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Topic: harmonic (Read 4621 times) |
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Grimbal
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Re: harmonic
« Reply #25 on: Oct 3rd, 2007, 8:21am » |
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Srn347,
it seems to me that you are very curious about mathematics, you want to explore it beyond what you are taught at school, and you already know much more than typical for your age.
But your approach is way too careless. You grab all the concepts you can get your hands on and apply them without any precaution. And in the process you take all the wrong turns.
The example you gave is one of the pitfalls of infinite series: sometimes, by reorganizing the terms you can change the result.
See Conditional Convergence on Mathworld.
I can understand that you want to explore and try more "exotic" concepts in mathematics. I do because I did the same at your age. But you have to know that each concept has a "user's manual" attached that tells you when and how the concept can be used. You cannot just pick it up and use it as your intuition tells you it should be used. To use it properly, you have to take the time to study the subject. If you don't, you are moving in a minefield. You will get funny results and you won't know why. If you are lucky, you will stumble over a paradox, like a "proof" that 1=0. These are a sure sign that you haven't been careful enough in one or the other concept you used. It is something you shouldn't ignore. Get back and understand what you did wrong.
I don't just want to criticize you. (In fact, yes, when you make bold claims about things you know only superficially.) But I am worried about your approach of mathematics. You cannot learn mathematics by just trying things and see if they works. It's not going to work in the long term. Eventually, you have to take a good book and learn the concepts thoroughly, chapter by chapter.
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towr
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Re: harmonic
« Reply #26 on: Oct 3rd, 2007, 8:32am » |
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on Oct 2nd, 2007, 11:38pm, srn347 wrote:| The first one is zero, because it is the harmonic series(infinity) minus itself. 1-1/2+1/3...=1-(1-1/2)+1/3-(1/2-1/4)...=(1-1)+(1/2-1/2)+(1/3)-1/3... |
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Roughly you can say  ni=1 1/i   + ln n (the approximation improves with n) where is the Euler constant.
And so the alternating harmonic series up to the nth term (for large n) gives + ln n - ( + ln n/2) = ln n - ln n/2 = ln n/[n/2] = ln 2
So as a formality we take n to infinity, and get lim n   ln 2 = ln 2
What you seem to be doing on the other hand, is
lim n (ln n - ln n/2) = lim n ln n - lim n ln n/2 = - , but you shouldn't haphazardly treat limits like that.
Who wants to try 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + ... ?
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| « Last Edit: Oct 3rd, 2007, 8:35am by towr » |
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Grimbal
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Re: harmonic
« Reply #27 on: Oct 3rd, 2007, 8:41am » |
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on Oct 3rd, 2007, 8:32am, towr wrote:| Who wants to try 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + ... ? |
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Easy! It is (1+1/2+1/3+...) - (2/3+2/6+2/9+...)
= (1+1/2+1/3+...) - 2/3·(1+1/2+1/3+...)
= infinity - 2/3·infinity = 1/3·infinity.
It is not infinity, but quite big anyway. 
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| « Last Edit: Oct 3rd, 2007, 8:41am by Grimbal » |
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pex
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Re: harmonic
« Reply #28 on: Oct 3rd, 2007, 8:53am » |
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on Oct 3rd, 2007, 8:41am, Grimbal wrote:
Easy! It is (1+1/2+1/3+...) - (2/3+2/6+2/9+...)
= (1+1/2+1/3+...) - 2/3·(1+1/2+1/3+...)
= infinity - 2/3·infinity = 1/3·infinity.
It is not infinity, but quite big anyway.
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LOL! 
More seriously, I find divergence to positive infinity.
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pex
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Re: harmonic
« Reply #29 on: Oct 3rd, 2007, 9:46am » |
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One that I just came up with, solved, and found interesting:
1/1 + 1/2 - 1/3 - 1/4 + 1/5 + 1/6 - 1/7 - 1/8 + 1/9 + 1/10 - 1/11 - 1/12 + ...
