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Math Help - rubber band

  1. #1
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    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.
    Last edited by Aquafina; June 10th 2009 at 05:51 AM.
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    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 4 inches along the band. It then jumps another inch, bringing its total distance to 5 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?
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    thanks for the reply

    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
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    Quote Originally Posted by Aquafina View Post
    thanks for the reply

    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.
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    how did you come to that solution?
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    Quote Originally Posted by Aquafina View Post
    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.
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    Hello,
    Quote Originally Posted by Opalg View Post
    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?
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    Quote Originally Posted by Moo View Post
    Hello,

    I'm quite surprised to read this from you
    it could even happen to Opalg! nobody is safe!
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    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
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  10. #10
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    Quote Originally Posted by Moo View Post
    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.)
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  11. #11
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    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?
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