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:
c = a - \frac{f(a) . (b-a)} {f(b) - f(a)}
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.
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
Initial interval: a=1, b=3
f(a) = 1^2 - 4 = -3
f(b) =3^2 - 4 = 5 Apply the formula:
c = a - \frac{f(a) \cdot (b - a)}{f(b) - f(a)} = 1 - \frac{(-3) \cdot (3 - 1)}{5 - (-3)} = 1 - \frac{-6}{8} = 1 + 0.75 = 1.75 So, the first approximation of the root is c=1.75.
Repeat: Now, use c=1.75 to update the interval. Sincef(1.75) = 1.75^2 - 4 = -0.9375 , 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/2 | xn+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 = −0.1481Since 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 = 0.1197Since 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 = −0.0061Since 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 = 0.0003Since 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 = −0.0001Since, 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−\frac{f(a)⋅(b−a)}{f(b)−f(a)}
⇒ c =0−\frac {1⋅(1−0)}{−0.4597−1}
⇒ c = 1/1.4597
⇒ c =0.684
and f(0.684) = cos(0.684) − 0.684 ≈ 0.0894Since 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 ≈ −0.0189Since 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 ≈ 0.0005Since 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 ≈ 0.0000Since, 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].
Read More,