Problem 69 - Calculus Early Transcendentals

Question

Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-x}$$

Step-by-Step Solution

Answer

Answer: The power series representation of the function \(f(x) = e^{-x}\) is \(\sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k!}\), and its interval of convergence is \(-\infty < x < \infty\).

Step 1: Recall the power series of the exponential function

The power series for the exponential function \(e^x\) is given by: $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}, \quad \text{for }-\infty < x < \infty$$

Step 2: Substitute \(-x\) into the power series of \(e^x\)

To find the power series for the function \(f(x) = e^{-x}\), we substitute \(-x\) into the power series of \(e^x\). $$f(x) = e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!}$$

Step 3: Simplify the power series for \(f(x)\)

Observe that \((-x)^k = (-1)^k x^k\). Thus, we can rewrite the power series for \(f(x)\) as: $$f(x) = e^{-x} = \sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k!}$$

Step 4: Identify the interval of convergence

The original interval of convergence for the \(e^x\) is \(-\infty < x < \infty\). Since the power series for \(f(x)\) is derived from the power series for \(e^x\), the interval of convergence should remain unchanged. Therefore, the power series for the function \(f(x) = e^{-x}\) is: $$f(x) = e^{-x} = \sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k!}$$ and the interval of convergence for the resulting series is \(-\infty < x < \infty\).

Key Concepts

Exponential Function

The exponential function, denoted as \( e^x \), is one of the most fundamental functions in mathematics. It's especially important because of its unique property of being its own derivative. This means that no matter how many times you differentiate \( e^x \), you will always get \( e^x \) again. The power series representation of the exponential function is incredibly handy, especially when you need to calculate values or solve problems involving \( e^x \). The power series for \( e^x \) is expressed as:

• \( e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \)

This series is an infinite sum where each term involves powers of \( x \) divided by the factorial of the term index \( k \). Factorials \( k! \) grow very rapidly, which means the terms of the series get smaller very quickly, making the series converge nicely.

Interval of Convergence

The interval of convergence is a crucial part of understanding power series. It tells us the range of values for \( x \), where the series converges to a definite value. For the exponential power series of \( e^x \), the interval of convergence is quite broad:

• \(-\infty < x < \infty\)

This means that for any real number \( x \), the series will converge. The interval of convergence is vital as it assures us that the series representation is valid across a wide range. This property is particularly useful when extending these concepts to other functions, like \( f(x) = e^{-x} \), because the interval remains the same.

Substitution

Substituting variables within a power series allows us to derive new series for related functions. In this exercise, we substitute \(-x\) into the power series for \( e^x \) to find the series for \( f(x) = e^{-x} \).Here's how the substitution works:

• Replace every instance of \( x \) in the original power series \( \sum_{k=0}^{\infty} \frac{x^k}{k!} \) with \(-x\).

This yields:

• \( f(x) = e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} \)

Substitution is a powerful tool that lets us build series for more complex functions based on simpler ones. It retains the characteristics of the original function, like the interval of convergence.

Simplification

Simplifying power series is often necessary to make them more practical for computation and analysis. In the given expression for \( f(x) = e^{-x} \), simplification involves handling the term \((-x)^k\). This can be done by noting:

• \((-x)^k = (-1)^k x^k\)

Thus, each term in the series can be written to separate the negative sign:

• \(f(x) = e^{-x} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!}\)

This transformation helps to visualize the series as alternating terms between positive and negative, dictated by \((-1)^k\). Simplification refines the power series for use in calculations and theoretical discussions.