/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Determine the radius of converge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 21

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum \frac{k^{2} x^{2 k}}{k !}$$

Short Answer

Expert verified
Question: What is the radius and interval of convergence for the power series: $$\sum \frac{k^{2} x^{2 k}}{k !}$$ Answer: The radius of convergence (R) for the given power series is 1. The interval of convergence is (-1, 1).

Step by step solution

01

Applying the Ratio Test

To apply the ratio test, we will find the limit of the ratio of the (k+1)th term to the kth term of the power series as k approaches infinity. $$\lim_{k \to \infty} \frac{a_{k+1}}{a_k}$$ Where, $$a_k = \frac{k^{2} x^{2 k}}{k !}$$. Now we will find \(a_{k+1}\) and the corresponding limit.
02

Calculate the Limit

To calculate the limit, first, we will find \(a_{k+1}\) by replacing k with k+1 in the expression of \(a_k\). $$a_{k+1} = \frac{(k+1)^{2} x^{2(k+1)}}{(k+1) !}$$ Now we will find the ratio and limit. $$\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} \frac{(k+1)^{2} x^{2(k+1)}}{(k+1) !} \cdot \frac{k!}{k^{2} x^{2 k}}$$
03

Simplify the Limit Expression

Now we will simplify the limit expression to easily calculate the limit. $$\lim_{k \to \infty} \frac{(k+1)^{2} x^{2(k+1)}}{(k+1) !} \cdot \frac{k!}{k^{2} x^{2 k}} = \lim_{k \to \infty} \frac{(k+1)^{2} x^{2}}{(k+1)} = x^2 \lim_{k \to \infty} \frac{(k+1)}{k+1}$$
04

Calculate the Limit Value

The limit expression can be further simplified to: $$x^2 \lim_{k \to \infty} \frac{(k+1)}{k+1}= x^2$$
05

Apply Ratio Test Criterion

For the ratio test, a series converges when the limit is less than one, meaning: $$x^2 < 1$$
06

Find the Radius of Convergence

To find the radius of convergence (R), we will look at the inequality above, $$|x^2| < 1 \Rightarrow |x| < R = 1$$ So, the radius of convergence is 1.
07

Test the Endpoints

Now test the endpoints by checking the series' convergence at x = 1 and x = -1. For \(x = 1\), we have: $$\sum \frac{k^{2}}{k !}$$ For \(x = -1\), we have: $$\sum \frac{k^{2} (-1)^{2 k}}{k !} = \sum \frac{k^{2}}{k !}$$ In both cases, the series is the same. By applying the ratio test for this modified series, we can see that the series diverges because the limit does not exist (k^2 divided by k! goes to infinity as k goes to infinity).
08

Determine the Interval of Convergence

Since the series diverges at both endpoints, the interval of convergence is: $$(-1, 1)$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Ratio Test
The Ratio Test is a powerful tool used in mathematics to determine the convergence or divergence of a series. It involves evaluating the limit of the ratio of successive terms in a sequence.
  • To apply the Ratio Test to a power series, you look at the ratio of the \( (k+1)^{th} \) term and the \( k^{th} \) term.
  • If the limit is less than 1, the series converges.
  • If the limit is greater than 1 or infinity, the series diverges.
  • If the limit equals 1, the test is inconclusive.
In our example, \[ a_k = \frac{k^{2} x^{2k}}{k!} \]We calculate \( \frac{a_{k+1}}{a_k} \)and find the limit as\( k \to \infty \).After simplifying, the limit becomes \(x^2\). So, according to the Ratio Test, for the series to converge, \(|x^2| < 1\).This tells us crucial information about where the series might be converging.
Power Series
A power series is an infinite series of the form:\[\sum_{k=0}^{\infty} a_k x^k\]where \(a_k\) is the coefficient of the \(k^{th}\) term, and \(x\) is the variable. Power series are highly useful in math because they function like polynomials but can handle more complex functions.
  • They can model functions analytically for values within their interval of convergence.
  • In our example, the power series is \(\sum \frac{k^{2} x^{2k}}{k!}\), which is centered at zero.
The terms of this series involve not just \(x\) but also factorials, making it converge based on complex interactions between powers and factorials. Understanding the nature of your power series helps in applying tests and finding where it behaves like a typical function.
Interval of Convergence
The Interval of Convergence is the set of all \(x\) values for which a power series converges. After finding the radius of convergence, determined by principles like the Ratio Test, it is key to identify where exactly the series holds true.
  • The radius gives a distance within which the series functions properly.
  • The interval must be tested at endpoints to see if they should be included.For our series,\[ |x| < R = 1 \]
  • It becomes crucial to check endpoints like \(x=1\) and \(x=-1\).
Since the series diverges at these points from further testing, as \( \sum \frac{k^{2}}{k!}\) fails to meet convergence criteria, the interval of convergence simplifies to \((-1,1)\). Knowing this interval lets us use the series correctly over defined boundaries.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Power series for sec \(x\) Use the identity \(\sec x=\frac{1}{\cos x}\) and long division to find the first three terms of the Maclaurin series for sec \(x.\)

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b. Determine the radius of convergence of the series. $$f(x)=\sqrt{1-x^{2}}$$

Suppose you wish to approximate \(e^{0.35}\) using Taylor polynomials. Is the approximation more accurate if you use Taylor polynomials centered at 0 or \(\ln 2 ?\) Use a calculator for numerical experiments and check for consistency with Theorem 9.2. Does the answer depend on the order of the polynomial?

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b. Determine the radius of convergence of the series. $$f(x)=\cos 2 x+2 \sin x$$

Let $$f(x)=\sum_{k=0}^{\infty} x^{k}=\frac{1}{1-x} \quad \text { and } \quad S_{n}(x)=\sum_{k=0}^{n-1} x^{k}$$ The remainder in truncating the power series after \(n\) terms is \(R_{n}(x)=f(x)-S_{n}(x),\) which depends on \(x\) a. Show that \(R_{n}(x)=x^{n} /(1-x)\) b. Graph the remainder function on the interval \(|x|<1\) for \(n=1,2,3 .\) Discuss and interpret the graph. Where on the interval is \(\left|R_{n}(x)\right|\) largest? Smallest? c. For fixed \(n,\) minimize \(\left|R_{n}(x)\right|\) with respect to \(x .\) Does the result agree with the observations in part (b)? d. Let \(N(x)\) be the number of terms required to reduce \(\left|R_{n}(x)\right|\) to less than \(10^{-6} .\) Graph the function \(N(x)\) on the interval \(|x|<1 .\) Discuss and interpret the graph.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.