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Topic: Impish Pixie revisited (Read 7754 times) |
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FredFnord
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Impish Pixie revisited
« on: May 30th, 2005, 3:03pm » |
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Suppose you have an infinite number of small balls, all of which have been uniquely numbered from 1 upwards. What is more, you have an infinitely large bucket, by which I mean it can be made as large as necessary. You decide to fill the bucket by throwing in all the balls, in order. Starting from 1, every minute you throw in two balls. But every minute, an impish pixie takes one ball back out. He always extracts the lowest-numbered ball in the bucket.
For example :
1st minute: You throw in Ball 1 and Ball 2.
Pixie extracts Ball 1.
2nd minute: You throw in Ball 3 and Ball 4.
Pixie extracts Ball 2.
3rd minute: You throw in Ball 5 and Ball 6.
Pixie extracts Ball 3.
and so on...
QUESTION: After an infinite amount of time has elapsed, how many balls are in the bucket?
Argument 1: There is an infinite number of balls in the bucket. After 1 minute there is 1 ball. After 2 minutes there are 2 balls. After 3 minutes there are 3 balls, etc.
Argument 2: There are no balls in the bucket. If there are some balls in the bucket, what is the lowest-numbered ball? It can't be Ball 1; that was extracted after 1 minute. Similarly, it can't be Ball 2; that was extracted after 2 minutes. It can't be Ball 3; that was extracted after 3 minutes, etc.
If the phrase 'after an infinite amount of time has elapsed' bothers you, then we can change the problem so that the 1st put-in-and-take-out operation is completed in 1/2 minute, the 2nd operation is completed in 1/4 minute, the 3rd in 1/8 minute, and so on. Now you can ask the question after 60 seconds, and "infinite time" is not longer an issue.
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I must admit to being somewhat unsatisfied with the answers to the Impish Pixie question. Unfortunately, I can't quite get a handle on the math that was posted in the original thread. (Which was apparently closed due to missing posts.) It might have been easier if we had access to real mathematical symbols... [omega] makes my head hurt. It also might have been easier if I'd gone to a real college. But oh well.
(Incidentally, I tried unicode but the math symbols still don't work, even if you're in a font that has them. Alas. Perhaps someone should revisit the forum software with a hacksaw and some glue?)
Anyway, here is the original thread:
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_har d;action=display;num=1049309591;start=0
I still can't make myself agree with the posted solution. Let me tell you why.
First, I think that everyone will agree with me that you have an equivalent problem if, instead of having one ball with every counting number, he has two. Two ones, two twos, two threes, etc, both placed into the bucket at the same time. Correct? Because the pixie must, by the same logic as before, have taken both of the ones by time 2, both of the twos by time 4, etc.
Now, a different question: imagine a person putting two balls in per turn, but one ball is always numbered '1' and the other balls are labeled with the counting numbers, including one. In this case, what will be left over is an infinite number of balls, labeled 1, 2, 3, etc. The pixie takes a '1' every time, and your other ball is safe. No? Yes?
Now, let's say you have two bowls. One has its own set of balls, labeled as above. (An infinite number of 1s, plus an infinite set of counting numbers.) The other is unlabeled, and likewise infinite. Each time, you put two balls into the first bowl as above, and you put two balls into the other bowl as well. The pixie takes one out of each, choosing the '1' ball in the first case and a random one in the second case. When you run out of numbered balls, both you and the pixie stop. Do the two bowls end up with the same number of balls in them? How can they not?
Yes, it's pretty obvious where this is going. Two bowls. Each round, I place two identically-numbered balls into the first bowl as in my first example. In the second bowl, each turn, I place a ball marked one and a ball marked with a counting number (the same number as on the first two identically-numbered balls), as in my second and third examples. The pixie takes the lowest-numbered ball out of each of the bowls each turn.
Do I, in fact, 'end up with' no balls at all (no balls at all, I married a man who had...) in the first bucket and an infinite number in the second bucket? This is clearly what is implied by the earlier answer, although if one were observing this from far enough away that one could not see the labels on the balls, the behavior for every step would be identical. Further, in theory, you run out of balls in both sets at the same time... there's a perfectly good one-to-one correspondence between them. Or one-to-two, depending on how you look at it.
