False Position (Regula Falsi) - Numerical Methods Step by Step

The false-position method refines a bracket [a,b] using a linear interpolant instead of the midpoint. The false position is xr=(a·f(b)-b·f(a))/(f(b)-f(a)). Applied to f(x)=x²-2 on [1,2]: after 3 steps the bracket tightens to [1.4118, 2.0000], width 0.5882.

false-position Like bisection, false position maintains a bracket where f changes sign. Unlike bisection, the new point xr is the x-intercept of the secant through (a,f(a)) and (b,f(b)). This typically converges faster than bisection.

Example

Add one false-position row per iteration using the secant x-intercept.

highlighted = computed this step

Step 1 — Set up

Start with the bracket from 1 to 2.

f(x)= x^{2}-2 \quad [ 1 , 2 ]

Step 2 — Iteration 1

Add false-position iteration 1 to the table.

\begin{array}{cccccc}\text{n} & \text{a} & \text{b} & \text{xr} & \text{f(xr)} & \text{update} \\ \hlmath{1} & \hlmath{1.0000} & \hlmath{2.0000} & \hlmath{1.3333} & \hlmath{-0.2222} & \hlmath{\text{a=xr}}\end{array}

Step 3 — Iteration 2

Add false-position iteration 2 to the table.

\begin{array}{cccccc}\text{n} & \text{a} & \text{b} & \text{xr} & \text{f(xr)} & \text{update} \\ 1 & 1.0000 & 2.0000 & 1.3333 & -0.2222 & \text{a=xr} \\ \hlmath{2} & \hlmath{1.3333} & \hlmath{2.0000} & \hlmath{1.4000} & \hlmath{-0.0400} & \hlmath{\text{a=xr}}\end{array}

Step 4 — Iteration 3

Add false-position iteration 3 to the table.

\begin{array}{cccccc}\text{n} & \text{a} & \text{b} & \text{xr} & \text{f(xr)} & \text{update} \\ 1 & 1.0000 & 2.0000 & 1.3333 & -0.2222 & \text{a=xr} \\ 2 & 1.3333 & 2.0000 & 1.4000 & -0.0400 & \text{a=xr} \\ \hlmath{3} & \hlmath{1.4000} & \hlmath{2.0000} & \hlmath{1.4118} & \hlmath{-0.0069} & \hlmath{\text{a=xr}}\end{array}

Step 5 — Bracketed root

The final bracket is 1.4118 to 2.0000 with width 0.5882.

[ \hlmath{1.4118} , \hlmath{2.0000} ]\quad w= \hlmath{0.5882}