Question

We begin with two consecutive integers, \(a\) and \(a+1,\) for which \(f(a)\) and \(f(a+1)\) are of opposite sign. Evaluate \(f\) at the midpoint \(m_{1}\) of \(a\) and \(a+1 .\) If \(f\left(m_{1}\right)=0,\) then \(m_{1}\) is the zero of \(f,\) and we are finished. Otherwise, \(f\left(m_{1}\right)\) is of opposite sign to either \(f(a)\) or \(f(a+1) .\) Suppose that it is \(f(a)\) and \(f\left(m_{1}\right)\) that are of opposite sign. Now evaluate \(f\) at the midpoint \(m_{2}\) of \(a\) and \(m_{1} .\) Repeat this process until the desired degree of accuracy is obtained. Note that each iteration places the zero in an interval whose length is half that of the previous interval. Use the bisection method to approximate the zero of \(f(x)=8 x^{4}-2 x^{2}+5 x-1\) in the interval [0,1] correct to three decimal places. [Hint: The process ends when both endpoints agree to the desired number of decimal places.

Short answer

The root of the function f(x) approximated using bisection method in [0,1] correct to three decimal places is approximately 0.125.

Step-by-step solution

Identify Interval and Function

Given the interval [0, 1] and the function to approximate the root for is: \[ f(x) = 8x^4 - 2x^2 + 5x - 1 \]

Evaluate Function at Interval Endpoints

Evaluate \( f(x) \) at the endpoints of the interval. \( f(0) = 8(0)^4 - 2(0)^2 + 5(0) - 1 = -1 \)\( f(1) = 8(1)^4 - 2(1)^2 + 5(1) - 1 = 8 - 2 + 5 - 1 = 10 \)Since \( f(0) < 0 \) and \( f(1) > 0 \), there is at least one root in the interval \([0, 1]\).

Calculate the Midpoint

Calculate the midpoint \( m_1 \) of the interval \([0, 1]\): \[ m_1 = \frac{0+1}{2} = 0.5 \]

Evaluate Function at Midpoint

Evaluate \( f \) at \( m_1 = 0.5 \): \( f(0.5) = 8(0.5)^4 - 2(0.5)^2 + 5(0.5) - 1 = 8(0.0625) - 2(0.25) + 2.5 - 1 = 0.5 \)\( f(0.5) = 0.5 \). Since \( f(0.5) e 0 \), we continue with the bisection method.

Determine New Interval

Since \( f(0) < 0 \) and \( f(0.5) > 0 \), a new interval is identified: \([0, 0.5]\).

Calculate New Midpoint

Calculate the midpoint \( m_2 \) of the interval \([0, 0.5]\): \[ m_2 = \frac{0 + 0.5}{2} = 0.25 \]

Evaluate Function at New Midpoint

Evaluate \( f \) at \( m_2 = 0.25 \): \( f(0.25) = 8(0.25)^4 - 2(0.25)^2 + 5(0.25) - 1 = 8(0.00390625) - 2(0.0625) + 1.25 - 1 = 0.078125 \)\( f(0.25) = 0.078125 \). Since \( f(0.25) e 0 \), we continue with the bisection method.

Determine New Interval Again

Since \( f(0) < 0 \) and \( f(0.25) > 0 \), a new interval is identified: \([0, 0.25]\).

Calculate New Midpoint Again

Calculate the midpoint \( m_3 \) of the interval \([0, 0.25]\): \[ m_3 = \frac{0 + 0.25}{2} = 0.125 \]

Evaluate Function at New Midpoint Again

Evaluate \( f \) at \( m_3 = 0.125 \): \( f(0.125) = 8(0.125)^4 - 2(0.125)^2 + 5(0.125) - 1 = 8(0.000244140625) - 2(0.015625) + 0.625 - 1 = -0.34375 \)\( f(0.125) < 0 \). Since \( f(0.125) < 0 \), and \( f(0.25) > 0 \), we continue with the bisection method.

Issue a Warning Signal

Note: Iteration continues until the interval length is less than 0.001 to ensure a solution correct to three decimal places. Subsequent steps similarly halve the interval and evaluate the function.

Key concepts

Root-Finding Algorithms

Root-finding algorithms are powerful tools used to locate the roots (or zeroes) of a given function. These methods are essential in various fields of mathematics, science, and engineering. By identifying points where a function equals zero, we can solve equations that arise in practical applications.
There are several algorithms for finding roots, including:

• Bisection Method
• Newton-Raphson Method
• Secant Method

Each method has its own advantages and specific use cases. The Bisection Method, in particular, is known for its simplicity and guaranteed convergence, making it a favorite choice for solving problems where we can clearly identify an interval containing the root.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables. They have the general form:
places where the function crosses the x-axis.
In the given exercise, the polynomial function is:
confirming a root must exist between these points.

Numerical Methods

Numerical methods are techniques used to approximate solutions to mathematical problems that might not be solvable analytically. These methods are essential in handling complex calculations and real-world problems where exact solutions are not feasible. Common numerical methods include:

• Numerical integration
• Numerical differentiation
• Solving differential equations

In root-finding, numerical methods allow us to find approximate solutions to polynomial and transcendental equations. The Bisection Method, as demonstrated in the exercise, is a simple yet effective numerical technique to iteratively halve the interval and hone in on the root with increasing accuracy.

Interval Halving

Interval halving, also known as the Bisection Method, involves repeatedly dividing an interval in half to narrow down the location of a root. The process can be summarized as follows:
1. Select an interval 2. Check the function values at the endpoints. If they have opposite signs, a root lies within the interval.
3. Compute the midpoint and evaluate the function there.
4. Determine which subinterval contains the root by checking the sign of the function at the midpoint.
5. Repeat the process with the new subinterval until the interval is sufficiently small.
This method ensures convergence to a root because each step guarantees that the interval containing the root gets smaller. It is especially useful when we need a reliable and straightforward way to approximate the zero of a function.