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**Septimus****Guest**

A caterpillar crawls along a very flexible 1 meter (100 cm) long bungee. It starts at on of the ends of the bungee. When the caterpillar has crawled 1 cm, someone stretches the bungee out to 2 meters. When the caterpillar then has crawled another cm, the bungee is streched out to 3 meters.

This goes on and on.

How long it would take before the caterpillar reached the other side if the caterpillar crawled 1 cm per second.

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Very good puzzle. It's a 1d version of what is happening to our universe.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Septimus****Guest**

Ricky wrote:

Very good puzzle. It's a 2d version of what is happening to our universe.

Excellent observation.

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

It would have been if I had written "1D". Changed it now.

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**coolcat23****Member**- Registered: 2006-06-21
- Posts: 553

very confusing puzzle.......................

The world revolves around me. Deal with it.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

This doesn't seem right. The bungie cord is expanding much faster then the catapillars speed. It seems to me it would never reach the other side. Sure you phrased the question right?

*Last edited by mikau (2006-09-18 15:23:25)*

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Remember that when the bunge cord expands, it's not every point which is expanding by an inch, but an inch overall. So 1/2 the bungee cord only expands by 1/2 an inch.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

This would probably be easier if we flipped around what happens. Instead of the bungee cord extending, let's say that the caterpillar moves less far each time.

So first it moves 1cm, then 1/2cm, then 1/3cm and so on.

You'd still get the same answer and it's easier to work out this way. There's probably an equation that will sum that series that you can use to work out how long it'll take.

Why did the vector cross the road?

It wanted to be normal.

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**Septimus****Guest**

Exactly, mathsyperson. And that's the harmonic series.

So, if we define the nth harmonic number as

then Hn grows about as fast as the natural logarithm of n.

And through some thinking, and some guidance from my math book, I got that there's little difference, so I went ahead and checked the difference.

After having compared various ones, I found out that there was a constant difference, which I later realized was Euler's constant (thanks wikipedia)

y or the greek letter gamma = ~0.577

SO!

ln (x) + y = 100 cm

ln (x) = 99.423

e^99.423 = 1,509594394 x 10^43 seconds

or

478 689 242 000 000 000 000 000 000 000 000 000 years

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I just wrote a BASIC program and ran it to 12% and noticed each percent the time increased by a factor of 4, then 3, then less than 3, and approached 2.718 about. I also agree hence between e up 99 and 3 up 99.

Excellent work, humans.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

That's about half of a trillion trillion trillions.

**igloo** **myrtilles** **fourmis**

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

Septimus wrote:

478 689 242 000 000 000 000 000 000 000 000 000 years

Hmm...I doubt the caterpillar would actually live that long, though.

My answer: The caterpillar stands no chance of reaching the end.

*Last edited by Devanté (2006-09-19 05:20:58)*

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**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

Devanté wrote:

Septimus wrote:478 689 242 000 000 000 000 000 000 000 000 000 years

Hmm...I doubt the caterpillar would actually live that long, though.

My answer: The caterpillar stands no chance of reaching the end.

I doubt the universe will live that long.

I could see that given enough time, the caterpillars progress each second would be greater than the rate at which the bungee lengthened, since the caterpillar's position with respect to time had a non-zero positive second derivative, while the second derivative of the bungee length was zero.

I wrote a C program to find the solution recursively, but after it ran for about twelve hours it was still not finished running.

*You can shear a sheep many times but skin him only once.*

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**Mr. Big****Guest**

The caterpillar will never reach the other end:

Let's assume there's a distance 'd' between the caterpillar and the end.

d = 100 cm.

The caterpillar travels 1 cm, d = 99 cm, the rope is stretched to 200 cm (twice as long), the distance scales by two, d = 198 cm.

The caterpillar travels another cm, d = 197, the rope is streched to 300 cm (3/2 as long), the distance scales by 3/2, d = 295.5 cm and so it goes on and on.

The distance only grows with time:

dt = (((((100 - 1) * 2) * 3/2) * (1 + 1/...)) * (1 + 1/t))

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Remeber that once the catipliar tavels X% of the rope, no matter how much the rope streches or shrinks, he will have always traveled X% of the rope.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's some interesting alternative spellings you've got there.

Mr. Big's reasoning is correct, but he forgets that each time the rope is stretched, the scale factor is less, which means the increase in distance is less. You can see it in his workings. First the caterpillar is 100cm away from the finish, then the finish gets 98cm further away, then 97.5cm and so on.

After a very long time, the increase in distance will become less than 1cm each time, meaning the caterpillar can start to gain ground again. Of course, it will still take ages to that because by that time the finish will be very far away indeed. But it will reach the finish.

Why did the vector cross the road?

It wanted to be normal.

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**Mr. Big****Guest**

The caterpillar will reach his target when dt = 0:

0 = (((((100 - 1) * 2) * 3/2) * ... * (1 + 1/t))

To make 'dt' zero, (1 + 1 / t) has to be zero, which is impossible.

Am I missing something?

**Mr. Big****Guest**

It's actually supposed to be:

0 = ((((((100 - 1) * 2 - 1) * 3/2 - 1) * 4/3 - 1) * ... - 1) * (1 + 1/t))

But that doesn't change much.

**Mr. Big****Guest**

The increase in distance is RELATIVELY less every time, but it is still the same amount of 100 cm!

Logically, how can the caterpillar ever finish if every time it passes one cm, the distance increases by 100?

**Mr. Big****Guest**

Ignore that last post.

You're right, the distance does decrease.

**Mr. Big****Guest**

Never mind. I'm wrong. '(((((100 - 1) * 2 - 1) * 3/2 - 1) * 4/3 - 1) * ... - 1)' will be zero at some point.

Sorry for all this posting.

Guests don't have an 'Edit' button.

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Well sign up then!

Don't worry about it, I thought exactly the same thing when I first saw the puzzle. It's very counter-intuitive.

Why did the vector cross the road?

It wanted to be normal.

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