Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the third-order Maclaurin polynomial for $$ (1+x)^{-1 / 2} $$ Ind bound the error \(R_{3}(x)\) if \(-0.05 \leq x \leq 0.05\).

Step-by-Step Solution

Answer

The third-order Maclaurin polynomial is \(P_3(x) = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3\). The error \(R_3(x)\) is at most approximately \(2.6 \times 10^{-6}\) for \(-0.05 \leq x \leq 0.05\).

Step 1: Understand the Function and its Derivatives

The function is \((1+x)^{-1/2}\). We will need the first few derivatives to build the third-order Maclaurin polynomial. Calculate the derivatives: 1. \(f(x) = (1+x)^{-1/2}\) 2. \(f'(x) = -\frac{1}{2}(1+x)^{-3/2}\) 3. \(f''(x) = \frac{3}{4}(1+x)^{-5/2}\) 4. \(f'''(x) = -\frac{15}{8}(1+x)^{-7/2}\)

Step 2: Calculate Maclaurin Coefficients

Evaluate the derivatives at \(x = 0\):- \(f(0) = 1\)- \(f'(0) = -\frac{1}{2}\)- \(f''(0) = \frac{3}{4}\)- \(f'''(0) = -\frac{15}{8}\)

Step 3: Form the Third-Order Maclaurin Polynomial

The Maclaurin polynomial is given by: \[ P_3(x) = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 \] This polynomial is constructed using the function value and its derivatives calculated at \(x=0\), along with factorial terms for each order.

Step 4: Calculate the Remainder Error Term

The remainder term \(R_3(x)\) is given by:\[ R_3(x) = \frac{f^{(4)}(c)}{4!} x^4 \]where \(c\) is a number in the interval between 0 and \(x\). To bound the error, we need to evaluate the 4th derivative.

Step 5: Evaluate the Fourth Derivative

Compute the 4th derivative:- \(f^{(4)}(x) = \frac{105}{16}(1+x)^{-9/2}\)For \(-0.05 \leq x \leq 0.05\), the maximum value of \(|f^{(4)}(x)|\) happens at \(x = -0.05\), which yields the maximum value: \(|f^{(4)}(-0.05)| \approx \frac{105}{16}(1-0.05)^{-9/2}\).

Step 6: Estimate the Error

Use the maximum value of the fourth derivative and the range of \(x\):\[ |R_3(x)| \leq \frac{105}{16 \cdot 4!} \times (0.05)^4 \]Calculate this to find the upper bound of the error \(R_3(x)\).

Key Concepts

Error Estimation in Polynomials

When we work with polynomials to approximate functions, error estimation helps us understand how close our approximation is. In the case of the Maclaurin polynomial, the error term, often called the remainder, shows the difference between the actual function value and the polynomial approximation.

The remainder for a Maclaurin polynomial of third order, denoted as \(R_3(x)\), is calculated using the formula:

• \( R_3(x) = \frac{f^{(4)}(c)}{4!} x^4 \)

Here, \(f^{(4)}(c)\) is the fourth derivative evaluated at some point \(c\) between 0 and \(x\).

In the original problem, the task was to bound this error when \(-0.05 \leq x \leq 0.05\). To find the maximum error, we take the largest absolute value of the fourth derivative in the given interval.

This approach helps to determine how accurately the Maclaurin polynomial represents the function within the specified range. Error estimation is crucial because it provides confidence in the polynomial's precision, especially when these approximations are used in calculus problems to solve real-world scenarios.

Taylor Series

Taylor series is a powerful tool in calculus, allowing us to express functions as infinite sums of their derivatives at a single point. The Maclaurin series is a special case of the Taylor series centered at zero.

For a function \(f(x)\), the Maclaurin series expansion is given by:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) \)

This series allows us to approximate functions using polynomials, which are easier to compute.

In the exercise, finding the third-order Maclaurin polynomial involved computing derivatives up to the third order. These derivatives help form the polynomial, which is used as an approximation of the function \((1+x)^{-1/2}\) around \(x = 0\).

Using a finite number of terms in the Taylor series, like in this example, provides a polynomial approximation. Although exact only if infinite terms are used, such approximations are practical and sufficient for small \(x\), especially when error estimation is also considered.

Calculus Problems

Calculus problems often involve understanding the behavior of functions and their derivatives. Approximations like Taylor and Maclaurin series are central in solving such problems when exact solutions are intractable.

In this particular problem, the aim was to approximate \((1+x)^{-1/2}\) using a polynomial. This strategy simplifies complex expressions into manageable polynomials. These approximations are not just about simplification, though; they are powerful in predicting behavior near a point, particularly when coupled with error estimates.

When tackling calculus problems involving function approximation:

• Identify the function and order of the polynomial needed.
• Find required derivatives at the expansion point.
• Construct the polynomial and calculate possible error bounds.

These steps ensure the simplification retains an acceptable level of precision, making calculus problems more approachable and easier to solve.