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Chapter 9: Problem 139

Write the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n} x^{n}}{n !}$$

Short Answer

Expert verified
The first five terms of the sequence are: \(a_1 = -x\), \(a_2 = \frac{x^2}{2}\), \(a_3 = -\frac{x^3}{6}\), \(a_4 = \frac{x^4}{24}\), \(a_5 = -\frac{x^5}{120}\)

Step by step solution

01

Understanding the formula

Here, \(a_n\) represents the nth term of the sequence. \(n\) stands for the position of the term in the sequence (like 1st term, 2nd term, etc.). The formula is \(a_{n}=\frac{(-1)^{n} x^{n}}{n !}\). It means, to find any term in the sequence, one has to substitute \(n\) with the position of the term in the sequence.
02

Applying the formula for n=1

Substitute \(n\) with 1 (to find the first term) in the formula: \(a_{1}=\frac{(-1)^{1} x^{1}}{1 !} = -x\)
03

Applying the formula for n=2

Substitute \(n\) with 2 (to find the second term) in the formula: \(a_{2}=\frac{(-1)^{2} x^{2}}{2 !} = \frac{x^2}{2}\)
04

Applying the formula for n=3

Substitute \(n\) with 3 (to find the third term) in the formula: \(a_{3}=\frac{(-1)^{3} x^{3}}{3 !} = -\frac{x^3}{6}\)
05

Applying the formula for n=4

Substitute \(n\) with 4 (to find the fourth term) in the formula: \(a_{4}=\frac{(-1)^{4} x^{4}}{4 !} = \frac{x^4}{24}\)
06

Applying the formula for n=5

Substitute \(n\) with 5 (to find the fifth term) in the formula: \(a_{5}=\frac{(-1)^{5} x^{5}}{5 !} = -\frac{x^5}{120}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
A factorial, represented by the symbol "!", is a fundamental concept in mathematics that arises in sequences, permutations, and various formulas. To compute the factorial of a non-negative integer \( n \), you multiply all the integers from 1 up to \( n \). Thus, \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \). This can be seen as a decorative way of stacking numbers. Here are some helpful pointers about factorials:
  • \( 0! = 1 \) by definition, which is an important starting point for many formulas.
  • Factorials grow very quickly; for example, \( 5! = 120 \) and \( 10! = 3,628,800 \).
  • Factorials are used extensively in combinatorics, calculus, and mathematical sequences, such as in our sequence formula \( a_n = \frac{(-1)^n x^n}{n!} \).
Understanding the factorial is crucial because it serves as the denominator in the given sequence formula, affecting each term's value as \( n \) increases.
Alternating Sequence
An alternating sequence is a type of sequence where the terms change signs in a regular pattern of positives and negatives. This pattern is commonly signified by the term \((-1)^n\) in a formula. Depending on \( n \), this ensures that some terms are positive and others are negative. Here鈥檚 why this concept is important:
  • The expression \((-1)^n\) results in alternating signs. If \( n \) is even, the result is \( +1 \); if \( n \) is odd, the result is \( -1 \).
  • Such sequences frequently appear in mathematical series because alternating signs help in convergence properties and balancing.
  • In our sequence, \( a_n = \frac{(-1)^n x^n}{n!} \), the alternation visually affects the calculated terms like \(-x\), \(\frac{x^2}{2}\), \(-\frac{x^3}{6}\), making the sequence more dynamic in behavior.
Grasping the idea of alternating sequences helps to predict the behavior of sequences and understand their patterns more clearly.
Algebraic Expressions
Algebraic expressions are mathematical statements comprising numbers, variables, and operators, like \(+ - \times \div \). They form the foundation of algebra and describe relationships between variables. In the sequence formula \( a_n = \frac{(-1)^n x^n}{n!} \), the expression is algebraic due to the variables and operations involved. Here's what to remember about them:
  • Variables, like \( x \), represent unknown or changeable values, critical when generalizing formulas.
  • Operations follow order precedence, meaning one must carefully consider parentheticals and the sequence of calculations (e.g., exponents before division).
  • Algebraic expressions can simplify complex problems, allowing us to express intricate relationships clearly and concisely.
Mastering algebraic expressions allows you to apply them to sequences and series effortlessly, as seen in the problem's solution involving terms like \(-x\) and \(-\frac{x^3}{6}\). They offer a way to model real-world and theoretical phenomena in mathematics effectively.

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Most popular questions from this chapter

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