Best ppt for regular falsi numerical method

1. REGULA FALSI METHOD
 Presented by:-
 Aditya raj (23bec004)
 Akshita sharma (23bec005)

2. Introduction:
 The Regula Falsi method, also known as the false
position method.
 It is a numerical technique used to find roots of a
continuous function.
 This method combines elements of the bisection
method and linear interpolation, making it both
straightforward and effective for root-finding
problems.

3. Convergence Criteria for this method:
 The Function f(x) should be continuous.
 The initial guesses a and b must bracket the root,
meaning they should satisfy f(a) f(b)<0 .This
⋅
ensures the method starts with two points on
either side of the root.

4. Advantages
 Guaranteed Convergence​
:
Under certain conditions, the method is
guaranteed to converge to a root. It
converges to the root if the function is
continuous and changes sign in the
initial interval.
 Faster Convergence​: ​
​
The Regula
Falsi Method often converges faster
than Bisection Method, especially when
the initial interval is large.​
 Suitable for Non-Differentiable
Functions ​:
Unlike methods like Newton-Raphson,
the Regula Falsi Method doesn't
require the function to be
differentiable. It can find roots of
functions that have sharp corners or
discontinuities

5. Disadvantages
 Slower Convergence: The Regula Falsi method may have a slower
convergence rate compared to other root-finding techniques like Newton-Raphson,
especially for certain types of functions.​
 Sensitivity to Initial Guesses: Poor initial guesses can lead to slow
convergence or even failure to converge.​
 Limited to Continuous Functions: It may not work well for
discontinuous or highly nonlinear functions.​
 Potential Oscillations: In some cases, the method may oscillate around
the root, leading to slower convergence or difficulty in determining the final
solution.​

6. Comparison between methods
Secant
method
• Similar to Regula
Falsi but less robust.
Can be faster but
might oscillate
around the root.​
Newton-
Raphson Method​
• Fastest convergence
rate, but requires
differentiable
function. Might fail if
the derivative is zero
or close to zero.​
Bisection
Method​
• Guaranteed
convergence but
slower convergence
rate compared to
Regula Falsi.
Requires continuous
function with a sign
change in the
interval.​

7. Applications and its Uses
Electrical Engineering : Finding the operating points of circuits or solving
equations involving impedances.​
Chemistry : Determining equilibrium constants or finding the concentration of
reactants and products in chemical reactions.​
Mechanical Engineering​: Analyzing stress and strain distributions in mechanical
systems or finding the optimal design parameters.​
Data Analysis​: Finding the optimal parameters in regression models or analyzing
data to identify trends.​

8. Mathematical Formulation:
 Initial Conditions: Choose a, b such that f(a) f(b)<0.
⋅
 Root Calculation: c = a f(b)− b f(a)
⋅ ⋅ / f(b) − f(a)
 Interval Update: Update ‘a’ or ‘b’ based on the sign of
f(c).
 Stopping Criteria: Continue until convergence is
achieved.

9. Procedure:
 Let a and b be two initial guesses such that: f(a) f(b)<0.
⋅
 Calculate the New Point : c = a f(b)− b f(a)
⋅ ⋅ / f(b) − f(a)
 Evaluate f(c) and Depending on the sign of f(c), update
either ‘a’ or ‘b’.
 Repeat Steps 2–3 until one of the stopping criteria is met ,
i.e :- f(c) < (desired tolerance).
∣ ∣ ϵ

10. Graphical Interpretation

11. Questions: Find a root for the equation 2ex
sin x = 3 using the
false position method and correct it to three
decimal places with three iterations.
Given equation: 2ex
sin x = 3
This can be written as: 2ex
sin x – 3 = 0
Let f(x) = 2ex
sin x – 3
So, f(0) = 2e0
sin 0 – 3 = 0 – 3
= -3 < 0
again,
f(1) = 2e1
sin 1 – 3
= 2e sin 1 – 3
= 1.5747 > 0
That means the root of f(x) lies between 0 and 1.

12. Let a = 0 and b = 1.
The first approximation = x1 = [a*f(b) – b*f(a)]/ [f(b) – f(a)]
= [0(1.5747) – 1(-3)]/ [1.5747 – (-3)]
= 3/4.5747
= 0.6557
Thus, x1 = 0.6557
Now, substitute x1 = 0.6557 in f(x).
So, f(0.6557) = 2e0.6557
sin(0.6557) – 3
= 2.3493 – 3
= -0.6507 < 0
As we know, f(1) > 0
That means a root lies between 0.6557 and 1.

13. Let a = 0.6557
The second approximation = x2 = [a*f(x1) – x1f(a)]/ [f(x1) – f(a)]
= [0.6557(1.5747) – 1(-0.6507)]/ [1.5747 – (-0.6507)]
= (1.0325 + 0.6507)/(2.2254)
= 1.6832/2.2254
= 0.7563
Therefore, x2 = 0.7563
Let us substitute 0.7563 in f(x).
So, f(0.7563) = 2e0.7563
sin(0.7563) – 3
= 2.9239 – 3
= -0.0761 < 0
We know that f(1) > 0
Thus, a root lies between 0.7563 and 1.

14. Hence, the third approximation = x3 = [a*f(x2) – x2f(a)]/ [f(x2) –
f(a)]
= [(0.7563)(1.5747) – 1(-0.0761)]/ [1.5747 – (-0.0761)]
= (1.1909 + 0.0761)/1.6508
= 1.2670/1.6508
= 0.7675
So,
x3 = 0.7675
Let us substitute 0.7675 in f(x).
So, f(0.7675) =2e0.7675
sin(0.7675) – 3
= -0.0082 < 0

15. We know that f(1) > 0
Thus, a root lies between a= 0.7675 and x3 = 1.
Hence, the fourth approximation =
x4 = [a*f(x3) – x3f(a)]/ [f(x3) – f(a)]
= [(0.7675)(1.5747) – 1(-0.0082)]/ [1.5747 – (-0.0082)]
=0.7687
Therefore, the best approximation of the root
up to three decimal places is 0.768

16. Graph

17. Algorithm:
 Define the function f(x).
Choose initial guesses a and b, such that (f(a) f(b)<0) .
⋅
Check the stopping criteria, such as a predefined tolerance ,
ϵ
or maximum number of iterations.
Compute the root approximation c using the false position formula
c = (a*f(b)−b*f(a)​
)/f(b)-f(a).
 Update the interval based on the sign of f(c)
If f(a) f(c)<0
⋅ , then the root lies in [a,c], so set b=c.
If f(b) f(c)<0
⋅ , then the root lies in [c,b], so set a=c.
 Check the convergence:
If f(c)∣is smaller than a predefined tolerance ϵ, then c is the
ϵ
root.
If not, repeat the process.

18. Program:

19. Result:

20. Thank You