By the way, I certainly do agree that things get weird and discontinuous and do not respond well to limits when you hit infinity. I'm just not yet convinced that that's the right way to explain this. And, frankly, I'm not sure that it's actually a rational question with a useful, unique answer.
Just as a bonus, what happens if you put in an infinite number of balls at each step, starting with the number of each step and ending at infinity? So first step you put in 1, 2, 3, ... and the pixie takes out 1. Second step you put in 2, 3, 4... and the pixie takes out one of the twos. Third step you put in 3, 4, 5... and the pixie takes out the other two. There are no twos left, and there will never be more than three threes, or four fours, so by the original logic there will be no more balls in the basket at time infinity.
Want even more? Double the number of marbles put down each time. 1, 2, 3, 4... on the first round, 2, 2, 3, 3, 4, 4... on the second, 3, 3, 3, 3, 4, 4, 4, 4 on the third, etc. There are still a finite, readily-calculable number of each digit, so the argument with the pixie still stands: for any counting number, you can point to a step and say 'that's when that number will be gone'.
You'll have to forgive me if I use my terms wrong here, because I never got as far in math as doing much with infinities. (Pity, it sounds like it would have been more fun than the math I did get to.) But isn't the set that you end up laying down if you do it this way some higher order than aleph null, but the pixie is compassing it using aleph null?
Now, admittedly, that's not the same as the original problem. But I don't see any reason that the original problem's logic is not just as applicable to it.
-fred
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Deedlit
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Re: Impish Pixie revisited
« Reply #1 on: May 30th, 2005, 3:58pm » |
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Quote:First, I think that everyone will agree with me that you have an equivalent problem if, instead of having one ball with every counting number, he has two. Two ones, two twos, two threes, etc, both placed into the bucket at the same time. Correct? Because the pixie must, by the same logic as before, have taken both of the ones by time 2, both of the twos by time 4, etc.
Now, a different question: imagine a person putting two balls in per turn, but one ball is always numbered '1' and the other balls are labeled with the counting numbers, including one. In this case, what will be left over is an infinite number of balls, labeled 1, 2, 3, etc. The pixie takes a '1' every time, and your other ball is safe. No? Yes? |
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Sure. You've just described a basic paradox of infinite sets: the difference between two infinite sets is not well-defined. Subtracting one countably infinite set from another can result in a set of any finite size, or of infinte size; it all depends on which elements are subtracted, not just on how many. Since our common sense has already failed, we shouldn't be surprised when things just get crazier from there.
Quote:| Now, let's say you have two bowls. One has its own set of balls, labeled as above. (An infinite number of 1s, plus an infinite set of counting numbers.) The other is unlabeled, and likewise infinite. Each time, you put two balls into the first bowl as above, and you put two balls into the other bowl as well. The pixie takes one out of each, choosing the '1' ball in the first case and a random one in the second case. When you run out of numbered balls, both you and the pixie stop. Do the two bowls end up with the same number of balls in them? How can they not? |
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Since the difference depends on which balls are removed, having the balls unlabelled means the result could be anything.
Quote:| Do I, in fact, 'end up with' no balls at all (no balls at all, I married a man who had...) in the first bucket and an infinite number in the second bucket? This is clearly what is implied by the earlier answer, although if one were observing this from far enough away that one could not see the labels on the balls, the behavior for every step would be identical. Further, in theory, you run out of balls in both sets at the same time... there's a perfectly good one-to-one correspondence between them. Or one-to-two, depending on how you look at it. |
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As we have discovered already, cardinality is not enough to describe the situation. One way to distinguish between the two cases is to look at the order types of the two sets of balls. The first set has order type [omega] - that is, it's a regular sequence:
1,1,2,2,3,3,4,4,5,5,...
The other has order type [omega] * 2:
1,1,1,1,1,1,1,1,1,1,...,
2,3,4,5,6,7,8,9,10,11...
If we allowed the imp two shots at it, it would remove the 1's during the first "infinite time" step, and the rest during the second.
I describe order types in more detail in the "Another (unusual) Number Game" thread.
Because we are only putting balls in the basket two at a time, the imp doesn't always take the smallest ball from the set, so that adds a further wrinkle to the problem.
