Problem 63 - Algebra and Trigonometry with Analytic Geometry

Question

The equation \(\frac{1}{3} \sqrt[3]{x}-x+2=0\) has a root near 2 . To approximate this root, rewrite the equation as \(x=\frac{1}{3} \sqrt[3]{x}+2\). Let \(x_{1}=2\) and find successive approximations \(x_{2}, x_{3}, \ldots\) by using the formulas $$ x_{2}=\frac{1}{3} \sqrt[3]{x_{1}}+2, \quad x_{3}=\frac{1}{3} \sqrt[3]{x_{2}}+2, \quad \ldots $$ until four-decimal-place accuracy is obtained.

Step-by-Step Solution

Answer

The root is approximately 2.46 with four-decimal-place accuracy.

Step 1: Understanding the Iterative Method

The original equation is \( \frac{1}{3} \sqrt[3]{x} - x + 2 = 0 \). You start with an initial guess of \( x_1 = 2 \) and rewrite the equation to \( x = \frac{1}{3} \sqrt[3]{x} + 2 \). This suggests an iterative approach for finding successive approximations.

Step 2: Calculate the First Approximation

Using the formula \( x_2 = \frac{1}{3} \sqrt[3]{x_1} + 2 \), substitute \( x_1 = 2 \). This gives \( x_2 = \frac{1}{3} \sqrt[3]{2} + 2 \). Calculate \( \sqrt[3]{2} \approx 1.26 \), thus \( x_2 = \frac{1}{3} \times 1.26 + 2 \approx 2.42 \).

Step 3: Calculate the Second Approximation

Now compute \( x_3 = \frac{1}{3} \sqrt[3]{x_2} + 2 \). Using \( x_2 \approx 2.42 \), calculate \( \sqrt[3]{2.42} \approx 1.34 \). Thus, \( x_3 = \frac{1}{3} \times 1.34 + 2 \approx 2.45 \).

Step 4: Check for Accuracy

Check if further iterations are required by calculating \( \sqrt[3]{2.45} \) and updating \( x_4 = \frac{1}{3} \sqrt[3]{x_3} + 2 \). Continue this process until the successive values differ by less than 0.0001.

Step 5: Further Approximation

Compute \( x_4 = \frac{1}{3} \sqrt[3]{x_3} + 2 \). If \( \sqrt[3]{2.45} \approx 1.35 \), then \( x_4 = \frac{1}{3} \times 1.35 + 2 \approx 2.45 \). Continue iterations: if \( \sqrt[3]{2.45} \approx 1.35 \) gives a new \( x \approx 2.46 \), keep iterating until desired accuracy.

Key Concepts

Cubic Roots

Cubic roots are numbers that, when multiplied by themselves three times, produce the original number. For example, if we have a number 8, the cubic root is 2 because \( 2 \times 2 \times 2 = 8 \). In mathematical terms, we denote the cubic root of \( x \) using the symbol \( \sqrt[3]{x} \). This concept is essential when solving equations that involve cubic terms or when approximating roots using methods like the one outlined in the exercise.
Understanding cubic roots helps us break down more complex problems into simpler parts. This knowledge proves crucial in the iterative process, as each step needs precise calculations of cubic roots to refine successive approximations. Calculating or estimating cubic roots accurately is fundamental to understanding how these values interact in solving equations.
Applying this in the iterative method presented in the exercise involves refining the approximation of the root of the equation \( \frac{1}{3} \sqrt[3]{x}-x+2=0 \). By finding the cubic root of \( x \) in each iteration step, you're gradually closing in on a more accurate solution.

Successive Approximations

Successive approximations are a method used to refine estimates of a solution through an iterative process. Starting with an initial guess, each iteration produces a new approximation that gets closer to the true value or root. This method is particularly helpful when an equation cannot be solved algebraically, allowing us to hone in on a solution numerically.
In the given exercise, we start with an initial guess \( x_1 = 2 \). Using the iterative formula \( x = \frac{1}{3} \sqrt[3]{x} + 2 \), we compute successive values \( x_2, x_3, \ldots \) until they stabilize to a desired precision level. Each new approximation uses the result from the previous one, progressively narrowing down the solution.

• Start with an initial estimate
• Apply the iterative formula
• Calculate the cubic root for the current iteration
• Update the approximation
• Repeat until the desired accuracy is achieved

This method demonstrates how iterative techniques refine and improve our estimates, honing closer to the true solution through each step.

Equation Solving

Equation solving involves finding the value or values that satisfy an equation—meaning the left-hand side equals the right-hand side. Equations can range from the simple, such as linear equations, to more complex types like quadratic, cubic, or systems of equations. Solving them often requires different strategies tailored to the specific form.
For equations that cannot be easily solved by straightforward algebraic manipulation, as the one in our exercise, numerical methods become very useful. Here, we encountered a transformation of the original equation \( \frac{1}{3} \sqrt[3]{x} - x + 2 = 0 \) into a form suitable for iterative approximation, \( x = \frac{1}{3} \sqrt[3]{x} + 2 \).
The process of equation solving using iterations involves:

• Identifying the suitable form for iteration
• Choosing an appropriate initial guess
• Applying the successive approximation method
• Checking the stability and convergence of the approximations

This iterative solving technique transforms challenging problems into simpler, approachable steps, allowing mathematical solutions where direct computation isn't feasible.

Numerical Methods

Numerical methods are algorithms used to solve mathematical problems using numerical approximation rather than analytical expressions. These techniques are invaluable in fields where exact solutions are difficult or impossible to obtain due to complex equations or large data sets.
The exercise utilizes a numerical method involving successive approximations to solve for the root of a non-linear equation. Such methods are crucial in engineering, physics, and computer graphics, where simulations rely on approximations of real-world phenomena.

• Provides solutions not possible by traditional algebraic means
• Useful for complex equations and large-scale computation
• Allows for iterative refinement of solutions
• Essential in applications where analytical solutions are impractical

In solving \( \frac{1}{3} \sqrt[3]{x} - x + 2 = 0 \), the iterative approach provides a practical avenue to reach solutions with desired precision, demonstrating how numerical methods bridge the gap between theoretical math and practical application.