Taylor Series ExpansionDate: 11/24/2001 at 15:14:59 From: Don Castaldini Subject: Taylor Series Expansion The distance from A to B is 1000 meters. As one traverses the distance from A to B, at 1 meter per second, the distance is instantaneously and uniformly stretched 1000 additional meters. How long does it take to get from A to B (in seconds)? What we've figured is that as we take one step toward B, the road stretches 999 meters in front and 1 meter behind. This happens each time we take a step toward B. What we've tried to do is come up with a formula for what fraction of the total distance has been covered after n seconds, but just before the distance stretches; however, we've been unable to come up with anything that isn't crazy looking! Date: 11/25/2001 at 15:23:38 From: Doctor Pete Subject: Re: Taylor Series Expansion Hi, This is an interesting problem, one that depends on whether you consider it the distance discrete or continuous. What do I mean by this? If the problem is discrete (as implied by the wording, and your comments), then each time we step 1 meter forward, the length stretches 1000 meters under our feet. On the other hand, if the problem is continuous, then you can imagine we're on a bicycle instead, moving at a constant rate of 1 m/s, and the road beneath us is stretching at a smooth, continuous rate of 1000 m/s. The important thing to realize is that the two situations I have thus described are different, and will give different answers. Let us consider the discrete case first. At time t, let x[t] be the distance from point A, and d[t] be the total distance from A to B. Then clearly, d[t] = 1000(t+1). The other thing to notice is that at time t+1, we will have traveled 1 meter farther than x[t], plus an amount corresponding to the stretching of the road beneath us. While the amount of stretch is the same (1000 m), how far it moves us from x[t] depends on how far we have traveled. In particular, this additional stretch moves us a distance of 1000(x[t]+1)/d[t], assuming that the road stretches after we step 1 meter forward from x[t]. Therefore, we find the distance x[t+1] = x[t] + 1 + 1000(x[t]+1)/d[t] = (x[t] + 1) + (x[t]+1)/(t+1) = (x[t] + 1)(1 + 1/(t+1)), or equivalently, x[t] = (x[t-1] + 1)(1 + 1/t), x[0] = 0. You may verify that at time t = 1, we have traveled a distance of x[1] = (x[0] + 1)(1 + 1/1) = 2. Now, the solution to this recursion relation is not easy. My best explanation is to say that the solution is x[t] = (t+1)H[t], where H[t] = 1 + 1/2 + 1/3 + 1/4 + ... + 1/t is the t(th) harmonic number. (For convenience, let H[0] = 0). To verify that this is the solution, we simply substitute it into the right-hand side of the recursion relation, giving (tH[t-1] + 1)(1 + 1/t) = (tH[t-1] + t/t)(1+1/t) = tH[t](1+1/t) = tH[t] + H[t] = (t+1)H[t], = x[t], as claimed. Finally, since x[t] gives the position from A at time t, and d[t] the total distance from A to B, we seek the value of t > 0 that gives x[t] = d[t], or (t+1)H[t] = 1000(t+1), or H[t] = 1000. This is not easy to compute, since t is very large. Using the relation lim [H[t] - ln[t]] = g, t->oo where g is the Euler-Mascheroni constant 0.5772156649015328606...., we have approximately 1000 - ln[t] = g, or t = Exp[1000 - g], or t = 1.106115110266 x 10^434, The total distance traveled is 1000(t+1), which is essentially the same as 1000t, which is 1.106 x 10^437 meters. The continuous case is actually quite a bit simpler. As before, we have d[t] = 1000(t+1), but now x[t] satisfies the differential equation x'[t] = 1 + x[t]/(t+1) (with initial condition x[0] = 0), because the speed at time t is 1 m/s plus the rate at which the ground behind us is stretching, which is 1000x[t]/d[t]. Since it is linear, the solution is simply x[t] = (t+1)(C + ln[t+1]); the initial condition x[0] = 0 implies C = 0, and hence x[t] = (t+1)ln[t+1]. Therefore, x[t] = d[t] implies ln[t+1] = 1000, or t = Exp[1000] - 1 which is approximately 1.970071114017 x 10^434. Note the similarity of the solutions. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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