Quote:| By the way, I certainly do agree that things get weird and discontinuous and do not respond well to limits when you hit infinity. I'm just not yet convinced that that's the right way to explain this. |
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Well, we're trying to explain why a certain abstract situation doesn't satisfy our common sense notions. There doesn't seem to be a good way to explain this to someone who doesn't want to believe it. Basically, our experience is based entirely in the finite, and conceptions of infinity simply don't fall in that realm. (and, some would argue that it's simply a mental fiction)
Quote:| And, frankly, I'm not sure that it's actually a rational question with a useful, unique answer. |
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Let's put it this way: We need to give a mathematical interpretation to any problem couched in non-mathematical terms. Obviously the answer we get will depend on which mathematical interpretation we give to the problem. This decision is necessarily nonmathematical and subjective, but most of the time we all agree on how to interpret the problem. There are certainly cases of ambiguity, however, and infinite structures in particular are goldmines for such ambiguities.
For the original problem, there seems to be a very natural mathematical interpretation. The balls in the basket at any particular time form a set, like S4 - {5,6,7,8}, and the "number of balls at infinity" clearly implies taking the limit of the sequence of sets. (Not all sequences of sets have well-defined limit, but this one does.) So this mathematical problem has an unambiguous answer of the null set.
An alternate mathematical interpretation is to just assign to xi the number of balls in the basket at time i, and take the limit as i goes to infinity. This gives an answer of infinity. It just seems, comparing the two models, that the first is "really what's happening" and the second fails because it's not given enough information. But, since the second works in all finite cases, our common sense tells us that it should be fine for the infinite case as well.
If the balls are unlabelled, the problem doesn't have a unique answer using the first model, while the second still gives infinity. So the temptation is to use the second model here. This gives a sort of beachhead through which we can start arguing about the labelled problem.
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| « Last Edit: May 30th, 2005, 4:04pm by Deedlit » |
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Ajax
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Re: Impish Pixie revisited
« Reply #2 on: May 31st, 2005, 12:16am » |
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First of all, I totally sympathise with Fred.
I believe that infinite is a concept and nothing more. How can we do something infinetely (I'd be infinetely earlier bored)? If we only put one ball at a time, which would be the number of the ball inserted at time oo:00? There would never be a ball having oo marked on it.
Even the "conventional" proposal of putting balls after 1/2' then 1/4' then 1/8' doesn't save us from "infinitism" (I'm creating new words  ) as we will end up putting balls every 1/oo minutes. Is there anyone or anything with such a speed?
I am not seeking for mathematical replies (I read too many in the original thread and didn't manage to understand that much). I fully comprehend the paradox, but my belief is that it's a trap that noone should fall into.
Let's change a bit Argument 2:
If there are some balls in the bucket, what is the biggest-numbered ball?
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towr
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Re: Impish Pixie revisited
« Reply #3 on: May 31st, 2005, 1:07am » |
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Working out a few of the proposed problem using (multi)sets
in: {2i-1, 2i | i [in] N} = N
out: {i | i [in] N} = N
N\N = 0
in: {i,i | i [in] N} = {1,1,2,2,3,3,..}
out: {ceil(i/2) | i [in] N} = {1,1,2,2,3,3,..}
{1,1,2,2,3,3,..}\{1,1,2,2,3,3,..} = 0
in: {1,i | i [in] N} = {1,1,1,..,1,2,3,..}
out: {1 | i [in] N} = {1,1,1,..}
{1,1,1,..,1,2,3,..}\{1,1,1,..} = {2,3,4,..}
in: M1={1}
Mn=Mn-1 U {n}n (where S1=S and Sn = S U Sn-1)
Minf={1,2,2,3,3,3,4,4,4,4,...}
out: {ceil(sqrt(1+8i)/2 - 1/2) | i [in] N} = {1,2,2,3,3,3,4,4,4,4,...}
{1,2,2,3,3,3,4,4,4,4,...}\{1,2,2,3,3,3,4,4,4,4,...}=0
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| « Last Edit: May 31st, 2005, 1:10am by towr » |
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Deedlit
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Re: Impish Pixie revisited
« Reply #4 on: May 31st, 2005, 1:52am » |
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on May 31st, 2005, 12:16am, Ajax wrote:First of all, I totally sympathise with Fred.
