Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

This exercise essentially utilizes various forms of Taylor's expansion and relies on expertise in calculus. (a) Prove that if \(f \in C^{2}[a, b]\) and there is a root \(x^{*}\) in \([a, b]\) such that \(f\left(x^{*}\right)=0, f^{\prime}\left(x^{*}\right) \neq 0\), then there is a number \(\delta\) such that, starting with \(x_{0}\) from anywhere in the neighborhood \(\left[x^{*}-\delta, x^{*}+\delta\right]\), Newton's method converges quadratically. (b) This is more challenging: prove the same conclusions under the same assumptions, except that now \(f\) is only assumed to have a first Lipschitz continuous derivative. Thus, while \(f^{\prime \prime}\) may not exist everywhere in \([a, b]\), there is a constant \(\gamma\) such that for any \(x, y \in[a, b]\) $$ \left|f^{\prime}(x)-f^{\prime}(y)\right| \leq \gamma|x-y| . $$

Short Answer

Expert verified
Answer: Newton's method converges quadratically for a root x* under the following two conditions: 1) The function f is twice continuously differentiable, f has a root x* such that f(x*) = 0 and f'(x*) ≠ 0. 2) The first derivative of f is Lipschitz continuous, f has a root x* such that f(x*) = 0 and f'(x*) ≠ 0.

Step by step solution

01

Explain Newton's Method

Newton's method is an iterative procedure to find the root of a given function, f. It starts from an initial guess, x₀, and proceeds iteratively by the formula: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, $$ where n = 0, 1, 2, 3, ...
02

Use Taylor's Expansion

To prove that Newton's method converges quadratically, we will use Taylor's expansion. Given that f is twice continuously differentiable on [a, b], we can express f(x) close to x* as $$ f(x) = f(x^{*}) + f'(x^{*})(x - x^{*}) + \frac{f''(\xi)(x - x^{*})^2}{2}, $$ where ξ is some point between x and x*. As we know that f(x*) = 0, this simplifies to $$ f(x) = f'(x^{*})(x - x^{*}) + \frac{f''(\xi)(x - x^{*})^2}{2}. $$
03

Determine x₁

Now apply Newton's method for one iteration starting from x₀, which gives us x₁ as $$ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. $$ Use the Taylor's expansion for f(x₀) to get $$ x_1 = x_0 - \frac{f'(x^{*})(x_0 - x^{*}) + \frac{f''(\xi)(x_0 - x^{*})^2}{2}}{f'(x_0)}. $$
04

Assess Quadratic Convergence

To show quadratic convergence, we need to show that $$ |x_1 - x^*| ≈ C|x_0 - x^*|^2, $$ with some constant C. Rewrite the above expression for x₁ in terms of the error term: $$ x_1 - x^{*} = (x_0 - x^{*}) - \frac{f'(x^{*})(x_0 - x^{*}) + \frac{f''(\xi)(x_0 - x^{*})^2}{2}}{f'(x_0)}. $$ Now, let's divide both sides by \((x_0 - x^{*})^2\) and rearrange to get $$ \frac{x_1 - x^{*}}{(x_0 - x^{*})^2} ≈ \frac{1}{(x_0 - x^*)f'(x_0)} - \frac{f''(\xi)(x_0 - x^{*})}{2f'(x_0)}. $$ As x₀ approaches x*, the left side of the equation approaches a constant C, proving quadratic convergence of Newton's method. #Part (b)#
05

Replace Assumptions

In this part of the exercise, we are given the same function f with root x* such that f(x*) = 0 and f'(x*) ≠ 0 but different assumptions. Now, it is only required that the first derivative is Lipschitz continuous i.e., there exists a constant γ such that $$ |f'(x) - f'(y)| \leq \gamma|x - y|. $$
06

Prove Same Conclusions

In order to prove the same conclusion as part (a), we will follow the same steps, but now using the Lipschitz continuous first derivative condition. First, rewrite the Newton's method iteration as $$ x_{n+1} - x^{*} = \frac{f(x_n)(x_n - x^{*})}{f'(x_n)}. $$ Once again, use Taylor's expansion for f(xₙ), but this time without the assumption of a continuous second derivative: $$ f(x_n) = f(x^{*}) + f'(x^{*})(x_n - x^{*}) + R_n(x_n - x^{*}), $$ with |Rₙ| ≤ γ. Now, plug the Taylor's expansion of f(xₙ) into the Newton's method iteration formula and divide both sides by (xₙ - x*)^2 to obtain $$ \frac{x_{n+1} - x^{*}}{(x_n - x^{*})^2} ≈ \frac{1}{(x_n - x^{*})f'(x_n)} - \frac{R_n}{f'(x_n)}. $$ As xₙ approaches x*, the left hand side approaches a constant C, verifying quadratic convergence of Newton's method under the Lipschitz continuous first derivative condition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Taylor's Expansion
Taylor's expansion is a powerful tool in calculus and numerical analysis for approximating functions near a specific point. It expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The idea is to represent the function as an infinite polynomial.

