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Hola Bo, You realize, of course, that I have been just itching to explain this. > Heh. I just read the post about infintesimals. It sounds like there is > another explanation. I always thought I understood the faultiness of > that paradox. Hmm... I know its off-topic, but if the group would be > willing, I certainly would be curious as to the explanation the previous Your reasoning is quite right -- the paradox is in your statement "eventually it approaches zero". The point is that it approaches zero only after an infinite number of iterations, i.e. clearly the distance between the runner and the tortoise keeps getting smaller, but it only reaches zero after an inifinite number of iterations. So the basic question posed by the paradox is: what is the sum of infinitely many infinitely small numbers (infinitesimals)? As a warm-up problem, consider the following: a physicist, a philosopher, and an engineer are all in a room with a nude woman at the other end. The philosopher says, "It will take me a certain amount of time to halve the distance to her. Then it will take a while longer to halve the distance again, and so on. Therefore it will take an inifinite time to reach the woman," and he leaves the room. The physicist, who is only here so there can be three people in the joke, says, "To first approximation the woman is a sphere," and goes off to ponder the problem. The engineer looks at the problem and says, "Well, I may never actually reach the woman, but I'll get close enough." Let's consider the poor philosopher's problem: It takes a certain amount of time to reduce the distance by half -- call it T/2. It then takes half as much time to halve the distance -- T/4. Continuing onwards, the total time is thus T/2 + T/4 + T/8 + T/16 + ... = T*(1/2 + 1/4 + 1/8 + 1/16 + ...) In binary, the sum on the right hand side is just 0.11111111..., which is certainly less than 1. In fact, if you calculate the infinite sum (using the method below), you get -- one! (You might also have heard in math class that 0.9999... repeating is equal to one -- same thing). Anyways, the total time to cross the floor is just T, as expected. Now, regarding the tortoise and the runner, let's first do it using the "physics" method -- if the tortoise starts at distance D, his position is given by xt = D + vt*t. Similarly, the runner's position is xr = vr*t. Setting these equal to each other, we find that the runner overtakes the tortoise at t = D/(vr-vt) -- that is, vr-vt is the closing velocity, and dividing the distance by the velocity gives the time (meters divided by meters/second gives seconds). Now let's be Greek about it. As above, vt = velocity tortoise, vr = velocity runner, and the tortoise starts at D. The time is takes the runner to reach D is t1 = D/vr In this amount of time, the tortoise has advanced by vt*t1 = vt*D/vr = dx1. That is, dx1 is the new distance between the two after the first iteration. The time it takes the runner to traverse this new distance is t2 = dx1/vr = vt*D/vr^2 and the tortoise advances by dx2 = vt*t2 = D*vt^2/vr^2. You can see that after n iterations, tn = D * vt^(n-1) / vr^n dxn = D * (vt/vr)^n = vt*tn and the total time and distance is given by T = t1 + t2 + t3 + ... X = dx1 + dx2 + dx3 + ... = D*(vt/vr) + D*(vt/vr)^2 + D*(vt/vr)^3 + ... = D*(r + r^2 + r^3 + ...) where r=vt/vr. At this point I have to recant on my earlier statement that the Greeks didn't know about infinitesimals, because I'm pretty sure the method for computing the above sum goes all the way back to the ancient Greeks: let S = r + r^2 + r^3 + ... = r * (1 + r + r^2 + ...) = r * (1 + S) Solving the above for S gives S = r/(1-r) = (vt/vr)/(1-vt/vr) = vt / (vr-vt) and hence X = D*S = D * vt/(vr-vt) T = X/vt = D / (vr-vt) which is exactly the result we got from the "physics" version: the initial distance divided by the closing velocity. The moral is: infinite sums can result in a finite number, and runners can indeed beat tortoises. Finally, it is worth pointing out that the full answer of the original question -- what is the sum of infinitely many infinitely small numbers -- is what leads to calculus: that's what an integral is. FINALLY... there's an old puzzle, where two trains are heading directly towards each other, each at a certain speed. A fly leaves one train and flies at a certain speed to the other train, whereupon he immediately turns around and flies back to the first train, and so on until the trains collide. How far did the fly travel? One way is to solve this with an infinite sum; the easy way is to compute the time it takes the trains to collide, and simply multiply by the fly's velocity. Supposedly this puzzle was given to John von Neumann, who quickly gave the right answer. The guy posing the question said, "Ah, you saw the trick!" To which von Neumann replied, "What trick?" Turns out he just did the sums in his head. We're having sum fun now. -Steve - This message was sent through the cbm-hackers mailing list. To unsubscribe: echo unsubscribe | mail cbm-hackers-request@dot.tcm.hut.fi.

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