I believe that infinite is a concept and nothing more.
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I'm not really following your position. If you believe that infinity is nothing more than a concept, then there are two ways to go. One way is to consider the problem the problem to be rubbish and refuse to discuss it. Let's not go that way.  The other way is to consider the problem, but purely as an abstract, mathematical problem, as that is really the only meaning we can attach to it. In that case, we should simply accept whatever our reasoning leads to, as that's all there is to do.
Quote:| How can we do something infinetely (I'd be infinetely earlier bored)? If we only put one ball at a time, which would be the number of the ball inserted at time oo:00? There would never be a ball having oo marked on it. |
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The idea is that all finite steps have occurred, and that's it. So we've added balls of every natural number, and the imp has removed balls of every natural number.
Quote:| Even the "conventional" proposal of putting balls after 1/2' then 1/4' then 1/8' doesn't save us from "infinitism" (I'm creating new words ) as we will end up putting balls every 1/oo minutes. Is there anyone or anything with such a speed? |
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It's not really a question of plausibility.
Quote:| I am not seeking for mathematical replies (I read too many in the original thread and didn't manage to understand that much). I fully comprehend the paradox, but my belief is that it's a trap that noone should fall into. |
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What is the trap exactly? Let me reiterate that the mathematical version is unambiguous - the limit of the sequence of sets is clearly the null set. Perhaps you are saying that the mathematical version is leading us astray. This would imply that there is some deeper reality to the problem that we are missing. But since you believe that infinity is just a concept, there should be no deeper reality.
Quote:Let's change a bit Argument 2:
If there are some balls in the bucket, what is the biggest-numbered ball? |
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Right - an infinite set doesn't necessarily have extrema. This is something we are more familiar with, though - it's not unusual to talk about, say, all numbers less than 1. Clearly there's no largest such number.
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| « Last Edit: May 31st, 2005, 1:55am by Deedlit » |
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rmsgrey
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Re: Impish Pixie revisited
« Reply #5 on: May 31st, 2005, 3:21am » |
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on May 31st, 2005, 1:52am, Deedlit wrote:
[...]all numbers less than 1. Clearly there's no largest such number. |
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For some value of clarity - the (now locked and replaced by a sticky) thread in the medium forum debating one candidate value was one of the longest threads on the forum.
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Ajax
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Re: Impish Pixie revisited
« Reply #6 on: May 31st, 2005, 3:43am » |
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on May 31st, 2005, 1:52am, Deedlit wrote:
The idea is that all finite steps have occurred, and that's it. So we've added balls of every natural number, and the imp has removed balls of every natural number. |
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But if there is always a bigger number than the last of "every natural number" there is no "has added" ; there is still "is adding", so we are still filling and the imp is still emptying; the process will never stop, hence the box will never be empty.
On the other hand, if we tried to do it in a finite period of time (already mentioned 1'), we would need infinite speed. Even if this could happen, and we were witnesses (even though we wouldn't be able to see the hands adding and subtracting), I can't think of finally seeing an empty box, but rather being instantly covered by infinate balls and finally dying either by the pressure of infinite weight on us or by suffocation, becoming thus unable to verify the outcome in the end of 1'. The universe would be filled by the infinite numbers of whatever the volume of the balls may be and everything would sink in an infinitely increasing sea . However, here we have a problem. The matter in universe is finite, so in order to aquire infinite amount of balls, they should be consisted of zero mass (and volume), therefore they will be practically invisible. So, the box will seem empty but we will have 0*oo indeterminateness.
Ok, don't take it all seriously 
Once again, I cannot judge or question the knowledge of so many, proving by maths the opposite. I'm only suggesting a more empirical way.
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| « Last Edit: May 31st, 2005, 3:46am by Ajax » |
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Deedlit
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Re: Impish Pixie revisited
« Reply #7 on: May 31st, 2005, 4:06am » |
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on May 31st, 2005, 3:43am, Ajax wrote:
But if there is always a bigger number than the last of "every natural number" there is no "has added" ; there is still "is adding", so we are still filling and the imp is still emptying; the process will never stop, hence the box will never be empty.