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Obob
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Re: harmonic
« Reply #30 on: Oct 3rd, 2007, 10:07am » |
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Without even knowing that the alternating harmonic series sums to ln 2, it is trivial to see that at least it is positive or divergent: regroup the terms like this:
(1-1/2)+(1/3-1/4)+(1/5-1/6)+...
Then the even partial sums S2, S4, S6, ... are increasing, since each of the parenthesized terms above is positive, and S2 is 1/2. So, if the series converges, its limit is at least 1/2.
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| « Last Edit: Oct 3rd, 2007, 10:07am by Obob » |
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pex
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Re: harmonic
« Reply #31 on: Oct 3rd, 2007, 10:26am » |
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on Oct 3rd, 2007, 9:46am, pex wrote:One that I just came up with, solved, and found interesting:
1/1 + 1/2 - 1/3 - 1/4 + 1/5 + 1/6 - 1/7 - 1/8 + 1/9 + 1/10 - 1/11 - 1/12 + ... |
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And, a bit more difficult:
1/1 + 1/2 + 1/3 - 1/4 - 1/5 - 1/6 + 1/7 + 1/8 + 1/9 - 1/10 - 1/11 - 1/12 + ...
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ThudnBlunder
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Re: harmonic
« Reply #32 on: Oct 3rd, 2007, 10:43am » |
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on Oct 2nd, 2007, 11:38pm, srn347 wrote:| The first one is zero, because it is the harmonic series(infinity) minus itself. 1-1/2+1/3...=1-(1-1/2)+1/3-(1/2-1/4)...=(1-1)+(1/2-1/2)+(1/3)-1/3... If you think it's arrogant, try taking ritalin. if you are offended, blame the one whom the phase originated from. |
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IIRC, one can re-bracket terms of an infinite series and get the right answer iff the series is absolutely convergent. Your series is conditionally convergent only. You simply don't know the answers to you own problems.
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| « Last Edit: Nov 9th, 2007, 2:43am by ThudnBlunder » |
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Eigenray
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Re: harmonic
« Reply #33 on: Oct 3rd, 2007, 11:08am » |
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1/1 + 1/2 + ... + 1/r - 1/(r+1) - 1/(r+2) - ... - 1/(2r) + 1/(2r+1) + ... + 1/(3r) - 1/(3r+1) - ... - 1/(4r) + 1/(4r+1) + ....
= 1/r * { log 2 + sin /(1-cos) * arctan[sin/(1-cos)] }
where the sum is over = (2k+1) /r, k going from 1 to r. The first few are
log(2)
1/2 [ log 2 + /2 ]
1/3 [ log 2 + 2/ 3 ]
1/4 [ log 2 + (1/2 + 2) ]
1/5 [ log 2 + 2{1+2/5} ]
1/6 [ log 2 + /6 (15 + 43) ]
1/7 [ log 2 + ?? ]
1/8 [ log 2 + (1/2 + 2 + 2{2+2})
...
1/r [ log 2 + *  r ].
Is r always algebraic?
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| « Last Edit: Oct 3rd, 2007, 2:51pm by Eigenray » |
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Obob
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Re: harmonic
« Reply #34 on: Oct 3rd, 2007, 11:09am » |
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If a series is convergent, then you can "rebracket" the terms however you want, in the sense that if your series is a0+a1+a2+... then it will equal say (a0+a1)+(a2+...). That is, you can add parentheses however you wish, so long as you don't rearrange terms. Adding parentheses corresponds to passing to a subsequence of the sequence of partial sums of the series. Since a subsequence of a convergent sequence is convergent and converges to the same limit, adding brackets doesn't change the limit.
However, you can rearrange the terms of the series if and only if the series is absolutely convergent. If the series converges, but not absolutely, then the terms of the series can be rearranged to make the sum come out to any value you want. So, by suitable rearrangement of the terms of the alternating harmonic series, we can make the limit be 0, say. This does not mean that the limit of the alternating harmonic series is 0, though! It is undeniably ln 2.
