Problem 1 - Calculus Early Transcendentals

Question

Approximate \(\sqrt{2}\) by applying Newton's Method to the equation \(x^{2}-2=0\)

Step-by-Step Solution

Answer

The approximation of \(\sqrt{2}\) using Newton's Method is approximately 1.4142.

Step 1: Define the Function and its Derivative

The equation we are considering is \(x^2 - 2 = 0\). For Newton's Method, we define a function \( f(x) = x^2 - 2 \). The derivative of this function, \( f'(x) \), is \( f'(x) = 2x \).

Step 2: Choose an Initial Guess

To begin using Newton's Method, we need an initial guess. A reasonable initial guess for \(\sqrt{2}\) is \(x_0 = 1.5\).

Step 3: Apply Newton's Iteration Formula

Newton's Method uses the iteration formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). We substitute \(f(x) = x^2 - 2\) and \(f'(x) = 2x\) into the iteration formula.

Step 4: Calculate the First Iteration

Using \(x_0 = 1.5\), we calculate:\[ x_1 = 1.5 - \frac{1.5^2 - 2}{2 \times 1.5} = 1.5 - \frac{0.25}{3} = 1.5 - 0.0833 = 1.4167 \]

Step 5: Calculate the Second Iteration

Using \(x_1 = 1.4167\), we calculate:\[ x_2 = 1.4167 - \frac{1.4167^2 - 2}{2 \times 1.4167} = 1.4167 - \frac{0.0069}{2.8334} = 1.4167 - 0.0024 = 1.4143 \]

Step 6: Calculate the Third Iteration

Using \(x_2 = 1.4143\), we calculate:\[ x_3 = 1.4143 - \frac{1.4143^2 - 2}{2 \times 1.4143} = 1.4143 - \frac{0.000006}{2.8286} = 1.4143 - 0.000002 = 1.4142 \]

Step 7: Convergence Check

Newton's Method has now provided a sufficiently accurate approximation of \(\sqrt{2}\): approximately 1.4142. We observe that \(x_3\) did not change significantly from \(x_2\), indicating convergence.

Key Concepts

Square Root Approximation

Approximating square roots is a common mathematical task. Using Newton's Method to find square roots is efficient and involves understanding the function's root.
\(\sqrt{2}\) specifically is a value often approximated in math exercises. To approximate \(\sqrt{2}\) using Newton's Method, we solve \(x^2 - 2 = 0\).
This equation roots to the square root of 2. The method provides successive approximations that get us closer to the true value.
The idea is to start from an initial guess, then iterate using Newton's Method to refine this guess. This approach is practical because the square root can’t be neatly expressed as a simple fraction or integer.
This iterative approach will help refine understanding of how closely we can get to the square root using algebraic techniques.

Iterative Methods

Iterative methods involve a repeated process to hone on a solution. In the case of Newton’s Method, you use an initial guess and improve it over multiple steps.
Newton’s Method formula is:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
This formula applies the initial guess through "iterations" to improve accuracy.

• First iteration: Start with a guess \(x_0\), calculate using the formula.
• Continued refinement: Each new \(x\) value \(x_{n+1}\) is more accurate.
• Stopping when changes are very small; indicating convergence.

This method is highly efficient for finding roots when analytical solutions are complex or nonexistent.
Every iteration brings us closer to the actual solution, in this case, \(\sqrt{2}\).

Derivative Calculation

Calculating the derivative is crucial in Newton's Method. It measures the slope of the tangent line to the function at a given point.
For the function \(f(x) = x^2 - 2\), the derivative is \(f'(x) = 2x\).
This derivative helps adjust the initial guess in Newton's iteration formula.

• Precisely, the derivative \(f'(x)\) guides the size of the step we take toward the root.
• A derivative that is zero or close to zero can make the method stall or diverge.
• In our exercise, as \(f'(x) = 2x\), it avoids these pitfalls near the actual root \(x = \sqrt{2}\).

Understanding and calculating the derivative accurately ensures that iterations are effective.
The quality of each step in Newton's Method hinges on this calculation, fine-tuning the path to the solution.