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| 25 MAR 2004 at 6:03pm | |
| Deleted User | Since the worm loses ground at a rate of 99 cm/min, he can never reach the end. |
| 25 MAR 2004 at 8:02pm | |
| Deleted User | Originally Posted By BacardiJim (25 MAR 2004 6:03pm) Are you sure that the rate of losing ground is really constant? If so, explain why. |
| 25 MAR 2004 at 8:09pm | |
| Deleted User | Actually, I'm wrong. I forgot to figure in the worm being carried along with the stretch. Back to the drawing board. |
| 25 MAR 2004 at 10:09pm | |
ElfstoneGuild Master Posts : 5892 Joined: 4 NOV 2002 Status : Online | The worm will travel 2cm/min because of the free ride it gets for each stretching. But the band will always be 50 times longer than the distance the worm has traveled. Which means it will never reach the end, but won't lose any ground, either. ? Edit: Well, in this case it won't lose any ground no matter how fast it travels. The ratio "distance from start" to "distance from end" will always be the same. [b]playing[/b]: Destination Treasure Island (done in two sittings, but it's nice), Syberia (ho-hum), Dracula: Last Sanctuary (on hold)&&[b]reading[/b]: even more study papers&&[b]listening to[/b]: [url=http://www.last.fm/user/Brax82/]this and that[/url], plus [url=http://www.musicovery.com/]Musicovery[/url]&&[b]TV favorites[/b]: (currently) Pushing Daisies, Chuck, Journeyman (cancelled! grrr...), Heroes&&all-time) 24, Stargate SG1, X-Files, Lost, House |
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| 26 MAR 2004 at 12:57am | |
MichalNGrand Inquisitor  Posts : 7058 Joined: 14 SEP 2003 Status : Online | Originally Posted By Elfstone (25 MAR 2004 10:08pm) Not true. The "free ride" gain will actually increase (in absolute terms, but not relative terms). Anyway here are my answers, all correct in some way: 1) No. Infinitely elastic rubber bands do not exist, hence it is impossible to reach the end of one, ever. 2) Yes. At some point there just won't be enough atoms in the universe to stretch the band any further and the worm will eventually catch up. 3) No. The worm will soon (relatively speaking) leave Earth's atmosphere, shrivel up and die. 4) Yes. Given infinite time, even at 1 cm/min, the worm will eventually cover infinite distance. I forgot my sig. |
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| 26 MAR 2004 at 1:13am | |
ElfstoneGuild MasterPosts : 5892 Joined: 4 NOV 2002 Status : Online | What makes you say the free ride gets longer? Any indications? ??? I hate it. It's a smartass problem! [b]playing[/b]: Destination Treasure Island (done in two sittings, but it's nice), Syberia (ho-hum), Dracula: Last Sanctuary (on hold)&&[b]reading[/b]: even more study papers&&[b]listening to[/b]: [url=http://www.last.fm/user/Brax82/]this and that[/url], plus [url=http://www.musicovery.com/]Musicovery[/url]&&[b]TV favorites[/b]: (currently) Pushing Daisies, Chuck, Journeyman (cancelled! grrr...), Heroes&&all-time) 24, Stargate SG1, X-Files, Lost, House |
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| 26 MAR 2004 at 1:34am | |
MichalNGrand InquisitorPosts : 7058 Joined: 14 SEP 2003 Status : Online | Originally Posted By Elfstone (26 MAR 2004 1:13am) Sure. Since the band stretches uniformly, the worm will get a free ride whose length is in proportion to worm's relative position on the band. After the worm crawled the first cm, the band will stretch from 100cm to 200cm. Since the worm is at 1/100 of the band's length, it'll get a free ride equal to 1/100 of the stretch, ie. 1cm. Thus after the first minute, the worm will be at 2cm of 200cm. Now after another minute of crawling, worm will be at 3cm out of 200 and will get a free ride equal to 3/200 of 100cm, ie. 1.5cm, ending up at 3 + 1.5 = 4.5cm. Here are the distances after the first few minutes: 2.0 4.5 7.333 10.416 13.700 17.150 20.742 24.460 28.289 32.218 36.238 You can see that in the last step listed above, the free ride was just over 3cm. I forgot my sig. |
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| 26 MAR 2004 at 3:27am | |
ElfstoneGuild MasterPosts : 5892 Joined: 4 NOV 2002 Status : Online | Relative position...damn! I didn't get that and regarded it as not important. : [b]playing[/b]: Destination Treasure Island (done in two sittings, but it's nice), Syberia (ho-hum), Dracula: Last Sanctuary (on hold)&&[b]reading[/b]: even more study papers&&[b]listening to[/b]: [url=http://www.last.fm/user/Brax82/]this and that[/url], plus [url=http://www.musicovery.com/]Musicovery[/url]&&[b]TV favorites[/b]: (currently) Pushing Daisies, Chuck, Journeyman (cancelled! grrr...), Heroes&&all-time) 24, Stargate SG1, X-Files, Lost, House |