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pex
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Re: harmonic
« Reply #35 on: Oct 3rd, 2007, 11:22am » |
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on Oct 3rd, 2007, 11:08am, Eigenray wrote:1/1 + 1/2 + ... + 1/(r-1) - 1/r - 1/(r+1) - ... - 1/(2r-1) + 1/(2r) + ... + 1/(3r-1) - 1/(3r) - ... - 1/(4r-1) + 1/(4r) + ....
= 1/r * { log 2 + sin/(1-cos) * arctan[sin/(1-cos)] }
where the sum is over = (2k+1)/r, k going from 1 to r. |
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Wow! Mind if I don't see where this comes from straight away? (Anyway, your first three values match the ones I found - so I guess my messy integrals were correct.)
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Eigenray
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Re: harmonic
« Reply #36 on: Oct 3rd, 2007, 11:26am » |
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If you already have them as integrals, try partial fractions.
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pex
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Re: harmonic
« Reply #37 on: Oct 3rd, 2007, 11:31am » |
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on Oct 3rd, 2007, 11:08am, Eigenray wrote:| Is r always algebraic? |
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If we let k go from 0 to r-1 (which makes no difference), so that all thetas are between zero and 2pi, and we recall sin(theta)/(1-cos(theta)) = cot(theta/2), we find that the summands equal
sin(theta)/(1-cos(theta)) * arctan(sin(theta)/(1-cos(theta))
= cot(theta/2) * arctan(cot(theta/2))
= cot(theta/2) * (pi - theta)/2.
Cotangents of such angles are always algebraic (IIRC) and (pi-theta) is a rational multiple of pi. So I'd say the answer is yes.
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| « Last Edit: Oct 3rd, 2007, 11:33am by pex » |
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pex
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Re: harmonic
« Reply #38 on: Oct 3rd, 2007, 11:32am » |
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on Oct 3rd, 2007, 11:26am, Eigenray wrote:| If you already have them as integrals, try partial fractions. |
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Yes, I did that before I posted the problems - I had found them in exactly the same forms you posted.
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Eigenray
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Re: harmonic
« Reply #39 on: Oct 3rd, 2007, 12:53pm » |
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on Oct 3rd, 2007, 11:31am, pex wrote:| we recall sin(theta)/(1-cos(theta)) = cot(theta/2) |
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Aha, yes. I suck at trig.
on Oct 3rd, 2007, 11:31am, pex wrote:| Cotangents of such angles are always algebraic (IIRC) and (pi-theta) is a rational multiple of pi. So I'd say the answer is yes. |
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Yes; they can be expressed in terms of a 2r-th root of unity, so r has degree no more than  (2r) over  . But it looks like we should be able to improve this bound to just (r).
on Oct 3rd, 2007, 11:32am, pex wrote:
Yes, I did that before I posted the problems - I had found them in exactly the same forms you posted. |
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Do you have the general solution as an integral?
We can show also that if
S(r) = 1/r [ log 2 + cot(/2)(-)/2 ]
is the sum in question, then
S(r) - log(2r/) - 0
as r . Do you see how?
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| « Last Edit: Oct 3rd, 2007, 12:58pm by Eigenray » |
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pex
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Re: harmonic
« Reply #40 on: Oct 3rd, 2007, 1:41pm » |
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on Oct 3rd, 2007, 12:53pm, Eigenray wrote:| Do you have the general solution as an integral? |
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Isn't it integral(0..1) (1-xr) / [(1-x)(1+xr)] dx? (Of course, we could cancel (1-x), but it just doesn't look as nice.)
