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September 7th, 2012, 08:33 PM  #1 
Senior Member Joined: Aug 2010 From: Germany Posts: 132 Thanks: 1  Dichotomy paradox reverse engineering
t = 0: move pen from 0 to 0.5 (taking 1/2 second), naming it segment #1 t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second), naming it segment #1 t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second), naming it segment #1 ... At which step should we stop using #1 as the segment's index in order to have enough segments of nonzero size left to assign the rest of the natural numbers? If we don't stop using #1 all segments in [0, 1) will be named #1 at t = 1, right? What about the size of all other segments (#2, #3, ...) at t = 1 then? I mean it like this: t = 0: move pen from 0 to 0.5, naming it segment #1, [0.5, 0.75] is #2, [0.75, 0.875] is #3, ... t = 0.5: move pen from 0.5 to 0.75, naming it segment #1, [0.75, 0.875] is #2, [0.875, 0.9375] is #3, ... t = 0.75: move pen from 0.75 to 0.875, naming it segment #1, [0.875, 0.9375] is #2, [0.9375, 0.96875] is #3, ... and so on. 
September 8th, 2012, 12:49 AM  #2  
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Re: Dichotomy paradox reverse engineering Quote:
f(n) = 1  (1/2^n) Clearly t = 1 is not in the range of f. So the behavior of your system at t = 1 is not defined. Most of these supertask problems can be resolved by noting that the task is not welldefined at the limit.  
September 8th, 2012, 01:53 PM  #3  
Senior Member Joined: Aug 2010 From: Germany Posts: 132 Thanks: 1  Re: Dichotomy paradox reverse engineering Quote:
The task is to design a process that enables us to draw a line from point 0 to point 1 in 1 second in infinitely many steps. Any suggestions how to accomplish that (i.e. which steps have to be defined)? The pencil is not allowed to change direction. Point 1 is the last point covered by the line. There is no last step in this process.  
September 8th, 2012, 05:07 PM  #4  
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Re: Dichotomy paradox reverse engineering Quote:
Continuous motion. What a concept!  
September 9th, 2012, 12:03 AM  #5  
Senior Member Joined: Aug 2010 From: Germany Posts: 132 Thanks: 1  Re: Dichotomy paradox reverse engineering Quote:
 
September 10th, 2012, 12:50 PM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Dichotomy paradox reverse engineering Quote:
 
September 10th, 2012, 07:23 PM  #7  
Senior Member Joined: Aug 2010 From: Germany Posts: 132 Thanks: 1  Re: Dichotomy paradox reverse engineering Quote:
t = 0: move pen from 0 to 0.5 (taking 1/2 second) t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second) t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second) t = 0.875: move pen from 0.875 to 0.9375 (taking 1/16 second) ... t = 1: move pen from 1 to 1 (taking 0 second) t = 1: move pen from 1 to 1 (taking 0 second) t = 1: move pen from 1 to 1 (taking 0 second) ... we are ending up in an endless loop which allows the last point 1 to be drawn without introducing a last step of the process. Even if we didn’t define a last step we can lift the pencil, since there is no difference between executing this endless loop and not executing it (the endless loop does do nothing in time and space). The problem is: This model is lacking a description of the process that enables us to adjoin the endless loop. That's why I think it might be useful to clarify this question: Quote:
Quote:
 
September 11th, 2012, 10:05 AM  #8  
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Re: Dichotomy paradox reverse engineering Quote:
You can take the millionth term of this sequence, or the billionth, or the googolth or the googolplextothegoogolplexth term, and none of them are 1. 1 is the limit of the sequence, but it's not attained by any element of the sequence. That's why the limit concept is so clever. It packages up this counterintuitive mystery of getting "arbitrarily close without attaining" in a concise definition that we can teach to calculus students. [Not that they understand it, but that's what Real Analysis is for ] But I don't understand what you're trying to do. The facts I outlined are wellunderstood and universally accepted in mathematics. No finite halving process will let you hit the endpoint. Another way to put this is that in binary, the sequence .1, .11, .111, .1111, ... has 1 as its limit, but none of the members of the sequence is 1. So in fact the binary number .111... is equal to 1; but that expression really denotes an infinite sum 1/2 + 1/4 + 1/8 + ... No finite sum of powers of 1/2 can be equal to 1.  
September 11th, 2012, 11:37 AM  #9  
Senior Member Joined: Aug 2010 From: Germany Posts: 132 Thanks: 1  Re: Dichotomy paradox reverse engineering
I think I can understand the concept of limits, but I am not that sure about it. If I try to apply the concept of limits to the task Quote:
 
September 11th, 2012, 03:14 PM  #10  
Senior Member Joined: Aug 2012 Posts: 2,342 Thanks: 731  Re: Dichotomy paradox reverse engineering Quote:
Let me repeat that. Your process is not defined at t = 1. So for all we know, at step n you have your pen moving to 1  1/2^n or whatever. And at t = 1 your pen turns upside down and whistles Dixie. Nobody knows what your pen does at t = 1 because you haven't taken the trouble to defined the behavior of the pen at t = 1.  

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