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) Not applicable (series diverges for x ≥ 1)

Step 1: Understand the function and Maclaurin series

The Maclaurin series is a Taylor series expansion of a function about 0. For the function \( f(x) = \frac{1}{1-x} \), the Maclaurin series is given by \( 1 + x + x^2 + x^3 + \cdots \) which can be generalized to a polynomial of order \( n \) as \( T_n(x) = 1 + x + x^2 + \cdots + x^n \).

Step 2: Find the 4th order Maclaurin polynomial

Using the general form from Step 1, the 4th order Maclaurin polynomial for \( f(x) = \frac{1}{1-x} \) is:\[ T_4(x) = 1 + x + x^2 + x^3 + x^4 \]

Step 3: Approximate f(0.1) using T_4(x)

Substitute \( x = 0.1 \) into the polynomial:\[ T_4(0.1) = 1 + 0.1 + 0.1^2 + 0.1^3 + 0.1^4 \]Calculate each term:- \( 0.1^2 = 0.01 \)- \( 0.1^3 = 0.001 \)- \( 0.1^4 = 0.0001 \)Thus:\[ T_4(0.1) = 1 + 0.1 + 0.01 + 0.001 + 0.0001 = 1.1111 \]

Step 4: Approximate f(0.5) using T_4(x)

Substitute \( x = 0.5 \) into the polynomial:\[ T_4(0.5) = 1 + 0.5 + 0.5^2 + 0.5^3 + 0.5^4 \]Calculate each term:- \( 0.5^2 = 0.25 \)- \( 0.5^3 = 0.125 \)- \( 0.5^4 = 0.0625 \)Thus:\[ T_4(0.5) = 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 1.9375 \]

Step 5: Approximate f(0.9) using T_4(x)

Substitute \( x = 0.9 \) into the polynomial:\[ T_4(0.9) = 1 + 0.9 + 0.9^2 + 0.9^3 + 0.9^4 \]Calculate each term:- \( 0.9^2 = 0.81 \)- \( 0.9^3 = 0.729 \)- \( 0.9^4 = 0.6561 \)Thus:\[ T_4(0.9) = 1 + 0.9 + 0.81 + 0.729 + 0.6561 = 4.0951 \]

Step 6: Understand shortcoming for approximating f(2)

The series \( 1 + x + x^2 + x^3 + \cdots \) converges for \( |x| < 1 \). Therefore, using this series to approximate \( f(2) \) results in divergence, as \( 2 > 1 \). The approximation is incorrect, and the polynomial doesn't approximate \( f(2) \) effectively.

Key Concepts

Polynomials

A polynomial is a mathematical expression made up of variables, coefficients, and exponents, combined using addition, subtraction, and multiplication. Polynomials are like algebraic recipes that can model a variety of behaviors. For instance:

• The function \(f(x) = 1 + x + x^2 + x^3\) is a polynomial.
• The expressions extend to any order (or degree), denoted by the highest power of \(x\).

The Maclaurin polynomial, which is a specific type of polynomial, approximates functions near zero. When developing the Maclaurin polynomial for \(f(x) = \frac{1}{1-x}\), as seen above, it becomes \(1 + x + x^2 + \ldots + x^n\). This showcases simplicity, a staple of polynomials.

Taylor Series

The Taylor series is a powerful tool in mathematics that represents functions as infinite sums of terms. Each term in the series adds more detail to the approximation of the function. The Maclaurin series is a special case of the Taylor series centered at zero. To illustrate:

• For \(f(x) = \frac{1}{1-x}\), the Maclaurin series expansion starts as \(1 + x + x^2 + \ldots\).
• This series can be manipulated to different orders depending on the desired accuracy, which, for our exercise, is \(n=4\) resulting in \(T_4(x) = 1 + x + x^2 + x^3 + x^4\).

Each additional term in the series function provides a refined approximation around zero, making it versatile for functions that are otherwise difficult to compute directly.

Convergence

Convergence is a crucial concept when working with series, such as the Maclaurin or Taylor series. It refers to whether the sum of the series approaches a definite value as more terms are added. Consider these key points:

• For a series like \(1 + x + x^2 + \ldots\), convergence occurs when \(|x| < 1\).
• Within this range, adding terms helps the polynomial approach the function’s actual value.
• Outside of this range, such as \(f(2)\), the series diverges and fails to produce a valid approximation.

In context, while \(T_4(x)\) performs well for values like \(f(0.1)\), \(f(0.5)\), and \(f(0.9)\), it struggles beyond \(x = 1\), highlighting both the power and limitation of convergence.