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Yes, after any finite number of steps. I think we can all agree that this problem doesn't work in physical reality. I said before that there were two ways to go from there; either call it rubbish and move on, or to fully accept the premise of "infinite time" or "limit" and discuss it further. It seems, though, that you want to discuss further the fact that it is rubbish.
Quote:
Once again, I cannot judge or question the knowledge of so many, proving by maths the opposite. I'm only suggesting a more empirical way. |
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I'm still confused by your position. You stated from the beginning that infinity, and therefore this problem, doesn't exist in physical reality, so we can only discuss it as an abstract problem. But you want to verify it emperically, via a physical experiment of some sort?
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Ajax
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Re: Impish Pixie revisited
« Reply #8 on: May 31st, 2005, 4:15am » |
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Forget my last sentence, just bls (isn't it for bullsh*t?)
However I'm still confused with the fact that we have successive additions and subtractions, preventing us from decreasing the sum.
You don't have to reply, as I fully understand the problem, but cannot accept it!!! 
Maybe such stuff is the fuel for the Infinite Improbability Drive...
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| « Last Edit: May 31st, 2005, 4:23am by Ajax » |
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towr
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Re: Impish Pixie revisited
« Reply #9 on: May 31st, 2005, 7:53am » |
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Taking the example where you compress the whole process into 1 minute.
Then for each number, you can give the exact moment it is put in, and the exact, later, moment it is pulled out.
And for any two numbered balls, you can give the time between them being put in or taken out.
So never do you have infinite speed, it's always finite. (The moment it would turn infinite you're finished.)
So basicly, if you can for each ball give the time it was put in, and the time it was taken out, and both ly within that 1 minute period, then afterwards the bucket must be empty.
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Grimbal
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Re: Impish Pixie revisited
« Reply #10 on: May 31st, 2005, 10:31am » |
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I believe the bucket ends up with infinity minus infinity balls. That is undefined.
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towr
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Re: Impish Pixie revisited
« Reply #11 on: May 31st, 2005, 11:59am » |
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on May 31st, 2005, 10:31am, Grimbal wrote:| I believe the bucket ends up with infinity minus infinity balls. That is undefined. |
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Only in infinitely many cases, but not so in infininitely many others.
Surely it wouldn't be a problem if you added 1 ball each step, and the imp removed it before you add the next. Still 'infinity minus infinity', but there's nothing left in the end.
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SMQ
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Re: Impish Pixie revisited
« Reply #12 on: May 31st, 2005, 2:29pm » |
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I thought, based on Icarus' arguments in the parent thread, that that case (where you put one in and the imp takes it back out) was still undefined too. After any finite number of steps there are clearly zero balls in the bucket, but the infinite case is unclear since there could be either zero or one balls in the bucket at any given point in continuous time.
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Re: Impish Pixie revisited
« Reply #13 on: May 31st, 2005, 2:59pm » |
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Well, it depends on whether they're numbered
(There's a little more to it, but I don't have the words to explain it now.. it's late..zzZZ)
Once you can say: 'no ball with any natural number can be left'; and then considering there only where balls with natural numbers to begin with, you're left to conclude they must all be gone.
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Deedlit
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Re: Impish Pixie revisited
« Reply #14 on: May 31st, 2005, 4:11pm » |
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on May 31st, 2005, 10:31am, Grimbal wrote:| I believe the bucket ends up with infinity minus infinity balls. That is undefined. |
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But the limit of the sequence of sets could certainly be defined.
Quote:| I thought, based on Icarus' arguments in the parent thread, that that case (where you put one in and the imp takes it back out) was still undefined too. After any finite number of steps there are clearly zero balls in the bucket, but the infinite case is unclear since there could be either zero or one balls in the bucket at any given point in continuous time. |
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I guess it depends on whether you define your sequence to be - if Sn is the set after the imp removes the nth ball, then all the Sn are null. If we include what's in the basket after we put balls in, but before the imp takes anything out, then it depends. (In the original problem it's the null set.)
Quote:Well, it depends on whether they're numbered
(There's a little more to it, but I don't have the words to explain it now.. it's late..zzZZ) |
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Why there should be a difference between numbered and unnumbered balls might be confusing, so let me explain. Of course, if we erase the numbers from the balls and do the exact same thing, we should expect the same result. The key is whether we consider each ball to have its own individual existence or not (and it seems natural that we should); then we can just mentally assign a name (like a natural number) to each ball, and ask what happens to that ball over time. So, it's not really about physical numbering, just whether the balls are distinguishable conceptually.