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| 26 MAR 2004 at 4:18am | |
adventuredogGuild Master Posts : 3255 Joined: 14 JAN 2003 Status : Offline | I would go for the yes answer. That is, the band will eventually stretch to infinity and given an infinite amount of time the worm will reach there. As if "infinity" is a place. Given infinite time the worm would travel an infinite distance. While it looks like the band is stretching to infinity "faster", infinite is still infinite (even though there are supposedly a range of infinities). So that will give the second answer: no, even though the worm will travel an infinite distance in infinite time, it will never reach the end of the band because the band continues infinitely to stretch! : A bit like one of Zeno's paradoxes (comparing discreet with continuous), or is it? Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 26 MAR 2004 at 4:47am | |
adventuredogGuild MasterPosts : 3255 Joined: 14 JAN 2003 Status : Offline | Maybe more along the lines of Cantor's infinities. If yes, how long would it take you ask? An infinite amount of time of course! There doesn't seem to be too much convergence short of an infinite amount of time. But, even in Michal's short sequence it moved from 2/100ths of the distance to just over 3/100ths. Hmmmmm........... Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 26 MAR 2004 at 8:12am | |
MichalNGrand InquisitorPosts : 7058 Joined: 14 SEP 2003 Status : Online | Another take (I really wish I hadn't forgotten all about differential equations!). Intuitively, one would say that the endpoint of the band always moves faster than the worm, hence the worm can never reach it. But this is deceptive - it's easy to see that if the worm had started out at 98th or 97th cm, it'd reach the end in short order. Maybe the key is the relative position - since that is not changed by the stretching(!). After the first minute, worm's position is 1/100 of the entire length. After second minute, it moves an additional 1/200. Next minute, another 1/300 and so on. So the relative postion after minute n can be expressed as the series 1/100 + 1/200 + 1/300 + 1/400 + ... + 1/(n*100) For sufficiently large n, the sum of this series will be 1 (because we keep adding a non-zero amount). I have a strong suspicion that the worm won't crawl to the end of the rubber band before this universe ceases to exist. I forgot my sig. |
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| 26 MAR 2004 at 2:57pm | |
ElfstoneGuild MasterPosts : 5892 Joined: 4 NOV 2002 Status : Online | I think you are right. The series will sum up to 1, that should be proof. [b]playing[/b]: Destination Treasure Island (done in two sittings, but it's nice), Syberia (ho-hum), Dracula: Last Sanctuary (on hold)&&[b]reading[/b]: even more study papers&&[b]listening to[/b]: [url=http://www.last.fm/user/Brax82/]this and that[/url], plus [url=http://www.musicovery.com/]Musicovery[/url]&&[b]TV favorites[/b]: (currently) Pushing Daisies, Chuck, Journeyman (cancelled! grrr...), Heroes&&all-time) 24, Stargate SG1, X-Files, Lost, House |
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| 26 MAR 2004 at 8:10pm | |
GreySorcerer Apprentice  Posts : 251 Joined: 3 SEP 2003 Status : Online | My answer is 100 minutes. The speed of the worm's "free ride" increases as it approaches the end of the band (if you decided to stretch it more time apon its arrival, it would be traveling at 101cm per second). Thus, the worm's average speed is sufficient to cover the 100 meters in 100 minutes. /\/\&&\/\/ |
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| 26 MAR 2004 at 10:48pm | |
MichalNGrand InquisitorPosts : 7058 Joined: 14 SEP 2003 Status : Online | Originally Posted By Grey (26 MAR 2004 8:09pm) I think that answer is off possibly by a few hundred orders of magnitude The speed of the worm's "free ride" increases as it approaches the end of the band. That's true - but I think you forgot that after the first minute, the worm has 198cm to travel instead of 100, and after 10 minutes it's almost 1068cm. I forgot my sig. |
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| 27 MAR 2004 at 12:49am | |
adventuredogGuild MasterPosts : 3255 Joined: 14 JAN 2003 Status : Offline | We really aren't fully answering the questions:Now the question is: Will the worm eventually reach the other end? If the answer is yes, how long will it approximately take? If the answer is no, show why. Sure the sum of the series will approach the limit of 1 for a sufficiently large n - is approach significant in this case, 'cause the band is still stretching. Or can you use Cantor's infinities type of proof by looking for a one-to-one correspondence between elements in the two sets: each distance for the worm and each length of the band? And yet, after each point in time, the band stretches another meter. If it never stops stretching, given an infinite universe, how can the poor worm ever reach the end? ??? Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 27 MAR 2004 at 1:07am | |
