# Math Help - rubber band

1. ## rubber band

Pretend the rubber band is a graph. It may be stretched horizontally, but the y-intercept will not change unless the graph is shifted (the y-intercept is equivalent to the center of the rubber band)
So the middle of the rubber band will remain in place as long as the rubber band itself is not moved or shifted other than the stretch, equal on both sides.

2. The key to this is to realise that once the flea is on the rubber band, as the band stretches it carries the flea with it.

After one jump, the kangaroo has moved 1 mile and the flea is 1 inch along the band. When the kangaroo jumps for the second time, the rubber band stretches from 1 mile to 2 miles, so the flea will then be 2 inches (not 1 inch) along the band. The flea then jumps another inch. So by the time they have both jumped twice, the flea is 3 inches along the band.

When the kangaroo makes its third jump the band stretches from 2 miles to 3 miles long, carry the flea from 3 inches along to 4½ inches along the band. It then jumps another inch, bringing its total distance to 5½ inches.

Work out what happens after the next few jumps, and see if you can guess a formula for how far the flea has moved after n jumps. Then prove by induction that your guess is correct.

The kangaroo moves a fixed distance of 1 mile for each jump. But the distance moved by the flea is increasing with each jump. After a miilion or so jumps, will the flea finally reach the kangaroo?

i managed to get that that the kangaroo travels: 1 mile, then 2, then 3.... n miles

and the flea travels correspondingly: n/(n-1) * (n-1th term) + 1
which holds for n ≥ 2

now how should i prove this using induction?

also, i know that ill have to convert between miles and inches so i can assume that the value of n miles = 63,360n inches

4. Originally Posted by Aquafina

i managed to get that that the kangaroo travels: 1 mile, then 2, then 3.... n miles

and the flea travels correspondingly: n/(n-1) * (n-1th term) + 1
which holds for n ≥ 2

now how should i prove this using induction?

also, i know that ill have to convert between miles and inches so i can assume that the value of n miles = 63,360n inches
My apologies: the recurrence relation for $x_n$ (the number of inches travelled by the flea after n jumps) is indeed $x_n = \tfrac n{n-1} x_{n-1}+1$, but the solution is not as evident as I first thought. In fact, the solution is $x_n = n\sum_{r=1}^n\frac1r$. This is approximately equal to $n\ln n$, so it grows at a rate slightly faster than a constant multiple of n.

However, $n\ln n$ only becomes greater than 63360n when n is greater than about $10^{37517}$, by which time flea and kangaroo will have travelled about $10^{37506}$ miles.

5. how did you come to that solution?

6. Originally Posted by Aquafina
how did you come to that solution?
I'm embarrassed to admit how long it took me to find the formula $x_n = n\sum_{r=1}^n\frac1r$, having previously said that you should be able to guess it after working out the first few terms. For n=1,2,3,4,5 these are $1,\ 3,\ 5\tfrac12,\ 8\tfrac13,\ 11\tfrac5{12}$. It seemed natural to write these as fractions with factorials as the denominators, namely $\frac1{0!},\ \frac3{1!},\ \frac{11}{2!},\ \frac{50}{3!},\ \frac{274}{4!}$. I still didn't recognise the numerators, so I searched for the sequence 3, 11, 50, 274 in the wonderful Online Encyclopedia of Integer Sequences (one of the most useful mathematical resources on the internet), and it pointed me in the direction of the harmonic series $\textstyle\sum\frac1r$.

7. Hello,
Originally Posted by Opalg
I'm embarrassed to admit how long it took me to find the formula $x_n = n\sum_{r=1}^n\frac1r$, having previously said that you should be able to guess it after working out the first few terms. For n=1,2,3,4,5 these are $1,\ 3,\ 5\tfrac12,\ 8\tfrac13,\ 11\tfrac5{12}$. It seemed natural to write these as fractions with factorials as the denominators, namely $\frac1{0!},\ \frac3{1!},\ \frac{11}{2!},\ \frac{50}{3!},\ \frac{274}{4!}$. I still didn't recognise the numerators, so I searched for the sequence 3, 11, 50, 274 in the wonderful Online Encyclopedia of Integer Sequences (one of the most useful mathematical resources on the internet), and it pointed me in the direction of the harmonic series $\textstyle\sum\frac1r$.
I don't want to hurt you, but I must say I'm quite surprised to read this from you
I must've got the correct idea accidentally... :s

$x_n=\frac{n}{n-1} \cdot x_{n-1}+1$

Then $\frac{x_n}{n}=\frac{x_{n-1}}{n-1}+\frac 1n$

If you don't see it, let $y_n=\frac{x_n}{n}$. We thus have :

$y_n=y_{n-1}+\frac 1n$

$y_n-y_{n-1}=\frac 1n$

So $y_n-y_1=\sum_{r=2}^n \{y_r-y_{r-1}\}=\sum_{r=2}^n \frac 1r$

$x_1=1 \Rightarrow y_1=1$

$\Rightarrow y_n=1+\sum_{r=2}^n \frac 1r=\sum_{r=1}^n \frac 1r$

$\Rightarrow \boxed{x_n=n \sum_{r=1}^n \frac 1r}$

huh?

8. Originally Posted by Moo
Hello,

I'm quite surprised to read this from you
it could even happen to Opalg! nobody is safe!

9. Originally Posted by NonCommAlg
it could even happen to Opalg! nobody is safe!
Sure ! But you know, Opalg is like an ideal/idol...

Anyway, don't take it badly, it isn't my intention

10. Originally Posted by Moo
Hello,

I don't want to hurt you, but I must say I'm quite surprised to read this from you
I must've got the correct idea accidentally... :s

$x_n=\frac{n}{n-1} \cdot x_{n-1}+1$

Then $\frac{x_n}{n}=\frac{x_{n-1}}{n-1}+\frac 1n$
Very nice!

Yes, I should have seen that. Clearly I need another vacation. (I'll be away for most of June and some of July. Maybe that will help to restore brain function.)

11. thanks oplag and moo!

how did you achieve the approximation to it being ?

also, what i had was basically:

saying that at a particular value, the increasing factor of the flea will equal 63360, i.e. it will increase by a mile.

so: n/n-1 + 1 = 63360

===> n = 63359/63358

what is wrong with this approach?