Once we have named the balls, we can hopefully answer the question by dividing it into two steps:
1) Look at ball i: Is that ball eventually left out of the basket permanently, or left in the basket permanently? In that case, we can say with certainty that the ball is out of the basket (or in) at "time infinity".
2) If we were able to decide all balls in part 1, then we know precisely the set of balls that are in the basket at time infinity.
I think many will accept part 1, but will balk when we get to part 2; they can't quite accept the notion that they can look at just one ball at a time, and ignore all other balls while they are doing that.
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| « Last Edit: May 31st, 2005, 4:15pm by Deedlit » |
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Icarus
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Re: Impish Pixie revisited
« Reply #15 on: May 31st, 2005, 4:32pm » |
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on May 30th, 2005, 3:03pm, FredFnord wrote:It might have been easier if we had access to real mathematical symbols... [omega] makes my head hurt. It also might have been easier if I'd gone to a real college. But oh well.
(Incidentally, I tried unicode but the math symbols still don't work, even if you're in a font that has them. Alas. Perhaps someone should revisit the forum software with a hacksaw and some glue?) |
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We HAD access to real mathematical symbols - that's why you see all those [omega]s and such in the other thread, and in many posts around here. William had managed to add a symbolry to the original YaBB software. But some growing resource management problems in the original software caused him to update to a newer version of YaBB. Unfortunately, the new version did not include any of the improvements he had added to the old. As he has time, he has been restoring this lost functionality, but the mathematical symbols have not yet been restored. So what you see in these posts are the tags we used to get the symbols to appear. (Apparently, unicode has security issues, so unicode symbolry has been disabled.)
I am aware that many of my posts in particular (including some in the original impish pixie thread), have been reduced to illegibility by this loss, but the only thing we can do is be patient, and wait for a very busy grad student to find the time to restore the capacity. 
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FredFnord
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Re: Impish Pixie revisited
« Reply #16 on: May 31st, 2005, 5:19pm » |
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I assume you mean the specific implementation of unicode that goes into the forum software has security issues? The entire product of the company that I work for is based around unicode, and I'm not aware of any security issues inherent to its design.
-fred
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Icarus
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Re: Impish Pixie revisited
« Reply #17 on: May 31st, 2005, 5:53pm » |
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That was just what I recalled from a conversation we had a long time ago here, starting with the struck-thru portion of the first post (the non-struck-thru text was added later, when the now disabled symbolry was added) and going through William's Reply #6. Since the struck-thru text is annoying to read, here is the pertinent part:
on Oct 21st, 2002, 6:19pm, william wu wrote:| (Note: So why can't you just type the HTML code for symbols that require ampersands? Well, YaBB will always parse your ampersands as ASCII text for security reasons: & is often used in HTML, and HTML is a security risk on any board. |
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After a discussion in which TimMann points out that not all browsers support the Symbol font, and wonders about the ampersand restriction, WWu replies with:
on Dec 15th, 2002, 12:31am, william wu wrote:| I'm not sure why ampersands are a hazard either, but that's what some supposed YaBB expert told me. |
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So if you want to find out more, I guess you'll have to contact the makers of YaBB.
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I wonder: Which is larger
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Icarus
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Re: Impish Pixie revisited
« Reply #18 on: May 31st, 2005, 6:27pm » |
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Concerning the subject of this thread, I agree with Deedlit, of course. But I would like to give my perspective on the same things:
No - the situation cannot occur in the real universe, for numerous reasons. Therefore the question of "what would actually happen if we tried this?" is meaningless. The only thing we can talk about is what happens in logically consistant models of the event. If we choose different models, of course the results may disagree.
There are consistant models for this puzzle for which the urn is not empty at the end (the "omega point"). However, the balls in the urn at the omega point are always "phantom balls" (i.e. balls that did not exist at the start), for logic still dictates that all the original balls have been removed. (Where would these phantom balls come from? Well, suppose that a particular position exists in the urn in which you place a ball whenever you find the spot empty. As the process continues, the relative amount of time this spot spends empty goes to zero. Since the position is filled up to the omega point, you could suppose that it remains filled afterwards.)
