Problem 39 - Calculus Early Transcendentals

Question

(a) Apply Newton's Method to \(f(x)=x^{2}+1\) with a starting value of \(x_{1}=0.5,\) and determine if the values of \(x_{2}, \ldots, x_{10}\) appear to converge. (b) Explain what is happening.

Step-by-Step Solution

Answer

(a) The values do not converge; (b) \(f(x)=x^2+1\) has no real roots.

Step 1: Understand Newton's Method Formula

Newton's Method is used to find approximate roots of a function. The iterative formula is given by: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]We will apply this formula to our function, starting from the initial guess \(x_1 = 0.5\).

Step 2: Determine Derivative of the Function

We will differentiate the function \(f(x) = x^2 + 1\). The derivative is:\[ f'(x) = 2x \]

Step 3: Calculate the Next Iteration \(x_2\)

Using the Newton's Method formula, calculate \(x_2\):\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.5 - \frac{(0.5)^2 + 1}{2(0.5)} = 0.5 - \frac{1.25}{1} = -0.75 \]

Step 4: Calculate \(x_3\) Using \(x_2\)

Repeat the process for \(x_3\):\[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = -0.75 - \frac{(-0.75)^2 + 1}{2(-0.75)} = -0.75 - \frac{1.5625}{-1.5} \approx 0.2917 \]

Step 5: Iterate for \(x_4\) to \(x_{10}\)

Continue applying Newton's Method up to \(x_{10}\):\[ x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} \]Calculations show that \[x_4 \approx 0.0496, \ x_5 \approx -0.0528, \ x_6 \approx -0.0007, \ x_7 \approx 0.00000024, \ x_8 \approx -0.00000000022 \]Observe that \(x_n\) alternates signs and gets smaller, not converging to a fixed number.

Step 6: Explain the Divergence

The function \(f(x) = x^2 + 1\) has no real roots, it only has imaginary roots, specifically \(i\) and \(-i\). Consequently, Newton's Method is not converging to a real number since the function does not cross the x-axis anywhere in the real number set.

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a function changes at any point. For the function given in the exercise, \(f(x) = x^2 + 1\), we seek its derivative to apply Newton's Method effectively. The derivative of this quadratic function can be found using the power rule. It essentially means taking the power of \(x\), multiplying by the coefficient, and then reducing the power by one. Hence, the derivative \(f'(x) = 2x\). This simple step is crucial in Newton's Method because, without the derivative, we cannot progress to calculating subsequent guesses or approximations. Just remember, in Newton's Method, this derivative helps determine how far from the x-axis your current point is moving.
Derivatives play a vital role in understanding the function's behavior and updating our guesses dynamically through the Newton's Method formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
So, when applying Newton's Method, always make sure you've correctly differentiated the function first.

Iterative Methods

Iterative methods like Newton's Method are algorithms used to find solutions by repeating certain processes. In this case, we focus on approximating the real roots of functions. The exciting part about these methods is their backbone of repeating cycles. Each step, or iteration, takes an approximation of the root and tries to improve it.
Newton's Method itself is particularly efficient and popular due to its quadratic rate of convergence. Given an initial guess, it uses the function and its derivative to quickly home in on a solution. Here, the term quadratic rate of convergence means the number of correct digits roughly doubles with each iteration, given a good initial guess and a function that behaves well.
Yet, not every application results in rapid convergence. If the function is not tractable in the expected way, like with our example function \(f(x) = x^2 + 1\), the iterative method can diverge or move toward complex numbers. Always recall that the success of iterative methods hinges on characteristics like the initial guess, function's behavior, and root's nature.

Real and Complex Roots

Roots of a function are the solutions for when the function equals zero, \(f(x) = 0\). These roots can either be real or complex numbers. Real roots are those that can be seen on a traditional x-y graph, intersecting the x-axis. Our function \(f(x) = x^2 + 1\), however, does not intersect the x-axis at any point in real numbers.
When the typical graph of a function does not show any intersections with the x-axis, this means that the function has no real roots, but it can still possess complex roots. These are numbers that involve the imaginary unit, \(i\), which is defined as \(\sqrt{-1}\). The roots of \(x^2 + 1 = 0\) are \(i\) and \(-i\), purely imaginary numbers.
Therefore, applying iterative methods like Newton's Method designed to find real roots might not work effectively when only complex roots exist. In such situations, the method may oscillate without finding a satisfactory root, or lead to complex values, which is an expected behavior when the function exclusively has complex roots.