Chapter 9.4, Problem 63E

To determine

To find: To find the radius of convergence.

Answer to Problem 63E

The radius of convergence is $2$ .

Explanation of Solution

Given:

The given series ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$ .

Calculation:

An expression of the form,

  ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}=c_0+c_1(x-a)+c_2{(x-a)}^2+c_n{(x-a)}^n+\mathrm{....}$

Is a power series centred at $x=a$ .

The convergence theorem for power series,

One possibility ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$ with respect to convergence. There is a positive number R such that the series diverges for $\left|x-a\right|>R$ and converges for $\left|x-a\right|<R$ . The series may or may not converges at either of the endpoints $x=a-R$ and $x=a+R$ . This R is called radius of convergence.

Now, students of my group make series like this students 1+=- radius of convergence is $1$ .

  $1+x+x^2+x^3+\mathrm{.....}$

Since, this is an infinite geometric series, hence it converges $\left|x\right|<2$ , hence radius of convergence is $2$ .

Other students made other series.

Students of the other group made series like this students 1- radius of convergence is $1$ .

  $1+(x-1)+{(x-1)}^2+{(x-1)}^3+\mathrm{..........}$

Since this is an infinite geometric series, hence it converges only for $\left|x-1\right|<1$ and hence radius of convergence is $1$ .

Student 2- radius of convergence is $2$ .

  $1+\frac{x-3}{2}+\frac{{(x-3)}^2}{4}+\frac{{(x-3)}^3}{8}+\mathrm{..................}$

Since this is an infinite geometric series, hence it converges only for $\left|x-3\right|<2$ and hence radius of convergence is $2$ .

Other students made the series.

The radius of convergence for the student was matched.