Problem 29 - University Calculus: Early Transcendentals

Question

Some curves are so flat that, in practice, Newton's method stops too far from the root to give a useful estimate. Try Newton's method on \(f(x)=(x-1)^{40}\) with a starting value of \(x_{0}=2\) to see how close your machine comes to the root \(x=1 .\) See the accompanying graph. (GRAPH CAN'T COPY)

Step-by-Step Solution

Answer

Newton's method starting from \(x_0 = 2\) yields \(x_1 = 1.975\), still far from the root at \(x = 1\) due to the curve's flatness.

Step 1: Understand Newton's Method

Newton's method is an iterative procedure to approximate the roots of a real-valued function. It uses the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]where \(x_n\) is the current approximation, \(f(x_n)\) is the function value at \(x_n\), and \(f'(x_n)\) is the derivative evaluated at \(x_n\).

Step 2: Compute the First Derivative

The function given is \( f(x) = (x-1)^{40} \). Differentiate it to get:\[ f'(x) = 40(x-1)^{39} \].

Step 3: Apply Newton's Formula for the First Iteration

Let the initial guess be \( x_0 = 2 \). Apply Newton's formula:\[ x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{(2-1)^{40}}{40(2-1)^{39}} \].Calculate:\[ x_{1} = 2 - \frac{1}{40} = 1.975 \].

Step 4: Second Iteration

Use \( x_1 = 1.975 \) for the next iteration:\[ x_{2} = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.975 - \frac{(1.975-1)^{40}}{40(1.975-1)^{39}} \].Due to the small values involved, the benefit to \( x_2 \) over \( x_1 \) will be negligible, emphasizing how flat the curve is.

Step 5: Analyze the Result

Despite two iterations, the value is still not close to the root at \( x = 1 \). The flatness of the curve \((x-1)^{40}\) near its root makes it challenging for Newton's method to converge efficiently to the root in few iterations.

Key Concepts

Newton's Method

Newton's Method is a powerful tool used in calculus to find approximate roots of a real-valued function. It is particularly useful when the function is too complex to solve directly. To begin, we choose an initial estimate, denoted as \( x_0 \). Newton's formula is then employed to iteratively bring this estimate closer to the actual root. The formula used is \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \), where \( f(x_n) \) is the value of the function at our current guess, and \( f'(x_n) \) is the derivative of the function at that same point.
This technique uses the idea that if the curve at a given point is approximated by a tangent line, the root of the tangent line provides a better approximation to the actual root than the current point.

• Start with an initial guess \( x_0 \).
• Calculate the function value and its derivative at \( x_0 \).
• Update the guess using the Newton's formula.
• Repeat until the change is insignificant.

Newton's Method is highly efficient, converging fast under favorable circumstances. However, it requires a good initial guess and depends greatly on the behavior of the function. If the function's derivative is zero or nearly zero at certain points, it may require many iterations or fail to converge altogether.

Root Approximation

Root approximation is a crucial calculus concept used to find where a function crosses the x-axis, known as roots. These roots can be actual solutions of an equation like \( f(x) = 0 \), or simply points where the function equals zero.
When using Newton's Method for root approximation, one's goal is to reach as close as possible to the true root within a reasonable number of iterations. However, the estimate may sometimes deviate significantly, particularly if the function is flat near the root, as with the example function \((x-1)^{40}\).
Flatness impairs convergence, meaning:

• Approximations may remain far from the true root even after many iterations.
• Changes in approximations may become vanishingly small.
• Additional iterations might not result in meaningful progress towards the root.

Thus, with flat functions, other techniques might be required or adjustments made to the starting value for better results.

Derivative Calculation

Calculating derivatives is foundational in using Newton's Method effectively. A derivative represents the instantaneous rate of change of a function, and it is essential for getting the slope of the tangent line at any point on the curve. In the given exercise, the function \( f(x) = (x-1)^{40} \) required calculating its derivative.
To find the derivative \( f'(x) \):

• Apply the power rule, where if \( f(x) = (x-a)^n \), then \( f'(x) = n(x-a)^{n-1} \).
• Thus, for our function, \( f'(x) = 40(x-1)^{39} \).

This derivative is used in each Newton iteration to compute the movement step size. If the derivative is too small, as it often is near the roots of very flat functions like \( (x-1)^{40} \), the method struggles to "jump" toward the root effectively. Knowing how to calculate derivatives accurately ensures that each estimated move is based on a reliable slope measurement, crucial for convergence.