Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Maclaurin polynomial of order \(n\) for \(f(x)=1 /(1-x)\). Then use it with \(n=4\) to approximate each of the following. (a) \(f(0.1)\) (b) \(f(0.5)\) (c) \(f(0.9)\) (d) \(f(2)\)

Step-by-Step Solution

Answer

(a) 1.1111, (b) 1.9375, (c) 4.0951, (d) 31.

Step 1: Define the Maclaurin Series Formula

The Maclaurin series for the function \( f(x) \) is given by the expression \( P_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots + \frac{f^n(0)}{n!}x^n \). This formula will be used to determine the polynomial for each required function to order \( n \).

Step 2: Derivatives of the Function

Calculate the derivatives of \( f(x) = \frac{1}{1-x} \) at \( x = 0 \):- \( f(x) = (1-x)^{-1} \), so \( f(0) = 1 \).- First derivative: \( f'(x) = (1-x)^{-2} \), so \( f'(0) = 1 \).- Second derivative: \( f''(x) = 2(1-x)^{-3} \), so \( f''(0) = 2 \).- Third derivative: \( f'''(x) = 6(1-x)^{-4} \), so \( f'''(0) = 6 \).- Fourth derivative: \( f^{(4)}(x) = 24(1-x)^{-5} \), so \( f^{(4)}(0) = 24 \).

Step 3: Construct the Maclaurin Polynomial

Substitute the derivatives into the Maclaurin series formula: \[ P_4(x) = 1 + x + \frac{2}{2}x^2 + \frac{6}{6}x^3 + \frac{24}{24}x^4 \]This simplifies to:\[ P_4(x) = 1 + x + x^2 + x^3 + x^4 \]

Step 4: Approximate f(0.1) using P_4

Plug \( x = 0.1 \) into \( P_4(x) \):\[ P_4(0.1) = 1 + 0.1 + 0.1^2 + 0.1^3 + 0.1^4 = 1 + 0.1 + 0.01 + 0.001 + 0.0001 = 1.1111 \]

Step 5: Approximate f(0.5) using P_4

Plug \( x = 0.5 \) into \( P_4(x) \):\[ P_4(0.5) = 1 + 0.5 + 0.5^2 + 0.5^3 + 0.5^4 = 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 1.9375 \]

Step 6: Approximate f(0.9) using P_4

Plug \( x = 0.9 \) into \( P_4(x) \):\[ P_4(0.9) = 1 + 0.9 + 0.9^2 + 0.9^3 + 0.9^4 = 1 + 0.9 + 0.81 + 0.729 + 0.6561 = 4.0951 \]

Step 7: Approximate f(2) using P_4

Plug \( x = 2 \) into \( P_4(x) \):\[ P_4(2) = 1 + 2 + 2^2 + 2^3 + 2^4 = 1 + 2 + 4 + 8 + 16 = 31 \]

Key Concepts

Polynomial Approximation

Polynomial approximation is a technique used in mathematics to approximate complex functions with polynomial expressions. A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. This method simplifies calculations and offers insights into the behavior of functions over certain intervals.
By using polynomial approximation:

• Complex functions become much easier to work with, especially for numerical computation.
• The approximating polynomial can mimic the behavior of the function at specific points or across intervals.

In our Maclaurin series example, we approximate the function \( f(x) = \frac{1}{1-x} \) using a polynomial of order 4. This approximation allows us to make predictions about the function's value at points like 0.1, 0.5, or even beyond using a simpler polynomial form.

Derivatives

Derivatives are fundamental tools in calculus, representing the rate at which a function changes. When computing a Maclaurin series, understanding how to find derivatives is crucial. It involves finding successive rates of change for the function at a specific point, usually at \( x = 0 \) for a Maclaurin series.
In the given exercise:

• The function \( f(x) = \frac{1}{1-x} \) was differentiated repeatedly to find the values of \( f(0) \), \( f'(0) \), and so forth.
• Each derivative measures how the output value of the function changes as \( x \) changes, with the first derivative being the most basic rate of change, and higher-order derivatives measuring more nuanced changes.

These derivatives are then used to form the coefficients of the polynomial in the Maclaurin series, giving us an approximate formula to work with.

Taylor Series

The Taylor series is a robust mathematical tool used to represent functions as infinite sums of terms calculated from the values of their derivatives at a single point. The Maclaurin series is a special case of the Taylor series, specifically centered at \( x = 0 \).
The general Taylor series formula is:

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

In the context of our problem:

• The Maclaurin series simplifies to using derivatives computed at \( x = 0 \), so the series takes the form \[ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \ldots \]

This makes it easier to compute values of the function at various points by considering a finite number of terms in the series. It’s a powerful method for tackling complex functions near the point of approximation.

Sequence and Series

Sequences and series are mathematical concepts that help us work with ordered lists of numbers and their sums, respectively. They provide a framework for analyzing how functions behave as their inputs progress or change.
Maclaurin series can be seen as a series - specifically, a series of terms derived from the derivatives of a function at a single point. For example, the series for \( f(x) \) starts with constant terms and adds on powers of \( x \) multiplied by the corresponding coefficients, which are the values of the derivatives divided by factorial.

• Each term in the series represents a part of the polynomial that approximates the original function.
• As more terms are added, the approximation becomes more refined, improving the closeness to the function over a larger interval.

Understanding sequences and series is vital for working with polynomial approximations and analyzing the convergence of series in calculus.