Bisection Method

The Bisection Method is a simple numerical technique used to find the root of a continuous function. It works by dividing an interval [a, b] into two halves and repeatedly narrowing down the interval where the root lies, based on the sign change of the function. The method guarantees convergence if the function is continuous and there’s a sign change at the endpoints. While reliable, the Bisection Method is slower compared to other root-finding methods, especially for high precision.

1.0What is the Bisection Method?

The Bisection Method is a numerical technique used to find the root (or zero) of a function. This method works on the principle of dividing an interval into two subintervals, checking whether the root lies in the first or second subinterval, and then narrowing down the interval repeatedly until the root is found with the desired level of accuracy.

2.0Bisection Method Formula

The formula used in the bisection method is:

$X_{new}=\frac{a+b}{2}$

Where:

• a and b are the endpoints of the current interval.
• $x_{new}$ is the midpoint of the interval [a, b].

The key idea is to check the function values at the midpoint and decide whether the root lies in the left half or the right half of the interval. The process is repeated until the interval becomes sufficiently small.

3.0Application Of The Bisection Method

The application of the bisection method spans various fields, from engineering to physics and beyond. It's widely used in solving equations that do not have an analytical solution or when finding an exact root is too complicated. The bisection method guarantees convergence if the function is continuous and the initial interval contains the root.

4.0Application Of Bisection Method In Real Life

In real-world applications, the bisection method plays a crucial role in solving engineering and physics problems. Some examples include:

1. Engineering Design: When designing systems that rely on the behavior of materials (such as structural analysis of beams or pressure systems), the bisection method can be used to find the correct design parameters that meet specific performance criteria.
2. Control Systems: The bisection method helps in tuning control systems, like those in automotive or robotics, to find optimal values for system stability.
3. Financial Modeling: In finance, the method can be used to find the interest rate or discount factor in situations where more complex models are impractical.
4. Electrical Engineering: The method is used to find impedance matching in circuit design, ensuring minimal energy loss between components.
5. Physics: In physics, problems like determining the equilibrium position of a system or finding the time when an object reaches a certain state (e.g., when it reaches a velocity or position) can be solved using the bisection method.

5.0Bisection Method Example

Let’s look at an example of the bisection method applied to a simple function:

Suppose we want to find the root of the equation:

$f(x)=x^2-4$

We know that f(x) has a root between 0 and 3, as:

• $f(0)=0^2-4=-4$ (negative)
• $f(3)=3^2-4=5$ (positive)

Thus, the root lies between 0 and 3. Using the bisection method:

1. Calculate the midpoint: $x_{mid}=\frac{0+3}{2}=1.5$
2. Evaluate the function at the midpoint:

$f(1.5)=(1.5{)}^2-4$

$f(1.5)=2.25-4$

$f(1.5)=-1.75$

Since f(1.5) is negative, the root lies between 1.5 and 3. 

Again, Calculate the mid-point

$x_{mid}=\frac{1.5+3}{2}=2.25$

$f(2.25)=(2.25{)}^2-4$

$f(2.25)=5.0625-4$

$f(2.25)=1.0625$

Since f(2.25) is positive, the root lies between 1.5 and 2.25.

Again, Calculate the mid-point

$x_{mid}=\frac{1.5+2.25}{2}=1.875$

$f(1.875)=(1.875{)}^2-4$

$f(1.875)=3.515625-4$

$f(1.875)=-0.484375$

Since f(1.875) is negative, the root lies between 1.875 and 2.25.

Again, Calculate the mid-point

$x_{mid}=\frac{1.875+2.25}{2}=2.0625$

$f(2.0625)=(2.0625{)}^2-4$

$f(2.0625)=(4.25390625-4$

$f(2.0626)=0.25390625$

Since f(2.0625) is positive, the root lies between 1.875 and 2.0625.

Again, Calculate the mid-point

$x_{mid}=\frac{1.875+2.0625}{2}=1.96875$

$f(1.96875)=(1.96875{)}^2-4$

$f(1.96875)=3.8759765625-4$

$f(1.96875)=-0.1240234375$

Since f(1.96875) is negative, the root lies between 1.96875 and 2.0625.

