Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use Newton's Method to calculate \(\sqrt[3]{6}\) to five decimal places. Hint: Solve \(x^{3}-6=0\).

Step-by-Step Solution

Answer

\(\sqrt[3]{6} \approx 1.81712\) to five decimal places.

Step 1: Identify the Function

To use Newton's Method, we first need to define the function related to the problem. Here, we need to find \(\sqrt[3]{6}\). This is equivalent to finding \(x\) such that \(x^3 = 6\). Hence, the function is \(f(x) = x^3 - 6\).

Step 2: Calculate the Derivative

Next, find the derivative of the function \(f(x) = x^3 - 6\). The derivative is \(f'(x) = 3x^2\).

Step 3: Initial Approximation

Choose a starting point \(x_0\) close to the actual root. Since \(\sqrt[3]{6}\) is between 1 and 2, we can start with \(x_0 = 1.5\).

Step 4: Apply Newton's Formula

Newton's Method requires the formula \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]Substitute the function and its derivative:\[ x_{n+1} = x_n - \frac{x_n^3 - 6}{3x_n^2} \]

Step 5: Iterate to Improve the Estimate

Compute successive approximations until the desired precision is reached. - \(x_1 = 1.5 - \frac{1.5^3 - 6}{3 \times 1.5^2} = 1.5 - 0.2963 = 1.2037\)- \(x_2 = 1.2037 - \frac{1.2037^3 - 6}{3 \times 1.2037^2} \approx 1.81756\)- \(x_3 = 1.81756 - \frac{1.81756^3 - 6}{3 \times 1.81756^2} \approx 1.81337\)- Continue this process until the result changes by less than \(0.00001\).

Step 6: Check Convergence

Calculate a few more iterations to ensure convergence. If the value doesn't change significantly (within the fifth decimal place), then the value is accurate enough.

Key Concepts

Numerical Methods

Numerical methods are mathematical tools used to solve problems that are difficult to answer analytically. These methods are crucial when working with equations that cannot be solved by algebraic means alone. By relying on a systematic approach, numerical methods allow us to obtain approximate solutions to complex problems.

They are extensively used in engineering, physics, finance, and many areas where an exact solution might be impossible to find. Even when exact solutions can be determined, numerical methods are often preferred because of their efficiency and simplicity in handling large datasets.

Key features of numerical methods include:

• Approximations: Providing close estimates instead of exact answers.
• Iterations: Continuous repetition to refine solutions.
• Error Reduction: Minimizing the difference between calculated and true values.

Cube Root Calculation

Calculating cube roots, such as finding \(\sqrt[3]{6}\) using numerical methods, involves identifying a value that, when multiplied by itself three times, equals the target number. Traditional analytical methods, like solving the equation \(x^3 = a\), where \(a\) is the number whose cube root we wish to find, can be algebraically challenging.

That is where numerical methods like Newton's Method come into play, offering a systematic way to approximate cube roots without directly solving the equation analytically. This is particularly useful for non-perfect cube numbers, which do not resolve neatly into whole numbers and might require iterative refinement for an accurate approximation.

Derivatives

Derivatives are fundamental components in calculus that measure how a function changes as its input changes. In simple terms, a derivative represents the slope or rate of change of a function at any given point.

When using Newton's Method, derivatives play a vital role because the process depends on calculating the slope of a tangent line to the function's curve. For the function \(f(x) = x^3 - 6\), the derivative is \(f'(x) = 3x^2\). This derivative helps to determine the direction and distance to adjust our estimates towards the root of the equation. The steeper the slope, the larger the adjustment.Newton's Method formula, \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \], specifically uses derivatives to modify guesses, ensuring each new approximation, \(x_{n+1}\), is closer to the actual root of the function than the previous.

Iterative Approximation

Iterative approximation is a technique used to improve the accuracy of a solution progressively. It involves repeatedly applying a mathematical process to refine an estimate.

In the context of Newton's Method, iterative approximation is core to narrowing down the cube root of a number like 6. Starting with an initial guess—such as 1.5 in our problem—the method iteratively applies the correction formula \[ x_{n+1} = x_n - \frac{x_n^3 - 6}{3x_n^2} \]. With each iteration, the approximation \(x\) becomes closer to \(\sqrt[3]{6}\).

The iterations continue until the difference between successive approximations is sufficiently small, ensuring the solution is accurate to the desired number of decimal places. This method is powerful and widely used because it offers precision and efficiency, making it suitable for computing solutions where exact answers are cumbersome to determine.