Problem 2 - Single Variable Calculus

Question

Use Newton's method to find the zero(s) of \(\bar{f}\) to four decimal places by solving the equation \(f(x)=2 x^{3}-15 x^{2}+36 x-20, \quad x_{0}=1\)

Step-by-Step Solution

Answer

Applying Newton's method with the iterative formula \(x_{n+1} = x_n - \frac{2x_n^3 - 15x_n^2 + 36x_n - 20}{6x_n^2 - 30x_n + 36}\) and initial guess \(x_0 = 1\), we get the approximated zero of the function \(f(x) = 2x^3 - 15x^2 + 36x - 20\) to be \(x \approx 1.4160\) to four decimal places.

Step 1: Find the Derivative of the Function

First, we need to find the derivative of the given function: \(f(x) = 2x^3 - 15x^2 + 36x - 20\). To do this, apply the power rule for each term: \[f'(x) = \frac{d}{dx} (2x^3 - 15x^2 + 36x - 20) = 6x^2 - 30x + 36\]

Step 2: Find the Iterative Formula for Newton's Method DependencyProperty

Now that we have the derivative of the function, we can use the formula for Newton's method: \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] In our case, the formula becomes: \[x_{n+1} = x_n - \frac{2x_n^3 - 15x_n^2 + 36x_n - 20}{6x_n^2 - 30x_n + 36}\]

Step 3: Apply the Iterative Formula

We're given the initial guess \(x_0 = 1\), and we need to find the zero(s) of \(f(x)\) to four decimal places. Apply the iterative formula until the difference between consecutive values of \(x_n\) is less than 0.0001. - First iteration using \(x_0=1\): \(x_1 = x_0 - \frac{2(1)^3 - 15(1)^2 + 36(1) - 20}{6(1)^2 - 30(1) + 36} = 1 + \frac{3}{12} \approx 1.2500\) Difference: \(|x_1 - x_0| = 0.2500\) - Second iteration using \(x_1 = 1.2500\): \(x_2 = x_1 - \frac{2(1.25)^3 - 15(1.25)^2 + 36(1.25) - 20}{6(1.25)^2 - 30(1.25) + 36} \approx 1.3695\) Difference: \(|x_2 - x_1| \approx 0.1195\) - Third iteration using \(x_2 = 1.3695\): \(x_3 = x_2 - \frac{2(1.3695)^3 - 15(1.3695)^2 + 36(1.3695) - 20}{6(1.3695)^2 - 30(1.3695) + 36} \approx 1.4055\) Difference: \(|x_3 - x_2| \approx 0.0360\) - Fourth iteration using \(x_3 = 1.4055\): \(x_4 = x_3 - \frac{2(1.4055)^3 - 15(1.4055)^2 + 36(1.4055) - 20}{6(1.4055)^2 - 30(1.4055) + 36} \approx 1.4149\) Difference: \(|x_4 - x_3| \approx 0.0094\) - Fifth iteration using \(x_4 = 1.4149\): \(x_5 = x_4 - \frac{2(1.4149)^3 - 15(1.4149)^2 + 36(1.4149) - 20}{6(1.4149)^2 - 30(1.4149) + 36} \approx 1.4159\) Difference: \(|x_5 - x_4| \approx 0.0010\) - Sixth iteration using \(x_5 = 1.4159\): \(x_6 = x_5 - \frac{2(1.4159)^3 - 15(1.4159)^2 + 36(1.4159) - 20}{6(1.4159)^2 - 30(1.4159) + 36} \approx 1.4160\) Difference: \(|x_6 - x_5| \approx 0.0001\) Since the difference between \(x_5\) and \(x_6\) is smaller than 0.0001, we can stop iterating and conclude that the zero of \(f(x) = 2x^3 - 15x^2 + 36x - 20\) is approximately \(x \approx 1.4160\) to four decimal places.

Key Concepts

Derivative

In calculus, the derivative represents the rate at which a function is changing at any given point. It is a fundamental concept necessary in Newton's Method. For the given problem, the function is\[f(x) = 2x^3 - 15x^2 + 36x - 20\]To find its derivative, apply the power rule. The power rule states that for any term of the form \(ax^n\), the derivative is \(n \cdot ax^{n-1}\). Therefore, the derivative of the function is:\[f'(x) = 6x^2 - 30x + 36\]This derivative, \(f'(x)\), is crucial as it helps us understand the steepness or slope of the function at any point \(x\). A higher positive derivative indicates that the function is rising steeply at that point, while a more negative derivative suggests a steep decline. Understanding the behavior of \(f(x)\) through its derivative \(f'(x)\) allows us to apply Newton's Method effectively.

Iterative Formula

Newton's Method is a powerful technique for finding the root, or zero, of a function. At the heart of this approach is the iterative formula:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]This formula allows you to progressively get closer to the root with each iteration. Here's how it works:

• Start with an initial guess, \(x_0\). For this problem, \(x_0 = 1\).
• Use the formula to calculate the next approximation, \(x_1\).
• Repeat the process until the difference between consecutive approximations is very small. This indicates you are close to the true root.

For the given function, our iterative formula becomes:\[x_{n+1} = x_n - \frac{2x_n^3 - 15x_n^2 + 36x_n - 20}{6x_n^2 - 30x_n + 36}\]This iterative process is essential for solving cases where a function does not have an easily obtainable analytical solution. By using repeated iterations, we can zero in on the solution to a high degree of precision.

Convergence of Iterations

The concept of convergence is vital to understanding when to stop iterating in Newton's Method. Convergence refers to how the repeated applications of the iterative formula bring us closer to the actual zero of the function. For continuous and differentiable functions, convergence can often be relied upon, as long as the initial guess is reasonably close to the true root.In our case, we look for when the change between two successive iterations, \(|x_{n+1} - x_n|\), is very small. A typical threshold is being less than 0.0001:

• The process continues from \(x_0 = 1\) until the change falls below this threshold.
• This ensures that the result, \(x \approx 1.4160\), is accurate to four decimal places.
• Each iteration brings the estimate \(x_n\) closer to the true root, demonstrating convergence.

Convergence makes Newton's Method a powerful tool since, unlike simple estimation, it methodically refines the guess to the required degree of accuracy. By observing the steadily decreasing differences between iterations, we gain confidence that we are approaching the true zero of the function.