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Chapter 9: Problem 41

Writing an Equivalent Series In Exercises \(41-44\) , write an equivalent series with the index of summation beginning at \(n=1 .\) $$ \sum_{n=0}^{\infty} \frac{x^{n}}{n !} $$

Short Answer

Expert verified
The equivalent series with the index of summation beginning at \(n=1\) is \(1 + \sum_{n=1}^{\infty} \frac{x^{n}}{n !}\).

Step by step solution

01

Understand the given series

The given series is an infinite series starting with \(n=0\): \(\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\). The pattern is \(\frac{x^n}{n!}\), where x is a constant, n is the variable, and n! is the factorial of n.
02

Rewrite the series starting from \(n=1\)

To rewrite the series to start at \(n=1\), all terms need to be shifted by one count in the index. This means \(n=0\) will become \(n=1\), \(n=1\) will become \(n=2\), and so forth. Therefore, the original series \(\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\) can be rewritten as \(\sum_{n=1}^{\infty} \frac{x^{(n-1)}}{(n-1) !}\). However this form still starts with n=0 when evaluated, so we must separate the first term and then rewrite the rest of the series.
03

Separate the first term and rewrite the rest of the series

Separate the first term from the rest of the given series and rewrite the remainder to begin at \(n=1\). \(\frac{x^{0}}{0!} + \sum_{n=1}^{\infty} \frac{x^{n}}{n !} = 1 + \sum_{n=1}^{\infty} \frac{x^{n}}{n !}\). Since the remaining series now starts at \(n=1\), this is the required equivalent series.

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Key Concepts

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

Summation Notation
Summation notation is a mathematical convention used to represent the sum of a sequence of numbers concisely. It involves the sigma symbol \( \Sigma \) to indicate summation and a formula inside the sum that determines each term in the series.
The basic structure of summation notation includes an expression for the general term, an index of summation which is the variable that sequentially takes on each value within the specified range of summation, and the upper and lower bounds of summation indicating where to start and end the sum.
Factorial

Understanding Factorial


The factorial of a non-negative integer \( n \) denoted as \( n! \) is the product of all positive integers less than or equal to \( n \). It's defined as \( n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1 \), with \( 0! \) being defined as 1 by convention.
  • Factorials grow at an extremely fast rate as \( n \) increases.
  • The concept of factorial is critical in permutations, combinations, and series expansion in calculus.
Series Convergence

What Is Series Convergence?


Series convergence is a property which determines whether a series approaches a finite limit as the number of terms grows indefinitely. For a series to be convergent, the terms should become infinitesimally small as the index goes to infinity, leading the sum to stabilize to a certain number.

Testing for Convergence


Several tests can be applied to determine the convergence of a series, including the ratio test, root test, and integral test. Infinite series that fail to converge are said to be divergent, meaning they do not sum up to a finite limit. The series involving factorials often converge due to the factorial's rapid growth rate which causes the terms to diminish quickly.
Index of Summation
The index of summation is the variable that is used to denote each individual term in a sum, functioning much like a loop variable in computer programming. It starts at the lower bound and increments in each term until it reaches the upper bound. Adjusting the index of summation can change how a series is expressed without altering its value. In the given exercise, the index started at \( n=0 \) and was redefined to start at \( n=1 \) by shifting the terms accordingly.
The careful manipulation of the index of summation is crucial for simplifying expressions and for mathematical proofs that involve infinite series.

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Most popular questions from this chapter

Area In Exercises 71 and \(72,\) use a power series to approximate the area of the region. Use a graphing utility to verify the result. $$ \int_{0}^{\pi / 2} \sqrt{x} \cos x d x $$

the terms of a series \(\sum_{n=1}^{\infty} a_{n}\) are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. \(a_{1}=2, a_{n+1}=\frac{2 n+1}{5 n-4} a_{n}\)

Use the Ratio Test or the Root Test to determine the convergence or divergence of the series. \(\begin{aligned} 1 &+\frac{1 \cdot 3}{1 \cdot 2 \cdot 3}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} \\ &+\frac{1 \cdot 3 \cdot 5 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}+\cdots \end{aligned}\)

Using a Binomial Series In Exercises \(17-26,\) use the binomial series to find the Maclaurin series for the function. $$ f(x)=\frac{1}{(2+x)^{3}} $$

Projectile Motion A projectile fired from the ground follows the trajectory given by $$y=\left(\tan \theta-\frac{g}{k v_{0} \cos \theta}\right) x-\frac{g}{k^{2}} \ln \left(1-\frac{k x}{v_{0} \cos \theta}\right)$$ where \(v_{0}\) is the initial speed, \(\theta\) is the angle of projection, \(g\) is the acceleration due to gravity, and \(k\) is the drag factor caused by air resistance. Using the power series representation $$\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots, \quad-1 < x < 1$$ verify that the trajectory can be rewritten as $$y=(\tan \theta) x+\frac{g x^{2}}{2 v_{0}^{2} \cos ^{2} \theta}+\frac{k g x^{3}}{3 v_{0}^{3} \cos ^{3} \theta}+\frac{k^{2} g x^{4}}{4 v_{0}^{4} \cos ^{4} \theta}+\cdots$$
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