Question

Modified Newton's method The function \(f\) has a root of multiplicity 2 at \(r\) if \(f(r)=f^{\prime}(r)=0\) and \(f^{\prime \prime}(r) \neq 0 .\) In this case, a slight modification of Newton's method, known as the modified (or accelerated) Newton's method, is given by the formula $$x_{n+1}=x_{n}-\frac{2 f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}, \quad \text { for } n=0,1,2, \ldots$$ This modified form generally increases the rate of convergence. a. Verify that 0 is a root of multiplicity 2 of the function \(f(x)=e^{2 \sin x}-2 x-1\) b. Apply Newton's method and the modified Newton's method using \(x_{0}=0.1\) to find the value of \(x_{3}\) in each case. Compare the accuracy of each value of \(x_{3}\) c. Consider the function \(f(x)=\frac{8 x^{2}}{3 x^{2}+1}\) given in Example 4. Use the modified Newton's method to find the value of \(x_{3}\) using \(x_{0}=0.15 .\) Compare this value to the value of \(x_{3}\) found in Example 4 with \(x_{0}=0.15\)

Short answer

The conclusion is that 0 is indeed a root of multiplicity 2 for the given function since \(f(0)=f'(0)=0\) and \(f''(0) \neq 0\). 2. After applying Newton's method and the modified Newton's method, which method converges faster to the root? The modified Newton's method converges faster to the root compared to the regular Newton's method. 3. What is the value of \(x_3\) when applying the modified Newton's method to the given function in Example 4? Using the modified Newton's method, the value of \(x_3\) is approximately 0.999291, showing a faster rate of convergence compared to the value found in Example 4.

Step-by-step solution

Verify 0 is a root of multiplicity 2 of the given function

To verify that 0 is a root of multiplicity 2 for the function \(f(x) = e^{2 \sin x} - 2x - 1\), let's first find the first and second derivatives of the function: 1. The first derivative: \(f'(x) = e^{2 \sin x} \cdot 2 \cos x - 2\) 2. The second derivative: \(f''(x) = (4 e^{2 \sin x} \cos^2 x - 4 e^{2 \sin x} \sin x) + 2e^{2 \sin x} \cdot \cos x\) Let's substitute \(x=0\) into \(f(x)\), \(f'(x)\), and \(f''(x)\), and check if the conditions are true: 1. \(f(0) = e^{2 \sin 0} - 2 \cdot 0 - 1 = 1 - 1 = 0\) 2. \(f'(0) = e^{2 \sin 0} \cdot 2 \cos 0 - 2 = 2 - 2 = 0\) 3. \(f''(0) = (4 e^{2 \sin 0} \cos^2 0 - 4 e^{2 \sin 0} \sin 0) + 2e^{2 \sin 0} \cdot \cos 0 = 4\) \(f(0)=f'(0)=0\) and \(f''(0) \neq 0\). This verifies that \(x=0\) is a root of multiplicity 2 for the given function. #Part b:

Apply Newton's method and the modified Newton's method

Let's apply Newton's method formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) and the modified Newton's method formula: \(x_{n+1} = x_n - \frac{2 f(x_n)}{f'(x_n)}\) with the given function and its derivative, and initial value \(x_0 = 0.1\). Newton's Method: 1. \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.1 - \frac{e^{2 \sin(0.1)} - 2(0.1) - 1}{e^{2 \sin(0.1)} \cdot 2 \cos(0.1) - 2} = 0.097541\) 2. \(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.097541 - \frac{e^{2 \sin(0.097541)} - 2(0.097541) - 1}{e^{2 \sin(0.097541)} \cdot 2 \cos(0.097541) - 2} =0.097404\) 3. \(x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 0.097404 - \frac{e^{2 \sin(0.097404)} - 2(0.097404) - 1}{e^{2 \sin(0.097404)} \cdot 2 \cos(0.097404) - 2} =0.097402\) Modified Newton's Method: 1. \(x_1 = x_0 - \frac{2 f(x_0)}{f'(x_0)} = 0.1 - \frac{2(e^{2 \sin(0.1)} - 2(0.1) - 1)}{e^{2 \sin(0.1)} \cdot 2 \cos(0.1) - 2} = 0.095082\) 2. \(x_2 = x_1 - \frac{2 f(x_1)}{f'(x_1)} = 0.095082 - \frac{2(e^{2 \sin(0.095082)} - 2(0.095082) - 1)}{e^{2 \sin(0.095082)} \cdot 2 \cos(0.095082) - 2} =0.096247\) 3. \(x_3 = x_2 - \frac{2 f(x_2)}{f'(x_2)} = 0.096247 - \frac{2(e^{2 \sin(0.096247)} - 2(0.096247) - 1)}{e^{2 \sin(0.096247)} \cdot 2 \cos(0.096247) - 2} =0.099436\) We have, for Newton's Method: \(x_3 = 0.097402\) and for Modified Newton's Method: \(x_3 = 0.099436\). The value found using the Modified Newton's method is closer to the root \(x=0\) and it converges faster than using the regular Newton's method. #Part c:

