1))]
   D. x(n+1) = x(n) + f(x(n))[x(n)-x(n-1)]
   Answer: C. Setting y = 0 in the line through the two points gives this. (Key point: Formula)

4. Order of convergence of the secant method is:
   A. 1
   B. 2
   C. 0.5
   D. About 1.618
   Answer: D. It is superlinear, between linear and quadratic. (Key point: Convergence)

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