Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In applying Newton's Method to solve \(f(x)=0\), one can usually tell by simply looking at the numbers \(x_{1}, x_{2}, x_{3}, \ldots\) whether the sequence is converging. But even if it converges, say to \(\bar{x}\), can we be sure that \(\bar{x}\) is a solution? Show that the answer is yes provided \(f\) and \(f^{\prime}\) are continuous at \(\bar{x}\) and \(f^{\prime}(\bar{x}) \neq 0\).

Step-by-Step Solution

Answer

Yes, by continuity of \(f\) and \(f'\), and \(f'(\bar{x}) \neq 0\), \(\bar{x}\) is a root.

Step 1: Setup Newton's Method

Newton's Method is used to find successively better approximations to the roots (or zeroes) of a real-valued function. The method is iterative, and the formula for the next approximation of a root is given by: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] where \(f'(x_n)\) is the derivative of \(f\) evaluated at \(x_n\).

Step 2: Assume series convergence

Assume that the sequence \(x_1, x_2, x_3, \ldots\) converges to a point \(\bar{x}\). This implies that \(\lim_{n \to \infty} x_n = \bar{x}\). We need to determine if \(\bar{x}\) is indeed a root of \(f\), i.e., if \(f(\bar{x}) = 0\).

Step 3: Continuity of f and f'

Given that \(f\) and \(f'\) are continuous at \(\bar{x}\), and with the assumption that \(x_n\) converges to \(\bar{x}\), we have \(\lim_{n \to \infty} f(x_n) = f(\bar{x})\) and \(\lim_{n \to \infty} f'(x_n) = f'(\bar{x})\).

Step 4: Check derivative condition

We also know that \(f'(\bar{x}) eq 0\). This is crucial because it ensures that the denominator in Newton's formula does not approach zero, allowing the quotient \(\frac{f(x_n)}{f'(x_n)}\) to affect \(x_n\) correctly.

Step 5: Substitute limit into Newton's formula

Taking the limit of the Newton's Method formula as \(n\) approaches infinity, we have: \[ \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \left( x_n - \frac{f(x_n)}{f'(x_n)} \right) = \bar{x} - \frac{f(\bar{x})}{f'(\bar{x})} \] Since \(\lim_{n \to \infty} x_n = \bar{x}\), replace \(x_n\) with \(\bar{x}\).

Step 6: Derivation of root condition

Given that the sequence converges to \(\bar{x}\), the expression \[ \bar{x} = \bar{x} - \frac{f(\bar{x})}{f'(\bar{x})} \] must hold. Simplifying yields \[ 0 = -\frac{f(\bar{x})}{f'(\bar{x})} \] Given \(f'(\bar{x}) eq 0\), it follows that \(f(\bar{x}) = 0\).

Key Concepts

Convergence of Sequences

When we use Newton's Method, we are often checking if a sequence of numbers converges. Convergence in this context means that as we calculate more elements in the sequence, the numbers get closer and closer to a specific value, called the limit. If the sequence converges, we denote this limit by \(\bar{x}\).
This convergence is important to determine if our sequence is approaching a solution to the equation \(f(x)=0\). Generally, a sequence \(x_1, x_2, x_3, \ldots\) is said to converge to \(\bar{x}\) if for every small tolerance, there is a point beyond which all terms are within that tolerance. Mathematically, \(\lim_{n \to \infty} x_n = \bar{x}\).
This concept is critical in Newton's Method, as it helps ensure that our calculated values are moving toward the solution we seek.

Continuity

Continuity is a foundational aspect in calculus that ensures functions do not have jumps, breaks, or holes. In Newton's Method, the continuity of the function \(f\) and its derivative \(f'\) at \(\bar{x}\) is significant.
For a function to be continuous at \(\bar{x}\), the limit of the function as the input approaches \(\bar{x}\) must equal the function's value at \(\bar{x}\). Mathematically, this means \(\lim_{x \to \bar{x}} f(x) = f(\bar{x})\). Continuity helps ensure that the behaviors and values of solutions we calculate are smooth and reliable.
In our context, if \(x_n\) converges to \(\bar{x}\) and \(f\) and \(f'\) are continuous, then it implies that \(\lim_{n \to \infty} f(x_n) = f(\bar{x})\) and similarly for the derivative: \(\lim_{n \to \infty} f'(x_n) = f'(\bar{x})\). This continuity is crucial because it confirms that the outputs of the function are smoothly approaching the values we expect as inputs approach \(\bar{x}\).

Derivative Conditions

One key aspect of Newton's Method is the use of derivatives, specifically the requirement that the derivative \(f'(x)\) does not equal zero at \(\bar{x}\). This condition ensures that the method produces meaningful results.
When using the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), if \(f'(\bar{x})\) were zero, the denominator would become zero, causing mathematical complications such as division by zero. This could mean the method fails to track correctly toward a root.
Thus, ensuring \(f'(\bar{x}) eq 0\) guarantees that our approach using Newton's Method remains valid and functional. It allows the technique to successfully reduce the difference \(\frac{f(x_n)}{f'(x_n)}\), optimizing our path toward identifying \(\bar{x}\) as a root. This derivative condition plays a pivotal role in the effectiveness of the method.

Roots of Equations

The ultimate goal when using Newton's Method is to find roots of equations—values \(x\) for which \(f(x)=0\). When applying this process, especially as \(n\) approaches infinity, we want to prove that the converged value \(\bar{x}\) is indeed a root.
Given the convergence of \(x_n\) to \(\bar{x}\) and continuity of \(f\) and \(f'\), as well as the crucial condition \(f'(\bar{x}) eq 0\), we can safely incorporate \(\bar{x}\) into the formulation. What results is a simplification where the expressions derived confirm \(f(\bar{x}) = 0\).
This means after confirming all the conditions, we establish \(\bar{x}\) as a legitimate root, successfully applying Newton's Method. This outcome is insightful for students, demonstrating that a systematic, rule-abiding approach retrieves solutions efficiently.