Problem 138 - Precalculus with Limits

Question

In Exercises 135-138, find the first five terms of the sequence. \( a_n = \dfrac{(-1)^n x^{2n + 1}}{(2n + 1)!} \)

Step-by-Step Solution

Answer

The first five terms of the sequence are \(a_1 = -\dfrac{x^3}{3!}\), \(a_2 = \dfrac{x^5}{5!}\), \(a_3 = -\dfrac{x^7}{7!}\), \(a_4 = \dfrac{x^9}{9!}\), \(a_5 = -\dfrac{x^{11}}{11!}\) respectively.

Step 1: Find the first term (\(a_1\))

Set \(n=1\) in the given equation, this gives: \(a_1 = \dfrac{(-1)^1 x^{2*1 + 1}}{(2*1 + 1)!} = -\dfrac{x^3}{3!}\)

Step 2: Find the second term (\(a_2\))

Set \(n=2\) in the given equation, which gives: \(a_2 = \dfrac{(-1)^2 x^{2*2 + 1}}{(2*2 + 1)!} = \dfrac{x^5}{5!}\)

Step 3: Find the third term (\(a_3\))

Set \(n=3\) in the given equation, which gives: \(a_3 = \dfrac{(-1)^3 x^{2*3 + 1}}{(2*3 + 1)!} = -\dfrac{x^7}{7!}\)

Step 4: Find the fourth term (\(a_4\))

Set \(n=4\) in the given equation, which gives: \(a_4 = \dfrac{(-1)^4 x^{2*4 + 1}}{(2*4 + 1)!} = \dfrac{x^9}{9!}\)

Step 5: Find the fifth term (\(a_5\))

Set \(n=5\) in the given equation, which gives: \(a_5 = \dfrac{(-1)^5 x^{2*5 + 1}}{(2*5 + 1)!} = -\dfrac{x^{11}}{11!}\)

Key Concepts

Series Expansion

A series expansion is a powerful technique in mathematics used to represent complex functions as infinite sums of simpler terms. It allows us to approximate functions to a desired level of accuracy, often simplifying problem-solving and analysis. These expansions are commonly seen in calculus, where functions are expressed as sums of polynomials.

• One of the most notable series expansion is the Taylor series, where a function is expanded around a specific point.
• The given sequence in the problem utilizes a specific pattern of series expansion involving powers of a variable and factorials in the denominator.
• This pattern is often seen in generating functions for trigonometric expressions, such as sine and cosine functions.

Series expansions are invaluable when dealing with problems in mathematical physics, engineering, and computational fields, where representing functions in a simplified form can drastically ease calculations and interpretations.

Factorial

The factorial of a number, denoted by \( n! \), is the product of all positive integers less than or equal to \( n \). It plays a crucial role in permutations, combinations, and series expansions. Factorials grow very rapidly with increasing values of \( n \).

• Factorials appear in series expansion as part of the denominator to normalize the terms, reducing their magnitude.
• For example, in the given sequence, terms like \(3!\) and \(5!\) appear in the denominators, emerging from the pattern \((2n + 1)!\).
• This division allows each term in a series to contribute more smoothly to the sum, especially in approximations or expansions.

Understanding factorials is essential in both basic and advanced mathematical problems, including binomial expansions, combinatorics, and solving differential equations.

Alternating Series

An alternating series is one where the signs of the terms alternate between positive and negative throughout the series. This pattern influences the convergence and behavior of the series. Alternating series often arise from functions that can be approximated or expressed as expansions.

• In the given sequence, the alternating sign is produced by \((-1)^n\), resulting in a swap between positive and negative terms depending on the value of \( n \).
• Alternating series can converge even when their non-alternating counterparts may diverge, due to the cancellation of terms with opposing signs.
• They are found frequently in expansions of trigonometric functions and are vital in creating accurate approximations of these functions.

Recognizing and understanding alternating series is fundamental to executing precise approximations and analyses in calculus and other mathematical fields.