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| « Last Edit: Oct 3rd, 2007, 1:41pm by pex » |
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pex
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Re: harmonic
« Reply #41 on: Oct 3rd, 2007, 1:57pm » |
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on Oct 3rd, 2007, 12:53pm, Eigenray wrote:We can show also that if
S(r) = 1/r [ log 2 + cot(/2)(-)/2 ]
is the sum in question, then
S(r) - log(2r/) - 0
as r . Do you see how? |
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Looking at it for a while: no, I don't. Unfortunately, I don't have any more time to spend on the problem now. 
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Eigenray
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Re: harmonic
« Reply #42 on: Oct 3rd, 2007, 2:49pm » |
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on Oct 3rd, 2007, 1:41pm, pex wrote:
Isn't it integral(0..1) (1-xr) / [(1-x)(1+xr)] dx? (Of course, we could cancel (1-x), but it just doesn't look as nice.) |
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And what happens when you expand this using partial fractions?
on Oct 3rd, 2007, 1:57pm, pex wrote:
Looking at it for a while: no, I don't. Unfortunately, I don't have any more time to spend on the problem now. |
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It can be done using the fact that
 02 { cot(t/2)(-t)/2 - [1/t+1/(2-t)] } dt = -2log(2).
(I just used Mathematica for this.) But maybe there is a nicer way.
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| « Last Edit: Oct 3rd, 2007, 2:53pm by Eigenray » |
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Sir Col
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Re: harmonic
« Reply #43 on: Oct 3rd, 2007, 3:30pm » |
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srn347, this is the first occasion I've had to say anything negative towards you, but your posts in this thread are entirely rude. You have received pertinent response to your original questions, so your subsequent responses are clear evidence that you are out of your depth with the mathematics. Moreover, in reading your responses here and in other threads I suspect that you have not quite appreciated how lucky you are to be a member on this forum. There are few other places on the planet where so many talented, knowledgeable, and skilled expositors gather. And what is more, they are always willing and entirely generous in sharing their knowledge.
I suggest you take serious stock of your actions and consider carefully if you wish to remain a member. Patience is wearing very thin, and in most cases you have already gone too far.
A sincere apology would be a starting point. From there I would strongly advise a more measured approach to any future posts. If you genuinely want help, then ask; if someone is explaining something and is using terms/concepts above your current level of mathematics then say so. Otherwise, if you have nothing useful to say, then please say nothing.
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ThudnBlunder
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Re: harmonic
« Reply #44 on: Oct 3rd, 2007, 7:30pm » |
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Well said, Grimbal and Sir Col!
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ima1trkpny
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Re: harmonic
« Reply #45 on: Oct 3rd, 2007, 7:35pm » |
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on Oct 3rd, 2007, 7:30pm, ThudanBlunder wrote:| Well said, Grimbal and Sir Col! |
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True. However it seems he is bent on being defiant and won’t take a hint even though he has been banned.
on Oct 3rd, 2007, 7:11pm, sm347 wrote:| But if you ban me, because i am so clever, I will find many more ways of showing you that I am smarter than u. |
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ima1trkpny
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Re: harmonic
« Reply #47 on: Jan 26th, 2008, 7:37pm » |
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on Jan 26th, 2008, 5:55pm, temporary wrote:
Prove it.
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temporary
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Re: harmonic
« Reply #48 on: Jan 27th, 2008, 11:25am » |
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Compare what they say.
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mikedagr8
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Re: harmonic
« Reply #49 on: Jan 27th, 2008, 3:40pm » |
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on Jan 27th, 2008, 11:25am, temporary wrote:
Look at how sm347 writes his age. Everyone who was used to srn347 (sad to say) knows he was obsessed with infinity. Maybe he was an imposter sm347. It really doesn't matter. He was booted before then. Someone was just having a joke. It hasn't harmed anyone as srn347 was already banned. I'm sorry you feel that way and you are certain they are not the same. I understand your opinions but I think this subject of srn347 should be put behind us. I don't want the problems he caused arrising again.
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| « Last Edit: Jan 27th, 2008, 7:05pm by mikedagr8 » |
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