Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the third-order Maclaurin polynomial for \((1+x)^{3 / 2}\) and bound the error \(R_{3}(x)\) if \(-0.1 \leq x \leq 0\).

Step-by-Step Solution

Answer

The third-order Maclaurin polynomial is \(1 + \frac{3}{2}x + \frac{3}{8}x^2 - \frac{1}{16}x^3\) with bounded error \(|R_3(x)|\).

Step 1: Identify the Function and Its Derivatives

The function we need the Maclaurin polynomial for is \((1+x)^{3/2}\). A Maclaurin polynomial is a Taylor polynomial centered at 0, so we first identify the function and its derivatives at \(x = 0\). The first few derivatives are needed up to the third order.

Step 2: Compute Derivatives

Calculate the derivatives:1. \(f(x) = (1+x)^{3/2}\)2. \(f'(x) = \frac{3}{2}(1+x)^{1/2}\)3. \(f''(x) = \frac{3}{4}(1+x)^{-1/2}\)4. \(f'''(x) = -\frac{3}{8}(1+x)^{-3/2}\).

Step 3: Evaluate Derivatives at Origin

Evaluate each of these derivatives at \(x = 0\):1. \(f(0) = (1+0)^{3/2} = 1\)2. \(f'(0) = \frac{3}{2}(1+0)^{1/2} = \frac{3}{2}\)3. \(f''(0) = \frac{3}{4}(1+0)^{-1/2} = \frac{3}{4}\)4. \(f'''(0) = -\frac{3}{8}(1+0)^{-3/2} = -\frac{3}{8}\).

Step 4: Form the Maclaurin Polynomial

Using the derivatives evaluated at 0, form the third-order Maclaurin polynomial:\(P_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!}\).Substituting the values:\(P_3(x) = 1 + \frac{3}{2}x + \frac{3}{8}x^2 - \frac{1}{16}x^3\).

Step 5: Estimating the Error Bound

The error \(R_3(x)\) of a third-order polynomial approximation is given by the next term's contribution in the Taylor series:\[ R_3(x) \leq \frac{|f^{(4)}(c)|}{4!} |x|^4 \]where \(c\) is in the interval between 0 and \(x\). We first find the fourth derivative:\(f^{(4)}(x) = \frac{15}{16}(1+x)^{-5/2}\).The maximum value in the interval \(-0.1\leq x \leq 0\) is \(f^{(4)}(-0.1) = \frac{15}{16}(0.9)^{-5/2}\). Calculate this numerical value, then plug into the error bound formula to find the maximum possible error.

Key Concepts

Taylor Series and Maclaurin Polynomials

Taylor series provide a way to represent functions as infinite sums of terms calculated from derivatives at a specific point. Each term in the series incorporates information about the value and rate of change of the function at that point. For example, the Taylor series of a function around a point 'a' is given by:

• \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots \]

A Maclaurin series is a special case of the Taylor series, centered at 0. This means that the formula simplifies because each term is evaluated at 0, resulting in:

• \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots \]

These series provide polynomial approximations of functions, helpful in calculating function values when they're difficult to compute exactly. As observed in the exercise, a third-order Maclaurin polynomial captures the primary behavior of the function \((1+x)^{3/2}\) with terms up to \(x^3\).

Derivative Calculation

Derivatives are a cornerstone of calculus representing the rate of change of a function. When it comes to building Taylor or Maclaurin polynomials, knowing how to calculate derivatives effectively is essential.
To construct the Maclaurin polynomial for \((1+x)^{3/2}\), the derivatives up to the third order are needed. Here's a quick overview of these calculations:

• First Derivative: The derivative of our function \(f(x) = (1+x)^{3/2}\) is \(f'(x) = \frac{3}{2}(1+x)^{1/2}\), indicating the initial rate of change.
• Second Derivative: The second derivative \(f''(x) = \frac{3}{4}(1+x)^{-1/2}\) shows how the rate of change itself is changing.
• Third Derivative: The third derivative \(f'''(x) = -\frac{3}{8}(1+x)^{-3/2}\) offers further insight into the behavior of the function.

Evaluating these derivatives at \(x = 0\) provides the coefficients of the terms in the Maclaurin polynomial. It's always crucial to compute derivatives accurately to ensure the reliability of your polynomial approximation. These calculations enable mathematicians and scientists to derive precise models of physical phenomena and theoretical problems.

Error Estimation in Taylor Series

Whenever we approximate a function using a Taylor or Maclaurin series, we introduce an error, known as the remainder or error term. This error quantifies how closely the polynomial approximation matches the actual function.
For a third-order polynomial, the error is represented by the next term in the series, ensuring we gauge how precise our approximation is. The formula for the error term in our scenario is:

• \[ R_3(x) \leq \frac{|f^{(4)}(c)|}{4!} |x|^4 \]

Here, \(f^{(4)}(c)\) is the fourth derivative evaluated at some point \(c\) between 0 and \(x\). The choice of \(c\), specifically the one that maximizes the derivative, provides a bound on the error.
In the exercise, it is crucial to evaluate \(f^{(4)}(x)\) within the interval \(-0.1 \leq x \leq 0\) to ensure error estimation remains valid. In practical terms, calculating such an error allows one to understand the potential deviation between a theoretical approximation and actual observations. Knowing and bounding this error is essential for confidence in mathematical modeling and practical implementations.