Problem 17 - Single Variable Calculus

Question

Use the power series representations of functions established in this section to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=x e^{-x}, \quad c=0\)

Step-by-Step Solution

Answer

The Taylor series of the function \(f(x)=xe^{-x}\) centered at \(c=0\) is given by: \(f(x) = 0 + x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + \cdots\) Using the ratio test, we determine that the radius of convergence is \(R=\infty\).

Step 1: Calculate the derivatives of \(f(x)\) with respect to \(x\)

Start by finding the first few derivatives of \(f(x) = xe^{-x}\) with respect to \(x\): \(f(x) = xe^{-x}\) \(f'(x) = e^{-x} - xe^{-x}\) \(f''(x) = -e^{-x} + 2e^{-x}x - e^{-x}x^2\) \(f^{(3)}(x) = e^{-x} - 6e^{-x}x + 6e^{-x}x^2 - e^{-x}x^3\)

Step 2: Evaluate the derivatives at \(c=0\)

Now, evaluate each of these derivatives at \(c=0\): \(f(0) = 0\cdot e^{0}=0\) \(f'(0) = e^{0} - 0\cdot e^{0}=1\) \(f''(0) = -e^{0} + 2e^{0}\cdot 0 - e^{0}\cdot 0^2= -1\) \(f^{(3)}(0) = e^{0} - 6e^{0}\cdot 0 + 6e^{0}\cdot 0^2 - e^{0}\cdot 0^3=1\)

Step 3: Write the Taylor series representation

Now we can write out the Taylor series representation based on the coefficients we found: \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\) \(f(x) = 0 + x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + \cdots\)

Step 4: Determine the radius of convergence

To find the radius of convergence, we will use the ratio test: \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n}\), where \(a_n = \frac{f^{(n)}(0)}{n!}\). For our particular case, this simplifies to: \(\lim_{n \to \infty} \frac{(-1)^{n+1}\frac{1}{(n+1)!}}{(-1)^n\frac{1}{n!}}\) Now, we cancel out the \((-1)^n\) terms and simplify: \(\lim_{n \to \infty} \frac{n!}{(n+1)!}\) \(\lim_{n \to \infty} \frac{1}{n+1}\) The limit goes to 0 as \(n\) goes to infinity, indicating that this power series converges for all values of \(x\). Therefore, the radius of convergence is \(R=\infty\).

Key Concepts

Radius of Convergence

Understanding the radius of convergence is essential in analyzing the behavior of power series. Imagine a power series as an infinite sum of functions, each with coefficients dependent on the terms of an underlying sequence. The radius of convergence represents the sphere within which this series converges, beyond which it may diverge or behave unpredictably.

The radius of convergence can be determined by various tests, of which the ratio test is particularly common. For a given power series \(\sum_{n=0}^{\text{\infty}} a_n(x - c)^n\), where \(a_n\) signifies the coefficients and \(c\) the center of the series, apply the ratio test as follows:

• Calculate \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
• If this limit is \(L\) and \(L < 1\), the series converges absolutely within a radius \(R = 1/L\).
• If \(L = 0\), the series converges for all \(x\), indicating an infinite radius of convergence, as in our example with \(f(x) = xe^{-x}\).
• If \(L > 1\) or \(L = \infty\), the series has no convergence within a nonzero radius around \(c\).

In our exercise, applying the ratio test simplifies to assessing the limit of \(\lim_{n \to \infty} 1/(n+1)\), which is 0, revealing that the Taylor series of \(f(x) = xe^{-x}\) converges for all \(x\), hence \(R = \infty\).

Power Series Representations

A power series representation is a form of expressing a function as an infinite sum of terms, each raised to successive power and multiplied by coefficients. These representations are crucial when dealing with complex functions as they can be used for approximation or to understand the behavior of functions over intervals.

The general form of a power series about a point \(c\) is given by \(\sum_{n=0}^{\infty} a_n(x - c)^n\) where \(a_n\) are the coefficients that tailor the series to the function it represents. To obtain these coefficients for a Taylor series, one method involves calculating the derivatives of the function at the center \(c\) and plugging them into the formula \(a_n = \frac{f^{(n)}(c)}{n!}\).
For the exercise at hand, we are provided with \(f(x) = xe^{-x}\) and asked to find its Taylor series around \(c = 0\). The coefficients obtained from the derivatives evaluated at zero are used in the series representation \(f(x) = \sum_{n=0}^{\text{\infty}} \frac{f^{(n)}(0)}{n!}x^n\), which yields the specific power series for the given function.

Derivatives in Calculus

Derivatives are a foundational concept in calculus, representing the rate at which a function changes at any given point. They are the building blocks for understanding motion, change, and a variety of quantitative relationships in physical and social sciences.

The derivative of a function \(f\) at a certain point \(x\) is denoted as \(f'(x)\) and physically represents the slope of the tangent line at that point on the function's graph. Higher-order derivatives, such as the second derivative \(f''(x)\), provide further insight into the function's behavior, such as concavity and inflection points.
In practice, to find higher-order derivatives for a Taylor series, we calculate \(f^{(n)}(x)\) successively. For \(f(x) = xe^{-x}\), we begin with the first derivative \(f'(x)\), and continue the process to obtain \(f''(x)\), \(f^{(3)}(x)\), and so on. These derivatives are then evaluated at the center point \(c\), in this case, zero, to find the series' coefficients. As illustrated in our example, finding the derivatives and evaluating them at \(c=0\) provides the necessary components to compose the Taylor series.