I do not care for "phantom ball" models, but admit they are logically consistant (they are not physically consistant, but since the problem isn't either, this is immaterial). If we deny the creation of phantom balls, only one logically consistant model remains: the urn is empty.
But, Fred, your complaint does not involve phantom balls. Instead, it makes the mistake of comparing apples to oranges. Metaphorically, it argues that since both are fruits, they should be exactly the same. You claim that the various scenarios you offer all "look the same" if you do not see the labels. But this is not true. In each case, the imp is removing a different collection of balls. For example, in the original, an observer would note that of the first two balls placed in the urn, the imp removes one on his first turn and the other on his second. In the situation with every other ball numbered 1, the observer sees that the imp removes one of the first two balls on his first turn, but never, ever, touches the second. Since the imp is behaving differently, it should not be surprising that the final result is different as well.
on May 31st, 2005, 12:16am, Ajax wrote:| I believe that infinite is a concept and nothing more. |
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I completely agree! Of course, 1 is also a concept, and nothing more. So it will take more than this to say that this problem is meaningless.
It is meaningless physically - something that was acknowleged right from the start. But it is not meaningless mathematically. In mathematics, infinity (actually "infinities", there are more than one) is just as well defined a concept as 1 is.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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jeffypop
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Re: Impish Pixie revisited
« Reply #19 on: Jul 20th, 2005, 9:29pm » |
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So, can't this be solved and understood mathematically?
f(n) = 2n - n = n
as n -> oo (infinity), f(n) does too.
For the lowest number ball at any given time,
g(n) = n+1
g(n->oo) = oo + 1 = oo
so even though we can't visualize this, we don't have to. That's why n approaches infinity... because it never actually reaches it.
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ThudnBlunder
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Re: Impish Pixie revisited
« Reply #20 on: Jul 22nd, 2005, 1:24pm » |
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on Jul 20th, 2005, 9:29pm, jeffypop wrote:So, can't this be solved and understood mathematically?
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Indeed it can.
But LIMIT f(x) = b does NOT imply f(a) = b
x -> a
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uMRod
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Re: Impish Pixie revisited
« Reply #21 on: Jul 22nd, 2005, 1:31pm » |
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on Jul 22nd, 2005, 1:24pm, THUDandBLUNDER wrote:
Indeed it can.
But LIMIT f(x) = b does NOT imply f(a) = b
x -> a |
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If f is continuous at a or if you are trying to determine if f is continuous at a, then f(a) = b is a necessary requirement.
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jeffypop
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Re: Impish Pixie revisited
« Reply #22 on: Jul 22nd, 2005, 2:51pm » |
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But can you actually have
f (infinity) = ??
I thought that was the purpose for the whole "limit as x->infinity" business was about, to circumvent that.
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uMRod
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Re: Impish Pixie revisited
« Reply #23 on: Jul 22nd, 2005, 3:25pm » |
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on Jul 22nd, 2005, 2:51pm, jeffypop wrote:But can you actually have
f (infinity) = ??
I thought that was the purpose for the whole "limit as x->infinity" business was about, to circumvent that. |
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I am not a PhD but I can tell you that the theory of limits is much deeper and richer that merely what you are saying. It is a lot more complex than what you might think.
The notation f(infinity) = ?? is not conventional. The closest to it is an indeterminate form and it is not that either.
To say that
Limit f(x) = b
x -> 00
means by, definition, given any positive real number, e, there exists a positive real number, M, such that
x > M implies |f(x) - b| < e.
This definition and others like it where x -> some finite number is/are the backbone of calculus.
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jeffypop
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Re: Impish Pixie revisited
« Reply #24 on: Jul 22nd, 2005, 3:59pm » |
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Thanks for the calc refresher, clearly mine is rusty.
But I wasn't not saying that limits are solely for use with cases of inifinity. What I was saying was that limits allows you to handle infinity, whereas otherwise you could not.
So the impish pixie is incorrect in asking for the number of the smallest ball after an infinite amount of time.
Essentially, he's asking for f(oo). And that's where the conundrumm comes from.
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