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Regula Falsi Method

Last Updated : 23 Jul, 2025

Regula Falsi Method, also known as the False Position Method, is a numerical technique used to find the roots of a non-linear equation of the form f(x)=0.

  • Based on the concept of bracketing, where two initial guesses, x0 and x1​, are chosen such that the function values at these points have opposite signs, indicating that a root lies between them
  • It operates on the fact that if a continuous function crosses zero over an interval, there exists a root within that interval.
  • When compared to other root finding algorithms, it has better convergence than bisection method, but slower convergence than Newton Raphson Method which has quadratic convergence.

Regula Falsi Method is a numerical technique used to find the roots of a non-linear equation of the form f(x) = 0. This method is particularly useful when the function is continuous and the root is located between two points.

Formula for Regula Falsi Method

The Regula Falsi method uses the following formula to approximate the root:

Where,

  • a and b are the endpoints of the interval [a, b].
  • f(a) and f(b) are the function values at points a and b.
  • c is the point where the linear interpolation intersects the x-axis.
👁 regular-falsi

How to Use Regula Falsi Method

To use Regula Falsi Method, we can use the following steps:

Step 1: Choose two initial points a and b such that function at those points have opposite sign i.e., f(a)⋅f(b) < 0.
Step 2: Calculate the point c where the linear approximation intersects the x-axis using the formula.
Step 3: Determine f(c).

  • If f(c) ⋅ f(a) < 0, then the root lies between a and c. Set b = c.
  • If f(c) ⋅ f(b) < 0, then the root lies between b and c. Set a = c.

Step 4: Repeat the steps until ∣f(c)∣ is less than a predefined tolerance level or the interval [a, b] is sufficiently small.

Example: the root of in the interval [1, 3].

Initial interval: a=1, b=3

f(b) =

Apply the formula:

So, the first approximation of the root is c=1.75.
Repeat: Now, use c=1.75 to update the interval. Since , we continue iterating to find more accurate roots.

This process would continue until f(c) is sufficiently close to 0.

Advantages of Regula Falsi Method

  • This method is very easy because it is very basic in its structure.
  • This bracketing method can guarantee the convergence if the function is continuous within the interval.
  • Unlike the Newton-Raphson method, the Regula Falsi method does not require the computation of derivatives.

Limitations of Regula Falsi Method

  • Regula Falsi, compared to other methods like Newton-Raphson may be slow to converge.
  • The speed of convergence depends heavily on the choice of initial interval.
  • The method can sometimes exhibit oscillatory behavior around the root, leading to slow convergence.

Regula Falsi vs Bisection Method vs Newton-Raphson Method

Feature

Regula Falsi Method

Bisection Method

Newton-Raphson Method

Initial Requirements

Requires two initial guesses a and b such that f(a).f(b)<0

Requires two initial guesses 𝑎 and 𝑏 such that f(a).f(b)<0

Requires one initial guess x0

Convergence Guarantee

Guaranteed if the function is continuous in [a, b]

Guaranteed if the function is continuous in [a, b]

Not guaranteed; depends on the choice of x0 and function behavior

Convergence Speed

Linear, generally faster than Bisection but slower than Newton-Raphson

Linear, generally slow

Quadratic, generally very fast

Formula

c = a− f(a)⋅(b−a)/f(b)−f(a)c = a+b/2xn+1 = xn − f(xn)/ f ′(xn)

Number of Function Evaluations

One per iteration

Two per iteration

One function and one derivative evaluation per iteration

Derivative Requirement

Not required

Not required

Requires first derivative

Application

Useful for non-differentiable functions

Useful for non-differentiable functions

Requires differentiable functions

Efficiency

More efficient than Bisection, less efficient than Newton-Raphson

Least efficient of the three

Most efficient if the derivative is available and well-behaved

Oscillatory Behavior

Can exhibit oscillatory behavior, slowing convergence

No oscillatory behavior

Can exhibit oscillatory behavior if initial guess is poor

Handling Multiple Roots

May converge to the nearest root

Will find a root within the interval but not necessarily the nearest

May fail to find multiple roots or diverge if poorly initialized

Numerical on Regula Falsi Method

Problem 1 : Find the root of the equation f(x)=x3−x−2 in the interval [1,2].