For a function that is twice continuously differentiable—often denoted as being in the class \(C^{2}[a, b]\)—the Taylor expansion around a point \(x^*\) within the interval [a, b] is given as:
\[ f(x) = f(x^{*}) + f'(x^{*})(x - x^{*}) + \frac{f''(\xi)(x - x^{*})^2}{2} + \ldots\]
In practice, we often truncate this series to only a few terms, which provides us a good approximation if \(x\) is sufficiently close to \(x^*\). This method is particularly useful for the quadratic convergence of iterative methods like Newton's Method, as it allows us to estimate the behavior of \(f\) around a root and subsequently refine our guesses for where the root lies.
Lipschitz Continuous Derivative
A Lipschitz continuous derivative is a concept that strengthens our understanding of function behavior beyond just knowing its differentiability. When we say a function \(f\) has a Lipschitz continuous derivative on an interval, it means there exists a constant, often denoted as \(\gamma\), such that for all \(x\) and \(y\) in the interval:
\[|f'(x) - f'(y)| \leq \gamma|x - y|.\]
This property provides a controlled rate at which the derivative can change. It comes in handy, especially when dealing with functions that may not be smooth enough to have higher-order derivatives at every point. In the context of Newton's Method, having a Lipschitz continuous derivative—even without a continuous second derivative—enables us to establish bounds on the error, which is crucial for proving quadratic convergence.
Iterative Methods
Iterative methods are procedures used to solve mathematical problems by repeating a sequence of operations to move closer to a desired result. They are central to numerical analysis, especially for problems that cannot be tackled analytically. Iterative methods often start with an initial guess and progressively improve this guess through a series of iterations.

Newton's Method, also known as the Newton-Raphson method, is a classic example of an iterative method used for finding roots of real-valued functions. By repeatedly applying the Newton iteration formula:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
we can converge to the true root from an initial approximation. The rate at which this process converges depends on the nature of the function and the quality of the initial guess. For certain well-behaved functions, Newton's Method can offer quadratic convergence, which means that the error decreases quadratically with each iteration. This is extremely efficient compared to many other methods that may offer only linear or sublinear convergence.
Numerical Root Finding
Numerical root finding involves methods for approximating the roots of equations when they cannot be solved analytically. The goal is to find values of \(x\) for which the function \(f(x)\) is zero, hence the term 'root'. These methods are indispensable for solving real-world problems where exact solutions are often unattainable.

Some of the commonly used numerical root finding methods include bisection, secant, and Newton's Method. Each of these methods brings a different approach to estimating roots:
  • The bisection method narrows down the interval where a root lies,
  • The secant method approximates the derivative by taking secants through points on the function's graph,
  • Newton's Method uses the function's derivative to predict where the function crosses the x-axis.

Newton's Method, in particular, is renowned for its potential to rapidly zero in on a root, providing a high speed of convergence under appropriate conditions, such as the presence of a continuous second derivative or a Lipschitz continuous first derivative for the function whose root is sought.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use your program from Exercise 21 to minimize (a) \(\phi(x)=\sin (x)\) and (b) \(\phi(x)=-\operatorname{sinc}(x)=-\frac{\sin (x)}{x}\) over the interval \([-10,10]\) with tol \(=10^{-8}\). You might find that determining the global minimum of the second function is significantly more challenging, even though the global minimum for the first one is not unique. Explain the source of difficulties and how you have addressed them. [Exercises 17 and 14 are relevant here.]

Apply the bisection routine bisect to find the root of the function $$ f(x)=\sqrt{x}-1.1 $$ starting from the interval \([0,2]\) (that is, \(a=0\) and \(b=2\) ), with atol \(=1 . \mathrm{e}-8\). (a) How many iterations are required? Does the iteration count match the expectations, based on our convergence analysis? (b) What is the resulting absolute error? Could this absolute error be predicted by our convergence analysis?

The derivative of the sinc function is given by $$ f(x)=\frac{x \cos (x)-\sin (x)}{x^{2}} $$ (a) Show that near \(x=0\), this function can be approximated by $$ f(x) \approx-x / 3 $$ The error in this approximation gets smaller as \(x\) approaches \(0 .\) (b) Find all the roots of \(f\) in the interval \([-10,10]\) for tol \(=10^{-8}\).

It is known that the order of convergence of the secant method is \(p=\frac{1+\sqrt{5}}{2}=1.618 \ldots\) and that of Newton's method is \(p=2\). Suppose that evaluating \(f^{\prime}\) costs approximately \(\alpha\) times the cost of approximating \(f\). Determine approximately for what values of \(\alpha\) Newton's method is more efficient (in terms of number of function evaluations) than the secant method. You may neglect the asymptotic error constants in your calculations. Assume that both methods are starting with initial guesses of a similar quality.

Suppose that the division button of your calculator has stopped working, and you have addition, subtraction, and multiplication only. Given a real number \(b \neq 0\), suggest a quadratically convergent iterative formula to compute \(\frac{1}{b}\), correct to a user-specified tolerance. Write a MATLAB routine that implements your algorithm, using \(\left|x_{k}-x_{k-1}\right|<10^{-10}\) as a convergence criterion, and apply your algorithm to \(b=\pi\) (that is, we compute \(\frac{1}{\pi}\) ), with two different initial guesses: (a) \(x_{0}=1\); and (b) \(x_{0}=0.1\). Explain your results.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free