Chapter 8.5, Problem 36E

a.

To determine

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

Answer to Problem 36E

The series having interval of convergence $(p,q)$ is ${{\displaystyle \sum _{n=0}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n$ .

Explanation of Solution

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

Calculation:

Let the series ${\displaystyle \sum f(x)}$ has interval of convergence $(a,b)$ .

Then the series ${\displaystyle \sum f\left(\frac{x}{c}\right)}$ has interval of convergence $(c.a,c.b)$ and the series ${\displaystyle \sum f(x-c)}$ has interval of convergence $(c+a,c+b)$ .

For the interval $(p,q)$ , the mid-point is $\frac{p+q}{2}$ and radius of convergence is $\frac{q-p}{2}$ .

Now, the interval of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }{x}^n}$ is $(-1,1)$ .

So, the interval of convergence of the series ${{\displaystyle \sum _{n=0}^{\infty }\left[\frac{x}{(q-p)/2}\right]}}^n$ is $\left(\frac{p-q}{2},\frac{q-p}{2}\right)$ and the interval of convergence of ${{\displaystyle \sum _{n=0}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n$ is $\left(\frac{p-q}{2}+\frac{p+q}{2},\frac{q-p}{2}+\frac{p+q}{2}\right)=(p,q)$ .

Simplifying, ${{\displaystyle \sum _{n=0}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n={{\displaystyle \sum _{n=0}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n$ .

Thus, the series having interval of convergence $(p,q)$ is ${{\displaystyle \sum _{n=0}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n$ .

b.

To determine

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

Answer to Problem 36E

The series having interval of convergence $(p,q]$ is ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{{(-1)}^n}{n}$ .

Explanation of Solution

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

Calculation:

Let the series ${\displaystyle \sum f(x)}$ has interval of convergence $(a,b]$ .

Then the series ${\displaystyle \sum f\left(\frac{x}{c}\right)}$ has interval of convergence $(c.a,c.b]$ and the series ${\displaystyle \sum f(x-c)}$ has interval of convergence $(c+a,c+b]$ .

For the interval $(p,q]$ , the mid-point is $\frac{p+q}{2}$ and radius of convergence is $\frac{q-p}{2}$ .

Now, the interval of convergence of the series ${\displaystyle \sum _{n=1}^{\infty }{x}^n}.\frac{{(-1)}^n}{n}$ is $(-1,1]$ .

So, the interval of convergence of the series ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x}{(q-p)/2}\right]}}^n.\frac{{(-1)}^n}{n}$ is $\left(\frac{p-q}{2},\frac{q-p}{2}\right]$ and the interval of convergence of ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n.\frac{{(-1)}^n}{n}$ is $\left(\frac{p-q}{2}+\frac{p+q}{2},\frac{q-p}{2}+\frac{p+q}{2}\right]=(p,q]$ .

Simplifying, ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n.\frac{{(-1)}^n}{n}={{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{{(-1)}^n}{n}$ .

Thus, the series having interval of convergence $(p,q]$ is ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{{(-1)}^n}{n}$ .

c.

To determine

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

Answer to Problem 36E

The series having interval of convergence $[p,q)$ is ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{1}{n}$ .

Explanation of Solution

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

Calculation:

Let the series ${\displaystyle \sum f(x)}$ has interval of convergence $[a,b)$ .

Then the series ${\displaystyle \sum f\left(\frac{x}{c}\right)}$ has interval of convergence $[c.a,c.b)$ and the series ${\displaystyle \sum f(x-c)}$ has interval of convergence $[c+a,c+b)$ .

For the interval $[p,q)$ , the mid-point is $\frac{p+q}{2}$ and radius of convergence is $\frac{q-p}{2}$ .

Now, the interval of convergence of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n}}$ is $[-1,1)$ .

So, the interval of convergence of the series ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x}{(q-p)/2}\right]}}^n.\frac{1}{n}$ is $\left[\frac{p-q}{2},\frac{q-p}{2}\right)$ and the interval of convergence of ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n.\frac{1}{n}$ is $\left[\frac{p-q}{2}+\frac{p+q}{2},\frac{q-p}{2}+\frac{p+q}{2}\right)=[p,q)$ .

Simplifying, ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n.\frac{1}{n}={{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{1}{n}$ .

Thus, the series having interval of convergence $[p,q)$ is ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{1}{n}$ .

d.

To determine

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

Answer to Problem 36E

The series having interval of convergence $[p,q]$ is ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{1}{n^2}$ .

Explanation of Solution

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

Calculation:

Let the series ${\displaystyle \sum f(x)}$ has interval of convergence $[a,b]$ .

Then the series ${\displaystyle \sum f\left(\frac{x}{c}\right)}$ has interval of convergence $[c.a,c.b]$ and the series ${\displaystyle \sum f(x-c)}$ has interval of convergence $[c+a,c+b]$ .

For the interval $[p,q]$ , the mid-point is $\frac{p+q}{2}$ and radius of convergence is $\frac{q-p}{2}$ .

Now, the interval of convergence of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^2}}$ is $[-1,1]$ .

So, the interval of convergence of the series ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x}{(q-p)/2}\right]}}^n.\frac{1}{n^2}$ is $\left[\frac{p-q}{2},\frac{q-p}{2}\right]$ and the interval of convergence of ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n.\frac{1}{n^2}$ is $\left[\frac{p-q}{2}+\frac{p+q}{2},\frac{q-p}{2}+\frac{p+q}{2}\right]=[p,q]$ .

Simplifying, ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{x-\frac{p+q}{2}}{(q-p)/2}\right]}}^n.\frac{1}{n^2}={{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{1}{n^2}$ .

Thus, the series having interval of convergence $[p,q]$ is ${{\displaystyle \sum _{n=1}^{\infty }\left[\frac{2x-(p+q)}{(q-p)}\right]}}^n.\frac{1}{n^2}$ .