`g(x)=5/(2x-3), c=-3` Find a power series for the function, centered at c and determine the interval of convergence.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

If x is a variable, then an infinite series of the form `sum_(n=0)^ooa_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+......+a_n(x-c)^n+.....` is a power series centered at x=c, where c is a constant.

Given `g(x)=5/(2x-3) , c=-3`

Let's write g(x) in the form `a/(1-r)`

`g(x)=5/(2x-3)`

`=(5/2)/(x-3/2)`

`=(5/2)/(x+3-3/2-3)`

`=(5/2)/(x+3-9/2)`

`=(5/2)/(-9/2(1-2/9(x+3)))`

`=((5/2)(-2/9))/(1-2/9(x+3))`

`=(-5/9)/(1-2/9(x+3))`

So a=`-5/9` and r=`2/9(x+3)`

So, the power series for g(x) is `sum_(n=0)^ooar^n`

`=sum_(n=0)^oo(-5/9)(2/9(x+3))^n`

`=-5sum_(n=0)^oo(2^n(x+3)^n)/9^(n+1)`

This power series is a geometric series and it converges if `|r|<1`

`|2/9(x+3)|<1`

`-1<2/9(x+3)<1`

`-9<(2x+6)<9`

`-9-6<2x<9-6`

`-15<2x<3`

`-15/2<x<3/2`

Interval of convergence is `(-15/2,3/2)`

 

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial