Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 2008-12-08 04:05:53
- GemmaJ1988
- Member
- Registered: 2008-10-08
- Posts: 39
power series
Q) for what values of Z is
Now im not sure if im going about this the right way, but so far ive done
so the radius of convergence is given by:
so we can give the circle of convergence as
can anyone tell me whether im going about this the right way?
thank you 
Last edited by GemmaJ1988 (2008-12-08 04:06:36)
Offline
#2 2008-12-08 07:45:33
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: power series
I'd do it by changing it into [Z/(1-Z)]^n.
I'm guessing you know that ∑ x^n converges if x is in [-1,1), so then you can find the answer by investigating which values of Z give Z/(1-Z) within that interval.
Why did the vector cross the road?
It wanted to be normal.
Offline
#3 2008-12-09 20:47:58
- GemmaJ1988
- Member
- Registered: 2008-10-08
- Posts: 39
Re: power series
ok i understand the bit about x^n converges when x is in [-1,1) but i dont know how to appy this to Z/(1-Z)
Offline
#4 2008-12-09 23:12:20
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: power series
You need to find the range of Z that solves -1 ≤ Z/(1-Z) < 1.
Why did the vector cross the road?
It wanted to be normal.
Offline
#5 2008-12-10 10:56:11
- GemmaJ1988
- Member
- Registered: 2008-10-08
- Posts: 39
Re: power series
im still confused how to do this 
Offline
#6 2008-12-10 22:33:17
- GemmaJ1988
- Member
- Registered: 2008-10-08
- Posts: 39
Re: power series
ive got the range of z that solves -1<=Z/(1-Z)<1 is when z<=0
Offline
#7 2008-12-11 02:24:10
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: power series
When I work it out, I get Z < 1/2 to be the solution.
Why did the vector cross the road?
It wanted to be normal.
Offline
Pages: 1