MichalNGrand InquisitorPosts : 7058 Joined: 14 SEP 2003 Status : Online | Originally Posted By adventuredog (27 MAR 2004 12:49am) No, the sum will exceed 1. Look at it this way - if the worm has 1cm to go, it will quite obviously reach the end (regardless of how long the band is). If it initially has 2cm to go out of 100cm, the positions after successive minutes will look like this: 98/100 198/200 298.5/300 399.333/400 500.417/500 and so on. Note that if the worm has 2cm to go out of, say, 10,000cm, it'll finish quicker since each stretch will add relatively shorter distance to go. If starting at 97th cm, it's even more interesting: 97/100 196/200 295.5/300 395.333/400 495.417/500 595.700/600 696.150/700 796.743/800 897.461/900 998.290/1000 1099.219/1100 1200.239/1200 Notice how after the first minute, the worm has 4cm to go instead of 3 - but it will catch up, because with each minute the distance added by stretching becomes smaller. I forgot my sig. |
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| 27 MAR 2004 at 1:18am | |
adventuredogGuild MasterPosts : 3255 Joined: 14 JAN 2003 Status : Offline | Very good idea to start with a simpler problem and look for a pattern. Will this pattern continue to hold? You have started from the back end of the problem and have come up with interesting results. Would it make any sense to start with more trivial examples? Keeping rate of worm constant at 1cm/minute. Band expands: 1cm/minute; 2cm/min; 3cm/min, 4cm/min, and then make ajump to the final 100cm/min Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 27 MAR 2004 at 1:19am | |
adventuredogGuild MasterPosts : 3255 Joined: 14 JAN 2003 Status : Offline | BTW - Are we working on one of Petter's class assignments? Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 27 MAR 2004 at 9:52pm | |
| Deleted User | An intriguing problem, isn't it? So simple to grasp the idea, so hard to figure out the answer... I've been reading your posts with great interest! Many good points there. However, before we get tangled up in discussions about infinities, let me clarify one thing. It is implied by the problem as it's stated, but maybe it's not so obvious. The assumption is that the band can stretch arbitrarily long, but its length will always be finite! The rubber band will never become infinitely long no matter how far you stretch it. If it was infinitely long it would not have any defined ends (like a rubber band before you cut/snap it, which would indeed leave the poor worm looping around it for infinity no matter how fast it is). Now, how to approach this problem? Well, a useful tactic imo would be to try and define the length of the rubber band and the covered distance of the worm at any given time in a mathematical way. Let's take it slowly and in easy steps: The rubber band is the easy part. Let t be the time (in minutes). To make it simple, we can limit ourselves to discrete times. (t always belongs to the set of natural numbers 0,1,2,...) Let f(t) be the length of the rubber band after t minutes. Clearly, the rubber band is 1 meter after 0 minutes, 2 meters after 1 minute and so on. It's always one meter longer than the amount of minutes passed. So we have: f(t)=t+1 With this function we can always compute the length of the rubber band at any given time t. If we now could only define a function for the worm's covered distance, also depending on t, we could compare these functions and see what happens with large values for t. Let g(t) be the distance covered by the worm (also in meters) after t minutes. Again, let's consider only positive integers (i.e. natural numbers) as valid t's. The movement is complicated because it consists of two components; one constant (the worm's 1 cm crawling) and one variable (the worm's movement due to stretching, depending on its current position). The variable bit of the function clearly depends on the function itself, so it seems easiest to define the function recursively (i.e. in terms of smaller instances of itself). Some of you probably know this way of reasoning very well, while others don't. It's not very complicated though. First, we can naively conclude that after 0 minutes (which is assumed to mean the time when the crawling starts) the worm has covered no ground at all. So we have: g(0) = 0 But what