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 ? and ? 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

Read More,

• Newton Raphson’s Method
• Difference Between Bisection Method and Regula Falsi Method
• Difference between Bisection Method and Newton Raphson Method 
• Difference between Newton Raphson Method and Regular Falsi Method
• Secant Method of Numerical analysis
• Root Finding Algorithm

Numerical on Regula Falsi Method

Problem: 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 ?(1.3333)⋅?(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 ?(1.3333)⋅?(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 ?(1.3513)⋅?(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: 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− f(a)⋅(b−a)/f(b)−f(a)

⇒ c =0− (1⋅(1−0))/−0.4597−1

⇒ 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) = x^2-4 in the interval [1,3] using the Regula Falsi method.

Problem 2: Solve f(x) = e^x – 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].

Conclusion

To a numerical analyst, The Regula Falsi method is a powerful tool in his toolbox as it provides a reliable way of solving nonlinear equations. Its simplicity and guaranteed convergence make it an invaluable method for engineering applications. Learning how to solve problems using these methodologies will enable students and practitioners tackle intricate challenges essential in their respective fields.”

FAQs: Regula Falsi Method

What is the Regula Falsi method?

Regula falsi method refers to the process of getting a root of a continuous function by iteratively shrinking an interval inside which the function changes sign using straight line interpolation.

How does the Regula Falsi method differ from the bisection method?

Whilst both are bracketing techniques, Regula Falsi estimates for example use linear interpolation roots while bisection just divides intervals.

Can all types of functions be handled by Regula Falsi?

This method finds its best use with continuous functions that can have an initial bracketing interval. It may fail to work well in situations where there is a discontinuity or more than one root occurs within a given interval.

How do you choose the initial interval for the Regula Falsi method?

The initial interval should be chosen such that the function changes sign over the interval, i.e.,

f(a) ⋅ f(b) < 0.