Again, Calculate the mid-point

$x_{mid}=\frac{1.96875+2.0625}{2}=2.015625$

$f(2.015625)=(2.015625{)}^2-4$

$f(2.015625)=4.062744140625-4$

$f(2.015625)=0.062744140625$

Since f(2.015625) is positive, the root lies between 1.96875 and 2.015625.

Again, Calculate the mid-point

$x_{\text{mid}}=\frac{1.96875+2.015625}{2}=1.9921875$

$f(1.9921875)=(1.9921875{)}^3-4$

$f(1.9921875)=3.9681103155625-4$

$f(1.9921875)=-0.03118896484375$

Since f(1.9921875) is negative, the root lies between 1.9921875 and 2.015625. 

Again, Calculate the mid-point

$x_{\text{mid}}=\frac{1.9921875+2.015625}{2}=2.00390625$

$f(2.00390625)=(2.00390625{)}^3-4$

$f(2.00390625)=4.015640258789063-4$

$f(2.00390625)=0.0156402587890625$

Since ( f(2.00390625) is positive, the root lies between 1.9921875 and 2.00390625.

Again, Calculate the mid-point

$x_{\text{mid}}=\frac{1.9921875+2.00390625}{2}=1.998046875$

$f(1.998046875)=(1.998046875{)}^3-4$

$f(1.998046875)=3.992171314697266-4$

$f(1.998046875)=-0.0078086853027344$

Since f(1.998046875) is negative, the root lies between 1.998046875 and 2.00390625.

Again, Calculate the mid-point

$x_{\text{mid}}=\frac{1.998046875+2.00390625}{2}=2.0009765625$

$f(2.0009765625)=(2.0009765625{)}^3-4$

$f(2.0009765625)=4.003907203674316-4$

$f(2.0009765625)=0.003907203674316$

Since f(2.0009765625) is positive, the root lies between 1.998046875 and 2.0009765625.

Again, Calculate the mid-point

$x_{\text{mid}}=\frac{1.998046875+2.0009765625}{2}=1.99951171875$

$f(1.99951171875)=(1.99951171875{)}^3-4$

$f(1.99951171875)=3.998047113418579-4$

$f(1.99951171875)=-0.0019528865814209$

Since f(1.99951171875) is negative, the root lies between 1.99951171875 and 2.0009765625.

Again, Calculate the mid-point

$x_{\text{mid}}=\frac{1.99951171875+2.0009765625}{2}=2.000244140625$

$f(2.000244140625)=(2.000244140625{)}^3-4$

$f(2.000244140625)=4.000976622104645-4$

$f(2.000244140625)=0.000976622104645$

Since f(2.000244140625 is positive, the root lies between 1.99951171875 and 2.000244140625.

We repeat this process, updating the interval in each iteration, and get closer to the root each time. After several steps, we’ll find that the root is approximately x = 2.

6.0Bisection Method of Finding Roots

The bisection method of finding roots is beneficial for cases where other methods (like Newton’s Method or Secant Method) might fail due to a lack of derivatives or initial guesses. The method’s strength lies in its reliability and simplicity, as it only requires that the function is continuous and that the root lies within the initial interval. However, it can be slower compared to other methods.

7.0Common Problems and Challenges

While the bisection method is reliable, it does come with certain limitations and challenges:

1. Slow Convergence: The method can be slower compared to other numerical methods, like Newton’s method, especially when high precision is required.
2. Initial Interval: The method requires that the initial interval [a, b] contains a root. If the function does not change sign between a and b, the method will fail.
3. Multiple Roots: If a function has more than one root in an interval, the bisection method can only find one of them, and typically only if it’s the one where the function changes sign.
4. Precision vs. Iterations: Achieving high precision requires a large number of iterations, which may be computationally expensive for complex functions.

8.0Bisection Method Problems

Here are a couple of bisection method problems to test your understanding:

1. Problem 1: Find the root of the function $f(x)=x^3-6x^2+11x-6$ between 1 and 3.
2. Problem 2: Use the bisection method to find the root of $f(x)=cos(x)-x$ between 0 and 1.