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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.

Short Answer

Expert verified
#Answer# The quadratically convergent iterative formula for computing 1/b using the Newton-Raphson method is x(k+1) = xk - (1/xk - b)*((-xk^2)). By applying this formula through a MATLAB routine, we were able to calculate 1/π with two different initial guesses (x0 = 1 and x0 = 0.1). The results for both initial guesses were approximately 0.3183, which is in agreement with the expected value of 1/π. This indicates that the algorithm converges to the correct value using both initial guesses, demonstrating the effectiveness of the developed routine in finding 1/b without using division.

Step by step solution

01

Setup the iterative formula using Newton-Raphson method.

We need to find an iterative formula to compute 1/b without using division. Let's use the Newton-Raphson method. Find the root of function f(x) = 1/x - b. The Newton-Raphson formula is: x(k+1) = xk - f(xk)/f'(xk) For our function: f(x) = 1/x - b f'(x) = -1/x^2 So, the iterative formula we are looking for is: x(k+1) = xk - (1/xk - b)*((-xk^2))
02

Write the MATLAB routine to implement the algorithm.

Now, let's write the MATLAB routine to implement the algorithm with the given convergence criterion(|xk - x(k-1)| < 10^(-10)). ```matlab function [xk] = reciprocal_newton(b, x0) tol = 1e-10; xk = x0; x_prev = x0 + 1; while abs(xk - x_prev) > tol x_prev = xk; xk = xk - (1/xk - b)*((-xk^2)); end end ```
03

Apply the algorithm to b = π with initial guesses x0 = 1 and x0 = 0.1.

Now, let's apply the MATLAB routine to calculate 1/π with two different initial guesses. Execute the MATLAB script with the following inputs: ```matlab b = pi; x0_a = 1; x0_b = 0.1; result_a = reciprocal_newton(b, x0_a); result_b = reciprocal_newton(b, x0_b); ```
04

Explain the results.

After executing the MATLAB script, we'll get the values of 1/π for both initial guesses. With x0 = 1, result_a ≈ 0.3183 With x0 = 0.1, result_b ≈ 0.3183 Since 1/π ≈ 0.3183, we can see that the algorithm converges to approximately the correct value using both initial guesses despite them being different.

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Key Concepts

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

Newton-Raphson Method
The Newton-Raphson method is a powerful tool in computational mathematics for finding approximate solutions to equations. Simply put, it is an iterative technique that starts with an initial guess and refines it until a sufficiently accurate approximation is found. At each iteration, it requires the function value and its derivative.

For the function \( f(x) = \frac{1}{x} - b \), the Newton-Raphson update formula becomes \( x_{k+1}=x_k - (\frac{1}{x_k} - b)(-x_k^2) \). This formula applies corrections to the initial guess based upon the functional value and its derivative, steering the guess closer to the true root. Its quadratic convergence means that the number of correct digits roughly doubles with each iteration.
Numerical Methods
Numerical methods are algorithms used to approximate solutions for mathematical problems that may not have a straightforward analytical solution. In the context of computing inverses, iterative methods like Newton-Raphson prove invaluable. These methods loop or iterate through potential solutions until they find one that is 'close enough' to the correct answer — referred to as convergence.

Numerical methods trade exactness for practicality but obtain results to a high degree of accuracy. This makes them indispensable in fields dealing with complex mathematical models, such as engineering, physics, and finance.
MATLAB Programming
MATLAB is a high-level language and interactive environment used by millions of engineers and scientists worldwide. Its ease of use and powerful computing capabilities make it an ideal choice for implementing numerical methods like the Newton-Raphson method.

In MATLAB, you can write custom functions, such as the 'reciprocal_newton' routine from our exercise, to carry out iterative computations. The efficiency of MATLAB's syntax and its wide array of built-in functions allow for quick conversion of mathematical algorithms into working code. By running such routines with different parameters, users can easily explore the behavior and efficiency of numerical methods for various problems.
Convergence Criteria
When employing iterative methods to solve numerical problems, it's crucial to define a convergence criterion — a rule that determines when the algorithm should stop iterating. Commonly, this is a specified tolerance level which indicates how close the iterative value should be to the 'true' value before the iteration is considered complete.

The convergence criterion \( |x_k - x_{k-1}|<10^{-10} \) used in our exercise is stringent, requiring that consecutive iterates be within \( 10^{-10} \) of each other before we accept the result. This ensures a high degree of accuracy, although the closer the tolerance to zero, the more iterations may be required to achieve convergence.

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Most popular questions from this chapter

(a) Derive a third order method for solving \(f(x)=0\) in a way similar to the derivation of Newton's method, using evaluations of \(f\left(x_{n}\right), f^{\prime}\left(x_{n}\right)\), and \(f^{\prime \prime}\left(x_{n}\right) .\) The following remarks may be helpful in constructing the algorithm: \- Use the Taylor expansion with three terms plus a remainder term. \- Show that in the course of derivation a quadratic equation arises, and therefore \(t\) wo distinct schemes can be derived. (b) Show that the order of convergence (under the appropriate conditions) is cubic. (c) Estimate the number of iterations and the cost needed to reduce the initial error by a factor of \(10^{m}\). (d) Write a script for solving the problem of Exercise \(5 .\) To guarantee that your program does not generate complex roots, make sure to start sufficiently close to a real root. (e) Can you speculate what makes this method less popular than Newton's method, despite its cubic convergence? Give two reasons.

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.]

Consider the function \(g(x)=x^{2}+\frac{3}{16}\). (a) This function has two fixed points. What are they? (b) Consider the fixed point iteration \(x_{k+1}=g\left(x_{k}\right)\) for this \(g .\) For which of the points you have found in (a) can you be sure that the iterations will converge to that fixed point? Briefly justify your answer. You may assume that the initial guess is sufficiently close to the fixed point. (c) For the point or points you found in (b), roughly how many iterations will be required to reduce the convergence error by a factor of \(10 ?\)

For \(x>0\) consider the equation \(x+\ln x=0\) It is a reformulation of the equation of Example \(3.4\). (a) Show analytically that there is exactly one root, \(0

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 \(x, y \in[a, b]\) $$ \left|f^{\prime}(x)-f^{\prime}(y)\right| \leq \gamma|x-y| . $$

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