Problem 137 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The first five terms of the sequence are \( a_1 = \dfrac{-x^2}{2} \), \( a_2 = \dfrac{x^4}{4!} \), \( a_3 = \dfrac{-x^6}{6!} \), \( a_4 = \dfrac{x^8}{8!} \), and \( a_5 = \dfrac{-x^{10}}{10!} \).

Step 1. Determine the First Term

Plug in \( n = 1 \) into the sequence formula. It becomes \( a_1 = \dfrac{(-1)^1 x^{2*1}}{(2*1)!} = \dfrac{-x^2}{2} \).

Step 2. Determine the Second Term

Next, substitute \( n = 2 \) into the sequence formula. This yields \( a_2 = \dfrac{(-1)^2 x^{2*2}}{(2*2)!} = \dfrac{x^4}{4!} \).

Step 3. Determine the Third Term

For the third term, substitute \( n = 3 \) into the sequence formula. So, \( a_3 = \dfrac{(-1)^3 x^{2*3}}{(2*3)!} = \dfrac{-x^6}{6!} \).

Step 4. Determine the Fourth Term

Next, substitute \( n = 4 \) into the equation. This gives \( a_4 = \dfrac{(-1)^4 x^{2*4}}{(2*4)!} = \dfrac{x^8}{8!} \).

Step 5. Determine the Fifth Term

Finally, substitute \( n = 5 \) into the equation. This gives \( a_5 = \dfrac{(-1)^5 x^{2*5}}{(2*5)!} = \dfrac{-x^{10}}{10!} \).

Key Concepts

Mathematical Sequences

Mathematical sequences are ordered lists of numbers that follow a specific pattern or rule. In precalculus and other branches of mathematics, we study sequences to understand how numbers behave collectively, how they progress and what kind of patterns they form. Sequences can be finite, having a limited number of terms, or infinite, extending indefinitely.

For example, in the given exercise, the sequence is defined by an explicit formula: \( a_n = \dfrac{(-1)^n x^{2n}}{(2n)!} \). To find the terms of this sequence, we apply the formula for successive values of \( n \) (where \( n \) starts at 1). This type of sequence is particularly interesting because it includes both polynomial expressions and factorial notation - a combination that we encounter in advanced mathematical concepts such as power series and the Taylor series.

Factorial Notation

Factorial notation is a mathematical operation that multiplies a series of descending natural numbers and is signified by an exclamation mark (\(!\)). For any non-negative integer \( n \), the factorial is expressed as \( n! \) and is the product of all positive integers less than or equal to \( n \). So, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

In the given problem, the factorial notation appears in the denominators of each term in the sequence. As an example, for the second term, when we substitute \( n = 2 \), the factorial notation yields to \( (2 \times 2)! = 4! \), which is \( 4 \times 3 \times 2 \times 1 = 24 \). Factorials grow extremely fast, which influences the convergence properties of sequences and series that contain them. Knowing how to work with factorial notation is an essential skill for studying mathematical sequences and calculus.

Sequence Convergence

Sequence convergence is a fundamental concept in calculus. A sequence is said to converge if it approaches a specific value as \( n \) approaches infinity. Essentially, the terms in the sequence become closer and closer to a single value, called the limit, as the sequence progresses. If a sequence does not approach a specific value, it's considered divergent.

Looking at our sequence \( a_n = \frac{(-1)^n x^{2n}}{(2n)!} \), it can be analyzed for convergence. One key aspect that affects convergence for such sequences is the factorial in the denominator. As we've seen, factorial values grow very rapidly, even faster than exponential functions like \( x^{2n} \), which suggests that the terms of the sequence diminish quickly as \( n \) increases. This is a hint that the sequence may converge to zero. In deeper studies, students will learn about tests for convergence that can formally prove whether sequences like this converge or not.