Problem 3 - Thomas Calculus in SI Units

Question

Use Newton's method to estimate the two zeros of the function \(f(x)=x^{4}+x-3 .\) Start with \(x_{0}=-1\) for the left-hand zero and with \(x_{0}=1\) for the zero on the right. Then, in each case, find \(x_{2}\).

Step-by-Step Solution

Answer

The approximations for the zeros are \(x_2 \approx -1.6452\) for the left and \(x_2 \approx 1.1654\) for the right.

Step 1: Understand Newton's Method Formula

Newton's method is an iterative process used to approximate the roots (or zeros) of a real-valued function. The formula is \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]where \(x_n\) is the current approximation, \(f(x_n)\) is the function value, and \(f'(x_n)\) is the derivative of the function at \(x_n\).

Step 2: Derive the Function and Its Derivative

Given the function \(f(x) = x^4 + x - 3\), first compute its derivative:\[ f'(x) = 4x^3 + 1 \]We will use this derivative in Newton's method formula.

Step 3: Compute the First Iteration (x1) for Left-Hand Zero

Starting with \(x_0 = -1\) for the left-hand zero, apply the formula:\[ x_{1} = -1 - \frac{(-1)^4 + (-1) - 3}{4(-1)^3 + 1} \]Calculate explicitly:\[ x_{1} = -1 - \frac{1 - 1 - 3}{-4 + 1} = -1 - \frac{-3}{-3} = -1 - 1 = -2 \]

Step 4: Compute the Second Iteration (x2) for Left-Hand Zero

Using \(x_1 = -2\), compute \(x_2\):\[ x_{2} = -2 - \frac{(-2)^4 + (-2) - 3}{4(-2)^3 + 1} \]Calculate explicitly:\[ x_{2} = -2 - \frac{16 - 2 - 3}{-32 + 1} = -2 - \frac{11}{-31} = -2 + 0.3548 \approx -1.6452 \]

Step 5: Compute the First Iteration (x1) for Right-Hand Zero

Starting with \(x_0 = 1\) for the right-hand zero, apply the formula:\[ x_{1} = 1 - \frac{1^4 + 1 - 3}{4(1)^3 + 1} \]Calculate explicitly:\[ x_{1} = 1 - \frac{1 + 1 - 3}{4 + 1} = 1 - \frac{-1}{5} = 1 + 0.2 = 1.2 \]

Step 6: Compute the Second Iteration (x2) for Right-Hand Zero

Using \(x_1 = 1.2\), compute \(x_2\):\[ x_{2} = 1.2 - \frac{(1.2)^4 + 1.2 - 3}{4(1.2)^3 + 1} \]Calculate explicitly:\[ x_{2} = 1.2 - \frac{2.0736 + 1.2 - 3}{6.912 + 1} \approx 1.2 - \frac{0.2736}{7.912} \approx 1.1654 \]

Key Concepts

Root Approximation

Root approximation is a valuable technique used to find values where a function equals zero. A function's zero, also known as a root, is any value of x for which the function f(x) equals zero. In calculus and many applied mathematics fields, finding these roots is essential. However, not all functions interact nicely to provide simple algebraic solutions for their roots. In these instances, approximation methods like Newton's method help. By starting with an initial guess of the root and iterating to improve this estimate, one can get increasingly closer to the true root value. Root approximation is particularly important for calculus problem-solving, where precise outcomes are needed but exact algebraic answers might be elusive.

Iterative Methods

Iterative methods are powerful techniques for solving equations where analytic solutions are difficult to obtain. Instead of finding a precise solution outright, they zero in on an approximate solution through continuous refinement. Newton's method is one such iterative method. It functions by repeatedly applying a formula that takes into account the current guess and adjusts it based on the function and its derivative. Each iteration brings the estimate closer to the actual root. These methods are especially useful for computational problem-solving, allowing computers to solve complex equations efficiently.

Calculus Problem-Solving

Calculus problem-solving often involves finding values where functions take extreme or zero values, and solving differential equations. Newton's method fits into this process by providing a way to calculate roots of a function using its derivative. By having a good initial guess, which can sometimes be informed by graphs or other analysis methods, Newton's method can quickly lead to satisfactory approximations. Calculus frequently extends into areas involving rates of change and slopes, where understanding and solving for derivatives are integral.

Derivatives in Calculus

Derivatives in calculus are fundamental for understanding how functions change. They provide information about the slope of a function at any given point. In Newton's method, the derivative of a function is crucial because it informs how large or small an adjustment should be to improve the approximation to a root. If the derivative is large, the function is changing quickly, and the adjustment might be more significant. Conversely, a small derivative indicates slower change, suggesting a smaller adjustment. Calculating derivatives correctly is essential, as it underpins many of the mechanical steps in calculus problem-solving, especially within iterative methods like Newton's.