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Sulaiman Khan
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False Position Method

The False Position Method, also known as the Regula Falsi Method, is a numerical technique used to find approximate roots of a real-valued continuous function. It is a bracketing method, meaning it starts with two initial points that bracket a root (i.e., the function changes sign between them) and iteratively refines this interval to approach the root.


Given a function:

f(x)=0f(x) = 0
C=a⋅f(b)−b⋅f(a)f(b)−f(a)C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}

How the False Position Method Works

  1. Root of a Function: A value C C such thatf(C)=0 f(C) = 0 .
  2. Initial Bracketing: Choose two initial points 𝑎 and 𝑏 such thatf(a).f(b)<0 f(a) . f(b) < 0 . This ensures that there is at least one root between 𝑎 and 𝑏.
  3. Calculate the False Position (Regula Falsi) Point:

    The next approximation C C is found using the formula:

    C=a⋅f(b)−b⋅f(a)f(b)−f(a)C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}

    This formula derives from the equation of the straight line (secant line) connecting (a,f(a)) and (b,f(b)) (a , f(a) ) \text{ and } ( b , f(b) ) and finding its intersection with the x-axis.

  4. Evaluate f(C)f(C):
    • If f(C)=0,C f(C) = 0 , C is the root.
    • If f(a).f(b)<0, f(a) . f(b) < 0 , the root lies between 𝑎 and C C . Set 𝑏 = C C .
    • If f(C).f(b)<0, f(C) . f(b) < 0 , the root lies between C C and 𝑏. Set 𝑎 = C C .
  5. Iterate:

    Repeat steps 2 and 3 until the approximate root C C converges to a desired level of accuracy.

Example

Given Function

x2=4 x^2 = 4

Error Margin is 0.01

Step 1: Initial Bracketing:

Choose  a=1,f(a)=12−4=−3\text{Choose} \; a = 1, f(a) = 1^2 - 4 = -3
Choose  b=3,f(b)=32−4=5\text{Choose} \; b = 3, f(b) = 3^2 - 4 = 5
Since f(a)⋅f(b)=−3⋅5<0,  a and b bracket the root.\text{Since} \ f(a) \cdot f(b) = -3 \cdot 5 < 0, \; a \text{ and } b \text{ bracket the root.}

Step 2: Calculate C:

C=a⋅f(b)−b⋅f(a)f(b)−f(a)C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}
C=1⋅5−3⋅(−3)5−(−3)=5+98=148=1.75C = \frac{1 \cdot 5 - 3 \cdot (-3)}{5 - (-3)} = \frac{5 + 9}{8} = \frac{14}{8} = 1.75
f(C)=f(1.75)=(1.75)2−4=3.0625−4=−0.9375f(C) = f(1.75) = (1.75)^2 - 4 = 3.0625 - 4 = -0.9375
Since  f(a)⋅f(C)=−3⋅(−0.9375)>0,  set a=1.75.\text{Since} \; f(a) \cdot f(C) = -3 \cdot (-0.9375) > 0, \; \text{set } a = 1.75.

Step 3: Second Iteration:

C=a⋅f(b)−b⋅f(a)f(b)−f(a)≈1.75⋅5−3⋅(−0.9375)5−(−0.9375)≈1.9474C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)} \approx \frac{1.75 \cdot 5 - 3 \cdot (-0.9375)}{5 - (-0.9375)} \approx 1.9474
f(1.9474)≈(1.9474)2−4≈3.7925−4=−0.2075f(1.9474) \approx (1.9474)^2 - 4 \approx 3.7925 - 4 = -0.2075
Since  f(a)⋅f(C)≈−0.9375⋅(−0.2075)>0,  set a=1.9474.\text{Since} \; f(a) \cdot f(C) \approx -0.9375 \cdot (-0.2075) > 0, \; \text{set } a = 1.9474.

Step 4: Third Iteration:

C=a⋅f(b)−b⋅f(a)f(b)−f(a)≈1.988C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)} \approx 1.988
f(1.988)≈(1.988)2−4≈3.952−4=−0.048f(1.988) \approx (1.988)^2 - 4 \approx 3.952 - 4 = -0.048
Since  f(a)⋅f(C)≈−0.2075⋅(−0.048)>0,  set a=1.988.\text{Since} \; f(a) \cdot f(C) \approx -0.2075 \cdot (-0.048) > 0, \; \text{set } a = 1.988.

Continue Iterating:

Repeating this process will yield increasingly accurate approximations of the root.

x≈2x \approx 2

Conclusion

The False Position Method is an efficient way to find the root of a function defined by a continuous equation. In this case, we demonstrated it for f(x)=x2−4 f(x) = x^2 - 4 and found that the root isx=2 x = 2 . The method guarantees convergence as long as you start with points that bracket the root, making it a reliable technique for root-finding problems.

False Position Method Solver