happens for larger values of t? Well, here the recursion comes in. We can clearly conclude that, (excluding t=0), just before the rubber band is stretched an additional meter, the worm will have covered the distance it was at one minute ago, plus an additional centimeter. So for t>0, we have that: g(t)=g(t-1)+0.01 Right? It's simply the accumulated distance from one minute ago plus the extra centimeter it covered this last minute! But we still haven't taken into account the stretching part. So what happens then? Well, it's not that complicated: Its current distance just gets boosted by the proportion the band is stretched at this time. So how big is that proportion? Well, we already have a function to compute the length of the rubber band at any given time. So at time t the band will become f(t)/f(t-1) times longer. The function is so simple that we can replace it with its defined form, and we get t+1/t. So after 1 minute, the band becomes 2/1=2 times longer, which you can easily see it does. After two minutes it gets 3/2=1.5 times longer and so on... So what happens to the worm is that its current position is moved forwards by this proportion. We represent this mathematically with a multiplication of the old function value: g(t)=(g(t-1)+0.01)*(t+1/t) We now have a valid functional description of the worm's movement, in the form of a recurrence: g(t)= 0, if t=0 (g(t-1)+0.01)*(t+1/t), otherwise The definition is awkward because it refers back to itself, but it's perfectly valid nonetheless. There are methods to solve recurrences so that we get a nicer functional description that is completely straightforward, but let's leave this for a moment. What happens for very large values of t? Well, f(n) just grows in a linear fashion. The rubber band will always be t+1 meters long. what about g(n) then? The stretching proportion is 2/1 the first time, then 3/2, then 4/3, then 5/4 and so on. It grows downwards towards 1, although it will never quite reach 1. You can say that as t grows towards infinity, t+1/t goes towards 1. So it seems that for large t's, the stretching of the rubber band will give the worm almost no boost at all! (Multiplication by 1). But does this prove that the worm will never reach the other end of the rubber band? No! We still aren't sure how much the worm's own crawling influences the distance. Looking at the solutions for the first few minutes gives a hint, but I'll leave the question open for some discussion without attempting to proceed... |
| 27 MAR 2004 at 9:54pm | |
| Deleted User | Originally Posted By adventuredog (27 MAR 2004 1:19am) Hehe, no you're not. This was just an interesting example assignment. I have different problems to solve and I really want to do them without help because I need to understand them properly. |
| 27 MAR 2004 at 11:12pm | |
adventuredogGuild MasterPosts : 3255 Joined: 14 JAN 2003 Status : Offline | I'll take a few minutes to digest this....... The first part is difficult to wrap my mind around - a band being able to infinitely stretch and still have finite length! Seems like an oxymoron..... Of course, I realize something is amiss if our band has a beginning and an end, but how could it "really" have an "end" if it has infinite stretchability? I do remember that we tacked one end to the table, but..... This is fun. Thanks for the response Petter and the PM - I hadn't realized I'd crossed another threshold. Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 27 MAR 2004 at 11:39pm | |
| Deleted User | Originally Posted By adventuredog (27 MAR 2004 11:12pm) I know! Infinities are always hard to deal with. : But the idea that there are finite but arbitrarily big things as compared to infinite things is very important. Take the set of natural numbers for instance. {0,1,2,3,...} This is an infinite set, because there is never a last natural number. But in a mathematical model you could easily define a finite set of natural numbers, say {0, 1, 2, 3, ..., n}, where n is some natural number. Clearly, this set is not infinite because it has a biggest number n. It is a subset of the infinite set of natural numbers as long as n is a member of that set. n can be any natural number no matter how big, so the size of the set can be as large as you like, but never infinitely large, because if it was, then it would include all natural numbers. This is a contradiction, because for any last number n in that set, you can immediately see that the numbers n+1, n+2 etc. are not in the set, and as those are natural numbers as well they should have been in the set in the first place! Confused yet? There are even infinities that are "bigger" than other infinities, although not simply by having more elements. The set of integers {..., -2, -1, 0, 1, 2, ...