Problem 11 - Calculus: Early Transcendentals

Question

Find the Maclaurin series for \( f(x) \) using the definition of a Maclaurin series. [ Assume that \( f \) has a power series expansion. Do not show that \( R_n (x) \to 0. \)] Also find the associated radius of convergence. \( f(x) = (1 - x)^2 \)

Step-by-Step Solution

Answer

The Maclaurin series is \( 1 - 2x + x^2 \), and the radius of convergence is \( \infty \).

Step 1: Write the Definition of a Maclaurin Series

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \). This series expands the function around \( x=0 \). We will apply this to find the Maclaurin series for \( f(x) = (1-x)^2 \).

Step 2: Calculate the Derivatives of the Function

First, calculate the derivatives of \( f(x) = (1-x)^2 \):- \( f(x) = (1-x)^2 \)- The first derivative: \( f'(x) = -2(1-x) \)- The second derivative: \( f''(x) = 2 \)- All higher derivatives \( f^{(n)}(x) \) for \( n \geq 3 \) are zero.

Step 3: Evaluate the Derivatives at x=0

Substitute \( x=0 \) into each derivative:- \( f(0) = (1-0)^2 = 1 \)- \( f'(0) = -2(1-0) = -2 \)- \( f''(0) = 2 \)- For \( n \geq 3 \), \( f^{(n)}(0) = 0 \).

Step 4: Write the Maclaurin Series Expansion

Plug the derivatives found above into the Maclaurin series formula:- \( f(x) = 1 + \frac{-2}{1!}x + \frac{2}{2!}x^2 + \frac{0}{3!}x^3 + \cdots \)- Simplifying, the series becomes \( f(x) = 1 - 2x + x^2 \).

Step 5: Find the Radius of Convergence

The Maclaurin series for a polynomial is the polynomial itself. Thus, the power series converges for all \( x \). Therefore, the radius of convergence is \( R = \infty \).

Key Concepts

Power Series Expansion

The concept of power series expansion is foundational when dealing with functions. It allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For the Maclaurin series, especially, we center the series at zero.

• The general form of a power series expansion is \[ f(x) = \sum_{n=0}^{\infty} c_n(x-a)^n \]where \( c_n \) represents the coefficients computed from the derivatives of the function at point \( a \). If \( a = 0 \), it becomes a Maclaurin series.

This powerful tool allows us to approximate complex functions with a polynomial, which can be easier to work with, especially near the center point. This is particularly useful for functions that cannot be represented in closed form. By expanding \( f(x) = (1 - x)^2 \), we obtain the series \( 1 - 2x + x^2 \).

Derivatives

Derivatives are crucial for finding the coefficients of a power series. The Maclaurin series requires evaluating the function and its derivatives at \( x = 0 \). In this problem, we calculate:

• The function itself at zero: \( f(0) = 1 \).
• The first derivative: \( f'(x) = -2(1-x) \), giving \( f'(0) = -2 \).
• The second derivative: \( f''(x) = 2 \), resulting in \( f''(0) = 2 \).
• Higher-order derivatives: since this is a quadratic polynomial function, derivatives of order three and higher are zero.

These derivatives provide the coefficients for each term in the series, shaping the overall polynomial approximation. Understanding how to compute these derivatives is essential for the power series expansion process.

Radius of Convergence

The radius of convergence is an essential aspect of a power series. It defines the interval within which the series converges to function.

• For a Maclaurin series of polynomial functions, the radius of convergence is always infinite.

This means our approximation through the Maclaurin series is valid for any \( x \) in the real number line. Unlike many functions that have a finite radius of convergence, polynomial functions inherently have no such limit, simplifying their analysis and expansion.

Polynomial Functions

Polynomial functions are equations involving terms raised to a power. They have a straightforward representation in terms of their degree, which determines their behavior and complexity.

• For instance, \( f(x) = (1-x)^2 \) is a quadratic polynomial, meaning its degree is 2.
• Polynomial functions have derivatives of descending degree until they reach zero.

When expanding polynomial functions as a series, they are somewhat unique because their power series (like the Maclaurin series) tend to reproduce the polynomial exactly. Thus, they are excellent candidates for learning and practicing with power series expansion because their derivatives are finite and manageable, eventually leading to zero for higher orders.