Inchworm

The worm will succeed. But it will take many billions of years...

Originally posted by PBE6
It's true that the rope is stretching faster than the worm can inch along, and at any point along the journey you can check the worm's position and see that it's falling behind. But "faster" only matters if there's a time limit.

The length of the rope will keep growing infinitely, and the worm will inch out an infinite distance. So the question is re ...[text shortened]... ough time. And since the longest rope length is no exception, the worm will eventually make it!

This is a great pair of explanations, PBE6.

I'm not sure it is infinity, though.

When I read that the rope is stretched uniformly, I took that to mean that the additional
stuff is added, not to one end, but evenly throughout.

So, after one second, he will have traveled not 1 cm of the now 2 meter rope, but 2 cm (because
99 cm of the new length was added in front of him, but 1 cm is added behind him).

I don't know enough about math to understand if this makes a difference, but intuitively I believe
that it should.

Nemesio

Originally posted by Nemesio
This is a great pair of explanations, PBE6.

I'm not sure it is infinity, though.

When I read that the rope is stretched [b]uniformly, I took that to mean that the additional
stuff is added, not to one end, but evenly throughout.

So, after one second, he will have traveled not 1 cm of the now 2 meter rope, but 2 cm (because
99 cm of the ...[text shortened]... to understand if this makes a difference, but intuitively I believe
that it should.

Nemesio[/b]

Oh! Now I get it!

Thanks Nemesio and PBE6!

i feel all enlightened!

Originally posted by davegage
However, whether or not the worm gets to the end is not independent of how much we stretch the rope each time. If we stretch it sufficiently far each time, then we can make it so that the worm can never make it to the end, even if he lived i ...[text shortened]... on of rope traversed by the worm converges to 1 as N goes to infinity.

I withdraw this follow up question on the grounds that I don't think it's as simple as I thought. I think actually, you might be able to show that if we stretch the rope always by the same finite amount, then it doesn't matter how long that amount is -- you will still end up with some modified version of the harmonic series and hence it must diverge, and hence the worm will always get to the end in finite time (though it may be an incredibly long time even on geological scale).

I think also, this may tie into what PBE6 was saying about surjective mapping, which I think is a good point.

Originally posted by davegage

The proof follows these lines:

Consider the fraction of rope that the worm has traversed after the Nth second. Since the stretching of the rope is uniform (ie, "affine" deformation), this fraction is preserved upon each stretch -- ...[text shortened]... nt is that he would only have to live a FINITE amount of time.

NOTE: THIS POST IS UNDER CONSTRUCTION. BEWARE POOR ARITHMETIC.

THE PREVIOUSLY POSTED MATERIAL HERE WAS GARBAGE. WIPE IT FROM YOUR MEMORY.

At time 0, the worm is at position 0 on a rope of 2 units. He has traversed 0/2=0% of the current rope length.

At time 1, the worm has crawled 1 unit, and since the rope has doubled in length, an extra unit has grown behind him. So he is at position 2 and the rope is now 4 units long. He has traversed 2/4=50% of the current rope length.

At time 2, the worm has crawled an additional 1 un ...[text shortened]... he worm has traversed over 100% of the rope. The quoted series encapsulates these iterations. [/b]

Originally posted by DoctorScribbles
Let us choose simpler quantities to show how the worm reaches the end of the rope. Suppose that the rope starts at 2 units and grows 2 units at each time step.

Can one figure out a rate of uniform growth that would make it such that the
worm would be traveling in place? I mean, if it grows (uniformly) at a billion
units at a time, would the worm still get to the end in some (long) finite time
or is there an upper bound?

If there is no upper bound, this boggles my mind.

Another question: instead of growing in spurts at discrete times, what if it
grew continuously (growing gradually 1 units of length per unit of time)? Does
this change the problem in an interesting way?

Nemesio

Originally posted by Nemesio

Another question: instead of growing in spurts at discrete times, what if it
grew continuously (growing gradually 1 units of length per unit of time)? Does
this change the problem in an interesting way?

