Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
In big-O notation is O((log n)^k) = O(log n), where k is some constant (e.g. the number of logarithmic for loops), true?
I was told by my professor that this statement was true, however he said it will be proved later in the course. I was wondering if any of you could demonstrate its validity or have a link where I could confirm if it is true.
(1) It is true that O(log(n^k)) = O(log n).
(2) It is false that O(log^k(n)) (also written O((log n)^k)) = O(log n).
Observation: (1) has been proven by nmjohn.
Exercise: prove (2). (Hint: f(n) = log^2 n is O(log^2 n). Is it O(log n)? What is a sufficiently large constant c such that, for all n greater than n0, c log n > log^2 n?)
EDIT:
On a related note, anybody who finds this question helpful and/or interesting should go show some love for the new “Computer Science” StackExchange site. Here’s a link. Go make this new place a reality!
http://area51.stackexchange.com/proposals/35636/computer-science-non-programming?referrer=rpnXA1_2BNYzXN85c5ibxQ2