Question

In Exercises \(15-18,\) use Newton's method to estimate all real solutions of the equation. Make your answers accurate to 6 decimal places. \(x^{3}+x-1=0\)

Short answer

The real solution of the equation \(x^{3}+x-1=0\) accurate up to six decimal places using Newton's method is approximatly \(0.682328\)

Step-by-step solution

Define the Function and its Derivative

We are given the equation \(x^{3}+x-1=0\). Let's denote the left hand side of the equation by \(f(x)\) so we have \(f(x) = x^{3}+x-1\). We need to find the derivative of \(f(x)\), \(f'(x) = 3x^{2}+1\).

Choose an Initial Approximation

We have to choose a seed or an initial approximation. A good initial approximation is essential for Newton's method to converge. A reasonable starting point for our problem could be \(x_{0} = 1\).

Apply Newton's Method with x0

To start the iteration, plug \(x_{0}\) into the Newton's method formula. Thus, we get: \(x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}\). Substitute \(x_{0} = 1\), \(f(x_{0}) = 1^{3}+1-1 = 1\), and \(f'(x_{0}) = 3*1^{2}+1 = 4\) into the equation you get: \(x_{1} = 1 - \frac{1}{4} = 0.75\).

Iteratively Apply Newton's Method

Repeat this process, each time plugging in the previous iteration's \(x\) value, until the result does not change, or changes very minimally (i.e., up to six decimal places). Continue until accuracy is reached, with the final value being the root of the equation.

Check Against the Original Equation

Once the root has been found, it is always a good practice to check if this is indeed a root by substituting it back into the original equation. Should the left and right side of the equation be equal (within a small tolerance), we can be sure that it is a root of the equation.

Key concepts

Numerical Solutions of Equations

When we encounter equations involving polynomials, trigonometry, or other complex functions, finding exact solutions can be quite challenging and sometimes impossible with algebra alone. This is where numerical solutions come into play. These methods allow us to approximate the solutions of an equation to a desired degree of accuracy.

Numerical approaches, like Newton's method, which is a root-finding technique, give us a practical way to tackle equations too tough for algebra. It is particularly useful because it uses the concept of iterations to hone in on a more precise answer with each step. However, to make sure these iterations lead to the correct answer, we start by carefully selecting an initial guess, then progressively improve our estimate.

Consider the equation from the exercise, \(x^{3} + x - 1 = 0\). The exact solution is not obvious, but through numerical methods, we can still arrive at a very precise approximation. Depending on the complexity of the equation and the initial guess, the number of iterations required can vary. Through this steadfast process, we can solve equations which are otherwise analytically unsolvable.

Derivative of a Function

The concept of the derivative is at the heart of calculus and plays a critical role in many numerical methods, including Newton's method. A derivative essentially measures how a function's output changes as its input changes. In technical terms, it represents the rate of change or the slope of the function at a particular point.

In the context of the given exercise, the function \( f(x) = x^{3} + x - 1 \) has a derivative \( f'(x) = 3x^{2} + 1 \). This derivative is crucial because, in Newton's method, it helps us understand how to adjust our current approximation to get closer to the actual root. When applying Newton's method, the value of \( f'(x) \) tells us the 'steepness' of the curve at our current guess, which influences the size and direction of the next step we take toward the root.

Understanding derivatives allows us to predict the behavior of functions and is the foundation for correctly applying iterative root-finding methods. Without this concept, methods like Newton's would not be possible.

Iteration for Root Finding

The process of iteration is a cornerstone of numerical methodologies for root finding. Iteration involves repeating a set of operations, with each repetition using the outcome of the previous one to move closer to the desired result. In our case, the mission is to identify a root of the equation—the value for which the function equals zero. Using Newton's method, we iterate by continually updating our guess of the root using a specific formula.

Each step of iteration involves two main activities: calculating the function value at the current guess, and evaluating its derivative. With these in hand, we apply the Newton's formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \), systematically nudging our guess closer to the actual root with each cycle. The beauty of this method lies in its speed and efficiency; typically, it converges to a highly accurate solution within just a few iterations.

However, the success of the iteration depends on the right choice of starting value and the nature of the function. Some functions may require more careful handling to ensure convergence. In the end, with a correct approach, iteration becomes a powerful tool in our mathematical arsenal to tackle equations that defy simpler, algebraic solutions.