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Secant Method

Computational Math
Convergence
Interpolation
Find a root like Newton’s method, but without the derivative.
Author

Apurva Nakade

Published

June 25, 2026

Both methods find a root of \(f(x)=0\) without a derivative, by interpolating the most recent iterates and taking the point where the interpolant crosses the \(x\)-axis as the next iterate. Write \(y_k = f(x_k)\).

The secant method uses the line through the last two points \((x_{n-1}, y_{n-1})\) and \((x_n, y_n)\), replacing Newton’s tangent line with a secant:

\[ x_{n+1}=x_n-y_n\,\frac{x_n-x_{n-1}}{y_n-y_{n-1}}. \]

Inverse quadratic interpolation (IQI) uses the last three points, and fits \(x\) as a quadratic function of \(y\), a sideways parabola \(x=q(y)\), rather than \(y\) as a function of \(x\). The next iterate is \(x_{n+1}=q(0)\):

\[ x_{n+1}=\frac{x_{n-2}\,y_{n-1}\,y_n}{(y_{n-2}-y_{n-1})(y_{n-2}-y_n)}+\frac{x_{n-1}\,y_{n-2}\,y_n}{(y_{n-1}-y_{n-2})(y_{n-1}-y_n)}+\frac{x_n\,y_{n-2}\,y_{n-1}}{(y_n-y_{n-2})(y_n-y_{n-1})}. \]

Because a sideways parabola crosses the \(x\)-axis exactly once, there is no need to solve a quadratic equation. (Read the same way, the secant line is \(x\) as a linear function of \(y\), evaluated at \(y=0\).)

Near a simple root the secant method converges with order \(\varphi=(1+\sqrt5)/2\approx 1.62\) and IQI with order about \(1.84\), both between bisection’s linear rate and Newton’s quadratic one. The slope of the log-log plot in “Convergence plots” estimates this order. Each method breaks down when two of the interpolated \(f\)-values coincide.

NoteConvergence plots