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91Ó°ÊÓ

Use a symbolic algebra utility to evaluate the summation. $$ \sum_{n=1}^{\infty} \frac{1}{(2 n) !} $$

Short Answer

Expert verified
The value of the summation \( \sum_{n=1}^{\infty} \frac{1}{(2n)!} \) is 1.

Step by step solution

01

Recognize the series

Recognize that \( \frac{1}{(2 n)!} \) is in the series expansion for \( \cos{x} \) at \( x=0 \), or more specifically just the even-indexed terms of the MacLaurin series expansion \( \cos{x} = \sum_{n=0}^{\infty}(-1)^n \cdot \frac{x^{2n}}{(2n)!} \). At \( x=0 \), the odd power terms vanish and we are left with \( \cos{0} = \sum_{n=1}^{\infty} \frac{1}{(2n)!} \).
02

Evaluate the recognized series

Given that the cosine of 0 is 1, the summation \( \sum_{n=1}^{\infty} \frac{1}{(2n)!} \) is also equivalent to 1.

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Key Concepts

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

Maclaurin Series
The Maclaurin series is a special type of Taylor series that represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point, specifically at zero. This series allows us to express complex functions in an easily manageable form.

For a general function \( f(x) \), the Maclaurin series is given by:
  • \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots \)
The series continues with the pattern, substituting higher-order derivatives and increasing powers of \( x \). This makes the concepts scalable and usable for approximating values of functions that are otherwise difficult to compute directly.

In the context of the exercise provided, we focus on deriving the series for the cosine function, which forms the base for expanding the function around zero. This is crucial for evaluating series that sum infinite terms, bypassing conventional computation methods.
Cosine Function
The cosine function, \( \cos(x) \), is a fundamental periodic function in mathematics, especially in trigonometry. It describes the relationship in a right-angled triangle between the length of the adjacent side to the angle and the length of the hypotenuse.

In analytic terms, cosine is commonly expanded using the Maclaurin series, which helps us to understand its behavior around \( x = 0 \). The expansion of \( \cos(x) \) through its Maclaurin series is:
  • \( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \)
The series alternates signs and contains only even powers of \( x \), which inherently shows how similar the structure of the cosine function is to other even-based polynomials.

The cosine function evaluated at \( x = 0 \) simplifies to 1, as all even power terms vanish beyond this point. This highlights how the function levels at zero and exemplifies the patterns within trigonometric identities, especially as they relate to infinite series expansions.
Symbolic Algebra
Symbolic algebra involves the manipulation and solution of equations and expressions using symbols and variables rather than numeric values. This branch of algebra is incredibly useful for automated computations and theoretical developments.

In problem-solving, symbolic algebra utilities, like software tools or mathematical solvers, allow us to evaluate series and infinite sums by representing functions in their generalized forms. This approach, as utilized in the exercise, provides an accurate means to identify the series expansion of functions such as the cosine function.

By recognizing the pattern in the Maclaurin series of \( \cos(x) \) and matching terms, symbolic algebra systems can derive expansions and evaluate limits without needing explicit computation of each term. This dramatically reduces the complexity involved in solving problems that use infinite series, making symbolic algebra an indispensable tool in both educational and real-world applications.

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Most popular questions from this chapter

Physical Science The ball in Exercise 47 takes the times listed below for each fall. ( \(t\) is measured in seconds.) $$ \begin{array}{rlrl}{s_{1}} & {=-16 t^{2}+16} & {s_{1}} & {=0 \text { if } t=1} \\\ {s_{2}} & {=-16 t^{2}+16(0.81)} & {s_{2}} & {=0 \text { if } t=0.9} \\\ {s_{3}} & {=-16 t^{2}+16(0.81)^{2}} & {s_{3}} & {=0 \text { if } t=(0.9)^{2}} \\\ {s_{4}} & {=-16 t^{2}+16(0.81)^{3}} & {s_{4}} & {=0 \text { if } t=(0.9)^{3}}\end{array} $$ $$ \begin{array}{cc}{\vdots} & {\vdots} \\ {s_{n}=-16 t^{2}+16(0.81)^{n-1}} & {s_{n}=0 \text { if } t=(0.9)^{n-1}}\end{array} $$ Beginning with \(s_{2},\) the ball takes the same amount of time to bounce up as it does to fall, and so the total time elapsed before it comes to rest is given by \(t=1+2 \sum_{n=1}^{\infty}(0.9)^{n}\) Find this total time.

Salary A person accepts a position with a company at a salary of 32,800 dollars for the first year. The person is guaranteed a raise of \(5 \%\) per year for the next 3 years. Determine the person's salary during the fourth year of employment.

Write the first five terms of the sequence of partial sums. $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots $$

Determine the convergence or divergence of the series. Use a symbolic algebra utility to verify your result. $$ \sum_{n=1}^{\infty} \frac{n !+1}{n !} $$

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