91Ó°ÊÓ

(Quadratic convergence of the Newton method) Let \(f:[a, b] \rightarrow \mathbb{R}\) be a differentiable function such that \(f^{\prime}\) is continuous on \([a, b], f^{\prime}(x) \neq 0\) for all \(x \in[a, b], f^{\prime \prime}\) exists and is bounded on \((a, b)\). If \(f(r)=0\) for some \(r \in[a, b]\), then show that \(r\) is the unique solution of \(f(x)=0\) in \([a, b]\) and there is a constant \(\alpha \in \mathbb{R}\) such that for any Newton sequence \(\left(x_{n}\right)\) for \(f\) with its initial point \(x_{0}\) in \([a, b]\), $$ \left|x_{n}-r\right| \leq \alpha\left|x_{n-1}-r\right|^{2}, \quad \text { provided } x_{n-1}, x_{n} \in[a, b] . $$ Deduce that \(\left|x_{n}-r\right| \leq \alpha^{2^{n}-1}\left|x_{0}-r\right|^{2^{n}}\), provided \(x_{1}, \ldots, x_{n} \in[a, b]\). [Note: In case \(\alpha<1\), the former inequality shows that the terms of the Newton sequence give a successively better approximation of the solution \(r\) of \(f(x)=0 .\) This inequality can also be interpreted to say that the Newton sequence converges quadratically.]

Step by step solution

01

Show unique solution of f(x) = 0 in [a, b]

Since f'(x) is continuous and non-zero on [a, b], it has the same sign throughout this interval. Suppose f'(x) > 0 on [a, b]. Then, f is strictly increasing on [a, b]. Given that f(r) = 0 for some r in [a, b], we can say that f(x) < 0 for x < r and f(x) > 0 for x > r. Therefore, the function f has only one root (r) in [a, b]. The case when f'(x) < 0 can be shown similarly.

02

Define the Newton sequence

The Newton sequence \((x_n)\) for the function f can be defined as follows: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \quad n \geq 0 $$ with the initial guess \(x_0\) in [a, b]. We will prove that this sequence converges quadratically to the unique root r.

03

Show that |x_n - r|

Define the constant α as: $$ \alpha = \max_{x \in (a, b)} \frac{1}{2}\left|\frac{f''(x)}{f'(x)}\right| $$ Since f''(x) is bounded on (a, b), α is finite. For any (n-1)-th term and n-th term of the Newton sequence, consider the Taylor expansion of f around \(x_{n-1}\): $$ f(x) \approx f(x_{n-1}) + f'(x_{n-1})(x - x_{n-1}) + \frac{1}{2}f''(c)(x - x_{n-1})^2 $$ where c is between x and \(x_{n-1}\). Now, let's replace x by r and analyze the term \(|x_n - r|\). We know that: $$ x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})} $$ So, $$ x_n - r = x_{n - 1} - r - \frac{f(x_{n-1})}{f'(x_{n-1})} $$ From the Taylor expansion, $$ f(x_{n-1}) = -f'(x_{n-1})(r - x_{n-1}) + \frac{1}{2}f''(c")(r - x_{n-1})^2 $$ where c" is between r and \(x_{n-1}\). Thus, $$ x_n - r = -\frac{1}{2}\frac{f''(c")}{f'(c")}(r - x_{n-1})^2 $$ and $$ \left|x_n - r\right| \leq \alpha \left|x_{n-1} - r\right|^2 $$ provided \(x_{n-1}, x_n \in [a, b]\).

04

Prove that |x_n - r|

We will prove this inequality by induction. Base case (n = 1): $$ \left|x_1 - r\right| \leq \alpha|x_0 - r|^2 $$ which is true from the previous step. Inductive step: Assume that the inequality holds for n = k. Now, we need to show it holds for n = k + 1. $$ \left|x_{k+1} - r\right| \leq \alpha\left|x_k - r\right|^2 $$ Using the inductive hypothesis, we can substitute the term \(\left|x_{k} - r\right|\): $$ \left|x_{k+1} - r\right| \leq \alpha\left(\alpha^{2^k - 1}\left|x_0 - r\right|^{2^k}\right)^2 $$ which simplifies to $$ \left|x_{k+1} - r\right| \leq \alpha^{2^{k+1} - 1}\left|x_0 - r\right|^{2^{k+1}} $$ Therefore, the inequality holds for all n, provided \(x_1, \ldots, x_n \in [a, b]\). The fact that \(\left|x_n - r\right|\) decreases quadratically at each step implies that the Newton sequence converges quadratically to the unique root r under the given conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Convergence

In the world of numerical methods, Newton's Method is well known for its fast convergence, specifically quadratic convergence. What does this mean? Simply put, if you start close enough to the root, Newton's method doubles the number of correct digits in each step. Let’s break this down further:

• For each iteration in Newton's sequence, the error (the distance between your guess and the actual root) gets squared.
• This means the error diminishes rapidly, leading to a highly accurate approximation in just a few steps.

Quadratic convergence is mathematically represented as \[ |x_{n} - r| \leq \alpha |x_{n-1} - r|^2 \] where \(x_{n}\) is the nth approximation, \(r\) is the actual root, and \(\alpha\) is a constant based on the function. This showcases the power of quadratic convergence in providing precise solutions efficiently.

Uniqueness of Solutions

The uniqueness of a solution refers to the condition where only one root exists in the given interval. For the function \( f(x) = 0 \) within the interval \([a, b]\), if \(f'(x)\) is continuous and non-zero, the solution is unique. Here's why:

• \(f'(x)\) as non-zero across the entire range implies the function is strictly monotonic, either strictly increasing or decreasing.
• Given \(f(r) = 0\), if \(f'(x) \gt 0\), then \(f(x)\) transitions from negative to positive as \(x\) passes through \(r\).
• This unique crossing indicates that \(r\) is the only point where \(f(x) = 0\) in the interval.

This concept guarantees that Newton’s method can reliably locate the correct root in the specified interval, highlighting its effectiveness in practical applications.

Taylor Expansion

Taylor expansion offers a way to approximate a function near a specific point using polynomials. It’s crucial for understanding why Newton's method works so well. Here's the process:

• The function \(f(x)\) can be approximated around a point \(x_{n-1}\) using the Taylor series: \[ f(x) \approx f(x_{n-1}) + f'(x_{n-1})(x - x_{n-1}) + \frac{1}{2}f''(c)(x - x_{n-1})^2 \]
• This formula breaks down complicated functions into simpler, easily analyzable pieces, using derivatives at \(x_{n-1}\).

In Newton’s method, we specifically exploit the linear part of this series, which simplifies finding roots effectively. The quadratic term contributes to the rapid convergence rate by squaring the error each step.

Inductive Proof

Inductive proof is a powerful technique to demonstrate that a property holds for every member of an infinite sequence, crucial for validating the performance of algorithms like Newton’s Method.

The proof structure consists of:

• **Base Case:** The statement is proven for the initial "n" (often \(n = 1\) or \(n = 0\)). In Newton’s Method, it involves showing the initial approximation satisfies the desired inequality for quadratic convergence.
• **Inductive Step:** Assuming the statement holds for \(n = k\), it must be shown true for \(n = k + 1\). For Newton’s Method, we assume \(|x_k - r|\) satisfies the inequality, showing it leads to the same inequality for \(|x_{k+1} - r|\).

This methodically proves that each subsequent approximation is increasingly close to the root, validating Newton’s Method's quadratic convergence property and ensuring confidence in its efficacy.