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Warp wrote:
2) In this situation, is there any way of determining whether you are in such a spacestation or on Earth?
I assume that in this situation there is also the limitation that you are stuck in a small room or some such, so you can't make obvious observations such as the behavior of the horizon and objects approaching it. Anyway, Einstein notwithstanding, there actually is a way to differentiate between gravitational acceleration and things such as fake centrifugal gravity and linear acceleration due to a rocket, or some such. Namely, the gravitational field around a massive body follows a gradient, which can be measured with precise enough instruments. Linear acceleration has no such gradient, while fake centrifugal gravity has a different gradient. (This doesn't actually contradict relativity since we're cheating by making certain assumptions and measuring at multiple points.)
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Pointless Boy wrote:
Anyway, Einstein notwithstanding, there actually is a way to differentiate between gravitational acceleration and things such as fake centrifugal gravity and linear acceleration due to a rocket, or some such. Namely, the gravitational field around a massive body follows a gradient, which can be measured with precise enough instruments. Linear acceleration has no such gradient, while fake centrifugal gravity has a different gradient. (This doesn't actually contradict relativity since we're cheating by making certain assumptions and measuring at multiple points.)
If you are in a small room, that gradient is probably so small that it would probably not be measurable (probably even physical limits come into play, such as the Planck length; although I'm not a physicist, so don't quote me on that).
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Warp wrote:
1) Assume that there's no physical limit (imposed by material strength, etc) to the size of the space station. To make the problem more difficult, let's build a space station which makes one full rotation in 24 hours, like Earth, and produces the same perceived gravity like that. Question: What would the radius of the space station have to be in order to achieve this (iow. it makes a full rotation in 24 hours and the perceived gravity at the outer edge is the same as on the surface of the Earth)?
1.6 Gm (gigametres)
2) In this situation, is there any way of determining whether you are in such a spacestation or on Earth?
Contrary to what you have said, a pendulum will NOT show rotation on a rotating space station, just as it will not show rotation at the earth's equator. A pendulum will rotate once per day at the poles, and slower at intermediate latitudes. This can be shown by a symmetry argument: how would the pendulum decide which direction to rotate in? There is nothing to show a preference for rotating left or right. Using this you can distinguish between [space station or close to earth's equator] and [sufficiently far from earth's equator], but not directly between earth's equator and the space station. A wheel mounted vertically will rotate once per day on the space station and at earth's equator, but will not rotate at all at the north pole. I would guess it rotates fractions of a complete rotation at other latitudes, similar to the pendulum but inverse. The other main difference I can think of is the difference in convergence points of gravitational field lines. On earth, field lines (or "down arrows") converge at the centre of the earth. On the space station, field lines ("up arrows") converge at the centre of the space station. You could put an upward-pointing laser in one corner of the room, place a large flat mirror perpendicular to it - so that the laser is reflected back along its original path - then place another upward-pointing laser in the opposite corner. If its reflection in the mirror goes towards the first laser, upwards arrows converge and you're on a space station. If its reflection in the mirror goes away from the first laser, upwards arrows diverge and you're on earth. This has the problem of accurately determining "up" using gravitational methods. Also if we have to consider local irregularities in earth's gravitation field, this method would not work, becuase locally the earth's field could be similar enough to a space station. Hmm. I'm stuck for now. Partial credit?
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rhebus wrote:
Contrary to what you have said, a pendulum will NOT show rotation on a rotating space station, just as it will not show rotation at the earth's equator.
Yeah, you are probably right. I got confused about the working principles of the Foucault pendulum. A rotating wheel can be used to make this measurement in any case, though.
A wheel mounted vertically will rotate once per day on the space station and at earth's equator, but will not rotate at all at the north pole.
It depends on the orientation of the secondary axis. If the secondary axis of the wheel allows it to reorient itself tangentially to the ground, then it will make a full rotation once per day at the poles. Optimally the wheel will have three axes of freedom, making it effectively a gyroscope. That ought to show the rotation of the Earth (or the spacestation) regardless of where it's located.
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so do you have a solution for distinguishing the space station from the earth's equator?
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henke37 wrote:
And a bonus one that some of you already know the answer to: You are a rude warrior that needs to pass two men. One always tells the truth and the other always lies. They wont let you pass until you have figured out who is the lying one. What is the fastest way to do so?
I suspect this question is inspired by the Order of the Stick. The wording suggests that the answer is to inflict harm on one of the men. Their reactions may reveal [1] who is the obligatory liar. Though it might not work, too.
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Your suspicion is correct.
