Question

Multiple Choice If Newton's method is used to find the zero of \(f(x)=x-x^{3}+2,\) what is the third estimate if the first estimate is 1\(?\) \((\mathbf{A})-\frac{3}{4} \quad(\mathbf{B}) \frac{3}{2} \quad(\mathbf{C}) \frac{8}{5} \quad(\mathbf{D}) \frac{18}{11}\) \((\mathbf{E}) 3\)

Short answer

The third estimate for the zero of the function \(f(x)=x-x^3+2\), starting with an initial estimate of 1, is \(\frac{18}{11}\), corresponding to choice (D).

Step-by-step solution

Newton's method formula

Newton's method uses the formula \(x_{n+1} = x_n - f(x_n)/f'(x_n)\) to generate the nth estimate from the (n-1)th estimate.

Derive \(f(x)\)

First we need to find the derivative of our function because it is needed in the Newton's method formula. The derivative of \(f(x) = x - x^3 + 2\) is \(f'(x) = 1 - 3x^2\).

Find the second estimate

Substitute \(x_1 = 1\) into the Newton's formula to get our second estimate \(x_2\). This gives us \(x_2 = x_1 - f(x_1)/f'(x_1) = 1 - (1 - 1^3 + 2) / (1 - 3 * 1^2) = 1 - 2 / -2 = 1 + 1 = 2\).

Find the third estimate

Now substitute \(x_2 = 2\) into our Newton's formula to give us our third estimate \(x_3\). This provides us with \(x_3 = x_2 - f(x_2)/f'(x_2) = 2 - (2 - 2^3 + 2) / (1 - 3 * 2^2) = 2 - 6 / -11 = 2 + 6/11 = \frac{18}{11}\).

Key concepts

Finding Zeros of Functions

Understanding how to find the zeros of a function is a fundamental skill in mathematics and in various applications across science and engineering. The zeros, or roots, are the points where the function crosses the x-axis, or simply put, the values of x for which the function equals zero.

For the function in our exercise, we see that the zero is the value of x for which the equation \(f(x) = x - x^3 + 2 = 0\). In an algebraic sense, this means we need to solve for x by factoring or other algebraic methods. However, many functions are too complex to solve with straightforward algebra, which is where numerical methods like Newton's method come into play.

By applying Newton's method, students can iteratively approach the function's zero, even when the function doesn't allow for a neat algebraic solution. This fusion of numerical methods with the concept of finding zeros greatly enhances problem-solving techniques.

Numerical Estimation

Numerical estimation is a critical concept that involves approximating a number when it's impossible or impractical to calculate the exact value. It is often applied in scenarios where direct calculation is either too complex or will take an impractical amount of time to compute.

Newton's method is a sterling example of numerical estimation in action. Through an iterative approach, we approximate the zero of a function to a desired level of accuracy. As demonstrated in our exercise, numerical estimation allows us to move progressively closer to the actual zero of the function by updating our estimates with each iteration.

For functions that have difficult-to-find exact solutions, numerical estimation tools are invaluable. They allow us to make educated guesses, improve those guesses iteratively, and reach a satisfactory solution efficiently.

Derivatives

The concept of derivatives is central to many fields within mathematics, including calculus, and plays a critical role in Newton's method. A derivative represents the rate at which a function is changing at any given point and is therefore a measure of its slope.

Practically speaking, knowing the derivative of a function allows us to understand how the function behaves—whether it's increasing or decreasing and how quickly at a particular point. In terms of Newton's method, the derivative helps to determine the direction and magnitude of the step we should take to approach the zero of the function.

In the exercise, we compute the derivative \(f'(x) = 1 - 3x^2\). This derivative is pivotal because Newton's method requires both the function value and its derivative value at the current estimate to calculate the next estimate.

Iterative Methods

Iterative methods, like Newton's method, use a sequence of improving approximations to converge on an accurate solution. These methods are iterative because the procedure is repeated multiple times, and with each repetition, the approximation becomes more precise.

Iteration is particularly useful for solving complex problems that cannot be easily solved using direct analytical methods. With each step, new values are computed based on the previous ones, and this process continues until the solution reaches an acceptable level of accuracy.

For instance, in our exercise, the iteration starts with an initial guess of 1. Then, based on the function and its derivative, subsequent estimates are calculated, getting closer and closer to the actual zero. Each step relies on the results of the last, showcasing the power and utility of iterative methods in numerical analysis and computational problem solving.