/*! 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 22 Determine the radius of converge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 22

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum k(x-1)^{k}$$

Short Answer

Expert verified
Answer: The radius of convergence is 1, and the interval of convergence is (0, 2).

Step by step solution

01

Use the Ratio Test for Convergence

To find the radius of convergence, we'll apply the Ratio Test to the power series. The Ratio Test states that for a series $$\sum a_k$$, the series converges if: $$\lim_{k\to\infty}\lvert\frac{a_{k+1}}{a_k}\rvert = L < 1$$ Our given power series is $$\sum k(x-1)^{k}$$, so let's set up the ratio test: $$\lim_{k\to\infty}\lvert\frac{(k+1)(x-1)^{(k+1)}}{k(x-1)^k}\rvert$$
02

Simplify the Limit Expression

Now, simplify the expression within the absolute value: $$\lim_{k\to\infty}\lvert\frac{(k+1)(x-1)}{k}\rvert$$
03

Determine the Limit

Calculate the limit as k goes to infinity: $$\lim_{k\to\infty}\lvert\frac{(k+1)(x-1)}{k}\rvert = |x-1|\lim_{k\to\infty}\lvert\frac{k+1}{k}\rvert = |x-1|$$
04

Apply the Ratio Test

For the series to converge using the Ratio Test, |x-1| < 1. Therefore, we get -1 < x-1 < 1, which means that the interval of convergence is: $$0 < x < 2$$ The radius of convergence, R, can be determined by half the length of the interval. As we have an interval between 0 and 2, the radius of convergence R is: $$R = \frac{2 - 0}{2} = 1$$
05

Test the Endpoints

Now, test the endpoints of the interval of convergence, x = 0 and x = 2: For x = 0, the series is $$\sum k(-1)^k$$, which is an alternating series and does not converge. For x = 2, the series is $$\sum k(1)^k$$ or $$\sum k$$, which is the harmonic series and also does not converge.
06

Determine the Interval of Convergence

Since neither endpoint converges, the interval of convergence remains unchanged: $$0 < x < 2$$ In summary, the radius of convergence is 1, and the interval of convergence is (0, 2).

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!

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

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=e^{-x}, a=0$$

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{k+1}}{4^{k}}$$

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$e^{2}$$

Choose a Taylor series and a center point a to approximate the following quantities with an error of \(10^{-4}\) or less. $$\sqrt[3]{83}$$

Tangent line is \(p_{1}\) Let \(f\) be differentiable at \(x=a\) a. Find the equation of the line tangent to the curve \(y=f(x)\) at \((a, f(a))\) b. Find the Taylor polynomial \(p_{1}\) centered at \(a\) and confirm that it describes the tangent line found in part (a).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

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.