Maths problem to solve

[Aractus]

This is asinine. If for every step you take, 99 more steps are added to the journey you will never complete the journey. If you present math that suggests otherwise, your math is wrong. Inescapable.

Well since a few of you now have suggested this I'll prove that it can complete the journey, and then leave it to you all to figure out how long it will take. We're not always adding 99 more steps as I just pointed out:

START:
0.
Elastic band: 1m
Ant walks 1cm
Distance in front of ant: 99cm
Distance behind: 1cm
Percentage of circumference so-far travelled (1/100): 1%
1.
Elastic band: 2m
Ant walks 1cm
Distance in front of ant: 197cm
Distance behind: 3cm
Percentage of circumference so-far travelled (3/200): 1.5%
2.
Elastic band: 3m
Ant walks 1cm
Distance in front of ant: 294.5cm
Distance behind: 5.5cm
Percentage of circumference so-far travelled (5.5/300): 1.8333... %
3.
Elastic band: 4m
Ant walks 1cm
Distance in front of ant: 391.666... cm
Distance behind: 8.333... cm
Percentage of circumference so-far travelled (8.333.../400): 2.08333... %
...

So you can see the ant is making progress, after 4 seconds he has travelled a little over 2% of the total circumference. That means that less distance is being added in front of him after each second than the previous second. That is, that only the first time was 99cm added in front of the ant, the next time it was stretched only 98.5cm was added, and then the next time only 98.1666... cm was added. So you can see the distance added to his journey each time gets smaller and smaller. So the ant can complete the journey. But you absolutely can't calculate it by hand second by second - if you try to do that believe me you will never find the answer, you need to generalise what's happening into a series and then solve for the nth term of the series (n being the number of seconds).

[Tiberius]

Ah I see, because the distance the ant has walked increases as you stretch the elastic band.

[TheRealJoeFish]

I am going to take a look at this tomorrow! (Not reading any comments but OP)

My first instinct is to try to create a function describing the % of the loop that's behind the ant after n seconds. But we'll see what happens.

[Whateverist]

I must not be understanding the question.  After one second the ant travels 1 cm but the distance left for him to travel increases from 100 cm to 199 cm.  After two seconds, he will have traveled 2 cm but now has 298 cm left to go.  How could the ant ever finish if there are no restraints on the capacity of the band to keep on stretching.  Since the length of band left for the ant to travel will always be greater than before, how can we talk of him ever completing the band?

Since no mention is made of the space available for the band to stretch I have no idea if that is a consideration, but I assume not.  What am I missing?

You're not missing anything, but you have incorrectly calculated the distance. After 1 second the ant travels 1cm, leaving him 99cm to go, and you stretch the elastic band to 2m. The distance in front of the ant is now 198cm (double), and the distance behind the ant is now 2cm. The he walks another cm, leaving him 197cm to go, and you stretch the elastic band to 3m. The distance in front of him is now 295.5cm, and the distance behind him now 4.5cm. So you can see the distance left to go is not quite as far as you thought (298cm).

Again please see my hint - construct a series that describes the distance the ant travels after n seconds accurately.

I see, it doesn't simply tack on another meter. Instead it stretches the entire band evenly, including the portion already traveled. Got it. Thanks.

[Alex K]

Ok, I have it apart from the numeric result

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n: length of band after last stretch, n=1,2,3,4,5...
f I use variable r denoting fraction of band travelled, this is invariant under stretching. I merely have to reduce the fraction walked in each step;

r_new=r_old + 0.01/n

where n is the length after last stretch, and we can start at n=1.
This is a harmonic series, and it diverges, i.e. the fraction will pass 1.

Test:

n=1: r1=0.01
n=2: r2=0.015
n=3: r3=0.0183333...

So the fraction is the harmonic sum 0.01 * sum_n 1/n

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Ok, so the ant will reach the end, but it will take her a looooooong time.

[Alex K]

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So, the harmonic sum approaches ln(n)+gamma, but we can neglect gamma when the log becomes >>1 which it will - so the ant approximately reaches the end when

1= 0.01 ln n

or

ln n =100

or

n=e^100

which is a very large rubber band indeed.

And by very large I mean a million billion times the size of the observable universe

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[Aractus]

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And by very large I mean a million billion times the size of the observable universe

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Well, that's very clearly understating it.

[BrianSoddingBoru4]

I'm surprised. The answer to math problems is usually 6.

Boru

[Aractus]

Sadly no. 6 minutes, six days, six months or even six years is in fact not enough time for the ant to complete his trip. Of course he is your pet, your favourite pet, so if after 6 years you feel it necessary to remove him from the elastic band and to cuddle him and show him affection, that's totally understandable.

[BrianSoddingBoru4]

Just to be clear...you're sure the answer isn't '6'?

Boru