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Convergent and Divergent series (2 Viewers)
- Thread starter KeypadSDM
- Start date
freaking_out
Saddam's new life
muhauhahaha- everyone's gonna spam affinity's e-mail now.
freaking_out
Saddam's new life
hehe, sif u can make a spammer pay!
turtle_2468
Member
Okay, does anyone want proof posted up
? I sent a copy off last night, so if ppl want it I don't have to do any more work
David
? I sent a copy off last night, so if ppl want it I don't have to do any more work
David
freaking_out
Saddam's new life
yeah, post it up here for all the maths amateurs!!
freaking_out
Saddam's new life
in that case put it up!!! we just wanna freak out!!
turtle_2468
Member
Umm... part of the proof omitted, I will put it in later tonight. Reason being that it's part of an assignment, and the deadline is today, so I don't want to do anything dodgy...
Okay.
Proof of convergent/divergent series...
I'm taking out the bits of Analysis that have been relevant so far. Most of them are also accessible on the website for MATH3610 on http://www.maths.unsw.edu.au/ForStudents/courses/math3610/index.shtml, I highly recommend you have at least chapter 1 open when you read this. (eg I really am too lazy to reproduce Cantor's argument, it's really nice tho.. go read that bit if nothing else)
Preliminary defns: |A|<=|B| for infinite sets means that there is a 1-1 (or onto) function from A to B ie one that doesn't have two function values equalling the same thing in B.
Countable: Can put in bijection with N. equivalent to "can put in a list containing every member"
In higher analysis I we proved that N^N (N being natural numbers, and if you take N^2 as pairs of ordered numbers, N^N is all sequences of natural numbers) is countable ie |N^N|=|N| and you can put it in a list as you can N.
(this is the bit I'll put up say a little later. It's an interesting exercise, not that easy though..)
Anyway, back to the question. The set of all series would then be R^N. By Cantor's diagonalisation argument, R is not countable (ie infinity of R is a step higher than infinity of N). It's quite easy to prove that if a subset of set X is not countable then X is not countable either. Therefore as R is uncountable, R^N is not countable either (take the n-tuple (x,0,0,0,0,....) which is basically same as R, and a subset of R^N... skipping a few slight definitonal problems here to save time)
BUT you can show that |R^N|=|R|. So the cardinality of a subset of R^N, ie set of divergent series (call this D) or convergent (call C) is at most that of R. (|D|<=|R|)
But now consider the first element of the series in D. By a function which takes a sequence (x_1, x_2,...) to the number x_1, f then has range all real numbers. Hence we have a one-to-one function f going to R, and so D must have cardinality at least that of R. (|D|>=|R|)
Taking the previous 2 facts, it would be obvious for finite sets that |D|=|R|, but actually you need a hard-to-prove (so I'm told) theorem to conclude that. Nevertheless it's true.
Hence you can find a bijection from D to R.
Similarly |C|=|R|, so there is a bijection from C to R and hence 1 from C to D (just put 2 bijections together).
Notice that divergence/convergence doesn't ultimately matter as we just take the first term
Okay.
Proof of convergent/divergent series...
I'm taking out the bits of Analysis that have been relevant so far. Most of them are also accessible on the website for MATH3610 on http://www.maths.unsw.edu.au/ForStudents/courses/math3610/index.shtml, I highly recommend you have at least chapter 1 open when you read this. (eg I really am too lazy to reproduce Cantor's argument, it's really nice tho.. go read that bit if nothing else)
Preliminary defns: |A|<=|B| for infinite sets means that there is a 1-1 (or onto) function from A to B ie one that doesn't have two function values equalling the same thing in B.
Countable: Can put in bijection with N. equivalent to "can put in a list containing every member"
In higher analysis I we proved that N^N (N being natural numbers, and if you take N^2 as pairs of ordered numbers, N^N is all sequences of natural numbers) is countable ie |N^N|=|N| and you can put it in a list as you can N.
(this is the bit I'll put up say a little later. It's an interesting exercise, not that easy though..)
