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Oh order comparsion

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Compare each pairs according to their respective orders. Classify these forms by the relationships between the indicated constants.

    Note: ki is a constant and all kis are mutually independent. (ki < kj where i < j),
    ki ≥1.0.

    1) n! Vs. K1^ n => Here its Cap K not little k.

    2) log(n^n ) Vs. log(k1^k2 ) => little k


    2. Relevant equations

    http://www.augustana.ca/~hackw/csc210/exhibit/chap04/bigOhRules.html

    3. The attempt at a solution

    Here I am doing Oh Comparison, and I am not sure how to say which greater than, less than, equal to. Say for problem 2:

    From the Log of a Power Rule (link above) the order would be O(log n) and O(log k1). Now n can be any number and k1 can be any number. So how would I know which is greater than, less than, or equal to? For all I know n=200 and k1 = 10, or maybe not?

    For problem 1 same thing. K1 can be any number as well as n. If n= 2, than 2! = 2, and K1^2 .
     
  2. jcsd
  3. Sep 22, 2008 #2

    HallsofIvy

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    I assume you are referring to the comparative orders as n goes to infinity.

    Look at the fractions.

    1) What is
    [tex]\frac{K1^n}{n!}[/tex]
    as n goes to infinity? Notice that numerator and denominator both have n terms but each factor in the numerator is K1 while factors in the denominator get larger and larger.

    2) is easy! the denominator log(k1k2) is a constant! Any unbounded function of n will eventually be larger than any constant as n goes to infinity!
     
    Last edited: Sep 23, 2008
  4. Sep 23, 2008 #3
    Thank You HallsofIvy for the help. I spoke to the teacher about this and apparently big K and little k are the same thing (K1=k1, K2 = k2), so would this change anything?
     
  5. Sep 23, 2008 #4

    HallsofIvy

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    Not at all. They are just arbitrary numbers.
     
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