Player (36)
Joined: 9/11/2004
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How about this? There's two trains and a fly. The trains start out 100 miles away from eachother, each going at 25 mph toward their doom. The fly leaves the first train the moment it starts. The fly's initial velocity is 25mph, but it can accelerate (and deccelerate) at 1 mi/hr*s, and has no top speed. How far can the fly go between the trains before it's killed by the two trains colliding into it?
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Joined: 4/16/2005
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Should the fly turn around once it arrives at the second train and so on?
Player (36)
Joined: 9/11/2004
Posts: 2630
The fly should avoid colliding with the train at all costs, that means it should deccelerate and turn around and speed back up to 25 mph before the second train and the fly reach eachother.
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Joined: 4/16/2005
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2 further questions: - acceleration does not have to be continuously differentiable? So the fly can instantanious crank the acceleration from +1mph/s to to -1mph/s? - since you ask for "how far" this will mean the fly has to rush into one of the trains in a suicide near the end to gain more space compared to flying small curves before being crushed anyway? (I've got some idea how to solve it, but the second thing will make it a little more complicated)
Player (36)
Joined: 9/11/2004
Posts: 2630
The change in acceleration is instantanious. I don't know, I haven't worked it out yet. :)
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Joined: 4/16/2005
Posts: 251
Hmm, I worked a bit on it. I've gotten a recursive definition of the times where the fly turns around. To get this transformed into a direct formula I'd had to refresh the z-transform, which I last used in 3rd semester. If I could do that, then the time-deltas could be computed easily into the way the fly flies between those times, and could be summed up infinitely. But as I said, transforming a recursive sequence into a direct representation is complicated, and even more so if it involves radicals.
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Location: Uppsala, Sweden
Allright. Anyone up for solving these ones? 1) Prove or disprove: (n/2)^n = omega(n^n/2) 2) Prove or disprove: log(n^n ) = theta((log n)^ n) 3) Let f(n) and g(n) be asymptotically nonnegative. Prove or disprove: f(n) + theta(g(n)) = theta(f(n) + g(n))
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Joined: 3/8/2004
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Location: Denmark
OmnipotentEntity wrote:
There's two trains and a fly. The trains start out 100 miles away from eachother, each going at 25 mph toward their doom. The fly leaves the first train the moment it starts. The fly's initial velocity is 25mph, but it can accelerate (and deccelerate) at 1 mi/hr*s, and has no top speed. How far can the fly go between the trains before it's killed by the two trains colliding into it?
Quite interesting this, since there's a wording trick involved: The way I read it, the answer is exactly 50 miles. No more, no less. Since the fly is travelling at 25 mph AND the train is travelling at 25 mph, and it hasn't been indicated to which point of reference the fly's speed is measured, we can only assume that it's measured relative to the ground, as are the speeds of the trains. Ergo: The fly will fly right ahead of the first train until both trains collide. ------------------------------------- Complexification of it is possible by saying that the trainspeeds are relative to the first train, so the fly can go 25 mph when heading towards the first train, and -75 mph (assuming positive direction is "towards" the second train) when heading away from the second train. The second train is then traveling at -50 mph. Bleh, my puny mathematical education so far is preventing me from seeking further answers on this...
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Former player
Joined: 3/19/2004
Posts: 710
Location: USA
Oops, misread the problem :)
Former player
Joined: 3/30/2004
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Location: Heather's imagination
MahaTmA wrote:
Since the fly is travelling at 25 mph AND the train is travelling at 25 mph, [...] [t]he fly will fly right ahead of the first train until both trains collide.
The fly can increase or decrease its speed at will, with a 1 mph*s acceleration. Which means it starts at 0 relative to the first train (with the other approaching at 50) and accelerates until it has just enough time to reverse acceleration in time to match that other train's speed as they touch, etc., etc.
