Snail!Date: 06/20/2002 at 12:06:22 From: Jean-Claude Desbordes Subject: Snail! Emergency: I need a good Doctor! I am involved in a treasure hunt in France and one of the riddles consists of a snail tring to climb a window-pane. :o) Would you please help me? I have done my best to translate it: One evening a snail stands on a window-pane at a height of e minus 1 meter from the base. Every night it slips slowly and loses 1 meter. On the following day it goes up, doubling its altitude of the morning. On the third day, it triples the altitude of the morning. On the fourth day, it quadruples its altitude, and so on, hoping some day to reach the top of the window. In doing this, what will be its altitude on the 51st day at dawn? Thank you very much for reading this to the end. I would be very pleased with the answer only, because I am afraid not to be able to understand the solution, my abilities in mathematics being very slightly superior to those of the snail of the riddle! ;o) Jean-Claude Desbordes Date: 06/24/2002 at 12:40:37 From: Doctor Ian Subject: Re: Snail! Hi Jean-Claude, Does the snail slip to e-2 meters before it starts crawling? Or does it begin the first dawn at e-1 meters? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 06/25/2002 at 11:21:09 From: Jean-Claude Desbordes Subject: Snail! Hello, Doctor Ian! First, I am very grateful for your answering so quickly. As you know, it's a riddle, so I can't get any additional details about it. In fact, my understanding of the problem is that the snail, having lost 1 meter during the first night, starts crawling on the second day at dawn at e-2 meters. But I could be wrong! Maybe you understand better now why we Frenchies hate those animals and eat them before they create problems! ;o)) Thank you very much indeed for your help, Jean-Claude Date: 06/26/2002 at 09:11:04 From: Doctor Ian Subject: Re: Snail! Hi Jean-Claude, At first I thought 'e' meant an unspecified distance, but I think it really means _e_, the base of the natural logarithm. I started fooling around with a spreadsheet, and when I use 2.718 as an approximation for e, I get a progression that looks like this: dawn 1: e - 2 = 0.718 dawn 2: 2(0.718) - 1 = 0.436 dawn 3: 3(0.436) - 1 = 0.308 dawn 4: 4(0.308) - 1 = 0.232 dawn 5: 5(0.232) - 1 = 0.160 dawn 6: 6(0.160) - 1 = -0.040 Obviously, from here on the altitude becomes increasingly negative. However, if I use the approximation 2.7183 for e, I get a different behavior: dawn 1: 0.7183 dawn 2: 0.4366 dawn 3: 0.3098 dawn 4: 0.2392 dawn 5: 0.1960 dawn 6: 0.1760 dawn 7: 0.2320 dawn 8: 0.8560 dawn 9: 6.7040 dawn 10: 66.4400 Again, obviously, from here on the altitude becomes increasingly positive. It looks to me as though if the exact value of e could be used, this would approach 0 as the number of days approaches infinity. Any lower approximation tends towards negative infinity; any higher approximation tends towards positive infinity. However, I can't see why this should be the case. One of our other math doctors, Dr. Peterson, had the following insight: This is pretty interesting! If we write the terms of the sequence exactly, we get a_1 = e - 2 a_2 = 2a_1 - 1 = 2e - (2*2 + 1) a_3 = 3a_2 - 1 = 3*2e - (3*2*2 + 3 + 1) a_4 = 4a_3 - 1 = 4*3*2e - (4*3*2*2 + 4*3 + 4 + 1) and so on; so a_n = n! e - (2*n! + n!/2! + n!/3! + n!/4! + ... + n!/n!) = n! (e - (1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!)) I've forgotten whatever I once knew about remainders of series approximations, so I don't know how to find this accurately enough that we can multiply by 51! and still get an accurate result for a_51. But I think we can show that the remainder is less than e/(n+1)!, so n! times that is less than e/(n+1), which does go to zero as you expected, and the answer to the problem is no more than 0.052. I guess it's good enough to answer "0". It's easy to see why, as your experiment showed, the result is highly dependent on the accuracy of the initial value, since if we use anything other than _exactly_ e, what we multiply by n! will not be approaching zero, and a_n will grow fast after some point. I've left the question active in our queue, in case anyone else comes up with an explanation. I hope this helps! - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 06/26/2002 at 13:23:29 From: Doctor Anthony Subject: Re: Snail! We can use difference equations to solve this. If u(n) = position at beginning of the nth day, then u(n) = n.u(n-1) - 1 u(n) - n.u(n-1) = -1 We now set up a chain of equations as shown: (1/2!)[u(2) - 2.u(1)] = -1/2! (1/3!)[u(3) - 3.u(2)] = -1/3! (1/4!)[u(4) - 4.u(3)] = -1/4! (1/5!)[u(5) - 5.u(4)] = -1/5! --------------------------------- add all the equations: u(5)/5! - u(1) = -[1/2! + 1/3! + 1/4! + 1/5!] u(5)/5! = u(1) - [1/2! + 1/3! + 1/4! + 1/5!] u(5) = 5![u(1) - [1/2! + 1/3! + 1/4! + 1/5!]] Note than on the left all the terms cancel BETWEEN rows except the first term on the last line and the second term on the first line. From this pattern, we see that the expression for u(n) will be u(n) = n![u(1) - [1/2! + 1/3! + 1/4! + ... +1/n!]] As n gets very large the series in brackets tends to e-2: u(n) = n![u(1) - (e-2)] and since u(1) = e-2 we obtain: u(n) = n! x 0 If n -> infinity this could be indeterminate but closer analysis or taking values of n of say 10, 11, 12 will show that u(n) -> 0 as n increases. We conclude that the snail will end up at the base of the window. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 06/27/2002 at 08:34:05 From: Jean-Claude Desbordes Subject: Thank you (Snail!) Thank you very much to Drs Ian, Peterson, and Anthony for helping me (and our team) in solving the snail problem. Really very kind of you to answer so quickly. As I told you, the snail problem is a riddle (the penultimate of 7) in a treasure hunt which started in November 2000, so it still could last for some time. I'll let you know as soon as we find the treasure! :o) If one of you comes to the Paris area one day, please let me know: I promise you a good meal. (What about a dozen snails with garlic butter as a starter? ;o)) Thank you again, Jean-Claude |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/