Can an Immortal Snail Ever Reach the End of a Stretching Rubber Band?

[Office_Shredder]

Similar to the previous two-part question, if you find part b to be an appropriate challenge please leave part a to those who are appropriately challenged by it.

This question is courtesy of mfb

An immortal snail is at one end of a perfect rubber band with a length of 1km. Every day, it travels 10cm towards the other side. Every night, the rubber band gets stretched uniformly by 1km. As an example, during the first day the snail will advance to x=10cm, then the rubber band gets stretched by a factor of 2, so the snail is now at x=20cm on a rubber band of 2km.

Will the snail ever arrive at the other side, and if yes, how long does it take approximately?

 

[cpscdave]

Hi everyone!
First crack at a math challenge. Not sure if I'm on the right track. Sadly all my relevant math notes and Ti-89 (which might as well be a third arm for me) is on the other side of the country.

But hoping someone can tell me if I'm barking up the right tree or if want I want to do isn't possible.

Spoiler

Calculating by hand I find the position of the snail to be:
(end of) Day 1 =0.0002
Day 2 = 0.00045
Day 3 = 0.000733
Day 4 = 0.001042 etc etc etc

This leads me to beliving the formula for the position of the snail is
$$x[n]=0.0002 \,\,n=1$$

$$x[n] = {\frac{(n+1)}{n}}*(x[n-1]+0.0001) \,\, n > 1$$

expanding this I get something like
$$x[n] = x[n-1] + 0.0001 + {\frac{1}{n}}*(x[n-1] + 0.0001)<br />$$

from here do the Z transform,
solve for x[z] and inverse it and you'll have a formula for the snails position. Subract n and figure out where the function goes positive??

I realize there is a bit of hand waving here :)

Specifically where I'm stuck is figuring out the z transform of

$${\frac{1}{n}}*(x[n-1] + 0.0001)$$

 

[hilbert2]

This challenge might be too easy for me, but here's my solution anyway...

Spoiler

After the first day, the fraction of rubber band's length traveled is $\frac{0.1m}{1000m}=\frac{1}{10000}$. Stretching the rubber band does not change this fraction. On the second day, the snail travels a fraction of $\frac{0.1m}{2000m}=\frac{1}{20000}$ along the rubber band. Therefore, after $n$ days, the fraction is $\frac{1}{10000}\sum_{k=1}^{n}\frac{1}{k}$. As the harmonic series $\sum_{k=1}^{\infty}\frac{1}{k}$ does not converge, the snail will eventually get to the other end of the rubber band. The number of days needed for this can be solved from the equation

$\frac{1}{10000}\sum_{k=1}^{n}\frac{1}{k}=1$.

To solve this, we make the approximation $\sum_{k=1}^{n}\frac{1}{k}\approx \ln n + \gamma$, where $\gamma$ is the Euler-Mascheroni constant. With this approximation, we get the result

$n \approx e^{10000-\gamma}$,

which is a very large number, about $4.945 \times 10^{4342}$.

 

[mfb]

@hilbert2: that is correct.