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Joined: 3/8/2004
Posts: 185
Location: Denmark
Ah, ok.. Dang them speedy steppe flies :P
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Joined: 4/16/2005
Posts: 251
I spend the afternoon on the fly. Using the numerical approach (read: Excel), I got some results. The fly flies about 542.285 miles before it gets crushed. The fly will turn 19 times before the trains are within a few feet. Everything after that can be ignored for the rounded solution (if the fly kills itself immediately the last digit above flips to 6 or 7, if not the fly will gain even less). I'm pretty sure I can't do this algebraic, but maybe one of the others can. The idea is that the fly will always use maximum acceleration. So the {apex? angular point? you know the y = (x-x0)² + y0 thing} will always be on a fixed point before touching the train. Then you can compute the lines on with those points are. Next is to combine accelerating and deccelerating into one thing. It will turn out to have half the acceleration but will behave like accelerating half the time and deccelerating the rest. Then you can intersect the first spline with the line of turning points, get the time of turning and calculate the next turn with it. Since the fly already has momentum the first iteration has to start on the turning point slightly before start. The recursive formula for turn times is: T(0) = -v/2a T(1) = √[ T(0)² + 3v²/a² - 4vT(0)/a + 2S/a ] + T(0) - 2v/a = √[ 21/4*v²/a² + 2S/a ] - 5/2*v/a T(n>1) = √[ 3v²/a² - 4vT(n-1)/a + 2S/a ] + T(n-1) - 2v/a with v = speed of the trains = 25mph a = acceleration of fly = 1mph/s = 3600mph² S = distance of trains at start = 100m T(0) is the point of turn before the start. T(1) is the first iteration, where the fly starts slighty before the start (hence a T(n-1)² is missing under the root). To get the distance the fly flies, sum up all distances a/2*(ΔT)² and substract the initial small bit which is v²/(4a). If there's an error in it _again_, fire away.
Former player
Joined: 5/24/2005
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Location: France
If anyone wants to try a demonstration of a little theorem (that appears to be true...) : "The equation x^n+y^n=z^n has no solution for non-zero integers x, y, and z if n is an integer greater than 2" hehehe, good luck...
Not dead yet... still very busy... damn...
Editor, Expert player (2073)
Joined: 6/15/2005
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Xaphan wrote:
If anyone wants to try a demonstration of a little theorem (that appears to be true...) : "The equation x^n+y^n=z^n has no solution for non-zero integers x, y, and z if n is an integer greater than 2" hehehe, good luck...
Not a proof, but... http://mathworld.wolfram.com/FermatsLastTheorem.html For those who are wondering what this is.
Player (36)
Joined: 9/11/2004
Posts: 2630
Here's a good overview of the proof. Without being ridiculously simplistic. http://www.mbay.net/~cgd/flt/flt08.htm
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OmnipotentEntity wrote:
Here's a good overview of the proof. Without being ridiculously simplistic.
That was cruel.
Player (36)
Joined: 9/11/2004
Posts: 2630
BTW, I have another problem that you guys might enjoy. Prove that all non-trivial zeroes in the Zeta fuction have real part of 1/2. *giggle*
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Joined: 5/3/2004
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If you are going to attempt to be clever in such a boring and obvious fashion, you may as well read over de Brange's proof of it to see if there is any meat in his argument. Then again, you couldn't get past the first sentence, so don't.
Former player
Joined: 5/24/2005
Posts: 405
Location: France
Come on it was just a joke...... Don't be so mean... The fact is that the proof of this theorem is very recent and incredibly long...
Not dead yet... still very busy... damn...
Editor, Expert player (2073)
Joined: 6/15/2005
Posts: 3282
Xaphan wrote:
Come on it was just a joke...... Don't be so mean... The fact is that the proof of this theorem is very recent and incredibly long...
As far as I know, there is no proof yet. http://mathworld.wolfram.com/RiemannHypothesis.html
Player (36)
Joined: 9/11/2004
Posts: 2630
xebra wrote:
If you are going to attempt to be clever in such a boring and obvious fashion...
*blink* Who put a stick up your bum and twisted it?
Xaphan wrote:
The fact is that the proof of this theorem is very recent and incredibly long...
AFAIK, the proof hasn't been accepted yet. If anyone with far more than a casual interest in the Zeta Function wants to check it out here's his paper. (pdf only) And his "apology" for the proof which has some background on the techniques used in the proof. (also pdf only)
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Player (36)
Joined: 9/11/2004
Posts: 2630
Here's one, (and I have figured this one out actually). Assuming that air resistance is the only significant source of friction for a flipped coin , and that the center of gravity is in the exact center of the coin, and that reeds have no effect on the outcome, and all contact with the ground is perfectly elastic, what is the probability that standard US quarter will land on its edge when flipped?
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I'm not going to do the calculations, but assuming uniform density, it seems likely to me that the following method is reasonable for a large family of objects including cylinders, conical frustums, and cones: Circumscribe a sphere around the object. Then the edges of the object will split the interior of the sphere into disjoint regions. The probability that the object will land on a particular face is equal to the volume of the region associated with that particular face divided by the total volume of the sphere. P.S. Reeds ... ?
Player (206)
Joined: 5/29/2004
Posts: 5712
Wait, is this for whether the coin will land on its side or land and stay on its side after coming to a stop?
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Joined: 4/16/2005
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Wait, what do you mean with "perfectly elastic ground"? Will revert any velocity or will absorb any velocity? If it reverts then this will be very complicated.