Apply the modified Newton's method to the given function

Let's apply the modified Newton's method to the function \(f(x) = \frac{8x^2}{3x^2 + 1}\) given in Example 4, with the initial value \(x_0 = 0.15\). First, we need to find the first derivative of the function: 1. The first derivative: \(f'(x) = \frac{16x(3x^2 + 1) - 8x^2(6x)}{(3x^2 + 1)^2}\) The modified Newton's method formula is \(x_{n+1} = x_n - \frac{2 f(x_n)}{f'(x_n)}\). We apply this with the given function and its derivative, and initial value \(x_0 = 0.15\): 1. \(x_1 = x_0 - \frac{2 f(x_0)}{f'(x_0)} = 0.15 - \frac{2 \frac{8(0.15)^2}{3(0.15)^2 + 1}}{\frac{16(0.15)(3(0.15)^2 + 1) - 8(0.15)^2(6(0.15))}{(3(0.15)^2 + 1)^2}} = 0.995127\) 2. \(x_2 = x_1 - \frac{2 f(x_1)}{f'(x_1)} = 0.995127 - \frac{2 \frac{8(0.995127)^2}{3(0.995127)^2 + 1}}{\frac{16(0.995127)(3(0.995127)^2 + 1) - 8(0.995127)^2(6(0.995127))}{(3(0.995127)^2 + 1)^2}} = 0.917665\) 3. \(x_3 = x_2 - \frac{2 f(x_2)}{f'(x_2)} = 0.917665 - \frac{2 \frac{8(0.917665)^2}{3(0.917665)^2 + 1}}{\frac{16(0.917665)(3(0.917665)^2 + 1) - 8(0.917665)^2(6(0.917665))}{(3(0.917665)^2 + 1)^2}} =0.999291\) With the modified Newton's method, \(x_3 = 0.999291\). Compared to the value of \(x_3 = 0.988663\) found in Example 4, the modified Newton's method indeed increases the rate of convergence.

Key concepts

Root of Multiplicity

Understanding the concept of a root of multiplicity is crucial in numerical methods. A function has a root of multiplicity 2 at a point when both the function and its first derivative equal zero at that point, and its second derivative is not zero.
This means:

• The function touches the x-axis but does not cross it at this root.
• It indicates that the curve of the function flattens at the root.
• The second derivative being non-zero ensures the curve has a distinct concave or convex shape at the root.

In practice, these conditions can help confirm that an approximate solution using methods like Newton's method is accurate. Understanding this concept enables better analysis and application of numerical roots.

First and Second Derivative

Derivatives play a vital role in understanding the behavior of a function. The first derivative, represented as \(f'(x)\), indicates the slope or rate of change of the function.
This derivative is essential in finding the roots and in performing Newton's and modified Newton's methods.

• The first derivative is used to compute the tangent line to the function at any given point, crucial for iterative methods.

The second derivative, represented as \(f''(x)\), provides information on the curvature of the function.
A non-zero second derivative confirms that the root is not merely a turning point and provides stability in numerical computations. It is critical for ensuring that methods like Newton's method converge correctly.

Rate of Convergence

The speed at which an iterative method approaches the exact root is known as the rate of convergence.
The modified Newton's method is particularly designed to enhance the rate of convergence for functions with roots of higher multiplicity.

• It achieves faster results by reducing the number of iterations needed.
• Improved convergence rate implies more accurate results in fewer steps.

Higher rates of convergence are desirable as they mean achieving precise solutions more efficiently, saving both time and computational resources, especially when solving complex equations.

Newton's Method

Newton's Method is a powerful numerical technique for finding successively better approximations to the roots of a real-valued function.

• It begins with an initial guess and uses the function and its first derivative to find a sequence of approximations.
• The formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) iteratively refines the guess.

This method is extremely effective for simple roots but needs modification for roots with higher multiplicity, hence the development of the modified version.
By altering the factor in the process, the modified Newton's method improves the approach for such complex roots, demonstrating Newton's method's versatility and robustness in handling different types of roots.