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Upperbounds of functions

  1. Jan 23, 2013 #1
    Lets say if g(n) is not an upper bound on f(n), then does that mean g(n) is a lower bound on f(n)?
    Can anyone help with this please?
     
    Last edited: Jan 23, 2013
  2. jcsd
  3. Jan 23, 2013 #2

    Simon Bridge

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    depends: how did you find g(n)? does f(n) have to be bounded at all?
     
  4. Jan 23, 2013 #3
    They are functions that map N to N (natural numbers).
    I guess f(n) is just an arbitrary function.
     
  5. Jan 24, 2013 #4

    Simon Bridge

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    ... good, what I figured, and the answers to the questions?

    Since f(n) is an arbitrary mapping N to N, does it have to be bounded? Can it not be unbounded in both directions? What does this say about g(n) as a bound?

    You did not say that g(n) is arbitrary - so how is it found? Is it selected from all possible N to N mappings to have some special relationship with f(n)?
     
  6. Jan 24, 2013 #5
    f(n) and g(n) is arbitrary. Both map from N to N.

    I think that the statement in the first post is true but I just don't understand how to show this...
     
  7. Jan 24, 2013 #6

    Simon Bridge

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    Line 2, post #4 remains unanswered. It is a repeat of a question asked in post #2.
    (If you did answer it, I missed it.)

    If you do not answer questions I cannot help you.
     
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