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Thread: The Worm

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    The Worm

    This is puzzle 59 from the 60 in The Age last week.

    Imagine a rubber rope one metre long. A worm starts at one end and travels along the rope at 1 centimeter per second.
    At the end of each second the rope is stretched so that it is one metre longer than it was before.
    The worm is carried along with the stretching.

    Does the worm ever reach the end of the rope?
    Regards
    John



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    Re: The Worm

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">No</span hide>

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    Re: The Worm

    Hans

    Did you work this out by writing the equations for this? or just solve it with 'intuition'. I ask because my intuitive answer was the same as yours, but the 'official' answer (Q59) is the opposite. This site will publish detail worked solutions in a few days.


    I am about to have a go at writing out the equations.
    Regards
    John



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    Re: The Worm

    My answer was intuitive (<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">the distance the worm still has to travel increases each second</span hide>)

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    Re: The Worm

    OK, I was right and wrong at the same time. The worm won't reach the end of the rope within its lifetime, or even within the lifetime of the universe. But in a theoretical, mathematical sense it will, eventually, if the whole thing could go on forever without spatial or temporal restrictions.

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">After n seconds, when the rope has just been stretched, its length is n+1 metres.
    If you write out the position of the worm, you'll find that it satisfies the formula (n+1)*(1+1/2+1/3+...+1/n) centimetres, or (n+1)*(1+1/2+1/3+...+1/n)/100 metres.
    So the question becomes: will 1+1/2+1/3+...+1/n ever become 100?
    It is well known (among mathematicians and similar nerds) that the sum 1+1/2+1/3+...+1/n grows like ln(n) where ln is the natural logarithm, and ln(n) grows slowly but without a limit, so eventually the sum will become greater than 100.
    For a very rough order-of-magnitude estimate: ln(n) = 100 when n = e^100. This is a number with 44 digits. For comparison, the current age of our universe is estimated by astronomers to be about 14.5 billion years, which translates to a number of seconds consisting of 18 digits. So the time needed to reach the end of the rope is completely outside our comprehension.
    Also, the rope would become extremely long - immeasurably longer than the current diameter of the universe.</span hide>

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    Re: The Worm

    So it is Schrödinger's worm
    Jerry

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    Re: The Worm

    My answer was arrived at purely by axiomatic deduction:

    - Hans is always correct.
    - The length of the hidden text in his answer seems, well, fairly short.
    - The correct answer must therefore be <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">NO</span hide>

    <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
    Alan

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    Re: The Worm

    Well as it is now a discussion, I disagree with Hans' first answer, it is in fact <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Yes</span hide>.

    This parable is a description of a Harmonic Series which is an infinite series

    <IMG SRC=http://upload.wikimedia.org/math/7/2/e/72eabbde965f1d21542db02dc261417c.png>

    To put this into context, the first 10^43 sum to less than 100!!!!!!!

    Eventually the worm will get there but it won't be in the lifetime of the Universe, so the answer is also <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">No</span hide>

    It could best be described as a Schrödinger worm
    Jerry

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    Re: The Worm

    I'll have to stick to NO I'm afraid. The XL sheet attached to the next post shows the process at work. The distance the worm has to travel to reach the rope's end is ever-increasing as far as I can see. These are the variables used:
    Attached Images Attached Images

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    Re: The Worm

    Also, well before the end of the universe, the thermodynamics of hyperelasticity would dictate that the rubber rope would break... and break again and again and again. So would the worm reach the (original) end in this case? <img src=/S/grin.gif border=0 alt=grin width=15 height=15>

    Alan
    Attached Files Attached Files

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    Re: The Worm

    Physically, the answer is clearly and unequivocally NO.
    I assume that the question was meant as a purely theoretical thought experiment: a dimensionless point sits on a straight line, etc.

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    Re: The Worm

    But even in the theoretical, idealized context, which was the basis for my analysis, the answer is still NO. I'm wondering what the YES explanation will be.

    Alan

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    Re: The Worm

    The purely theoretical answer is Yes - see <post:=745,126>post 745,126</post:> and <post:=745,219>post 745,219</post:>.

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    Re: The Worm

    Each second the worm travels 1 cm, plus a proportion of the stretching. It will only get closer to the end , if it travels more than 1 metre in the second.
    Does it ever do this? Yes. Once it has traveled more than 99% of the length of the rope, the stretch component of its movement will be more than 99 cm.

    How can it ever get to the 99% mark if the distance to travel increases each second?
    Even though the distance still to travel each second is (initially) increasing, the proportion of the rope traveled is also increasing.

    The proportion of the rope traveled is given by Harmonic series previously referred to, so eventually it reaches 99.

    The 'official' solution is now published here, and uses the same series.
    Regards
    John



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    Re: The Worm

    I arrived at the same result myself (in terms of the % of rope "consumed" with each second). But I also saw the ever-increasing amount of rope still to be consumed. For some reason I discounted the former in favour of the latter. I can now see that the situation changes at the point you describe.

    Alan

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