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Power series; 2

by arl146
Tags: power, series
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Mark44
#19
Feb23-12, 01:58 PM
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Um I don't know, that if it's less than 1 it converges and if greater than 1 it diverges
Be specific. If what is less than 1. Don't use "it". Write mathematics equations/inequalities.
arl146
#20
Feb23-12, 03:46 PM
P: 343
|x-4|<1 converge
|x-4|>1 diverge
Mark44
#21
Feb23-12, 03:51 PM
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|x-4|<1 converge
|x-4|>1 diverge
OK, what interval are we talking about for convergence.

What about if |x - 4| = 1? The Ratio test doesn't cover this situation, so you have to investigate it as a special case.
arl146
#22
Feb23-12, 08:00 PM
P: 343
interval? .... i dont know O_O
Mark44
#23
Feb23-12, 08:06 PM
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Quote Quote by arl146 View Post
interval? .... i dont know O_O
Solve |x - 4| < 1.
arl146
#24
Feb23-12, 08:07 PM
P: 343
oh, x < 5 ... so just (-infinity, 5) ?
Mark44
#25
Feb23-12, 08:48 PM
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oh, x < 5 ... so just (-infinity, 5) ?
No. You should get a finite length interval. According to you, -300 would satisfy |x - 4| < 1. Does it?

You should review working with absolute values in equations and inequalities.
arl146
#26
Feb23-12, 08:51 PM
P: 343
ohhhhhhhh, i dont know why I KEEP just looking over the absolute values. sooo .... (4, 5) ?
Mark44
#27
Feb23-12, 08:54 PM
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What about 3.5? Doesn't it satisfy |x - 4| < 1?
arl146
#28
Feb23-12, 09:08 PM
P: 343
ohh right .... so (3, 5) because since it's not a square bracket it doesn't include the 3 right
but anything after 3 and before 4 would really never actually equal 1 which satisfies the less than sign
Mark44
#29
Feb24-12, 12:13 AM
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Quote Quote by arl146 View Post
ohh right .... so (3, 5) because since it's not a square bracket it doesn't include the 3 right
but anything after 3 and before 4 would really never actually equal 1 which satisfies the less than sign
Between 3 and 5, not 4.
OK, this establishes that the radius of convergence is 1. In the interval (3, 5), the series converges absolutely. If you aren't clear on what this means, look up the definition of absolute convergence in your book.

To finish the problem you need to check the two endpoints: x = 3 and x = 5. Substitute these numbers in your formula for the series and see what you get. Obviously (??) you won't be able to use the Ratio Test, but you should be able to use what other facts you know about series to say whether the series converges (either absolutely or conditionally) or diverges, at each of these values. If you aren't clear on conditional convergence, look up that term as well.
arl146
#30
Feb26-12, 01:17 AM
P: 343
I know that, I was explaining how I understood why it is (3,5) instead of (4,5).

I think at x=5 it absolutely converges. It seems to get closer and closer to one and theyre all positive values so Looking for absolute convergence with the absolute values doesn't change that.

For x=3 I think it coverages just conditionally. Not absolute because the values don't seem to get closer to a Specific value when you add the absolute value signs... And I don't think it diverges because The values are -,+,-,+ and so on an they seem to end up going towards a specific number

Although, I've never really understood when something converges or diverges
vela
#31
Feb26-12, 01:52 AM
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You need to be able to prove your conclusions. You can't just look at a few of the partial sums and guess that it's going to converge or not.

What series do you get when you set x=5?
arl146
#32
Feb26-12, 07:04 PM
P: 343
i got summation [n(1)^n] / [n^3 + 1] for x=5. and the same for x=3 just with the negative, [n(-1)^n] / [n^3 + 1]

and thats when n=1 to infinity
Mark44
#33
Feb26-12, 07:12 PM
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Quote Quote by arl146 View Post
i got summation [n(1)^n] / [n^3 + 1] for x=5. and the same for x=3 just with the negative, [n(-1)^n] / [n^3 + 1]

and thats when n=1 to infinity
So when x = 5, the series is
[tex]\sum_{n = 1}^{\infty}\frac{n\cdot 1^n}{n^3 + 1}[/tex]
Does that series converge or diverge? (You should simplify it first.)
Why?

What about when x = 3? Same questions.
arl146
#34
Feb26-12, 07:20 PM
P: 343
no, not n^3 + 3, its n^3 + 1 ... right

i said that it absolutely converges. but the other person said i need to prove it. i dont know how i do that
Mark44
#35
Feb26-12, 08:53 PM
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Quote Quote by arl146 View Post
no, not n^3 + 3, its n^3 + 1 ... right
That was a typo, which I have now fixed.
Quote Quote by arl146 View Post

i said that it absolutely converges. but the other person said i need to prove it. i dont know how i do that
You must have worked similar problems a little earlier in your course. What techniques do you have to determine whether a series converges?
arl146
#36
Feb28-12, 05:31 PM
P: 343
you mean like, root test, alternating series test, integral test .. those kinds? wont the root test work for this?


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