Question

The roots of \(f(x)\) are known or are easily found. Use 5 iterations of Newton's Method with the given initial approximation to approximate the root. Compare it to the known value of the root. $$ f(x)=x^{2}-2, x_{0}=1.5 $$

Short answer

The approximation after 5 iterations is 1.4142135, closely matching the known root \(\sqrt{2}\).

Step-by-step solution

Understand the Problem

We need to use Newton's Method to approximate the root of the function \(f(x) = x^2 - 2\) starting from an initial approximation of \(x_0 = 1.5\). We will perform 5 iterations of the method and then compare our approximation to the known root.

Identify the Known Root

Since \(f(x) = x^2 - 2\), we can solve \(x^2 = 2\), giving the roots \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). For this problem, given the initial approximation of \(x_0 = 1.5\), we are likely interested in \(x = \sqrt{2}\), approximately 1.414.

Derive the Formula for Newton's Method

Newton's Method uses the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). For the function \(f(x) = x^2 - 2\), the derivative \(f'(x) = 2x\).

First Iteration of Newton's Method

Start with \(x_0 = 1.5\), then compute: \[ x_1 = x_0 - \frac{x_0^2 - 2}{2x_0} = 1.5 - \frac{1.5^2 - 2}{2*1.5} = 1.5 - \frac{0.25}{3} = 1.5 - 0.08333 = 1.41667 \]

Second Iteration

Using \(x_1 = 1.41667\), compute: \[ x_2 = x_1 - \frac{x_1^2 - 2}{2x_1} = 1.41667 - \frac{1.41667^2 - 2}{2*1.41667}\] Calculating gives \(x_2 \approx 1.4142157\).

Third Iteration

Using \(x_2 = 1.4142157\), compute: \[ x_3 = x_2 - \frac{x_2^2 - 2}{2x_2} \] After calculation, \(x_3 \approx 1.4142135\).

Fourth Iteration

Using \(x_3 = 1.4142135\), compute: \[ x_4 = x_3 - \frac{x_3^2 - 2}{2x_3} \] This yields \(x_4 \approx 1.4142135\), indicating convergence.

Fifth Iteration

Using \(x_4 = 1.4142135\), compute: \[ x_5 = x_4 - \frac{x_4^2 - 2}{2x_4} \] Calculation results in \(x_5 \approx 1.4142135\).

Compare to the Known Root

The known root is \(\sqrt{2} \approx 1.4142135\). The approximation \(x_5 = 1.4142135\) matches the known root very closely, showing that Newton's Method converges quickly for this problem.

Key concepts

Roots of Equations

In mathematics, the term "roots of equations" refers to the solutions of an equation where the function equals zero. For example, if we have a function \( f(x) = x^2 - 2 \), the roots are the values of \( x \) that satisfy the equation \( f(x) = 0 \), or \( x^2 - 2 = 0 \). Solving \( x^2 = 2 \) gives the roots \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). These roots are the points where the graph of the function crosses the x-axis.
Understanding the roots of equations is essential in calculus and algebra, as they help us find where something equals nothing, making them vital in problem-solving. Knowing the roots can greatly aid in graph plotting and analysing real-world scenarios where variables achieve specific values.

Numerical Methods

Numerical methods are techniques used to find approximate solutions to equations or mathematical problems, especially when an exact form is difficult or impossible to obtain. These methods are a powerful tool for engineers and scientists.
Among these methods, Newton's Method stands out due to its efficiency in finding roots of equations. It works by starting with an initial guess and iteratively improving that guess with the formula:

• \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)

Here, \( f'(x_n) \) represents the derivative of the function at \( x_n \).
For \( f(x) = x^2 - 2 \), the derivative \( f'(x) = 2x \). Starting from an initial guess like \( x_0 = 1.5 \), applying the formula yields increasingly accurate approximations of \( \sqrt{2} \).
Numerical methods are crucial as they let us perform calculations that are otherwise very difficult with traditional algebraic techniques.

Iterative Approximations

Iterative approximations refer to the process of repeated calculations to get closer to the desired result. Each step in the iteration uses the output from the previous step to refine the solution. This is central to many numerical methods, including Newton's Method.
The idea is to start with a guess and apply a set formula to improve it. In our example, starting with \( x_0 = 1.5 \), we apply:

• \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)

By iterating this process, we get consecutive approximations \( x_1 \), \( x_2 \), etc., inching closer to the true value of \( \sqrt{2} \).
In this example with five iterations, the approximations quickly converge towards the root 1.4142135. This demonstrates the power and speed of iterative methods in calculus, where a method like Newton's can rapidly solve otherwise complex problems.