Problem 108 - College Algebra with Modeling and Visualization

Question

The Natural Exponential Function The following series can be used to estimate the value of \(e^{a}\) for any real number a: $$e^{a} \approx 1+a+\frac{a^{2}}{2 !}+\frac{a^{3}}{3 !}+\dots+\frac{a^{n}}{n !}$$ where \(n !=1 \cdot 2 \cdot 3 \cdot 4 \cdot \cdots \cdot n\). Use the first eight terms of this series to approximate the given expression. Compare this estimate with the actual value. $$ e^{-1} $$

Step-by-Step Solution

Answer

The series approximation for \( e^{-1} \) is 0.3679, which matches its actual value.

Step 1: Identify the Series Terms

To approximate the value of \( e^{-1} \), we will use the first eight terms of the series:\[ e^a = 1 + a + \frac{a^2}{2!} + \frac{a^3}{3!} + \frac{a^4}{4!} + \cdots + \frac{a^n}{n!} \]For \( a = -1 \), the series becomes:\[ e^{-1} \approx 1 - 1 + \frac{(-1)^2}{2!} - \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} - \frac{(-1)^5}{5!} + \frac{(-1)^6}{6!} - \frac{(-1)^7}{7!} \]

Step 2: Calculate Each Term

Calculate each of the terms one by one:1. \( 1 = 1 \)2. \( -1 = -1 \)3. \( \frac{(-1)^2}{2!} = \frac{1}{2} = 0.5 \)4. \( \frac{(-1)^3}{3!} = \frac{-1}{6} = -0.1667 \)5. \( \frac{(-1)^4}{4!} = \frac{1}{24} = 0.0417 \)6. \( \frac{(-1)^5}{5!} = \frac{-1}{120} = -0.0083 \)7. \( \frac{(-1)^6}{6!} = \frac{1}{720} = 0.0014 \)8. \( \frac{(-1)^7}{7!} = \frac{-1}{5040} = -0.0002 \)

Step 3: Sum the Terms

Sum the calculated terms to find the approximation:\[1 - 1 + 0.5 - 0.1667 + 0.0417 - 0.0083 + 0.0014 - 0.0002 \]This simplifies to:\[0.3679\]

Step 4: Compare with Actual Value

The actual value of \( e^{-1} \) is approximately 0.3679. Comparing this with our approximation of 0.3679, they are very close, indicating the series provides a good approximation using the first eight terms.

Key Concepts

Exponential Series

An exponential series is a way to express an exponential function as a sum of terms. These terms are infinite and involve powers and factorials. For any number \( e^a \), the series is written as: \[ e^a = 1 + a + \frac{a^2}{2!} + \frac{a^3}{3!} + \frac{a^4}{4!} + \cdots + \frac{a^n}{n!} \] This series is especially useful when calculating \( e^a \) without the immediate availability of a calculator. - **Power of \(a\):** Every subsequent term increases the power of \(a\). For example, the next term in the series after \( a^3/3! \) is \( a^4/4! \). - **Factorials in Denominators:** The factorials are used to balance the growth of the powers, ensuring convergence of the series.

Taylor Series

The Taylor series is an important concept for approximating functions. It represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. For the exponential function \( e^a \), the Taylor series is centered at \( a = 0 \) or the origin, which can be written as: \[ e^a = \sum_{n=0}^{\infty} \frac{a^n}{n!} \] This is essentially the exponential series for \( e^a \), making it a key tool in calculus. - **Infinite Terms:** While Taylor series ideally have infinite terms, practical applications often use a finite number of terms for approximation. - **Derivatives Role:** Each term in the Taylor series involves a derivative of the function, which captures the function's behavior accurately around the expansion point.

Factorials

Factorials are at the heart of both exponential and Taylor series. They are denoted by an exclamation mark \( n! \), which reads as 'n factorial.' A factorial is the product of all positive integers up to \( n \). For instance: - \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) - \( 0! = 1 \) by definition. Factorials grow extremely fast as \( n \) increases, which plays a crucial role in ensuring the convergence of series like exponential series. - **Balancing Large Powers:** In series, factorials in the denominator help to balance out the large powers in the numerator. This helps to prevent divergence. - **Computation Simplicity:** Factorials simplify the computation of series terms, making processes like approximating functions much more manageable.