Problem 4 - Thomas Calculus in SI Units

Question

Use Newton's method to estimate the two zeros of the function \(f(x)=2 x-x^{2}+1 .\) Start with \(x_{0}=0\) for the left-hand zero and with \(x_{0}=2\) for the zero on the right. Then, in each case, find \(x_{2}\).

Step-by-Step Solution

Answer

The two approximated zeros are \( x_2 \approx -0.5833 \) and \( x_2 \approx 2.4167 \).

Step 1: Compute Derivative of the Function

First, we compute the derivative of the function. Given \( f(x) = 2x - x^2 + 1 \), the derivative is found using the standard rules of differentiation: \[ f'(x) = 2 - 2x \]

Step 2: Apply Newton's Method Formula

Newton's method iteratively finds approximate solutions to the equation \( f(x) = 0 \). The formula is given by: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

Step 3: Compute First Iteration for Left-Hand Zero

Start with \( x_0 = 0 \). Compute \( x_1 \) using Newton's formula:\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \] Since \( f(0) = 1 \) and \( f'(0) = 2 \), \[ x_1 = 0 - \frac{1}{2} = -0.5 \]

Step 4: Compute Second Iteration for Left-Hand Zero

Now use \( x_1 = -0.5 \) to find \( x_2 \):\[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \] Calculate \( f(-0.5) = 2(-0.5) - (-0.5)^2 + 1 = -1 + 0.25 + 1 = 0.25 \) And \( f'(-0.5) = 2 - 2(-0.5) = 3 \), \[ x_2 = -0.5 - \frac{0.25}{3} = -0.5 - 0.0833 \approx -0.5833 \]

Step 5: Compute First Iteration for Right-Hand Zero

Start with \( x_0 = 2 \) and compute \( x_1 \): \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \] Here, \( f(2) = 4 - 4 + 1 = 1 \) and \( f'(2) = 2 - 4 = -2 \), \[ x_1 = 2 - \frac{1}{-2} = 2.5 \]

Step 6: Compute Second Iteration for Right-Hand Zero

Now use \( x_1 = 2.5 \) to compute \( x_2 \): \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \] Calculate \( f(2.5) = 5 - 6.25 + 1 = -0.25 \) And \( f'(2.5) = 2 - 5 = -3 \), \[ x_2 = 2.5 - \frac{-0.25}{-3} = 2.5 + 0.0833 \approx 2.4167 \]

Key Concepts

Derivative of a function

The derivative of a function is a fundamental concept in calculus focused on understanding how a function changes as its input changes. It gives the slope of the function at any point, essentially indicating the function's rate of change. For a polynomial function like the one used in the exercise, the power rule is commonly applied to find the derivative. This rule states that if you have a term of the form \(ax^n\), its derivative will be \(nax^{n-1}\).
The function in this exercise is \(f(x) = 2x - x^2 + 1\). The derivative is calculated by applying the power rule to each term individually:

• The derivative of \(2x\) is \(2\), because the exponent of \(x\) is 1, and \(1 \times 2 = 2\).
• The derivative of \(-x^2\) is \(-2x\), since the coefficient of \(x^2\) is -1, and applying the power rule gives \(-2 \times 1 \times x^{2-1} = -2x\).
• The constant term \(+1\) becomes \(0\), as constants do not change and thus have a derivative of zero.

By combining these derivatives, we obtain \(f'(x) = 2 - 2x\), which is used in Newton's method to adjust guesses towards the function's zeros.

Zeros of a function

Zeros of a function, also referred to as roots, are the input values where the function outputs zero. That is, they satisfy the equation \(f(x) = 0\). Identifying zeros is essential as they often represent crucial points of interest in mathematical problems, such as intersections with the x-axis in graphical terms.
Understanding zeros is not only important mathematically but also in practical applications where these points might represent break-even points, equilibrium in systems, or other pivotal states.
For the given function \(f(x) = 2x - x^2 + 1\), zeros are the values of \(x\) that make the expression equal to zero. Newton's Method, which was applied in the original step-by-step solution, is a powerful tool for approximating these values when they cannot be found easily by algebraic manipulation.
The exercise uses Newton's method to find two zeros by iterative estimation, beginning with trial points \(x_0 = 0\) and \(x_0 = 2\).

Iterative methods in calculus

Iterative methods in calculus involve applying a formula repeatedly to get closer to an answer. They are particularly useful for problems where exact solutions are difficult to calculate analytically, or even impossible with simple algebra. Newton's Method is a widely used iterative method to find zeros of a function.
Here's how it works:

• Start with an initial guess, often denoted as \(x_0\) for the function's zero.
• Use the iterative formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) to get a better approximation. This formula involves both the function and its derivative, using the concept of local linear approximation to better zero in on an actual zero.
• Repeat this process until the value converges to a constant, meaning changes are negligible. This indicates a close approximation to the zero is found.

In practice, as we saw in the exercise, executing the method generally involves a couple of iterations to substantially narrow down the estimates. Starting with rough guesses such as \(x_0 = 0\) and \(x_0 = 2\), the process quickly homes in on the true zeros of the function, illustrating the precision and power of iterative calculus methods like Newton's Method.