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Chapter 9: Problem 45

Writing an Equivalent Series In Exercises 45-48 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 to \(\sum_{n=0}^{\infty} \frac{x^{n}}{n!}\) with the summation index starting at \(n=1\) is \(\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}\).

Step by step solution

01

Analyzing the Given Series

The given series is \(\sum_{n=0}^{\infty} \frac{x^{n}}{n!}\). This is an infinite series, with the index (n) starting at 0.
02

Noting the Starting Point

The exercise requires you to write the series starting from \(n=1\). Note that the current series starts from \(n=0\).
03

Reshaping the Series

Due to the requirement of starting the series from \(n=1\), you need to subtract 1 from each index, resulting in \(\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}\). This is the required series with the index of summation starting at 1.

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Key Concepts

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

Index of Summation
In mathematics, the index of summation is a variable that helps us iterate over terms in a series. When dealing with an infinite series like the given example, the index of summation specifies the starting point and progression of our calculation.
In the original series, the index of summation is \(n=0\). This means each term in the series is calculated starting from when \(n\) equals zero.
  • This index indicates at which value the series begins summarizing.
  • The task specifies changing this starting index to \(n=1\), meaning each term of our newly written series begins from when \(n\) equals one.
  • Adjusting the index often requires altering the terms accordingly, such as substituting variables or adjusting the power and factorial, to maintain the series' integrity.
Altering the index may be a common requirement in exercises to practice re-framing mathematical problems, ensuring comprehension of how summation index affects series calculation.
Equivalent Series
An equivalent series is a new series that's mathematically identical to the original, but it has been rewritten to fulfill certain conditions.
The goal is to transform the expression while maintaining its value over any range of summation.
  • Typically, forming an equivalent series involves finding a new expression that provides the same sum at every stage.
  • In this problem, we transform the series starting index from \(n=0\) to \(n=1\).
  • To do this, generally, you adjust the formula of the series terms accordingly, respecting the new index framework.
For the provided series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), converting it to begin at \(n=1\) resulted in \( \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} \). This transformation maintains the series' equivalency in terms of result and function.
Factorial in Series
Factorials often appear in series, especially those related to expansions of exponential functions, as they serve to scale the terms in increasing order, contributing to convergence properties.
Each term in the series \( \frac{x^n}{n!} \) is marked by \(n!\):
  • Factorials are the product of all positive integers up to a specified number \( n \).
  • In each term \( \frac{x^n}{n!} \), the factorial \(n!\) acts as a divisor, significantly impacting the magnitude of each term.
  • As \(n\) increases, \(n!\) grows much faster than \(x^n\), typically leading to smaller individual term values and aiding in series convergence.
When the index is altered from \(n=0\) to \(n=1\), as in the exercise, \( \frac{x^{n-1}}{(n-1)!} \) preserves the factorial sequence, ensuring the series equates correctly when the index starts at 1. Factorials in series like this help manage term sizes, ensuring the series converges neatly.

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Most popular questions from this chapter

Finding the Sum of a Series In Exercises 47-52, find the sum of the convergent series by using a well-known function. Identify the function and explain how you obtained the sum. $$\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{3^{n} n}$$

Finding Intervals of Convergence In Exercises \(49-52\) , find the intervals of convergence of $$f(x),\left(\text { b) } f^{\prime}(x),\left(\text { c) } f^{\prime \prime}(x), \text { and }\left(\text { d) } \int f(x) d x\right.\right.\right.$$ (Be sure to include a check for convergence at the endpoints of the intervals.) $$f(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-2)^{n}}{n}$$

Identifying a Function In Exercises \(69-72,\) the series represents a well- known function. Use a computer algebra system to graph the partial sum \(S_{10}\) and identify the function from the graph. $$f(x)=\sum_{n=0}^{\infty}(3 x)^{n}$$

Bessel Function The Bessel function of order 0 is $$J_{0}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{2 k}}{2^{2 k}(k !)^{2}}$$ $$\begin{array}{l}{\text { (a) Show that the series converges for all } x \text { . }} \\ {\text { (b) Show that the series is a solution of the differential }} \\ {\text { equation } x^{2} J_{0}^{\prime \prime}+x J_{0}^{\prime}+x^{2} J_{0}=0}\end{array}$$ $$\begin{array}{l}{\text { (c) Use a graphing utility to graph the polynomial composed }} \\ {\text { of the first four terms of } J_{0} \text { . }} \\\ {\text { (d) Approximate } \int_{0}^{1} J_{0} d x \text { accurate to two decimal places. }}\end{array}$$

$$\begin{array}{l}{\text { Using a Power Series } \text { Let }} \\ {g(x)=1+2 x+x^{2}+2 x^{3}+x^{4}+\cdots} \\ {\text { where the coefficients are } c_{2 n}=1 \text { and } c_{2 n+1}=2 \text { for } n \geq 0 \text { . }} \\ {\text { (a) Find the interval of convergence of the series. }} \\ {\text { (b) Find an explicit formula for } g(x) \text { . }}\end{array}$$
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