/*! 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 18 Find the radius of convergence \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 18

Find the radius of convergence \(R\) and interval of convergence for \(\sum a_{n} x^{n}\) with the given coefficients \(a_{n}\). $$ \sum_{k=1}^{\infty} \frac{k^{e} x^{k}}{e^{k}} $$

Short Answer

Expert verified
The radius of convergence is \( R = e \), and the interval of convergence is \(-e < x < e\) without including the endpoints.

Step by step solution

01

Identify the Power Series

The given series is \( \sum_{k=1}^{\infty} \frac{k^e x^k}{e^k} \). This is a power series in the form \( \sum_{k=1}^{\infty} a_k x^k \) where \( a_k = \frac{k^e}{e^k} \).
02

Apply the Ratio Test

To find the radius of convergence \( R \), use the ratio test. Consider the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1} x^{k+1}}{a_k x^k} \right| \).
03

Simplify the Ratio

Calculating the ratio gives:\[L = \lim_{k \to \infty} \left| \frac{\frac{(k+1)^e}{e^{k+1}} x^{k+1}}{\frac{k^e}{e^k} x^k} \right| = \lim_{k \to \infty} \left| \frac{(k+1)^e}{k^e} \cdot \frac{x}{e} \right|.\]
04

Evaluate the Limit

Simplifying the expression gives:\[L = \frac{|x|}{e} \lim_{k \to \infty} \left( 1 + \frac{1}{k} \right)^e.\]As \(k\) approaches infinity, the term \(\left( 1 + \frac{1}{k} \right)^e\) converges to 1. Thus, \(L = \frac{|x|}{e}\).
05

Find the Radius of Convergence

For the ratio test, convergence requires \(L < 1\). So we have:\[\frac{|x|}{e} < 1 \Rightarrow |x| < e.\]Therefore, the radius of convergence \( R \) is \( e \).
06

Determine the Interval of Convergence

The interval of convergence is determined by \(|x| < e\), meaning the series converges for \(-e < x < e\). Investigate endpoints \( x = -e \) and \( x = e \) separately using the series test to confirm if they converge or not.

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 to determine the convergence of an infinite series. It's particularly useful for power series, where each term involves a base raised to a power of a variable. The Ratio Test involves taking the limit of the absolute value of the ratio of successive terms.

Let's consider a series \[ \sum a_k x^k \]where we are testing for convergence.
  • Calculate the ratio of successive terms:\[L = \lim_{k \to \infty} \left| \frac{a_{k+1} x^{k+1}}{a_k x^k} \right|\]

  • Next, simplify this ratio to find the limit \( L \).
  • If \( L < 1 \), the series converges.
  • If \( L > 1 \) or \( L \) is infinite, the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In our example, the ratio test helped determine that when \( \frac{|x|}{e} < 1 \), the series converges, aiding in finding the radius of convergence.
Power Series
A power series is a series of the form:\[ \sum_{k=0}^{\infty} a_k x^k \]where \( a_k \) are coefficients, and \( x \) is a variable. Essentially, it's a function expressed as an infinite sum of terms involving powers of \( x \).

  • The coefficients \( a_k \) determine the scale of each term in the series.
  • The power series converges for certain values of \( x \), which dictates its domain.
  • Understanding a power series helps in analyzing functions and can help approximate complex functions with a series of simpler terms.
In our problem, the series \[ \sum_{k=1}^{\infty} \frac{k^e x^k}{e^k} \]was identified as a power series with coefficients \( a_k = \frac{k^e}{e^k} \), and finding its convergence characteristics was the main goal.
Convergence of Series
The convergence of a series is about whether the sum of its terms approaches a finite number as more terms are added. In essence, a series converges if its terms settle into a particular value rather than growing indefinitely.

  • For a series \( \sum a_k \), if the partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approach a finite limit as \( n \to \infty \), the series converges.
  • If the partial sums grow without bound, the series diverges.
  • A series can be tested for convergence using various methods like the Ratio Test, Integral Test, or Comparison Test.
  • Knowing if a series converges is crucial for understanding the function it represents.
In our exercise, convergence was confirmed by ensuring the ratio \( \frac{|x|}{e} < 1 \), which was derived using the Ratio Test, indicating a finite limit to the series' sum.
Interval of Convergence
The interval of convergence for a power series is the set of values of \( x \) for which the series converges. It is not just about finding a central radius but also includes checking whether the series converges at the endpoints of this interval.

  • The series converges within an interval \(-R < x < R\), determined by the radius of convergence \( R \).
  • To discover the interval of convergence, evaluate \( |x| < R \), and for endpoints \( x = R \) and \( x = -R \), use other tests for convergence, such as the Alternating Series Test or p-Series Test.
  • The interval may include both, either, or neither endpoint, resulting in open, closed, or half-open intervals.
In the example of our exercise, the interval of convergence was initially found as \(-e < x < e\) through the ratio criterion but required further investigation of endpoints.

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

\(\quad f(x)=\sin ^{2} x \quad\) using the identity \(\sin ^{2} x=\frac{1}{2}-\frac{1}{2} \cos (2 x)\).

Find the Maclaurin series of \(F(x)=\int_{0}^{x} f(t) d t\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero. $$ F(x)=\int_{0}^{x} e^{-t^{2}} d t ; f(t)=e^{-t^{2}}=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{2 n}}{n !} $$

Use the binomial approximation \(\sqrt{1-x} \approx 1-\frac{x}{2}-\frac{x^{2}}{8}-\frac{x^{3}}{16}-\frac{5 x^{4}}{128}-\frac{7 x^{5}}{256}\) for \(|x|<1\) to approximate each number. Compare this value to the value given by a scientific calculator. \(\frac{1}{\sqrt{2}}\) using \(x=\frac{1}{2}\) in \((1-x)^{1 / 2}\)

Evaluate \(\int_{0}^{\pi / 2} \sin ^{4} \theta d \theta\) in the approximation \(T=4 \sqrt{\frac{L}{g}} \int_{0}^{\pi / 2}\left(1+\frac{1}{2} k^{2} \sin ^{2} \theta+\frac{3}{8} k^{4} \sin ^{4} \theta+\cdots\right) d \theta \quad\) to obtain an improved estimate for \(T\).

An equivalent formula for the period of a pendulum with amplitude \(\theta_{\max } \quad\) is \(T\left(\theta_{\max }\right)=2 \sqrt{2} \sqrt{\frac{L}{g}} \int_{0}^{\theta_{\max }} \frac{d \theta}{\sqrt{\cos \theta}-\cos \left(\theta_{\max }\right)}\) where \(L\) is the pendulum length and \(g\) is the gravitational acceleration constant. When \(\theta_{\max }=\frac{\pi}{3} \quad\) we \(\frac{1}{\sqrt{\cos t-1 / 2}} \approx \sqrt{2}\left(1+\frac{t^{2}}{2}+\frac{t^{4}}{3}+\frac{181 t^{6}}{720}\right)\). Integrate this approximation to estimate \(T\left(\frac{\pi}{3}\right)\) in terms of \(L\) and \(g\). Assuming \(g=9.806\) meters per second squared, find an approximate length \(L\) such that \(T\left(\frac{\pi}{3}\right)=2\) seconds.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.