Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 8.5, Problem 36E

To determine

To find: A power series whose interval of convergence is $(p,q)$ .

Expert Solution

Answer to Problem 36E

The series having interval of convergence $(p,q)$ is $∑n=0∞[2x−(p+q)(q−p)]n$ .

Explanation of Solution

Given Information: $p$ and $q$ are real numbers with $p<q$ .

Calculation:

Let the series $∑f(x)$ has interval of convergence $(a,b)$ .

Then the series $∑f(xc)$ has interval of convergence $(c.a,c.b)$ and the series $∑f(x−c)$ has interval of convergence $(c+a,c+b)$ .

For the interval $(p,q)$ , the mid-point is $p+q2$ and radius of convergence is $q−p2$ .

Now, the interval of convergence of the series $∑n=0∞xn$ is $(−1,1)$ .

So, the interval of convergence of the series $∑n=0∞[x(q−p)/2]n$ is $(p−q2,q−p2)$ and the interval of convergence of $∑n=0∞[x−p+q2(q−p)/2]n$ is $(p−q2+p+q2,q−p2+p+q2)=(p,q)$ .

Simplifying, $∑n=0∞[x−p+q2(q−p)/2]n=∑n=0∞[2x−(p+q)(q−p)]n$ .

Thus, the series having interval of convergence $(p,q)$ is $∑n=0∞[2x−(p+q)(q−p)]n$ .

To determine

To find: A power series whose interval of convergence is $(p,q]$ .

Expert Solution

Answer to Problem 36E

The series having interval of convergence $(p,q]$ is $∑n=1∞[2x−(p+q)(q−p)]n.(−1)nn$ .

Explanation of Solution

Given Information: $p$ and $q$ are real numbers with $p<q$ .

Calculation:

Let the series $∑f(x)$ has interval of convergence $(a,b]$ .

Then the series $∑f(xc)$ has interval of convergence $(c.a,c.b]$ and the series $∑f(x−c)$ has interval of convergence $(c+a,c+b]$ .

For the interval $(p,q]$ , the mid-point is $p+q2$ and radius of convergence is $q−p2$ .

Now, the interval of convergence of the series $∑n=1∞xn.(−1)nn$ is $(−1,1]$ .

So, the interval of convergence of the series $∑n=1∞[x(q−p)/2]n.(−1)nn$ is $(p−q2,q−p2]$ and the interval of convergence of $∑n=1∞[x−p+q2(q−p)/2]n.(−1)nn$ is $(p−q2+p+q2,q−p2+p+q2]=(p,q]$ .

Simplifying, $∑n=1∞[x−p+q2(q−p)/2]n.(−1)nn=∑n=1∞[2x−(p+q)(q−p)]n.(−1)nn$ .

Thus, the series having interval of convergence $(p,q]$ is $∑n=1∞[2x−(p+q)(q−p)]n.(−1)nn$ .

To determine

To find: A power series whose interval of convergence is $[p,q)$ .

Expert Solution

Answer to Problem 36E

The series having interval of convergence $[p,q)$ is $∑n=1∞[2x−(p+q)(q−p)]n.1n$ .

Explanation of Solution

Given Information: $p$ and $q$ are real numbers with $p<q$ .

Calculation:

Let the series $∑f(x)$ has interval of convergence $[a,b)$ .

Then the series $∑f(xc)$ has interval of convergence $[c.a,c.b)$ and the series $∑f(x−c)$ has interval of convergence $[c+a,c+b)$ .

For the interval $[p,q)$ , the mid-point is $p+q2$ and radius of convergence is $q−p2$ .

Now, the interval of convergence of the series $∑n=1∞xnn$ is $[−1,1)$ .

So, the interval of convergence of the series $∑n=1∞[x(q−p)/2]n.1n$ is $[p−q2,q−p2)$ and the interval of convergence of $∑n=1∞[x−p+q2(q−p)/2]n.1n$ is $[p−q2+p+q2,q−p2+p+q2)=[p,q)$ .

