/*! 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 4 Find the radius of convergence a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 4

Find the radius of convergence and interval of convergence for the given power series. $$ \sum_{k=0}^{\infty} k !(x-1)^{k} $$

Short Answer

Expert verified
The radius of convergence is 0, and the interval of convergence is \( x = 1 \).

Step by step solution

01

Understanding the Problem

We need to find the radius of convergence and interval of convergence for the power series \( \sum_{k=0}^{\infty} k !(x-1)^{k} \). This involves determining the values of \( x \) for which this series converges.
02

Using the Ratio Test for Convergence

To find the radius of convergence, we can apply the Ratio Test to the series \( a_k = k! (x-1)^k \). The Ratio Test requires that we compute the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).
03

Compute the Ratio

Calculate \( \frac{a_{k+1}}{a_k} = \frac{(k+1)! (x-1)^{k+1}}{k! (x-1)^k} = (k+1)|x-1| \).
04

Finding the Limit for the Ratio Test

The limit therefore becomes \( \lim_{k \to \infty} (k+1)|x-1| \). As \( k \to \infty \), this limit tends to infinity for any non-zero \( |x-1| \).
05

Conclusion of the Ratio Test

Since the limit tends to infinity whenever \( x eq 1 \), the series converges only for \( x = 1 \). The radius of convergence \( R \) is therefore 0.
06

Interval of Convergence

Given that the series converges only at the point \( x = 1 \), it follows that the interval of convergence is \( x = 1 \).

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.

Understanding Power Series
A power series is incredibly useful in mathematics and engineering. It's a series of the form:\[\sum_{k=0}^{\infty} c_k (x-a)^{k}\]where \(c_k\) are coefficients and \(a\) is the center of the series. It represents a function similar to a polynomial but with infinitely many terms. The power series is centered at \(a\), meaning all terms are in the form \((x-a)^k\). This centering allows us to model functions as if they are polynomials near that specific point \(a\).
  • Coefficients \(c_k\): Each term of the power series has a coefficient which influences the shape and convergence of the series.
  • Center \(a\): This is the point around which the series is expanded. For our exercise's series, the center is \(x=1\).
Understanding the structure of a power series helps simplify complex functions into manageable algebraic expressions.
Discovering the Interval of Convergence
The interval of convergence is the set of values for which a power series converges. For our purposes, finding this interval involves determining the values of \(x\) such that the series converges to a finite sum. For series centered at \(x=1\), we identify points where changing \(x\) leads to the series converging or diverging.

To check convergence at different \(x\) values, we calculate the radius of convergence \(R\):
  • Inside the interval where \(|x-a| < R\), the series converges.
  • Outside, where \(|x-a| > R\), it diverges.
In the given series from the exercise:
  • The series only converges at \(x=1\), meaning the radius of convergence is \(R=0\).
  • This results in a convergent interval of just the single point \(x=1\).
Understanding the interval helps determine how effectively we can use series to approximate functions in certain regions.
Utilizing the Ratio Test for Convergence
The Ratio Test is a potent method for assessing the convergence of series, especially power series. To employ the test, we evaluate the limit:\[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|\]Here \(a_k\) is the general term of the power series. This test helps understand how quickly the terms of the series are approaching zero.
  • Convergence: If \(L < 1\), the series converges.
  • Divergence: If \(L > 1\), the series diverges.
  • If \(L = 1\), the test is inconclusive; other methods might be necessary.
For the exercise, the limit \( L = \lim_{k \to \infty} (k+1)|x-1|\) was obtained. As \(k\) approaches infinity, this limit tends to infinity unless \(x = 1\). Therefore, the radius of convergence \(R\) is \(0\), leading to a convergence at \(x=1\) only. This practical approach simplifies solving convergence problems.

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

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. $$ \left(x^{3}-2 x^{2}+3 x\right)^{2} y^{\prime \prime}+x(x-3)^{2} y^{\prime}-(x+1) y=0 $$

$$ y^{\prime \prime}+x y^{\prime}+2 y=0, y(0)=3, y^{\prime}(0)=-2 $$

The differential equation $$ \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+\alpha^{2} y=0 $$ where \(\alpha\) is a parameter, is known as Chebyshev's equation after the Russian mathematician Pafnuty Chebyshev (18211894). Find the general solution \(y(x)=c_{0} y_{1}(x)+c_{1} y_{2}(x)\) of the equation, where \(y_{1}(x)\) and \(y_{2}(x)\) are power series solutions centered at the ordinary point 0 and containing only even powers of \(x\) and odd powers of \(x\), respectively.

$$ x^{2} y^{\prime \prime}+\left(\frac{5}{3} x+x^{2}\right) y^{\prime}-\frac{1}{3} y=0 $$

Without actually solving the differential equation \((\cos x) y^{\prime \prime}+\) \(y^{\prime}+5 y=0\), find a lower bound for the radius of convergence of power series solutions about \(x=0 .\) About \(x=1\).
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Scientific Method Physics

Read Explanation

Thermodynamics

Read Explanation

Electricity and Magnetism

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.