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Chapter 3: Problem 7

Consider Steffensen's method $$ x_{k+1}=x_{k}-\frac{f\left(x_{k}\right)}{g\left(x_{k}\right)}, \quad k=0,1, \ldots, $$ where $$ g(x)=\frac{f(x+f(x))-f(x)}{f(x)} $$ (a) Show that in general the method converges quadratically to a root of \(f(x)\). (b) Compare the method's efficiency to the efficiency of the secant method.

Short Answer

Expert verified
Question: Show that the Steffensen's method converges quadratically to a root of \(f(x)\) and compare the efficiency of Steffensen's method to the efficiency of the secant method. Answer: Steffensen's method converges quadratically to a root of \(f(x)\), with a limit of the form \(\lim_{k \to \infty} \frac{E_{k+1}}{E_{k}^2} = C\), where \(E_k\) is the error at the \(k\)-th step and \(C\) is a constant. In comparison, the secant method converges with an order of approximately \(1.618\), which is slower than the quadratic convergence of Steffensen's method. Thus, Steffensen's method is more efficient than the secant method.

Step by step solution

01

Understand Steffensen's Method Iteration Formula

Steffensen's method has an iteration formula as follows: $$ x_{k+1}=x_{k}-\frac{f\left(x_{k}\right)}{g\left(x_{k}\right)}, \quad k=0,1,\ldots $$ This iteration formula is used to find the successive approximations of the root of the function \(f(x)\). We know that in general, if an iterative method converges quadratically, its error will satisfy the following condition: $$ \lim_{k \to \infty} \frac{E_{k+1}}{E_{k}^2} = C $$ where \(E_k\) is the error at the \(k\)-th step and \(C\) is a constant.
02

Rewriting the Error Equation using Steffensen's Method Iteration Formula

Let \(r\) be the root of \(f(x)\), which means \(f(r) = 0\). Then the error \(E_k\) at the \(k\)-th step is given by $$ E_k = r - x_k $$ Now substituting this into the iteration formula and rearranging to write an equation for \(E_{k+1}\) we get: $$ E_{k+1} = r - x_{k+1} = r - \left(x_k - \frac{f(x_k)}{g(x_k)}\right) = \frac{f(x_k)}{g(x_k)} - E_k $$
03

Derive the Convergence of Steffensen's Method

To show that the method converges quadratically, we need to find the limit: $$ \lim_{k \to \infty} \frac{E_{k+1}}{E_{k}^2} $$ Using the equation from Step 2, we have $$ \lim_{k \to \infty} \frac{\frac{f(x_k)}{g(x_k)} - E_k}{E_{k}^2} $$ Now, since \(x_k \to r\) as \(k \to \infty\), we can use L'Hopital's rule to evaluate this limit. Differentiate the numerator and denominator with respect to \(k\) to get $$ \lim_{k \to \infty} \frac{\frac{f'(x_k) - g'(x_k)}{g(x_k)}}{2E_k} = \lim_{k \to \infty} \frac{f'(x_k) - g'(x_k)}{2E_kg(x_k)} $$ As \(x_k \to r\), we have \(E_k \to 0\) and \(g(x_k) \to g(r)\). Thus, the limit becomes: $$ \lim_{k \to \infty} \frac{f'(r) - g'(r)}{2E_kg(r)} $$ This limit exists and is finite since \(f'(r), g'(r), g(r)\) are finite. Hence, the Steffensen's method converges quadratically to a root of \(f(x)\).
04

Comparing Efficiency with the Secant Method

Now let's compare the efficiency of the Steffensen's method to the secant method. The error at the \(k\)-th step in the secant method satisfies: $$ \lim_{k \to \infty} \frac{E_{k+1}}{E_{k}^{\varphi}} = C' $$ where \(\varphi \approx 1.618\) (golden ratio) and \(C'\) is a constant. Comparing the convergence rates, we notice that Steffensen's method converges quadratically, with an order of 2, whereas the secant method has a convergence order of approximately \(1.618\). As quadratic convergence is faster than the secant's convergence, we can conclude that Steffensen's method is more efficient than the secant method.

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Key Concepts

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

Numerical Methods
Numerical methods are techniques used to obtain numerical solutions to various problems in mathematics and physics when analytical or exact solutions are difficult or impossible to achieve. They are particularly valuable for solving problems such as integrating a complex function, solving differential equations, or finding roots of equations.

Numerical methods often involve iterative processes, where a sequence of increasingly accurate approximations is generated. One common application is root finding, where we seek the value for which a function outputs zero. Steffensen's method is an example of these techniques and is specifically designed for root finding with certain properties like quadratic convergence.
Root Finding Algorithms
Root finding algorithms are a subset of numerical methods geared towards locating the zeros of functions. These algorithms are fundamental in computer science, engineering, applied mathematics, and scientific computing.

In essence, the goal is to find the value of the variable, say 'x', such that when plugged into function 'f(x)', it returns zero. There are many different algorithms for this purpose, but they generally fall into two categories: bracketing methods, like bisection, which require an initial interval containing the root; and open methods, like Newton-Raphson and Steffensen's Method, which start from a single initial guess and iterate towards the solution. Open methods can offer faster convergence but might require a closer initial guess and can be prone to diverge if the guess is poor.
Quadratic Convergence
Quadratic convergence refers to the speed at which an iterative method approaches the solution. For a method to exhibit quadratic convergence, the error of successive iterations should square each time, rapidly decreasing towards zero.

In the context of Steffensen's method, the error term at each step approximates to the square of the previous step's error, providing us with a very efficient convergence. Mathematically, this is expressed as the limit ratio of the error in the subsequent iteration to the square of the current error becoming a constant. This level of acceleration in finding the accurate root makes methods with quadratic convergence, such as Steffensen's, highly sought after for problems where a high degree of precision is required quickly.
Secant Method Efficiency
The secant method is another iteration-based root-finding algorithm that is known for its efficiency and simplicity. It does not require the derivative of the function, unlike Newton's method, which makes it advantageous for functions where the derivative is not readily available or difficult to compute.

However, while the secant method is efficient, it converges at the rate of the golden ratio, \( \varphi \approx 1.618 \) per iteration, which is slower than quadratic convergence. In contrast, Steffensen's Method, which converges quadratically, refines its approximation more rapidly with every step taken. While the secant method is straightforward and versatile, Steffensen's Method can often reach a similar level of accuracy with fewer iterations, making it more efficient for certain problems.

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Most popular questions from this chapter

Find all the roots of the function $$ f(x)=\left\\{\begin{array}{ll} \frac{\sin (x)}{x}, & x \neq 0 \\ 1, & x=0 \end{array}\right. $$ in the interval \([-10,10]\) for tol \(=10^{-7}\). [This function is special enough to have a name. It is called the sinc function.]

(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.

Consider the fixed point iteration \(x_{k+1}=g\left(x_{k}\right), k=0,1, \ldots\), and let all the assumptions of the Fixed Point Theorem hold. Use a Taylor's series expansion to show that the order of convergence depends on how many of the derivatives of \(g\) vanish at \(x=x^{*}\). Use your result to state how fast (at least) a fixed point iteration is expected to converge if \(g^{\prime}\left(x^{*}\right)=\cdots=\) \(g^{(r)}\left(x^{*}\right)=0\), where the integer \(r \geq 1\) is given.

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?

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.
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