} is known as a countable infinite set, as you can put a label (a counter) on the elements in it. The obvious way is to just label them by their numerical values. It is also infinite in both directions. But consider the set of real numbers, that is all possible decimal numbers and not just the integers. Of course, this set includes all the integers, so that set is a subset of the set of real numbers. But there are so many more real numbers than integers, even if there are infinitely many integers. Only between two integers, say 3 and 4, there are infinitely many real numbers (e.g. 3.1, pi, 3.5, 3.9999 etc.) and they are so "densely packed" that you cannot even clearly define what number is the immediate follower of the previous one. This is known as an uncountable infinite set, and it is mightier than the set of integers, which means that it's in a sense much, much bigger. |
| 27 MAR 2004 at 11:48pm | |
adventuredogGuild MasterPosts : 3255 Joined: 14 JAN 2003 Status : Offline | But consider the set of real numbers, that is all possible decimal numbers and not just the integers. Of course, this set includes all the integers, so that set is a subset of the set of real numbers. But there are so many more real numbers than integers, even if there are infinitely many integers. Only between two integers, say 3 and 4, there are infinitely many real numbers (e.g. 3.1, pi, 3.5, 3.9999 etc.) and they are so "densely packed" that you cannot even clearly define what number is the immediate follower of the previous one. This is known as an uncountable infinite set, and it is mightier than the set of integers, which means that it's in a sense much, much bigger.But I thought that Cantor proved that these two sets share a one-to-one correspondence, so that there is no such thing as larger and smaller infinities. Isn't that the trick with infinity? - intuitively we would think that some infinities are larger than others, but that is not true. Has someone else proved otherwise recently? Still adventuring after all these years! Patiently awaiting The Last Crown: Haunting of Hallowed Isle, and Bracken Tor... ... and Asylum if it's not tooooo scary... |
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| 28 MAR 2004 at 1:24am | |
| Deleted User | Originally Posted By adventuredog (27 MAR 2004 11:48pm) I'll return to this later. Have to look up some reference material. This is confusing to me as well. About the worm problem, it really is harder than it looks at first glance... I tried some things with interesting results. Solving the recurrence is not very easy it turns out. At least not for me. But when it gets hard it's important to remember your goals. We can do some simplification and see if that is enough. The problem lies in solving the recursive function g(t)=g(t-1)+0.01)*(t+1/t). But let's just see what neighbourhood we are in... The multiplying factor due to stretching goes from 2 to 1 and never strays out of these borders. So let's compare it to a simpler function. Clearly the following function will always give smaller values than g(t): h(t)=h(t-1)+0.01 This is simply g(t) with the stretching factor set to 1 (i.e. no stretching effect at all). How can we compute the value of this without recursion? Well, let's iterate on the definition and see where we go: We have the original function: h(t)=h(t-1)+0.01 Now, let's use the definition of h(t) to replace the occurance of h(t-1) within itself. In other words, we pick out the h(t-1) bit above and just replace it with the full function, with t-1 where we had t before: h(t)=h(t-1)+0.01= =h(t-1-1)+0.01+0.01 = h(t-2)+0.02 Let's do it one more time, now replacing the h(t-2) bit: h(t)=h(t-2)+0.02= =h(t-2-1)+0.01+0.02 = h(t-3)+0.03 You can see the pattern emerging clearly: Every time we iterate, we get a new recursion for an input one step smaller and a 0.01 addition at the end. Since we know initially that h(0)=0 and we know that after t iterations we will have an expression of the form h(t)=h(t-t)+0.01t we can end the iteration by simply replacing h(t-t) with its value, 0: h(t)=h(t-t)+0.01t=h(0)+0.01t=0+0.01t=0.01t What does it tell us? Well, it shows that the function h(t) grows in a linear fashion, just like f(t), and it has a smaller value. This is the worm crawling along without getting a boost from the stretching of the rubber band, and obviously it will never reach its goal then either, losing 99 cm each minute. Not so exciting perhaps. We only know now that the worm moves faster than a speed where it would certainly never reach its goal. |
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