No, it does not really change the problem. The same sort of discrete analysis holds, for you can simply choose to observe the continuous system at discrete points in time. In fact, I interpreted the problem to mean that the rope grows continuously.

Originally posted by DoctorScribbles
NOTE: THIS POST IS UNDER CONSTRUCTION. BEWARE POOR ARITHMETIC.

THE PREVIOUSLY POSTED MATERIAL HERE WAS GARBAGE. WIPE IT FROM YOUR MEMORY.

Crap. I thought I understood it.

😞

Originally posted by Nemesio
If there is no upper bound, this boggles my mind.
Nemesio

Prepare for some mind-boggling:

To be general, suppose that the rope is initially x cm long, and suppose that we stretch it an extra x cm after each second (so this is just like the initial case posted if you substitute x = 100 cm, but this will be more general). We could probably also be completely general with the speed of the worm, too, but let's just say he is moving 1 cm/sec.

Then you can show that after the first second, the fraction of rope traversed is 1/x; after the second second, it is 1/x + 1/(2x). After the third it is 1/x + 1/(2x) + 1/(3x), and generally, after the Nth second, it is Summation (i=1 to i=N)[1/(ix)].

Thus the fraction of rope traversed is simply (1/x) times the harmonic series carried out to the Nth term. But since the harmonic series diverges, you can always find some number M such that the harmonic series carried out to this Mth term is greater than x (and hence the fraction traversed exceeds 1), no matter what finite value x is.

In other words, there is no upper bound for how large x can be. As long as x is finite, the worm will always traverse in some finite amount of time.

So even if we choose, say, x = 10^79839, the worm will still traverse in some (really long, but) finite time.

If that boggles your mind, then I'm right there with you. I think the analysis is correct though....

Originally posted by DoctorScribbles
No, it does not really change the problem. The same sort of discrete analysis holds, for you can simply choose to observe the continuous system at discrete points in time. In fact, I interpreted the problem to mean that the rope grows continuously.

I agree that it shouldn't change the problem whether the rope grows continuously or in discrete spurts. So I think in the above post x could also represent a rate of rope growth as in x cm/sec.

EDIT: an important point, however, is that the rope is not growing, per se, but stretching and the worm moves also with the stretching process relative to some fixed axes. But keeping this in mind, you could think of it as a continuous stretching process.

Originally posted by Nemesio
Crap. I thought I understood it.

😞

Try this. My edit time expired on my last post.

------------------------------------

I concur with davegage's conclusion and proof. I'll attempt to post an illustration of it for those unfamiliar with sequences, series, and divergence.

In the stated problem, the 100 unit rope grows 100 units uniformly each time period, and the worm moves one unit each time period. Let us choose simpler quantities to show how the worm reaches the end of the rope. Suppose that the rope starts at 2 units and grows 2 units at each time step.

Let L(t) denote the total length of the rope at time t.
Let C(t) denote the total distance that the worm has crawled using his own legs at time t.
Let S(t) denote the total distance behind the worm that he did not physically crawl at time t.
Let R(t) denote the ratio between the worm's remaining distance on the current rope length and the total rope length at time t.

Note that R(t) = [L(t) - [C(t)+S(t)]]/L(t).
Note further that L(t) = 2 + 2*t.
Note further that C(t) = t.
Thus R(t)=[2+2*t - t - S(t)]/[t + 2*t]=[2+t-S(t)]/[t+ 2*t].

So, the problem reduces to asking whether there exists a time t such that
2+t-S(t) <= 0, or equivalently,
S(t)>=t+2

So, let us examine the sequence of S values.

S(0) = 0, since the rope hasn't strected. So S(0) = 0 < 0+2.

S(1) = 1, since the rope stretched 2 units during the first time unit, and since half of that stretch was distribted behind the worm, who had crawled half of the rope length before the stretch. So, S(1) = 1 < 3.

S(2) = S(1)+2*(3/4)=2.5, since the rope stretched 2 units during the second time unit, and since 3/4 of that stretch was distribted behind the worm, who had crawled 3/4 of the rope length before the stretch. So, S(2) = 2.5 < 4.

