Question

Approximating reciprocals To approximate the reciprocal of a number \(a\) without using division, we can apply Newton's method to the function \(f(x)=\frac{1}{x}-a\) a. Verify that Newton's method gives the formula \(x_{n+1}=\left(2-a x_{n}\right) x_{n}\) b. Apply Newton's method with \(a=7\) using a starting value of your choice. Compute an approximation with eight digits of accuracy. What number does Newton's method approximate in this case?

Short answer

Question: Use Newton's method to approximate the reciprocal of 7 with eight digits of accuracy. Answer: By applying Newton's method iteratively, we find that the reciprocal of 7 can be approximated with eight digits of accuracy as 0.14285714, which is the same as $\frac{1}{7}$.

Step-by-step solution

Understand Newton's Method

Newton's method is an iterative method for finding the roots of a real-valued function \(f(x)\). It works by choosing an initial guess, \(x_0\), and successively computing better approximations of the root using the following formula: \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] In our case, the function is \(f(x) = \frac{1}{x} - a\). We need to find the derivative of this function to apply Newton's method.

Find the derivative of \(f(x)\)

\(f(x) = \frac{1}{x} - a\) has the following derivative: \(f'(x)=-\frac{1}{x^2}\) Now we can plug \(f(x)\) and \(f'(x)\) back into Newton's method formula.

Verify the Newton's method formula

For our function \(f(x) = \frac{1}{x} - a\) and its derivative \(f'(x) = -\frac{1}{x^2}\), the Newton's method formula gives: \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] Substitute \(f(x_n)\) and \(f'(x_n)\): \[x_{n+1} = x_n - \frac{\frac{1}{x_n} - a}{-\frac{1}{x_n^2}}\] Simplify the term: \[x_{n+1} = x_n - \left(-x_n^2\left(\frac{1}{x_n} - a\right)\right)\] \[x_{n+1} = x_n + x_n^2\left(1 - ax_n\right)\] \[x_{n+1} = x_n\left(1 + x_n\left(1 - ax_n\right)\right)\] Finally, we get: \[x_{n+1} = \left(2 - ax_n\right)x_n\] Thus, we have verified the Newton's method formula for our function.

Apply Newton's method with \(a = 7\)

Now, with \(a = 7\), we have the function \(f(x) = \frac{1}{x} - 7\). We will apply Newton's method to approximate the reciprocal of \(a\). Let's choose a starting value, \(x_0 = 0.1\). Use the formula \(x_{n+1} = \left(2 - ax_n\right)x_n\) to compute subsequent approximations until we achieve eight digits of accuracy: \(x_1 = \left(2 - 7\cdot 0.1\right)(0.1) = (2 - 0.7)(0.1) = 1.3(0.1) = 0.13\) \(x_2 = \left(2 - 7\cdot 0.13\right)(0.13) = (2 - 0.91)(0.13) = 1.09(0.13) \approx 0.1417\) Continue this process iteratively until the desired accuracy is reached: \(x_3 \approx 0.1429\) \(x_4 \approx 0.1429\) We have reached eight digits of accuracy with \(x_4 \approx \frac{1}{7} \approx 0.14285714\). In this case, Newton's method approximates the reciprocal of 7, which is \(\frac{1}{7}\).

Key concepts

Iterative Method

Newton's Method is a powerful iterative method used for finding roots of real-valued functions. An iterative method means you start with an initial guess and repeatedly apply a formula to get closer to the desired solution.

For example, if you want to find the root of a function \( f(x) \), you start with an initial guess \( x_0 \). You then use the Newton's Method formula to iteratively calculate better approximations. The formula you use is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. \]As you apply this step multiple times, the values \( x_1, x_2, \) and so on, get closer to the actual root. It's crucial to have a good starting point because it affects the speed and accuracy of convergence.

This iterative approach is especially useful because it's usually faster and requires fewer function evaluations compared to other methods.

Reciprocals Approximation

Approximating reciprocals without division can be achieved using Newton's Method on the function \( f(x) = \frac{1}{x} - a \). The goal is to find a value of \( x \) such that \( f(x) = 0 \), which implies \( \frac{1}{x} = a \), or equivalently, \( x = \frac{1}{a} \).

This clever setup means when you find the root of the function \( f(x) \), you find the reciprocal of \( a \). Newton's Method provides a way to update the initial guess \( x_0 \) using the formula:\[ x_{n+1} = (2 - ax_n)x_n. \]By plugging in the values and iterating, you gradually get closer to the exact value of the reciprocal \( \frac{1}{a} \). For example, to approximate \( \frac{1}{7} \), begin with a starting point like \( x_0 = 0.1 \).

Step by step, you refine this approximation with each iteration, until you achieve sufficient accuracy.

Function Derivative

Finding the right function derivative is key to accurately applying Newton's Method. The derivative helps determine how much to change your current guess to move closer to the actual root.

If you have a function \( f(x) = \frac{1}{x} - a \), its derivative is calculated as \( f'(x) = -\frac{1}{x^2} \). This derivative indicates how steep the function is at any point \( x \), allowing you to adjust your next guess accordingly.

The adjustment is calculated by dividing \( f(x_n) \) by \( f'(x_n) \), where \( f(x_n) \) is the value of the function at your current guess. Incorporating the derivative facilitates informed corrections.

This not only helps converge more rapidly to the root but also ensures stability and effectiveness of the iterative method, especially when the initial guess is not very close to the actual root.

Root Finding

The essence of Newton's Method is rooted in finding the value of \( x \) such that \( f(x) = 0 \). This process is known as root finding. Root finding is a common problem in mathematics, particularly in solving equations.

By iteratively applying Newton's Method, you can identify where the function crosses the x-axis, which is the root. For the function \( f(x) = \frac{1}{x} - a \), solving for a root effectively finds the reciprocal of \( a \).

In these scenarios, ensuring the chosen function and its derivative are accurately evaluated at each step of the iteration is crucial. This ensures convergence to the true root within a practical number of steps. If done correctly, the method refines your approximation of the root, or reciprocal in this case, with each step. This systematic approach effectively solves the initial mathematical challenge, providing solutions that are usable and accurate.