Secant Method Calculator
The secant method replaces the derivative with the slope through the last two points: x² − 612 = 0 converges to 24.7386337537, √612.
Use x, ^ for powers, and sin, cos, tan, exp, ln, log, sqrt, abs, pi, e. 2x and xsin(x) are understood.
Stop when successive values or |f(x)| fall below this.
24.738633753706
Full answer, then cut to fewer decimals (not rounded):
- 24.738633753706
- 24.73863375370
- 24.7386337537
- 24.738633753
- 24.73863375
- 24.7386337
- 24.738633
- 24.73863
- 24.7386
- 24.738
- 24.73
- 24.7
- f(root)
- 1.13687 × 10⁻¹³
- Iterations
- 6
| n | xₙ | f(xₙ) |
|---|---|---|
| 0 | 10 | -512 |
| 1 | 30 | 288 |
| 2 | 22.8 | -92.16 |
| 3 | 24.5454545455 | -9.52066 |
| 4 | 24.7465437788 | 0.391429 |
| 5 | 24.7386027537 | -0.00153379 |
| 6 | 24.7386337488 | -2.45172 × 10⁻⁷ |
| 7 | 24.7386337537 | 1.13687 × 10⁻¹³ |
Step-by-step solution
- xₙ₊₁ = xₙ − f(xₙ)·(xₙ − xₙ₋₁) ÷ (f(xₙ) − f(xₙ₋₁))
- Converged to 24.7386337537 after 6 iterations
Secant Method Formula
xₙ₊₁ = xₙ − f(xₙ)(xₙ − xₙ₋₁) ÷ (f(xₙ) − f(xₙ₋₁))
order of convergence ≈ 1.618
Worked Example
f(x) = x² − 612 starting from 10 and 30:
- xₙ₊₁ = xₙ − f(xₙ)·(xₙ − xₙ₋₁) ÷ (f(xₙ) − f(xₙ₋₁))
- Converged to 24.7386337537 after 6 iterations
Iterations: 6
| n | xₙ | f(xₙ) |
|---|---|---|
| 0 | 10 | -512 |
| 1 | 30 | 288 |
| 2 | 22.8 | -92.16 |
| 3 | 24.5454545455 | -9.52066 |
| 4 | 24.7465437788 | 0.391429 |
| 5 | 24.7386027537 | -0.00153379 |
| 6 | 24.7386337488 | -2.45172 × 10⁻⁷ |
| 7 | 24.7386337537 | 1.13687 × 10⁻¹³ |
GATE tip
Order of convergence is a favourite one-mark question: bisection 1 (linear), regula falsi 1, secant ≈ 1.618, Newton-Raphson 2.
More Numerical Methods Calculators
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Newton-Raphson Method Calculator
xₙ₊₁ = xₙ − f(xₙ) ÷ f′(xₙ)
quadratic convergence near a simple root
Example: f(x) = x³ − 2x − 5 starting at x₀ = 2: Root x ≈ 2.09455148154.
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Bisection Method Calculator
c = (a + b)/2; regula falsi c = (a·f(b) − b·f(a)) ÷ (f(b) − f(a))
keep the half where f changes sign
Example: f(x) = x³ − x − 2 on [1, 2]: Root x ≈ 1.5213804245.
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Lagrange Interpolation Calculator
P(x) = Σ yᵢ Π (x − xⱼ)/(xᵢ − xⱼ)
degree ≤ n − 1 through n points
Example: The points (1, 1), (2, 8), (3, 27), (4, 64) at x = 2.5: P(2.5) = 15.625.
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Newton Interpolation Calculator
P(x) = f[x₀] + f[x₀,x₁](x − x₀) + f[x₀,x₁,x₂](x − x₀)(x − x₁) + …
f[xᵢ,…,xᵢ₊ₖ] = (f[xᵢ₊₁,…] − f[xᵢ,…]) ÷ (xᵢ₊ₖ − xᵢ)
Example: The points (1, 1), (2, 8), (3, 27), (4, 64) at x = 2.5: P(2.5) = 15.625.
Frequently asked questions
How does the secant method differ from Newton-Raphson?
It uses the slope between the last two iterates instead of f′(x), so no derivative is needed; it converges with order about 1.618 instead of 2.
Do the two starting points need to bracket the root?
No, unlike bisection and regula falsi. That is also why the secant method can diverge from poor starting points.
Sources
Last reviewed: . Factors are stored exactly and computed with 50-digit decimal arithmetic. See how we verify conversion factors.
Cite this page
APA
Scientific Converters. (2026, October 15). Secant Method Calculator. Retrieved October 25, 2026, from https://scientificconverters.com/calculators/secant-method-calculator/
IEEE
Scientific Converters, "Secant Method Calculator," scientificconverters.com. Accessed: Oct. 25, 2026. [Online]. Available: https://scientificconverters.com/calculators/secant-method-calculator/