送交者: lakeblue 于 December 17, 2009 23:43:44:
回答: 借用Chi版主的话：湖蓝仁兄/贤弟/贤侄 由 老六 于 December 17, 2009 21:54:44:
Assume they start from the same point. When YP catches up with lala, they should have travelled the same distance, i.e.
10^2+10^4+....+10^(2N) = 1!+2!+...+N!
Here dt = 1 second
On the left it is a power series sum for which there is a formula, but there is no formula for the sum on the right. If N is large, the last term on both sides is overwhelmingly larger than the rest, and the above equation can be approximated as
10^(2N) = N!
same as what lala has given. Taking natural log and using the Stirling approximation,lnN!=NlnN-N, we have
2Nln10 = NlnN - N
Solving for N yields that N is approximately 271. For such a large N, the approximation of reducing the first equation to the second one is justified.
A speed of 10^(2x271) is an astronomical number no matter how small your units are. lala should have reached the speed of light much earlier than 271 s. So the answer is no, YP can never catch up on lala.