Problem 24 - Single Variable Calculus

Question

Estimate the value of the radical accurate to four decimal places by using three iterations of Newton's method to solve the equation \(f(x)=0\) with initial estimate \(x_{0}\). \sqrt{3} ; \quad f(x)=x^{2}-3 ; x_{0}=1.5

Step-by-Step Solution

Answer

By applying three iterations of Newton's method to the equation \(f(x) = x^2 - 3\) with an initial estimate of \(x_0 = 1.5\), we can estimate the value of \(\sqrt{3}\) to be approximately \(1.7321\) accurate to four decimal places.

Step 1: Find the derivative of the function

First, we need to find the derivative of the function \(f(x) = x^2 - 3\). The derivative of this function is: \(f'(x) = 2x\).

Step 2: Apply Newton's method formula

Now that we have both the function \(f(x)\) and its derivative \(f'(x)\), we can apply Newton's formula for the three iterations: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\)

Step 3: Perform the first iteration

For the first iteration, we have \(x_0 = 1.5\). Let's plug this value into the formula: \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1.5 - \frac{(1.5^2 - 3)}{2(1.5)} = 1.5 - \frac{-0.75}{3} \approx 1.7500\)

Step 4: Perform the second iteration

Now, we will perform the second iteration using the value of \(x_1\): \(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.7500 - \frac{(1.7500^2 - 3)}{2(1.7500)} \approx 1.7321\)

Step 5: Perform the third iteration

Finally, we will perform the third iteration using the value of \(x_2\): \(x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.7321 - \frac{(1.7321^2 - 3)}{2(1.7321)} \approx 1.7321\) After three iterations of Newton's method, our approximation of \(\sqrt{3}\) is \(1.7321\), accurate to four decimal places.

Key Concepts

Numerical Approximation

Numerical approximation is a fundamental technique in mathematics and computational science, allowing us to find approximate solutions to equations that may not be solvable analytically. It is especially useful when we want to estimate the value of irrational numbers, complex roots, or the solutions to transcendental equations.

Numerical methods, like Newton's method, help bridge the gap between mathematical theory and practical applications. By starting with an initial guess and improving that estimate iteratively, numerical approximation provides a systematic way to hone in on a more accurate solution. In the case of approximating a radical, such as \(\sqrt{3}\), Newton's method uses both the function values and their derivatives to converge on an estimate that becomes more precise with each iteration. This method of approximation is invaluable in fields like engineering, physics, and finance, where precise calculations are crucial.

Derivatives

Derivatives are a core concept in calculus, expressing how a function changes as its input changes. Technically, the derivative of a function at a certain point is the slope of the tangent to the function's curve at that point.

In the context of Newton's method, derivatives play a critical role. They tell us the rate at which our function \(f(x)\) is changing at our current estimate. By understanding this rate of change, we can make an educated guess about where the function's curve crosses the x-axis—essentially where it reaches zero. In our exercise, the derivative of the function \(f(x) = x^2 - 3\) is given by \(f'(x) = 2x\). This simple derivative informs the adjustments made to the initial estimate \(x_0\) during each iteration of the process.

Iteration

Iteration is a process of repeating steps to get closer to a desired outcome or to refine an approximation. It is fundamental to many algorithms in numerical analysis, including Newton's method. Each iteration takes us from one estimate of the solution (\(x_n\)) to a more refined estimate (\(x_{n+1}\)).

The beauty of iteration in numerical methods is its simplicity and power. With each cycle, using the function and its derivative, we adjust our guess based on the outcome of a defined formula. As in the exercise, you'll see \(x_1\), \(x_2\), and \(x_3\) each represent the current estimate, honed by the information gathered from the previous step. This iterative process continues until we achieve a sufficiently accurate result, or until the estimates converge to a stable value.

Radical Estimation

Radical estimation is the process of finding the approximate value of a root, like \(\sqrt{3}\). These roots are often irrational numbers that cannot be expressed exactly as fractions or decimals. To deal with such numbers, we use numerical methods to obtain a decimal approximation that is accurate to a required number of decimal places.

Using Newton's method for radical estimation involves an initial guess, which is then refined through iterations to yield a more accurate approximation. In our example, we aim to estimate \(\sqrt{3}\) by solving the equation \(x^2 - 3 = 0\). Newton's method is particularly useful here, as it quickly converges to a precise value, allowing us to approximate radicals with the desired accuracy. After the iterations, we achieved an estimation of \(\sqrt{3}\) accurate to four decimal places as \(1.7321\), showcasing the method's effectiveness in radical estimation.