Bisection Method

The bisection method is a simple and effective technique used in numerical analysis to find the root of an equation. It works by repeatedly dividing an interval into halves and narrowing down the section where the root lies. This method is particularly useful for solving transcendental equations and polynomial equations, making it one of the fundamental techniques in root-finding methods.

If you are looking for resources like bisection method PDF, or want to implement the bisection method in MATLAB, this article covers everything, including the bisection method formula, step-by-step explanation, and solved examples.

Bisection Method Definition

The bisection method is a numerical approach for solving equations by repeatedly dividing an interval into two halves and selecting the subinterval where the root exists. This method is based on the Intermediate Value Theorem, which states that if a function is continuous on an interval [a, b] and f(a) and f(b) have opposite signs, then there exists at least one root in that interval.

Other names for the bisection method include:

• Interval Halving Method
• Binary Search Method
• Dichotomy Method

How Does the Bisection Method Work?

1. Choose an interval [a, b] where f(a) and f(b) have opposite signs.
2. Compute the midpoint, t = (a + b)/2.
3. Check if f(t) = 0. If yes, t is the root. If not, continue.
4. Determine the new interval:
   • If f(a) * f(t) < 0, set b = t.
   • If f(t) * f(b) < 0, set a = t.
5. Repeat the process until the interval is sufficiently small or f(t) is close to zero.

The bisection method in MATLAB can be implemented using simple code, which we will discuss later. You can also download a bisection method PDF for further practice.

Bisection Method Example – Solved Problem

Example 1: Find the root of the equation $x^2-3=0$$x$$2$$-$$3$$=$$0$ in the interval [1, 2] using the bisection method.

Solution:

Given: $f(x)=x^2-3$$2$

Find f(a) and f(b):

$2$ 
$$f(2) = 2^2 - 3 = 1 > 0$$$2$

Since f(1) and f(2) have opposite signs, a root exists in [1, 2]. Now, apply the bisection method formula:

The root is approximately 1.7344 after seven iterations.

You can also try implementing this bisection method in MATLAB using simple code.

Advantages and Limitations of the Bisection Method

The bisection method in MATLAB is widely used due to its straightforward approach, but it also has some drawbacks. The table below summarizes the key advantages and limitations of the method.

Advantages of the Bisection Method

Limitations of the Bisection Method

Application of the Bisection Method

The bisection method in MATLAB and other programming languages is widely used in real-world applications. It plays a crucial role in different scientific and engineering fields.

Common Applications of the Bisection Method

Many textbooks and research papers provide detailed explanations, and you can find a bisection method PDF online with solved examples and theoretical concepts.

Bisection Method Formula

The bisection method formula follows these simple steps:

1. Select an initial interval $[a, b]$ where the function changes sign (i.e., $f(a) \cdot f(b) < 0$).
2. Compute the midpoint: $$c = \frac{a + b}{2}$$
3. Check the function value at $c$:
   • If $f(c) = 0$, then $c$ is the root.
   • If $f(c)$ has the same sign as $f(a)$, set $a = c$; otherwise, set $b = c$.
4. Repeat the process until the difference between $a$ and $b$ is smaller than the desired accuracy.

This step-by-step approach is easy to follow, and you can implement it using MATLAB code for the bisection method to automate calculations.

Tips and Tricks for the Bisection Method

• Select a proper interval to minimize iterations.
• Use high precision to avoid floating-point errors.
• Implement an effective stopping condition to balance accuracy and speed.
• Compare with other methods, like the False Position Method, for better convergence.

By following these steps, you can effectively use the bisection method in MATLAB and apply it to real-world problems.

Bisection Method Questions for Practice

Here are some questions to practice:

1. Find the root of $x^3 - 4x - 9 = 0$$3$ using the bisection method in the interval [2, 3].
2. Solve $x^3 - 4 = 0$$3$ in the interval [1, 2] using the bisection method in numerical analysis.
3. Determine the root of $x^3 - 3$$3$ using the bisection method formula.
4. Solve $x^2 - 5 = 0$$2$ in [2, 3] using the bisection method example given above.
5. Find the root of $x^3 - 3x - 5 = 0$$3$ using the bisection method.