Problem 87 - A Graphical Approach to College Algebra

Question

Solve each problem. The series $$e^{a} \approx 1+a+\frac{a^{2}}{2 !}+\frac{a^{3}}{3 !}+\cdots+\frac{a^{n}}{n !}$$ where $$n !=1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot n$$. can be used to estimate the value of \(e^{a}\) for any real number \(a\). Use the first eight terms of this series to approximate each expression. Compare this estimate with the actual value. Give values to six decimal places. (a) \(e\) (b) \(e^{-1}\) (c) \(\sqrt{e}\)

Step-by-Step Solution

Answer

Approximately \( e = 2.718254 \), \( e^{-1} = 0.367881 \), \( \sqrt{e} = 1.648721 \); close to known values.

Step 1: Understanding the Series

The series given is the Taylor series expansion of the exponential function \( e^a \) around \( a = 0 \). It can be used to approximate the value of \( e^a \) by evaluating the sum of the first few terms of the series.

Step 2: Approximating \( e \)

Set \( a = 1 \). Compute the approximate value of \( e^1 = e \) using the first eight terms of the series:\[1 + 1 + \frac{1^2}{2!} + \frac{1^3}{3!} + \frac{1^4}{4!} + \frac{1^5}{5!} + \frac{1^6}{6!} + \frac{1^7}{7!} = 1 + 1 + 0.5 + 0.166667 + 0.041667 + 0.008333 + 0.001389 + 0.000198 = 2.718254\]

Step 3: Approximating \( e^{-1} \)

Set \( a = -1 \). Compute the approximate value of \( e^{-1} \) using the first eight terms:\[1 - 1 + \frac{(-1)^2}{2!} - \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} - \frac{(-1)^5}{5!} + \frac{(-1)^6}{6!} - \frac{(-1)^7}{7!} = 1 - 1 + 0.5 - 0.166667 + 0.041667 - 0.008333 + 0.001389 - 0.000198 = 0.367881\]

Step 4: Approximating \( \sqrt{e} \)

Set \( a = 0.5 \). Compute the approximate value of \( e^{0.5} = \sqrt{e} \) using the first eight terms:\[1 + 0.5 + \frac{0.5^2}{2!} + \frac{0.5^3}{3!} + \frac{0.5^4}{4!} + \frac{0.5^5}{5!} + \frac{0.5^6}{6!} + \frac{0.5^7}{7!} = 1 + 0.5 + 0.125 + 0.020833 + 0.002604 + 0.000325 + 0.000027 + 0.000002 = 1.648721\]

Step 5: Comparing with Actual Values

Compute the actual values (rounded to six decimal places):- \( e = 2.718282 \)- \( e^{-1} = 0.367879 \)- \( \sqrt{e} = 1.648721 \).Compare the approximations:- Approximated \( e = 2.718254 \)- Approximated \( e^{-1} = 0.367881 \)- Approximated \( \sqrt{e} = 1.648721 \).

Key Concepts

Exponential Function

An exponential function is a mathematical expression in the form of \( e^x \), where \( e \) is a constant approximately equal to 2.71828. The exponential function has significant applications in various fields such as mathematics, finance, and natural sciences. It describes growth and decay processes, like population growth or radioactive decay. In our context, the exponential function is used to model continuous growth at a constant rate. The function is unique because its derivative is equal to itself, making it very useful in calculus. Moreover, the exponential function is closely linked with the concept of natural logarithms, which serve as its inverse.

Approximation

Approximation is the process of finding values that are close to the exact answer. When dealing with complex functions or calculations, such genuine representations are unattainable or unnecessary. Therefore, using approximations can simplify and facilitate calculations. In mathematics, particularly when dealing with functions like \( e^a \) from a Taylor series, approximations become extremely handy.

The purpose of using approximations in the Taylor series is to simplify the computation. Instead of evaluating an infinite number of terms to get the exact value, we use a finite number, which gives a value very close to the actual one. In this exercise, the first eight terms of the series provide reasonably accurate approximations for \( e \), \( e^{-1} \), and \( \sqrt{e} \). This approach is common in areas of science and engineering where manageable accuracy is acceptable.

Factorial

The factorial function, denoted as \( n! \), is the product of all positive integers up to \( n \). For instance:

• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)

Factorials play a crucial role in the Taylor series as they serve as denominators for the terms, influencing their magnitude. When evaluating a Taylor series, the factorials are used to divide the powers of \( a \), producing smaller terms that help achieve a more precise approximation of a function's value. As \( n \) grows larger, \( n! \) grows rapidly, causing the contributions of terms in a Taylor series to diminish significantly after a few initial terms. This property is why only a few terms are often needed for practical approximations.

Math Series Evaluation

Math series evaluation involves breaking down complex functions into an infinite sum of simpler parts or terms. In a Taylor series, like the one used to approximate \( e^a \), each term adds a layer of precision, bringing the total sum closer to the desired function's value. The series is defined as

• \( 1 + a + \frac{a^2}{2!} + \frac{a^3}{3!} + \cdots \)

This approach allows mathematicians and scientists to evaluate functions that are difficult to compute exactly using simple arithmetic and algebraic operations.

Evaluating mathematical series requires careful consideration of how many terms to include. The more terms you include, the higher the precision but also the greater the computational effort. In the original exercise, using eight terms provides a balance between computational feasibility and accuracy. Students should appreciate the underlying beauty and efficiency in using such a powerful mathematical tool to closely approximate complex expressions.