Simplifying, $∑n=1∞[x−p+q2(q−p)/2]n.1n=∑n=1∞[2x−(p+q)(q−p)]n.1n$ .

Thus, the series having interval of convergence $[p,q)$ is $∑n=1∞[2x−(p+q)(q−p)]n.1n$ .

To determine

To find: A power series whose interval of convergence is $[p,q]$ .

Expert Solution

Answer to Problem 36E

The series having interval of convergence $[p,q]$ is $∑n=1∞[2x−(p+q)(q−p)]n.1n2$ .

Explanation of Solution

Given Information: $p$ and $q$ are real numbers with $p<q$ .

Calculation:

Let the series $∑f(x)$ has interval of convergence $[a,b]$ .

Then the series $∑f(xc)$ has interval of convergence $[c.a,c.b]$ and the series $∑f(x−c)$ has interval of convergence $[c+a,c+b]$ .

For the interval $[p,q]$ , the mid-point is $p+q2$ and radius of convergence is $q−p2$ .

Now, the interval of convergence of the series $∑n=1∞xnn2$ is $[−1,1]$ .

So, the interval of convergence of the series $∑n=1∞[x(q−p)/2]n.1n2$ is $[p−q2,q−p2]$ and the interval of convergence of $∑n=1∞[x−p+q2(q−p)/2]n.1n2$ is $[p−q2+p+q2,q−p2+p+q2]=[p,q]$ .

Simplifying, $∑n=1∞[x−p+q2(q−p)/2]n.1n2=∑n=1∞[2x−(p+q)(q−p)]n.1n2$ .

Thus, the series having interval of convergence $[p,q]$ is $∑n=1∞[2x−(p+q)(q−p)]n.1n2$ .

Chapter 8.5, Problem 35E

Chapter 8.6, Problem 1E

Chapter 8 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

Ch. 8.1 - Prob. 11E Ch. 8.1 - Prob. 12E Ch. 8.1 - Prob. 13E Ch. 8.1 - Prob. 14E Ch. 8.1 - Prob. 15E Ch. 8.1 - Prob. 16E Ch. 8.1 - Prob. 17E Ch. 8.1 - Prob. 18E Ch. 8.1 - Prob. 19E Ch. 8.1 - Prob. 20E Ch. 8.1 - Prob. 21E Ch. 8.1 - Prob. 22E Ch. 8.1 - Prob. 23E Ch. 8.1 - Prob. 24E Ch. 8.1 - Prob. 25E Ch. 8.1 - Prob. 26E Ch. 8.1 - Prob. 27E Ch. 8.1 - Prob. 28E Ch. 8.1 - Prob. 29E Ch. 8.1 - Prob. 30E Ch. 8.1 - Prob. 31E Ch. 8.1 - Prob. 32E Ch. 8.1 - Prob. 33E Ch. 8.1 - Prob. 34E Ch. 8.1 - Prob. 35E Ch. 8.1 - Prob. 36E Ch. 8.1 - Prob. 37E Ch. 8.1 - Prob. 38E Ch. 8.1 - Prob. 39E Ch. 8.1 - Prob. 40E Ch. 8.1 - Prob. 41E Ch. 8.1 - Prob. 42E Ch. 8.1 - Prob. 43E Ch. 8.1 - Prob. 44E Ch. 8.1 - Prob. 45E Ch. 8.1 - Prob. 46E Ch. 8.1 - Prob. 47E Ch. 8.1 - Prob. 48E Ch. 8.1 - Prob. 49E Ch. 8.1 - Prob. 50E Ch. 8.1 - Prob. 51E Ch. 8.1 - Prob. 52E Ch. 8.1 - Prob. 53E Ch. 8.1 - Prob. 54E Ch. 8.1 - Prob. 55E Ch. 8.1 - Prob. 56E Ch. 8.1 - Prob. 57E Ch. 8.1 - Prob. 58E Ch. 8.1 - Prob. 59E Ch. 8.1 - Prob. 60E Ch. 8.2 - 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