The Bisection Method

If a function f(x) is continuous between a and b, and f(a) and f(b) are of opposite signs, then there exists at least one root between a and b. For definiteness, let f(a) be negative and f(b) be positive. Then the root lies between a and b and let its approximate value be given by x₀= (a+b)/2. If f(x₀)=0, we conclude that x₀ is a root of the equation f(x) =0. Otherwise, the root lies either between x₀ and b, or between x₀ and a depending on whether f(x₀) is negative or positive.

The method is shown graphically on the below

Example: - Find a real root of the equation f(x) = x³ – x – 1 = 0.

Solⁿ: -

Given, f(x) = x³ – x – 1

And f(1) = -1

f(2) = 5

Since f(1) is negative and f(2) is positive, a root lies between 1 and 2,

Step-1:

Here, a = 1 and b = 2

So, xo =(1+2)/2=1.5

f(x_o) = 0.875

x_o =1.5

Step-2:

a = 1 , b = 1.5

x₁ = (1+1.5)/2 = 1.25

f (x₁) = -0.296875

x₁ = 1.25

Step-3:

a = 1.25 , b = 1.5

x₂ = 1.375

f(x₂) = 0.2246

x₂ =1.375

The procedure is repeated and the successive approximations are

x₃ = 1.3125, x₄ = 1.34375, x₅ = 1.328125, etc.

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Numerical Analysis

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