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henke37 wrote:
And a bonus one that some of you already know the answer to: You are a rude warrior that needs to pass two men. One always tells the truth and the other always lies. They wont let you pass until you have figured out who is the lying one. What is the fastest way to do so?
This should work, and should be pretty quick: Turn to one of them and simply ask "am I speaking to you right now?". If he is the one who is always telling the truth he will of course say yes. If he is lying, he will answer no. This question does not require any real knowledge from either one of them, opposite to any question like "is the grass green?" or something like that.
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Warp wrote:
If you are in a small room, that gradient is probably so small that it would probably not be measurable (probably even physical limits come into play, such as the Planck length; although I'm not a physicist, so don't quote me on that).
Oh, a number of systems based on MEMS pendulums or laser-cooled atomic interferometry accurate enough and small enough for your purposes exist today. You can also use superconducting sensors (which require some cooling apparatus, but those aren't so big nowadays) to measure test-mass displacement to very fine degrees on something akin to a conventional spring scale. I also remember a long time ago in university I wrote a paper about how you could use the resonance frequency of an exceptionally tiny "springboard" to measure field gradients of all sorts (assuming the springboard had a known something on its tip that interacted with the field being measured) and as I recall I calculated such a method should be precise enough for your purposes. Given that no one has ever seriously considered something like my obvious design (as far as I know), doubtless the other methods are even better.
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I'm looking for a general expression for a combinatorics problem, it would be great if someone could provide an expression and a motivation for it. The problem is: You have n integers a(1), a(2), ..., a(n) where a(i)>=1 for all i, a(i)>=a(i-1) (i.e. they are sorted in descending order) and a(1)<=b where b is a positive integer. How many possible combinations of a(1),...,a(n) can you create? If possible, express it as an explicit function of n and b.
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Randil wrote:
I'm looking for a general expression for a combinatorics problem, it would be great if someone could provide an expression and a motivation for it. The problem is: You have n integers a(1), a(2), ..., a(n) where a(i)>=1 for all i, a(i)>=a(i-1) (i.e. they are sorted in descending order) and a(1)<=b where b is a positive integer. How many possible combinations of a(1),...,a(n) can you create? If possible, express it as an explicit function of n and b.
Maybe I didn't understand the problem, but sounds to me like http://en.wikipedia.org/wiki/Combination
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I've read that page, but it doesn't seem to answer my problem. To give an example of what I mean: Let's say you have n=5 and b=9. This means you have 5 integers, a(1) to a(5), all between 1 and 9. Without the condition that a(j)>=a(j-1), we have 9^5 possible combinations, since all a(j) can be anything between 1 and 9. However, the tweak to this problem is that the list is sorted in descending order, so we get much less than 9^5 possible combinations. For example, the combination 2,1,9,8,7 isn't allowed because 9>=1, breaking the a(j)>=a(j-1) condition. The combination 1,3,3,3,5 is allowed, since the list is non-decreasing. What I'm looking for is, for arbitrary values of n and b, how many non-decreasing combinations you can create.
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I think I have an ugly solution. instead of all integers being greater than 1, let's say they are all greater than 0. This does not affect the problem at all. We can define the function f(n,b), the combinations of all the numbers, implicitly with f(n,b) = f(n-1,b) + f(n-1,b-1) + f(n-1,b-2) ... f(n-1,2) + f(n-1,1) + f(n-1,0) We want an explicit solution, so we will continue, until everything is of the form f(0,x), since f(0,x) = 1. f(n,b) = f(n-2,b) + 2*f(n-2,b-1) + 3*f(n-2,b-2) ... (b-1)*f(n-2,2) + b*f(n-2,1) + (b+1)*f(n-2,0) so the general case is: f(n,b) = ((i-1)C(0)f(n-i,b) + ((i)C(1))*f(n-i,b-1) +((i+1)C(2))*f(n-i,b-2) ... ((b+i-3)C(b-2))*f(n-i,2) + ((b+i-2)C(b-1))*f(n-i,1) + ((b+i-1)C(b)*f(n-i,0) so we now extend this until i=n, and we have out explicit formula: f(n,b) = ((n-1)C(0))*(1) + ((n)C(1))*(1) +((n+1)C(2))*(1) ... ((b+n-3)C(b-2))*(1) + ((b+n-2)C(b-1))*(1) + ((b+n-1)C(b))*(1) cleaned up a bit: f(n,b) = (n-1)C(0) + (n)C(1) +(n+1)C(2) ... (b+n-3)C(b-2) + (b+n-2)C(b-1) + (b+n-1)C(b) This can probably be simplified even further, and I've probably made a fatal mistake somewhere. (it's not f(0,x)=0, I'm pretty sure of that one, even though you can have a hard time rearranging 0 elements)
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Randil wrote:
I'm looking for a general expression for a combinatorics problem, it would be great if someone could provide an expression and a motivation for it. The problem is: You have n integers a(1), a(2), ..., a(n) where a(i)>=1 for all i, a(i)>=a(i-1) (i.e. they are sorted in descending order) and a(1)<=b where b is a positive integer. How many possible combinations of a(1),...,a(n) can you create? If possible, express it as an explicit function of n and b.
