False Position Method

In mathematics, an ancient method of solving an equation in one variable is the false position method (method of false position) or regula falsi method. In simple words, the method is described as the trial and error approach of using “false” or “test” values for the variable and then altering the test value according to the result. In this article, you will learn how to solve an equation in one variable using the false position method. Also, get solved examples on the regula falsi method here.

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False Position Method (or) Regula Falsi Method

Consider an equation f(x) = 0, which contains only one variable, i.e. x.

To find the real root of the equation f(x) = 0, we consider a sufficiently small interval (a, b) where a < b such that f(a) and f(b) will have opposite signs. According to the intermediate value theorem, this implies a root lies between a and b.

Also, the curve y = f(x) will meet the x-axis at a certain point between A[a, f(a)] and B[b, f(b)].

Now, the equation of the chord joining A[a, f(a)] and B[b, f(b)] is given by:

Let y = 0 be the point of intersection of the chord equation (given above) with the x-axis. Then,

This can be simplified as:

Thus, the first approximation is x₁ = [af(b) – bf(a)]/ [f(b) – f(a)]

Also, x₁ is the root of f(x) if f(x₁) = 0.

If f(x₁) ≠ 0 and if f(x₁) and f(a) have opposite signs, then we can write the second approximation as:

x₂ = [af(x₁) – x₁f(a)]/ [f(x₁) – f(a)]

Similarly, we can estimate x₃, x₄, x₅, and so on.

Geometrical representation of the roots of the equation f(x) = 0 can be shown as:

False Position Method Solved Example

Question:

Find a root for the equation 2e^x sin x = 3 using the false position method and correct it to three decimal places with three iterations.

Solution:

Given equation: 2e^x sin x = 3

This can be written as: 2e^x sin x – 3 = 0

Let f(x) = 2e^x sin x – 3

So, f(0) = 2e⁰ sin 0 – 3

= 0 – 3

= -3 < 0

And

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.

Let a = 0 and b = 1.

The first approximation = x₁ = [af(b) – bf(a)]/ [f(b) – f(a)]

= [0(1.5747) – 1(-3)]/ [1.5747 – (-3)]

= 3/4.5747

= 0.6557

Thus, x₁ = 0.6557

Now, substitute x = 0.6557 in f(x).

So, f(0.6557) = 2e^0.6557 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.

Let a = 0.6557

The second approximation = x₂ = [af(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) = 2e^0.7563 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.

Hence, the third approximation = x₃ = [af(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

Therefore, the best approximation of the root up to three decimal places is 0.768 (up to three decimal places).

Practice Problems

1. Find a root for the equation x³ – 3x + 1 = 0 using the false position method and correct it to three decimal places with three iterations.
2. Find the root correct to two decimal places of the equation xe^x = cos x, using the regula falsi method.
3. Find the roots of f(x) = √12.