Question

Given \(a>0\), we wish to compute \(x=\ln a\) using addition, subtraction, multiplication, division, and the exponential function, \(e^{x}\). (a) Suggest an iterative formula based on Newton's method, and write it in a way suitable for numerical computation. (b) Show that your formula converges quadratically. (c) Write down an iterative formula based on the secant method. (d) State which of the secant and Newton's methods is expected to perform better in this case in terms of overall number of exponential function evaluations. Assume a fair comparison, i.e., same floating point system, "same quality" initial guesses, and identical convergence criterion.

Short answer

In conclusion, both Newton's and secant methods are effective ways to solve numerical problems, but Newton's method has the advantage of faster convergence and fewer evaluations of the exponential function if the initial guesses are of the same quality. The secant method, on the other hand, requires two initial guesses and more function evaluations per iteration. However, it might be preferred in cases where the derivative of the function is not available, or its computation is more time-consuming than the function itself.

Step-by-step solution

Part (a): Newton's method iterative formula

To find the iterative formula based on Newton's method, we first need the derivative of the function \(f(x) = e^x - a\). The derivative is: \(f'(x) = \frac{\text{d}}{\text{d}x}(e^x - a) = e^x\) Now, the Newton's method formula is given by: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) Substituting \(f(x_n)\) and \(f'(x_n)\), we get: \(x_{n+1} = x_n - \frac{e^{x_n} - a}{e^{x_n}}\) Simplifying, we get the iterative formula: \(x_{n+1} = x_n - \frac{1}{a}({e^{x_n}} - a)\)

Part (b): Convergence of the formula

To show the convergence of this formula, we will use the following steps: 1. Define the error at the nth step, \(e_n = x_n - x_0\), where \(x_0\) is the exact value we want to compute (\(x = \ln a\)). 2. Compute the error at the \((n+1)^\text{th}\) step, \(e_{n+1} = x_{n+1} - x_0\). 3. Show that the relation between \(e_n\) and \(e_{n+1}\) indicates quadratic convergence. First, let's compute the error at the \((n+1)^\text{th}\) step: \(e_{n+1} = x_{n+1} - x_0 = (x_n - \frac{1}{a}(e^{x_n} - a)) - x_0\) Now, let's express \(e_n\) in terms of \(x_n\) and \(x_0\): \(e_n = x_n - x_0 \implies x_n = e_n + x_0\) Substituting this into the expression for \(e_{n+1}\), we get: \(e_{n+1} = (e_n + x_0 - \frac{1}{a}(e^{e_n + x_0} - a)) - x_0\) Simplifying, we get: \(e_{n+1} = e_n - \frac{1}{a}(e^{e_n + x_0} - a)\) Since we want to find the relation between \(e_n\) and \(e_{n+1}\), we can observe that \(e_{n+1}\) is proportional to the square of \(e_n\) (as evident from the exponential term): \(e_{n+1} \propto e_n^2\) Hence, the iterative formula converges quadratically.

Part (c): Iterative formula based on the secant method

The secant method requires two starting points, \(x_{n-1}\) and \(x_n\). The iterative formula for the secant method is given by: \(x_{n+1} = x_n - \frac{f(x_n)(x_n-x_{n-1})}{f(x_n) - f(x_{n-1})}\) Substituting \(f(x_n)\), we get: \(x_{n+1} = x_n - \frac{(e^{x_n} - a)(x_n-x_{n-1})}{e^{x_n} - a - (e^{x_{n-1}} - a)}\) Simplifying, we get the iterative formula based on the secant method: \(x_{n+1} = x_n - \frac{(e^{x_n} - a)(x_n-x_{n-1})}{e^{x_n} - e^{x_{n-1}}}\)

Part (d): Comparison between Newton's and secant methods

The main goal is to compare the number of exponential function evaluations in both methods. In Newton's method, we need to calculate \(e^x\) once in each iteration, while in the secant method, we need to calculate \(e^x\) twice in each iteration. Given that both methods have "same quality" initial guesses and identical convergence criterion, plus the fact that Newton's method has quadratic convergence, we can conclude that Newton's method will perform better in terms of overall number of exponential function evaluations.

Key concepts

Newton's Method

Newton's Method is a powerful numerical technique for finding better approximations of the roots of a real-valued function. The method involves an iterative process that starts from an initial guess and improves it using a formula derived from the function and its derivative.

To find the natural logarithm, \( x = \ln a \), we consider the function \( f(x) = e^x - a \). The root of this function, where \( f(x) = 0 \), is precisely our required logarithm. Thus, applying Newton's Method, we use the iterative formula:

• Identify \( f(x) = e^x - a \)
• Calculate its derivative: \( f'(x) = e^x \)
• Newton's formula becomes: \( x_{n+1} = x_n - \frac{e^{x_n} - a}{e^{x_n}} \)

This method is particularly efficient, as it shows quadratic convergence, meaning the approximation quickly becomes more accurate with each iteration.
By recognizing the simplicity of \( e^x \) in the derivative, this approach shines for exponential functions with accessible first derivatives.

Secant Method

The Secant Method offers another way to approximate the roots of a function, similar to Newton's Method but without the use of derivatives. Instead, it leverages two initial guesses to form a linear approximation, iterating to hone in on the root. This setup is useful when the derivative is complex or not readily available.

For calculating \( x = \ln a \), the method uses the function \( f(x) = e^x - a \) and two initial guesses \( x_{n-1} \) and \( x_n \). Unlike Newton's Method, it replaces the derivative with a secant line defined by these points:

• Utilize two current approximations: \( x_{n-1} \) and \( x_n \)
• Reformulate using secant slope: \( x_{n+1} = x_n - \frac{(e^{x_n} - a)(x_n - x_{n-1})}{e^{x_n} - e^{x_{n-1}}} \)

The Secant Method converges more slowly, typically only linearly, but without needing derivatives it can be more versatile in certain contexts, where such calculations are intensive.

Exponential Function

The Exponential Function, denoted as \( e^x \), is fundamental in calculus and many areas of math and science. It describes growth processes where the rate of change is proportional to the value itself, such as populations or radioactive decay.

Key properties of the exponential function include:

• Always positive and rapidly increasing
• Derivative is itself: \( \frac{d}{dx} e^x = e^x \)
• Inverse is the natural logarithm: \( x = \ln e^x \)

In numerical methods like Newton's and Secant methods, \( e^x \) plays a crucial role due to its simplicity and properties making derivations and iterative processes straightforward.

Understanding this function's behavior helps when calculating logarithms and similar operations, as misjudging its rapid growth can lead to errors in initial guesses or calculations, which are critical in achieving convergence in numerical methods.

Logarithm Calculation

Logarithms specifically address the inverse of exponential functions, helping to solve equations of the form \( e^x = a \). Calculating logarithms, such as the natural log \( \ln a \), involves finding \( x \) such that \( e^x = a \).

This inverse relationship provides a strong foundation for applying numerical methods to solve for \( x \). For instance:

• Express the original equation in a more workable form: \( f(x) = e^x - a \)
• Use iterative methods like Newton's or Secant to refine guesses of \( x \)

The applications of accurate logarithm calculations span computational and theoretical fields. Efficiency in these calculations is often essential, particularly in computer algorithms that rely on precision and speed.
The key to effective logarithm calculation lies in choosing "good" initial guesses and understanding the dynamics between exponential growth and its logarithmic counterpart.