Problem 3 - Thomas Calculus

Question

Use Newton's method to estimate the two zeros of the function \(f(x)=x^{4}+x-3 .\) Start with \(x_{0}=-1\) for the left-hand zero and with \(x_{0}=1\) for the zero on the right. Then, in each case, find \(x_{2}\)

Step-by-Step Solution

Answer

For the left-hand zero, \( x_2 \) is 3. For the right-hand zero, \( x_2 \) is approximately 1.121.

Step 1: Understanding Newton's Method

Newton's Method is used to find successively better approximations to the roots (or zeroes) of a real-valued function. It uses the formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to iteratively find closer approximations.

Step 2: Calculate the derivative of the function

Given the function \( f(x) = x^4 + x - 3 \), we need to find its derivative to use in Newton's formula. The derivative is \( f'(x) = 4x^3 + 1 \).

Step 3: First Iteration for Left-hand Zero

Starting with \( x_0 = -1 \), calculate \( f(-1) \ = (-1)^4 + (-1) - 3 = 1 - 1 - 3 = -3 \ \) and \( f'(-1) = 4(-1)^3 + 1 = -4 + 1 = -3 \). Use Newton's formula: \( x_1 = -1 - \frac{-3}{-3} = 0 \).

Step 4: Second Iteration for Left-hand Zero

Use \( x_1 = 0 \) to find \( x_2 \). Calculate \( f(0) = 0^4 + 0 - 3 = -3 \) and \( f'(0) = 4(0)^3 + 1 = 1 \). Use Newton's formula: \( x_2 = 0 - \frac{-3}{1} = 3 \).

Step 5: First Iteration for Right-hand Zero

Starting with \( x_0 = 1 \), calculate \( f(1) = 1^4 + 1 - 3 = -1 \) and \( f'(1) = 4(1)^3 + 1 = 5 \). Use Newton's formula: \( x_1 = 1 - \frac{-1}{5} = 1.2 \).

Step 6: Second Iteration for Right-hand Zero

Use \( x_1 = 1.2 \) to find \( x_2 \). Calculate \( f(1.2) = (1.2)^4 + 1.2 - 3 \approx 0.6896 \) and \( f'(1.2) = 4(1.2)^3 + 1 \approx 8.728 \). Use Newton's formula: \( x_2 = 1.2 - \frac{0.6896}{8.728} \approx 1.121 \).

Key Concepts

Numerical Analysis

Numerical analysis is a branch of mathematics that deals with algorithms for solving mathematical problems numerically. This means instead of finding a solution through algebraic manipulation, numerical methods use approximations and iterations to find solutions as close as possible to the real answer.
A common application of numerical analysis is finding roots of functions, which means solving for x values that make the function equal zero.

• Newton's Method is one such algorithm; it iteratively finds successively better approximations of roots.
• This method is particularly useful when dealing with complex functions, which might be difficult to solve using direct algebraic methods.

Each step involves calculations using current estimates and improving on them based on the derivative of the function. By applying Newton’s Method multiple times, we converge closer to the actual root with each iteration.

Root-Finding Algorithms

Root-finding algorithms are techniques used for determining the zeros of a function, where the function crosses the x-axis. These points are important because they demonstrate where the function reaches a value of zero, often reflecting equilibrium in real-world applications.
Newton's Method is one of the more efficient algorithms for root-finding due to its rapid convergence under certain conditions.

• The method starts with an initial guess, which is crucial for its success. A good initial guess leads to quicker convergence.
• The formula for Newton's Method is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

Here, \( x_n \) is the current approximation, and \( x_{n+1} \) is the next approximation. By repeatedly applying this formula, we use the tangent line at the current point to find a better guess, honing in on the true zero of the function.

Calculus

Calculus is the mathematical study of change and is integral to Newton's Method. It provides the toolkit for understanding how a function behaves, notably through derivatives that indicate how the function's rate of change or slope varies at any given point.
To apply Newton's Method, it is essential to compute the derivative of the function first.

• In the exercise, the function \( f(x) = x^4 + x - 3 \) has a derivative \( f'(x) = 4x^3 + 1 \).
• This derivative is crucial because it is used in the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).

By knowing the rate at which the function changes, we can adjust our guesses more accurately towards the root. Thus, calculus enables us to use derivatives to guide our iterative process, making each step more precise.