Problem 460
An ant on the move
On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.
In each step, the ant at point (x0, y0) chooses one of the lattice points (x1, y1) which satisfy x1 ≥ 0 and y1 ≥ 1 and goes straight to (x1, y1) at a constant velocity v. The value of v depends on y0 and y1 as follows:
- If y0 = y1, the value of v equals y0.
- If y0 ≠ y1, the value of v equals (y1 - y0) / (ln(y1) - ln(y0)).
The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P1(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
From P1(1, 3) to P2(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 ≈ 0.6667. From P2(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required time is sqrt(5) / 1.8205 ≈ 1.2283.
Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233.
The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4.
Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) ≈ 2.960516287.
We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.
Find F(10000). Give your answer rounded to nine decimal places.
蚂蚁旅行
在欧几里德平面上，一只蚂蚁打算从点A(0, 1)运动到点B(d, 1)，其中d是一个整数。
每一步，位于点(x0, y0)的蚂蚁会选择一个格点(x1, y1)，满足x1 ≥ 0及y1 ≥ 1，然后以恒定速率v径直移动过去。v的值取决于y0和y1，如下所示：
- 若y0 = y1，则v的值为y0。
- 若y0 ≠ y1，则v的值为(y1 - y0) / (ln(y1) - ln(y0))。
左图是当d = 4时的其中一条路径。首先蚂蚁从点A(0, 1)移动到点P1(1, 3)，速率为(3 - 1) / (ln(3) - ln(1)) ≈ 1.8205，所需时间为sqrt(5) / 1.8205 ≈ 1.2283。
再从点P1(1, 3)移动到点P2(3, 3)，蚂蚁的速率为3，所需时间为2 / 3 ≈ 0.6667。从点P2(3, 3)移动到点B(4, 1)，蚂蚁的速率是(1 - 3) / (ln(1) - ln(3)) ≈ 1.8205，所需时间为sqrt(5) / 1.8205 ≈ 1.2283。
因此，总耗时为1.2283 + 0.6667 + 1.2283 = 3.1233。
右图是另一条可行路径，总耗时为0.98026 + 1 + 0.98026 = 2.96052，可以看出这是d = 4时的最快路径。
记F(d)是蚂蚁选择最快路径时所需的耗时，例如F(4) ≈ 2.960516287。
我们可以验证F(10) ≈ 4.668187834以及F(100) ≈ 9.217221972。
求F(10000)，保留9位小数。