username970964
Badges: 14
Rep:
?
#1
Report Thread starter 5 years ago
#1
I think I've done this wrong, but I don't know how to start it. Name:  20150411_102525.jpg
Views: 406
Size:  344.6 KB
0
reply
rayquaza17
Badges: 17
Rep:
?
#2
Report 5 years ago
#2
Are you trying to find the radius of convergence?
0
reply
DFranklin
Badges: 18
Rep:
?
#3
Report 5 years ago
#3
(Original post by Airess3)
I think I've done this wrong, but I don't know how to start it. Name:  20150411_102525.jpg
Views: 406
Size:  344.6 KB
This looks like a radius of convergence question. It is usually best to solve these questions using the ratio test (possibly with a bit of tweaking/common sense, but this is not necessary here).

I'll also note that you have some very messed up manipulation of exponents in your working. I know this is basic AS material, but if you can't do it well and reliably you're setting yourself up for failure with questions like this.
0
reply
michaelxukong
Badges: 0
Rep:
?
#4
Report 5 years ago
#4
I'm not sure what you mean by finding the convergence, but I have spotted a working mistake if that helps. When you took out the 3 from the denominator, the (-1) power should be on the 3 taken out, not (x^n/3^2n)
0
reply
0x2a
Badges: 2
Rep:
?
#5
Report 5 years ago
#5
If you're trying to find the radius of convergence, then use the ratio test/root test as suggested above.

Say you're using the root test, then to find the radius of convergence, you want to just look at the sequence (a_n) itself where  \sum a_nx^n is your power series.

So perform the root test as such:  \displaystyle \limsup_{n \rightarrow \infty} \sqrt[n]{|a_n|} = \alpha. Then your radius of convergence should be  R = \dfrac{1}{\alpha} and  |x| < R.

Or with the ratio test, let  \displaystyle \alpha = \limsup_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right|, then your radius of convergence should then be R = \dfrac{1}{\alpha} again.
0
reply
username970964
Badges: 14
Rep:
?
#6
Report Thread starter 5 years ago
#6
(Original post by 0x2a)
If you're trying to find the radius of convergence, then use the ratio test/root test as suggested above.

Say you're using the root test, then to find the radius of convergence, you want to just look at the sequence (a_n) itself where  \sum a_nx^n is your power series.

So perform the root test as such:  \displaystyle \limsup_{n \rightarrow \infty} \sqrt[n]{|a_n|} = \alpha. Then your radius of convergence should be  R = \dfrac{1}{\alpha} and  |x| < R.

Or with the ratio test, let  \displaystyle \alpha = \limsup_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right|, then your radius of convergence should then be R = \dfrac{1}{\alpha} again.
Name:  20150411_173535.jpg
Views: 176
Size:  379.3 KB Name:  20150411_173548.jpg
Views: 182
Size:  346.8 KB

For the first paper working, the third last line, I meant to write 1^2 instead of 1^-1.

Tried both, but have no idea how to do the second one. The professor said the answer is R=9 because of 1/3 x 1/3. And IxI is greater than or equal to 9. And it converges absolutely. My answers are nowhere near that.
0
reply
rayquaza17
Badges: 17
Rep:
?
#7
Report 5 years ago
#7
(Original post by Airess3)
Image Image

For the first paper working, the third last line, I meant to write 1^2 instead of 1^-1.

Tried both, but have no idea how to do the second one. The professor said the answer is R=9 because of 1/3 x 1/3. And IxI is greater than or equal to 9. And it converges absolutely. My answers are nowhere near that.
\frac{3^{2n+1}}{3^{2n+3}}=\frac{  1}{9}
0
reply
username970964
Badges: 14
Rep:
?
#8
Report Thread starter 5 years ago
#8
(Original post by rayquaza17)
\frac{3^{2n+1}}{3^{2n+3}}=\frac{  1}{9}
Could I ask you how you did that? I thought if you did subtracted the powers away from each other then you'd get 1^(-2). And if you substitute infinity into the n's then the answer would be 1.
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#9
Report 5 years ago
#9
(Original post by Airess3)
Could I ask you how you did that? I thought if you did subtracted the powers away from each other then you'd get 1^(-2). And if you substitute infinity into the n's then the answer would be 1.
\dfrac{3^{2n+1}}{3^{2n+3}} = 3^{(2n+1) - (2n+3)} = 3^{-2} = \dfrac{1}{9}

Is that what you were having trouble with, or was it some other part of the process?
0
reply
username970964
Badges: 14
Rep:
?
#10
Report Thread starter 5 years ago
#10
(Original post by davros)
\dfrac{3^{2n+1}}{3^{2n+3}} = 3^{(2n+1) - (2n+3)} = 3^{-2} = \dfrac{1}{9}

Is that what you were having trouble with, or was it some other part of the process?
Yes, I understand it now, thanks! The question also asked about the range of values of real numbers x for which the series converges. The answer for that is IxI is greater than or equal to 9. But I don't know why it's "greater than or equal to".
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#11
Report 5 years ago
#11
(Original post by Airess3)
Yes, I understand it now, thanks! The question also asked about the range of values of real numbers x for which the series converges. The answer for that is IxI is greater than or equal to 9. But I don't know why it's "greater than or equal to".
Are you sure it's not |x| \leq 9 ??
0
reply
username970964
Badges: 14
Rep:
?
#12
Report Thread starter 5 years ago
#12
(Original post by davros)
Are you sure it's not |x| \leq 9 ??
Yeah, sorry that's what I meant.
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#13
Report 5 years ago
#13
(Original post by Airess3)
Yeah, sorry that's what I meant.
So are you happy with the question now?

Note that your series is basically just a multiple of a GP with common ratio x/9, so you'd expect the answer to be |x/9| < 1 or |x| < 9
0
reply
username970964
Badges: 14
Rep:
?
#14
Report Thread starter 5 years ago
#14
(Original post by davros)
So are you happy with the question now?

Note that your series is basically just a multiple of a GP with common ratio x/9, so you'd expect the answer to be |x/9| < 1 or |x| < 9
Yeah, I understand it now.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (147)
14.54%
I'm not sure (43)
4.25%
No, I'm going to stick it out for now (303)
29.97%
I have already dropped out (26)
2.57%
I'm not a current university student (492)
48.66%

Watched Threads

View All