How do I compare orders of forms with different constants?

  • Thread starter Thread starter Ad2d
  • Start date Start date
Ad2d
Messages
9
Reaction score
0

Homework Statement



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


Homework Equations



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

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 .
 
Physics news on Phys.org
I assume you are referring to the comparative orders as n goes to infinity.

Look at the fractions.

1) What is
\frac{K1^n}{n!}
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 by a moderator:
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?
 
Ad2d said:
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?

Not at all. They are just arbitrary numbers.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top