Problem 47 - Calculus: Early Transcendentals

Question

Find the Maclaurin series of \( f \) (by any method) and its radius of convergence. Graph \( f \) and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and \( f? \) \( f(x) = xe^{-x} \)

Step-by-Step Solution

Answer

The Maclaurin series for \( xe^{-x} \) is \( x - \frac{1}{2} x^2 + \frac{1}{6} x^3 + \cdots \), with an infinite radius of convergence. Graph shows Taylor polynomials approximate \( f \) closely near 0.

Step 1: Find the Derivatives

To find the Maclaurin series, we need the derivatives of the function at 0. Start with the function:\[ f(x) = xe^{-x} \]Calculate derivatives:- First derivative: \( f'(x) = e^{-x} - xe^{-x} \)- Second derivative: \( f''(x) = -e^{-x} + 2xe^{-x} \)- Third derivative: \( f'''(x) = e^{-x} - 3xe^{-x} \)Evaluate these derivatives at \( x = 0 \).

Step 2: Evaluate at Zero

Evaluate the derivatives at \( x = 0 \).- \( f(0) = 0 \cdot e^{0} = 0 \)- \( f'(0) = e^{0} - 0 = 1 \)- \( f''(0) = -e^{0} + 0 = -1 \)- \( f'''(0) = e^{0} - 0 = 1 \)These will be used in the Maclaurin series formula.

Step 3: Write the Maclaurin Series

The Maclaurin series is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^n(0)}{n!} x^n \]Using the values:- \( f(0) = 0 \)- \( f'(0) = 1 \)- \( f''(0) = -1 \)- \( f'''(0) = 1 \) Write the series:\[ f(x) = x - \frac{1}{2} x^2 + \frac{1}{6} x^3 + \cdots \]

Step 4: Determine the Radius of Convergence

The radius of convergence can be found using the ratio test on the series:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = R \]Here, all coefficients are dictated by the factorial in the denominator:\[ \lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n} \right| = \lim_{n \to \infty} \frac{|x|}{n+1} = 0 \]The series converges for all \( x \), so the radius of convergence is \( \infty \).

Step 5: Graph the Functions

Graph \( f(x) = xe^{-x} \) and the first few polynomials:- Zeroth: \( P_0(x) = 0 \)- First: \( P_1(x) = x \)- Second: \( P_2(x) = x - \frac{1}{2} x^2 \)- Third: \( P_3(x) = x - \frac{1}{2} x^2 + \frac{1}{6} x^3 \)Use a graphing tool to visualize. Notice that as we include more terms, the polynomial approximates more of the curve.

Step 6: Analyze the Relationship

Notice how the approximations (Taylor polynomials) become more accurate as more terms are included. For smaller values of \( x \), fewer terms are usually needed for a good approximation of \( f(x) = xe^{-x} \). This is because the power series is centered at 0, aligning with the true function near this point.

Key Concepts

Radius of Convergence

The radius of convergence is an essential concept when dealing with power series, like the Maclaurin series. It tells us the range of values, or 'radius', where the series will converge to the actual function. When we use the power series formula for a function, it may not necessarily converge for every value of the variable. That's why determining the radius of convergence is crucial. In this exercise, the radius is found using the ratio test, which involves calculating the limit of the ratio of successive terms of the series. Specifically, this is expressed as:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]For the function \(f(x) = xe^{-x}\), the series converges for all real numbers. This results in a radius of convergence of \(\infty\), meaning the series is valid and converges everywhere along the number line.

Taylor Polynomial

A Taylor polynomial is a polynomial used to approximate a function. By assuming the polynomial closely resembles the function near a point, we sum the derivatives of the function at that particular point to it. The Maclaurin series is a special case of the Taylor series when this point is zero. In the exercise, we're looking at the Maclaurin series for \(f(x) = xe^{-x}\). To find the series, you evaluate the derivatives of the function at \(x = 0\) and use those to construct the polynomial.

• The zeroth polynomial: \(P_0(x) = 0\)
• The first polynomial: \(P_1(x) = x\)
• The second polynomial: \(P_2(x) = x - \frac{1}{2}x^2\)
• The third polynomial: \(P_3(x) = x - \frac{1}{2}x^2 + \frac{1}{6}x^3\)

These polynomials successively approximate the function more accurately as more terms are included. This is particularly useful in graphing since adding more terms usually results in a closer fit to the actual function \(f(x)\).

Derivatives

Derivatives play a crucial role in constructing the Taylor and Maclaurin series. They measure how a function changes as its input changes, essentially describing the slope at any given point.For the function \(f(x) = xe^{-x}\), you'll need to find its derivatives and evaluate them at \(x=0\) to build the series:

• First derivative: \( f'(x) = e^{-x} - xe^{-x} \). Evaluated, \(f'(0) = 1\).
• Second derivative: \( f''(x) = -e^{-x} + 2xe^{-x} \). Evaluated, \(f''(0) = -1\).
• Third derivative: \( f'''(x) = e^{-x} - 3xe^{-x} \). Evaluated, \(f'''(0) = 1\).

These values are critical for constructing each polynomial in the series. They dictate the coefficients of the Taylor polynomials and therefore, how closely the polynomial aligns with the function at the center point. The more derivatives you include, the closer the approximation will be over a wider interval.

Exponential Function

The exponential function, often expressed as \(e^x\), is a fundamental mathematical function with unique properties. It grows rapidly and is one of the few functions that equals its own derivative. In the context of our problem, the function \(f(x) = xe^{-x}\) combines a linear component \((x)\) with the decaying exponential \((e^{-x})\). This mixture results in a function that rises initially alongside \(x\), before being overtaken by the exponential decay part, causing it to turn and tail off.This distinctive shape makes the series expansion interesting, as lower-degree polynomials capture the initial linear rise, while higher-degree terms reflect the later decline.By understanding how the exponential component impacts the function, students can better appreciate how the series approximations work. The exponential part ensures convergence across all real numbers, which explains why x-values extending into positive and negative infinity can be accounted for.