Secant Method Calculator
Find a root of f(x)=0 using the secant method — no derivative needed.
Inputs
Function whose root you seek.
Root
2.094551481542
Iterations
7
|f(root)|
3.5527 × 10⁻¹⁵
Status
Converged
Step by step
Values used
f(x) = x^3 - 2*x - 5; x₀ (first guess) = 2; x₁ (second guess) = 3; Tolerance = 0.0000; Max iterations = 100
Secant iteration
x(n+1) = x(n) - f(x(n)) · (x(n) - x(n-1)) / (f(x(n)) - f(x(n-1)))
Root
= 2.094551481542
Iterations
= 7
|f(root)|
= 0.0000
Status
= Converged
How it works
The secant method approximates Newton's method by replacing the derivative with a finite difference: f'(xₙ) ≈ [f(xₙ) − f(xₙ₋₁)]/(xₙ − xₙ₋₁). It requires two initial guesses rather than one, and converges at rate φ ≈ 1.618 (superlinear) — faster than bisection but slightly slower than Newton's quadratic rate. No derivative computation is needed.
Formula
Secant iteration
x(n+1) = x(n) - f(x(n)) · (x(n) - x(n-1)) / (f(x(n)) - f(x(n-1)))
- x(n)
- Current approximation
- x(n-1)
- Previous approximation
Frequently Asked Questions
When should I use secant vs Newton?
Use secant when the derivative is expensive or unavailable. It needs one extra initial point but avoids derivative computation entirely.
Can the secant method fail?
Yes, if the two function values are nearly equal (horizontal secant line) or if the guesses are too far from the root. Bisection is safer as a fallback.