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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
Associated Topics:
College Calculus
High School Calculus
High School Puzzles

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