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Analysis: Least Upper Bound

  1. May 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Find, with proof, the least upper bound of the set of real numbers E given by:
    E ={14n + 9/16n + 13: n [itex]\in[/itex] N}
    
    :

    2. Relevant equations



    3. The attempt at a solution

    So I said that 16n+13>14n+9 for all N

    From this I get n>-2

    What do I do with this? I understand that as n increase E will decrease but I don't know how to answer the question. Any help would be much appreciated.
     
  2. jcsd
  3. May 15, 2014 #2

    pasmith

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    As written you have [tex]
    E = \{ 14n + \frac{9}{16n} + 13 : n \in \mathbb{N}\}
    [/tex] which has no upper bound.

    I infer from this that you actually meant [tex]
    E = \{ \frac{14n + 9}{16n + 13} : n \in \mathbb{N} \}.
    [/tex]

    You should get that 1 is an upper bound for E. But is it the least?

    Let [tex]
    y = \frac{14n + 9}{16n + 13}
    [/tex] and solve for [itex]n[/itex]. You need [itex]n[/itex] to be positive, so that gives you a bound [itex]y_0[/itex] on [itex]y[/itex].

    You then need to see whether you can make [tex]y_0 - \frac{14n + 9}{16n + 13}[/tex] arbitrarily small by suitable choice of [itex]n[/itex].
     
  4. May 15, 2014 #3
    Hey pasmith,

    You're right, that is what I meant (long day).

    y = (14n + 9)/(16n + 13)

    And you said solve for n so:

    n= (9-13y)/(16y-14)

    However n would not be postive here?
     
    Last edited: May 15, 2014
  5. May 15, 2014 #4

    Ray Vickson

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    Does N mean all integers (negative and positive) or just the non-negative positive integers?
     
  6. May 16, 2014 #5
    Hey Ray Vickson, N is natural numbers so all positive integers.
     
  7. May 16, 2014 #6

    pasmith

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    You can see that the numerator is positive when [itex]y < 9/13[/itex] and negative when [itex]y > 9/13[/itex] and the denominator is positive when [itex]y > 16/14[/itex] and negative when [itex]y < 16/14[/itex]. For the quotient to be positive, the numerator and denominator must have the same sign.
     
  8. May 16, 2014 #7

    HallsofIvy

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    I don't see any point in solving for n. You are concerned with values of the fraction, not values of n. Instead, divide both numerator and denominator by n:
    [tex]\frac{14+ \frac{9}{n}}{16+ \frac{13}{n}}[/tex]
    Now, it is obvious what happens as n goes to infinity. Is y ever larger than that number for finite n?
     
  9. May 16, 2014 #8
    So the limit is 14/16 and as n goes to infinity.

    As n increases E decreases so away from L so its divergent?
     
    Last edited: May 16, 2014
  10. May 16, 2014 #9

    HallsofIvy

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    I am no longer sure what you are talking about. Saying "the limit is 14/16" means the sequence is convergent, doesn't it? It has a limit. Clearly the fraction approaches 14/16= 7/8 as closely as we please. The only question left is "is it ever, for some finite value of n, larger than 7/8:

    Can you solve [itex]\frac{14n+ 9}{16n+ 13}> \frac{7}{8}[/itex]?
     
  11. May 16, 2014 #10
    Ok I understand that bit now.

    Solving the inequality:

    8(14n+9)>7(16n+13)
    112n+72>112n+91
    72>91

    Which doesn't make sense so 7/8 is the least upper bound?
     
  12. May 16, 2014 #11

    HallsofIvy

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    Yes, trying to solve [itex]\frac{14n+ 9}{16n+ 13}> \frac{7}{8}[/itex] leads to a statement that is false for all n (I would not say "doesn't makes sense"- just "false") so the inequality is never true- 7/8 is an upper bound on the fraction. But we also know that it comes arbitrarily close to 7/8 so 7/8 is the "least upper bound".
     
  13. May 16, 2014 #12

    Ray Vickson

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    Alternatively, use calculus to examine the behavior of the function
    [tex] f(x) = \frac{9 + 14x}{13+16x}, \; x \geq 0[/tex]
     
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