In general, we see that there is a pattern: S(t) = S(t-1) + 2*[S(t-1)+C(t)]/2*t].
So, we can simply compute the S values.
S(3)=4.3< 5
S(4)=6.4 > 6

Thus, sometime before time 4, the worm reaches the end of the rope.

A similar analysis applies for the original problem with the 100 unit rope.

Correction:

Thus R(t)=[2+2*t - t - S(t)]/[t + 2*t]=[2+t-S(t)]/[t+ 2*t].

should instead read

Thus R(t)=[2+2*t - t - S(t)]/[2 + 2*t]=[2+t-S(t)]/[2+ 2*t].

This is a fortunate typo in that in does not affect anything that follows, as the denominator of the fraction has no bearing on the analysis.

I was trying to come up with an analytical solution for the worm's position and the length of the rope with respect to time, and I think I have one. Unfortunately, it's not even close to my discrete model. One or both is wrong. How does this look to y'all?

let W = worm's position = coordinate on the x-axis
let L = rope length = coordinate of the rope's end on the x-axis
let L0 = rope's initial length
let k = rope stretch rate (in this case 100 cm/s)
let w = worm's crawl rate (in this case 1 cm/s)

Now, the change in the worm's position with time is just the rope stretch rate "k", so:

dL/dt = k

The change in the worm's position is its crawl rate plus a function of the rope stretch rate. Since the rope is stretched uniformly, the speed boost the worm gets is proportional to its position along the rope, so:

dW/dt = w + k(W/L) = w + (dL/dt)(W/L)

To solve this equation, we note that:

d(W/L)/dt = [ L(dW/dt) - W(dL/dt) ]/L^2

L(dW/dt) - W(dL/dt) = L^2[d(W/L)/dt]

by the quotient rule. Rearranging our worm equation, we get:

L(dW/dt) = wL + W(dL/dt)

L(dW/dt) - W(dL/dt) = wL

and subbing in we get:

L^2[d(W/L)/dt] = wL

d(W/L)/dt = w/L

Integrating both sides, we get

(W/L)-(W0/L0) = int(w/L) = int(w/(L0 + kt)) = (w/k)*ln(L/L0)

Now W0 = 0, so we end up with:

(W/L) = (w/k)*ln(L/L0)

Solving for W, we get:

W = L*(w/k)*ln(L/L0)


Can anyone spot a logical mistake, or does it look OK?

Originally posted by PBE6
I was trying to come up with an analytical solution for the worm's position and the length of the rope with respect to time, and I think I have one. Unfortunately, it's not even close to my discrete model. One or both is wrong. How does this look to y'all?

let W = worm's position = coordinate on the x-axis
let L = rope length = coordinate of the rope ...[text shortened]... r W, we get:

W = L*(w/k)*ln(L/L0)


Can anyone spot a logical mistake, or does it look OK?

I can't find anything wrong with your analysis, but I find the result a bit perplexing.

I was under the impression that there was no difference between whether we stretch the rope in discrete increments or whether the rope is stretched continuously, but your result above does not match the discrete case.

For example, consider the original case where the original lenth is 1 meter, and we stretch it another meter after each second, and the speed of the worm is 1 cm/sec. Then the fraction of rope traversed after 1 second is clearly 1/100 = 0.01 (it's this value right before and right after the rope is stretched).

But if we use w = 1 cm/sec, k = 100 cm/sec, and L0 = 100 cm in your answer above, then your continuous model says the fraction traversed (W/L) after 1 second is only 0.01*ln2 = 0.000693....

This confuses me greatly...I don't see why, in general, the two cases should yield different results. Regardless, I cannot find anything wrong with your analysis.

With your continuous model, we can solve for the time when the worm will reach the end by simply setting W/L = 1 and by using L = L0 + kt.

When I do that for the original case of L0 = 1 meter, w = 1 cm/sec, and k = 100 cm/sec, i get that it should take the worm about 8.5 X 10^35 years! This is considerably longer, I think, than what the discrete model predicts.

i'm very confused...