Solution:

Lets assume Initial Points i.e., a=1, b=2
Now,

  • f(1) = 13−1−2 = −2
  • f(2) = 23−2−2 = 4

Since f(1) ⋅ f(2) < 0, the root lies between 1 and 2.

Iteration 1: c = a− [f(a)⋅(b−a)/f(b)−f(a)]
⇒ c = 1−[(−2)⋅(2−1)/4−(−2) ]
⇒ c = 1− (−2/6) = 4/3
⇒ c = 1.3333
and f(1.3333) = 1.33333 − 1.3333 − 2 = −0.1481

Since f(2)⋅ f(1.3333) <0, update the interval to [4/3, 2].

Iteration 2: c = 1.3333− (−0.1481)⋅(2−1.3333)/4−(−0.1481)
⇒ c =1.3333−(−0.1481⋅0.6667/4.1481)
⇒ c = 1.3672
and f(1.3672) = 1.36723 − 1.3672 − 2 = 0.1197

Since 𝑓(1.3333)⋅𝑓(1.3672)<0, update the interval to [1.3333,1.3672].

Iteration 3: c = 1.3333− (−0.1481)⋅(1.3672−1.3333)/0.1197−(−0.1481)
⇒ c =1.3513
and f(1.3513) = 1.35133 − 1.3513 − 2 = −0.0061

Since 𝑓(1.3333)⋅𝑓(1.3513)<0, update the interval to [1.3513,1.3672].

Iteration 4: c = 1.3513− (−0.0061)⋅(1.3672−1.3513)/0.1197−(−0.0061)
⇒ c =1.3535
and f(1.3535) = 1.35353 − 1.3535 − 2 = 0.0003

Since 𝑓(1.3513)⋅𝑓(1.3535)<0, update the interval to [1.3513,1.3535].

Iteration 5: c = 1.3513− (−0.0061)⋅(1.3535−1.3513)/0.0003−(−0.0061)
⇒ c =1.3520
and f(1.3520) = 1.35203 − 1.3520 − 2 = −0.0001

Since, there are no significant changes in the value of approximate root.

Thus, the root is approximately x = 1.3522.

Problem 2: Find the root of the equation f(x) = cos(x) - x in the interval [0,1].

Solution:

Let the initial points be a = 0, b = 1

  • f(0) = cos(0) − 0 = 1
  • f(1) = cos(1) − 1 ≈ −0.4597

Since f(0)⋅ f(1) < 0, the root lies between 0 and 1.

Iteration 1: c = a−
⇒ c =0−
⇒ c = 1/1.4597
⇒ c =0.684
and f(0.684) = cos(0.684) − 0.684 ≈ 0.0894

Since f(0)⋅f(0.684) < 0, update the interval to [0,0.684].

Iteration 2: c = 0− [1⋅(0.684−0)/0.0894−1]
⇒ c = 0.684/0.9106
⇒ c = 0.751
and f(0.751) = cos(0.751) − 0.751 ≈ −0.0189

Since f(0.684)⋅f(0.751) < 0, update the interval to [0.684,0.751].

Iteration 3:

c = 0.684− (0.0894⋅(0.751−0.684))/(−0.0189−0.0894)
⇒ c = 0.739
and f(0.739) = cos(0.739) − 0.739 ≈ 0.0005

Since f(0.684)⋅f(0.739)<0, update the interval to [0.684,0.739].

Iteration 4: c = 0.684− 0.0894⋅(0.739−0.684)/0.0005−0.0894
⇒ c =0.7391
and f(0.7391) = cos(0.7391) − 0.7391 ≈ 0.0000

Since, there are no significant changes in the value of approximate root.

The root is approximately x = 0.7391.

Practice Problems: Regula Falsi Method

Problem 1: Find the root of the equation f(x) = x2-4 in the interval [1,3] using the Regula Falsi method.

Problem 2: Solve f(x) = ex - 3x in the interval [0,1] using the Regula Falsi method.

Problem 3: Determine the root of f(x) = sin(x)−0.5x in the interval [1,2] using the Regula Falsi method.

Problem 4: Apply the Regula Falsi method to find the root of f(x) = log(x)+x−5 in the interval [1,3].

Problem 5: Use the Regula Falsi method to solve f(x) = x3+3x2−1 in the interval [0,1].

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