Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the third-order Maclaurin polynomial for \((1+x)^{1 / 2}\) and bound the error \(R_{3}(x)\) for \(-0.5 \leq x \leq 0.5\).

Step-by-Step Solution

Answer

The third-order Maclaurin polynomial is \(1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3\) with an error bound of \(0.0098\) for \(-0.5 \leq x \leq 0.5\).

Step 1: Find the Function and its Derivatives

We start with the function \( f(x) = (1+x)^{1/2} \). To find the Maclaurin series, we need to compute the first few derivatives of \( f(x) \) and evaluate them at \( x = 0 \):- \( f(x) = (1+x)^{1/2} \)- \( f'(x) = \frac{1}{2}(1+x)^{-1/2} \)- \( f''(x) = -\frac{1}{4}(1+x)^{-3/2} \)- \( f'''(x) = \frac{3}{8}(1+x)^{-5/2} \) Evaluating at \( x = 0 \):- \( f(0) = 1 \)- \( f'(0) = \frac{1}{2} \)- \( f''(0) = -\frac{1}{4} \)- \( f'''(0) = \frac{3}{8} \).

Step 2: Construct the Maclaurin Polynomial

The Maclaurin series is given by:\[ P_n(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... \]For a third-order polynomial, plug in the values from Step 1:\[ P_3(x) = 1 + \frac{1}{2}x - \frac{1}{4\cdot2}x^2 + \frac{3}{8\cdot6}x^3 \]\[ P_3(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 \].

Step 3: Determine the Error Bound

The remainder (error) term for a Maclaurin series of order \( n \) is:\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \]where \( c \) is between 0 and \( x \). For the third-order polynomial, the fourth derivative is \( f^{(4)}(x) = -\frac{15}{16}(1+x)^{-7/2} \).The maximum of \( |f^{(4)}(x)| \) in \(-0.5 \leq x \leq 0.5\) will be \( f^{(4)}(-0.5) = -\frac{15}{16}(1-0.5)^{-7/2} = -\frac{60}{16} = 3.75 \).So the error is bounded by:\[ |R_3(x)| \leq \frac{3.75}{4!}|x|^4 \]\[ |R_3(x)| \leq \frac{3.75}{24}|x|^4 \]\[ |R_3(x)| \leq \frac{0.15625}{2}|x|^4 \].

Step 4: Evaluate the Error Bound

Evaluate the error bound specifically for \( x \) in \(-0.5 \leq x \leq 0.5 \):- The maximum value of \( |x| \) in the given range is 0.5.- Substitute into the error bound:\[ |R_3(x)| \leq 0.15625 \times (0.5)^4 \]\[ |R_3(x)| \leq 0.15625 \times \frac{1}{16} \]\[ |R_3(x)| \leq 0.009765625 \].

Key Concepts

Polynomial Approximation

A polynomial approximation aims to simplify complex functions using polynomials. These are easier to compute and analyze, especially for small values of variables. The Maclaurin series is one such polynomial approximation technique that's particularly useful when functions can be complex or transcendental.

The task of creating a polynomial approximation involves finding a function's Maclaurin series. For the function \( f(x) = (1+x)^{1/2} \), we compute a sequence of derivatives to determine the Maclaurin polynomial. These derivatives give us insights into the function's behavior at zero, which is the base point of a Maclaurin series.

For the third-order Maclaurin polynomial, we take the function and compute up to its third derivative. The polynomial then becomes a simple expression that approximates the original function near zero. In this case, the third-order polynomial is \( P_3(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 \). By using only a few terms from the series, we can approximate \( (1+x)^{1/2} \) quite closely near \( x = 0 \).

Remainder Term

The remainder term is crucial in understanding the accuracy of a polynomial approximation. It measures the difference between the actual function and its polynomial approximation.

In a Maclaurin series, the remainder term is often denoted as \( R_n(x) \). For a function approximated by a polynomial of order \( n \), \( R_n(x) \) represents the error due to ignoring higher-order terms beyond \( n \).

The remainder term is mathematically expressed using higher derivatives of the function evaluated at some point \( c \), where \( 0 < c < x \). Specifically, for a third-order Maclaurin polynomial, the expression for the remainder term is:

• \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \)

This means the accuracy of the polynomial hinges on this term, and with proper calculation, we can ensure the approximation is as precise as necessary for a given range of \( x \).

Derivative

Derivatives are the backbone of polynomial approximations like the Maclaurin series. In this context, they provide us with coefficients for each term of the series.

The process begins with finding derivatives of the function of interest. For \( f(x) = (1+x)^{1/2} \), derivatives are computed successively to find their values at \( x = 0 \).

• \( f'(x) = \frac{1}{2}(1+x)^{-1/2} \)
• \( f''(x) = -\frac{1}{4}(1+x)^{-3/2} \)
• \( f'''(x) = \frac{3}{8}(1+x)^{-5/2} \)

By evaluating these derivatives at zero, we obtain:

• \( f(0) = 1 \)
• \( f'(0) = \frac{1}{2} \)
• \( f''(0) = -\frac{1}{4} \)
• \( f'''(0) = \frac{3}{8} \)

These values form the coefficients of the polynomial, determining how the approximation behaves as \( x \) increases from zero.

Error Bound

The error bound provides a quantitative measure of the maximum possible error in a polynomial approximation. It ensures that users of the approximation understand its range of validity.

Calculating the error bound involves determining the maximum of the absolute value of the remainder term across the intended interval of \( x \). For our Maclaurin polynomial approximation, we need the error bound within \(-0.5 \leq x \leq 0.5\).

Using the fourth derivative of the original function, we identify that:

• \( f^{(4)}(x) = -\frac{15}{16}(1+x)^{-7/2} \)

To find the maximal \( |f^{(4)}(x)| \), we evaluate it at the endpoints of our interval. This guides us in calculating the error bound as:

• \( |R_3(x)| \leq \frac{3.75}{24}|x|^4 \)

Substituting \( |x| = 0.5 \), the bound becomes \( |R_3(x)| \leq 0.009765625 \). This boundary informs us that within the set interval, the third-order polynomial approximation will not deviate from the function by more than this calculated amount.