In other words, how many monotone sequences of length n are there consisting of integers from 1 to b? Here is a way of thinking about it that may help you: A monotone sequence is given by having k1 1's, k2 2's, k3 3's, ..., kb b's, such that the sum of the k's is n. (The k's may be 0.) So consider placing k1 white balls, then a black ball, then k2 white balls, then a black ball, etc., remembering that you can have 0 white balls between 2 black balls. Each way of ordering the balls uniquely determines a sequence, and each sequence uniquely determines a way of ordering the balls. Then you have a sequence of n+b-1 balls, b-1 of which you choose to be black. Thus, (n+b-1)C(b-1) is the answer. This combinatorial trick commonly goes under the stupid name of stars and bars.
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Pointless Boy: Thank you, I understand your solution. I did not consider writing it on the form 1*k1 + 2*k2 + ... I am familiar with the stars and bars trick, and I've been sniffing on it a little when looking for a solution to this problem, but I didn't find a way to apply it. Well, thanks for the solution!
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Four positive integers a,b,c,d satisfying ab = cd Prove that a+b+c+d cannot be a prime.
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ab = cd a = cd/b a + b + c + d cd/b +b +c +d (b^2 +b(c +d) + cd)/b which we can factorise (b + c)(b + d)/b Although a rigourous proof would go past this step, I see no real reason to continue. even though you divide by b, and as long as none of the numbers are zero, it should be pretty clearby now that there are other factors.
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Actually, it's not clear for me that there are other factors. :P I think you should continue ^^
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p4wn3r wrote:
Actually, it's not clear for me that there are other factors. :P I think you should continue ^^
If (b+c)(b+d)/b divides evenly, then 2 positive integers exist x and y such that b = xy, (b+c)/x divides evenly and (b+d)/y divides evenly. Neither (b+c)/x nor (b+d)/y can be 1 because b+c>x and b+d>y; therefore a+b+c+d = [(b+c)/x][(b+d)/y] is a composite number.
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(b+c)(b+d)/b. Well this is my reasoning for why this is composite.There will be HCF (highest common factor) of b and b+c. Now whatever that HCF is, whether it be 1 (i.e b+c, and b are coprime) or b (i.e, b+c is a multiple of b), (b+c)/(HCF) is going to be a factor of ((b+c)(b+d)/b. I like to think of it as (b+c)/(HCF) is the part of (b+c) that isn't divided by b. This is similar to what rhebus posted. basically instead of b = xy, we use b=HCF*(b/HCF), so x = HCF, and y = b/HCF. However x and y don't need to be the HCF and b/HCF, they can be other values as well.
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Yeah, I understand what you guys mean. It's an interesting solution. I solved it like this: ab=cd implies a/c = d/b. Since the fractions are equal, they can be simplified to an irreductible fraction p/q where p and q are positive integers. So, there are two numbers m and n where a=mp, c=mq, d=np and b=nq a+b+c+d=mp+nq+mq+np=(m+n)(p+q) Since m,n,p,q >=1 m+n,p+q>=2 and we've factored the number.
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Here's one I just can't do just did: I spend half my money. The amount I have remaining in "pounds and pence" is the same as the "pence and pounds" I had originally. How much did I start with? Feel free to change the currency
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One that I can do! There are infinitely many solutions to the "pounds and pence" problem, but here is one: Start with 199.98 pounds. Half is 99.99 pounds, which is also 98 pounds and 199 pence. Any integer solution where 98 a = 199 b will work, where a is the pounds and b is the pence you start with.
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Well, this one is fairly well known, but is fun and interesting. Prove that EDIT: Heh, sorry, it seems the LaTeX generator I used changes image links a lot. What I want you to prove is: \binom{2n}{n} = \sum_{k=0}{n} {\binom{n}{k}}^2
HHS
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That's easy. Assume that is false. Then, exists. Because , the difference does not exists, and there is a contradiction. I have proved that the expression is not false, therefore it must be true.