Problem 11 - Calculus: Early Transcendentals

Question

Find the \(3^{\text {rd }}\) degree Taylor polynomial for \(f(x)=\frac{1}{1-x}-1\) around \(a=0 .\) Explain why this approximation would not be useful for calculating \(f(5)\).

Step-by-Step Solution

Answer

The 3rd degree Taylor polynomial for \( f(x) \) is \( P_3(x) = x + x^2 + x^3 \). This approximation isn't useful for calculating \( f(5) \) because it's outside the convergence interval \(|x| < 1\).

Step 1: Understand the Function

The function given is \( f(x) = \frac{1}{1-x} - 1 \). This is a simplified form of the geometric series expansion, which we'll use to find the Taylor polynomial.

Step 2: Recognize Geometric Series

The series \( \frac{1}{1-x} \) can be expanded as a geometric series: \( 1 + x + x^2 + x^3 + \cdots \). By subtracting 1 from this series, we have: \( x + x^2 + x^3 + \cdots \).

Step 3: Identify Taylor Polynomial Terms

Since we want the 3rd degree Taylor polynomial, we need the terms up to \(x^3\) from the series for \( f(x) \). These terms are \( x, x^2, \text{ and } x^3 \).

Step 4: Write the Taylor Polynomial

The 3rd degree Taylor polynomial for \( f(x) \) is \( P_3(x) = x + x^2 + x^3 \). This polynomial includes terms up to \( x^3 \).

Step 5: Consider Convergence Range

The series expansion \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \) converges for \( |x|<1 \). This means the Taylor polynomial approximation is accurate only for values within this range.

Step 6: Evaluate Approximation for f(5)

The value \( f(5) \) is outside the convergence interval \(|x| < 1\). Therefore, using this Taylor polynomial to approximate \( f(5) \) would not be valid or useful.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. In the context of the problem, the function \( f(x) = \frac{1}{1-x} - 1 \) can be expanded into a series using this idea. The geometric series expansion of \( \frac{1}{1-x} \) takes the form:

• \( 1 + x + x^2 + x^3 + \cdots \)

By subtracting 1, we simplify it to \( x + x^2 + x^3 + \cdots \).
This is crucial in identifying the Taylor polynomial terms as the problem asks to find the Taylor series centered at a particular point. The geometric nature of this series provides a systematic way to understand and compute each term of the polynomial approximation.These series are particularly useful in mathematical analysis as they offer a simple form to represent more complex functions, thus making calculation easier.

Degree of Polynomial

The degree of a polynomial refers to the highest power of the variable in the polynomial expression. In this example, we seek the 3rd degree Taylor polynomial.The problem specifies the function:\( f(x) = \frac{1}{1-x} - 1 \).From the geometric series expansion, the terms contributing to the Taylor polynomial are \( x, x^2,\) and\( x^3\).The 3rd degree Taylor polynomial, thus, takes the form:

• \( P_3(x) = x + x^2 + x^3 \)

When we talk about the 'degree of a polynomial,' it directly affects the polynomial's ability to approximate the original function over a certain range. The higher the degree, the better the approximation, especially close to the center of expansion, which in this problem is \( a = 0 \).This polynomial accurately reflects the behavior of the function up to the cubic term, making it a powerful tool for approximation in the right conditions.

Convergence Radius

Understanding the concept of convergence is vital when working with series, as it defines the interval over which a series accurately represents a function.For a Taylor series, the convergence radius determines the interval within which the approximation is valid. In the example problem, the geometric series expansion of \( \frac{1}{1-x} \)converges when \(|x| < 1 \).For \( f(x) = \frac{1}{1-x} - 1 \), the same convergence constraints apply. This means that the Taylor polynomial \( P_3(x) = x + x^2 + x^3 \)accurately represents \( f(x) \)only within \(|x| < 1 \).This convergence radius highlights why, for \( f(5) \), the approximation would fail; \( x = 5 \)falls outside \(|x| < 1 \).When considering the radius of convergence,

• Always evaluate if the value of interest falls within the range.
• Use alternative methods for approximation if outside the convergence radius.

Understanding this concept helps ensure that mathematical models and approximations are both valid and reliable.