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Really stumped with a limit proof question

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    I have the following question that I'm struggling with quite a bit:

    aNI7p.png

    Our instructor gave us the hint that we could break it up by proving

    a) Proving that 0 <= an < 1 which I managed to do by induction

    b) Proving that 0 <= an =< k^(n-1) for n > 0

    and then using the squeeze theorem.


    2. Relevant equations

    None that I know of?

    3. The attempt at a solution

    I'm having a lot of trouble with part b, I had the thought that I could try to again do it by induction but I'm hitting a bit of a brick wall. I'll show my working so far.

    I thought that if I could prove that a(n+1) =< an then it would be trivial to prove by induction that an =< k^(n-1) because I could show a1 =< k^(1-1) and then claim an =< k^(n-1) and then it'd follow that a(n+1) =< k^(n-1) because a(n+1) =< an. Is this a legitimate way of proving it?

    So I was thinking I needed to prove that an >= a(n+1) and I tried doing it like this:

    so k = a(n+1)/(an(1 - an))

    then for an =\= 0:

    0 <= a(n+1)/(an(1 - an)) < 1
    0 <= a(n+1)/an < (1 - an)
    0 <= a(n+1)/an < 1
    0 <= a(n+1) < an

    So as long as an isn't 0 then it's always larger than a(n+1). This is great and helps me with the k^(n-1) bit.

    but what about when an is 0? How on earth do I factor that in? I've been scratching my head for ages but haven't really come up with a way to go about doing it.

    My best thought so far is to prove that if an is ever 0 then it has always been 0 and it always will be which means that it will always be equal or less than k^(n - 1). I know an will only be 0 if a0 is equal to 0 or 1 or if k is 0 and I think I can show fairly easily if any of those conditions are true then an will stay 0 forever but how do I go back the other way and prove that if an is 0 it means it always has been 0? (and therefore always less than k^(n-1)?) Really pulling out my hair here.

    Or am I going about it completely wrong and there's an easy way of doing this that I'm completely missing?

    I get the feeling I might be a very very small nudge away from getting there so if anyone has any thoughts they'd really be appreciated!
     
  2. jcsd
  3. Mar 9, 2012 #2

    Dick

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    If you look at the function f(x)=x*(1-x) then what's the maximum of that function for x=[0,1]?
     
  4. Mar 9, 2012 #3
    It has a maximum between 0 and 1 so at x = 0.5 so 0.25. Sorry, wish I was smart enough to see where you were going with this but I'm not sure what to do with this information.
     
    Last edited: Mar 9, 2012
  5. Mar 9, 2012 #4

    Dick

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    Actually, you probably don't even need that. If you have already figured out that 0 <= an < 1. Then isn't an*(1-an)<an? So [itex]a_{n+1}<=k a_n[/itex].
     
  6. Mar 9, 2012 #5
    I'm trying to prove that an*(1-an)<an so that I can prove [itex]a_{n+1}<=k a_n[/itex] but I'm having trouble with the former proof.
     
  7. Mar 9, 2012 #6

    Dick

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    (1-an)<=1. So (1-an)*an<=an. Isn't that enough?
     
  8. Mar 10, 2012 #7
    What am I doing wrong here?

    0 <= 1 - an <= 1
    0 <= k*(1 - an) < 1 (because 0 <= k < 1)
    0 <= k*an(1 - an) < an
    0 <= a(n + 1) < an

    I know a(n + 1) can be equal to an so I shouldn't be getting a strictly less than there should I? But there's definitely no way for k*(1 - an) = 1, because k can't be 1.
     
  9. Mar 10, 2012 #8

    Dick

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    I think the only way a(n+1) can equal a(n) is if a(n)=0. I don't think you are doing anything wrong.
     
  10. Mar 10, 2012 #9
    I agree but shouldn't my inequality reflect that? For me to say that a(n+1) is less than an is correct isn't it? Or do I say that a(n+1) is less than an unless an and a(n+1) are both 0? Wouldn't I need to prove that?
     
  11. Mar 10, 2012 #10

    Dick

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    If you have the strict inequality a<b and c>=0, then you only know ac<=bc. You can't conclude ac<bc because c could be 0. Is that what you are asking about?
     
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