Anyway, back to the question. The set of all series would then be R^N. By Cantor's diagonalisation argument, R is not countable (ie infinity of R is a step higher than infinity of N). It's quite easy to prove that if a subset of set X is not countable then X is not countable either. Therefore as R is uncountable, R^N is not countable either (take the n-tuple (x,0,0,0,0,....) which is basically same as R, and a subset of R^N... skipping a few slight definitonal problems here to save time)
BUT you can show that |R^N|=|R|. So the cardinality of a subset of R^N, ie set of divergent series (call this D) or convergent (call C) is at most that of R. (|D|<=|R|)
But now consider the first element of the series in D. By a function which takes a sequence (x_1, x_2,...) to the number x_1, f then has range all real numbers. Hence we have a one-to-one function f going to R, and so D must have cardinality at least that of R. (|D|>=|R|)
Taking the previous 2 facts, it would be obvious for finite sets that |D|=|R|, but actually you need a hard-to-prove (so I'm told) theorem to conclude that. Nevertheless it's true.
Hence you can find a bijection from D to R.
Similarly |C|=|R|, so there is a bijection from C to R and hence 1 from C to D (just put 2 bijections together).
Notice that divergence/convergence doesn't ultimately matter as we just take the first term
freaking_out
Saddam's new life
hey there turtle, ya link (http://www.maths.unsw.edu.au/ForStudents/courses/math3610/index.shtml ) seems to b a dead link.
Xayma
Lacking creativity
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Works for me.Originally posted by freaking_out
hey there turtle, ya link (http://www.maths.unsw.edu.au/ForStudents/courses/math3610/index.shtml ) seems to b a dead link.
freaking_out
Saddam's new life
haha, it works now...i guess it got fixed up.
KeypadSDM
B4nn3d
Wow, I didn't expect my question to be answered...
"In higher analysis I we proved that N^N is countable ie |N^N|=|N| and you can put it in a list as you can N.
(this is the bit I'll put up say a little later. It's an interesting exercise, not that easy though..)"
Hmm... I claimed it was uncountable since 2^N is a subset of N^N and we know that |2^N|>|N|... of course if you found a bijection that would probably do the trick, though I fail to see that there would be one under the circumstances!
edit:
In fact, it's obviously not countable!
Apply Cantor's diagonalization arguement to the thing!
(this is the bit I'll put up say a little later. It's an interesting exercise, not that easy though..)"
Hmm... I claimed it was uncountable since 2^N is a subset of N^N and we know that |2^N|>|N|... of course if you found a bijection that would probably do the trick, though I fail to see that there would be one under the circumstances!
edit:
In fact, it's obviously not countable!
Apply Cantor's diagonalization arguement to the thing!
Last edited:
turtle_2468
Member
*cries*
don't want to think about it... grr, there goes my 100
although, there is still a lot more to come
Anyway, I guess marks don't matter for me anyway. But I don't like being wrong... grr
I'll think about that one. But I think my argument still works... you can't apply Cantor's because it's not exactly the same... The sequences don't have to be infinite. ie for the place you choose it could be a finite sequence, and things would stuff up..
ie (5,4) is a member of N^N as I take it..
that's a sequence right?
don't want to think about it... grr, there goes my 100
although, there is still a lot more to come
Anyway, I guess marks don't matter for me anyway. But I don't like being wrong... grr
I'll think about that one. But I think my argument still works... you can't apply Cantor's because it's not exactly the same... The sequences don't have to be infinite. ie for the place you choose it could be a finite sequence, and things would stuff up..
ie (5,4) is a member of N^N as I take it..
that's a sequence right?
Yes, it's a sequence, but the 2^N argument still works, even in that case.
And in any event you can apply Cantor's as follows:
Let c_k = 2 if d_kk = 1
Let c_k = 1 if d_kk = anything other than 1, including if d_kk does not exist.
This still generates a sequence of natural numbers, which hence must lie in N^N
And in any event you can apply Cantor's as follows:
Let c_k = 2 if d_kk = 1
Let c_k = 1 if d_kk = anything other than 1, including if d_kk does not exist.
This still generates a sequence of natural numbers, which hence must lie in N^N
turtle_2468
Member
umm... I'll go and take a rest.
I think you're right...
I think you're right...
freaking_out
Saddam's new life
oooh maniac guy trying to teach turtle a thing or two...hmm...what has the world come to.
Grey Council
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hah! both of them could teach YOU a thing or two.