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Monotonic and Bounded

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine if the sequence is Monotonic and Bounded.


    2. Relevant equations
    an = 2 - (3/n)


    3. The attempt at a solution
    Depending on the domain: Ex: a1, a2, a3 ... n=1 ; n=2 it would be bounded by [1,2]

    however, if we have negative n values and values as fractions we have no bounds and it can be increasing or decreasing.

    2 - (3/n) n=1,2,3 (range = 1 to 2)
    2 - (3/n) n=-1/10, -1/100, -1/600... (range = +∞)
    2 - (3/n) n=1/10, 1/100, 1/600... (range = -∞)
     
  2. jcsd
  3. Mar 26, 2013 #2

    mfb

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    It is a sequence, if nothing else is specified n is limited to the natural numbers.
     
  4. Mar 26, 2013 #3
    for the bounds would it be: [-1,2) or [-1,2] ??
     
  5. Mar 26, 2013 #4

    mfb

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    Both are intervals which bound the sequence, the first one is a bit better than the second one.
    [-2,2] or similar is fine, too - you just have to determine that it is bounded.
     
  6. Mar 26, 2013 #5

    micromass

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    That is an interval, not a bound. A bound consists of one number.

    For example, an upper bound is just a number that is larger or equal than every element of the sequence. So for the sequence 0,1,0,1,0,1,... we can say that an upper bound is given by 100. This is not the best upper bound however. The best upper bound is 1. This is the lower possible upper bound and it has a special name: a supremum.

    So, if I were to prove that 0,1,0,1,0,1,... is bounded. I would need to find an upper bound and a lower bound. I could say that an upper bound is given by 1 (or a higher number) and that a lower bound is given by 0 (or a lower number).
     
  7. Mar 26, 2013 #6
    also this sequence is increasing, correct?

    as n->inf an goes from -1 to 2.

    also, derivative is positive: f'(n) = 3/n^2 > 0
     
  8. Mar 26, 2013 #7

    micromass

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    Yes, the sequence is increasing.

    Watch out. The derivative of a sequence doesn't make any sense. A derivative is only defined for functions (on a suitable domain of definition). So if you take a sequence ##f(n) = 2- 3/n##, then it doesn't make any sense to talk about the derivative.

    What you want to say is to take the following function ##f:\mathbb{R}^+\rightarrow \mathbb{R}:x\rightarrow 2-3/n##. This function agrees with the sequence in the sense that ##f(n) = a_n##. The crucial point is that the function is now defined on entire ##\mathbb{R}^+##. Thus the derivative is something that makes sense.
     
  9. Mar 26, 2013 #8
    so should i just say that the sequence is represented in terms of a function f(x) then set f(x) = 2-3/x

    then take the derivative as you mentioned. or should i just not take the derivative at all; then how do i prove it is increasing?
     
  10. Mar 26, 2013 #9

    micromass

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    Taking the derivative is fine. But you should watch out with differentiating a sequence, that doesn't make any sense. You should say that there is a function ##f(x)=2-3/x## such that ##f(n) = a_n## for all natural numbers ##n##.

    There are other such functions however!! For example, take ##f(x) = (2 - 3x)\cos(2\pi x)##. This function is not increasing, but it does coincide with ##a_n## for natural numbers ##n##. But it is not because this function is not increasing, that the original sequence is not increasing. However, it is true that if you can find an increasing function that extends the sequence, then the original sequence is increasing.
     
  11. Mar 26, 2013 #10
    what do you mean by "extends the sequence." ?

    would f(x) = 2 - 3/x work?
     
  12. Mar 26, 2013 #11

    mfb

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    It extends the definition from the natural to the (positive) real numbers.
    Yes, it would work.
     
  13. Mar 26, 2013 #12
    I would show [tex]a_{n+1}-a_{n}>0[/tex] for all n which is pretty obvious in your case.
     
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