Is O(n^2 * 2^(log n)) == O(n^2) ?Why I think that this could be the case:In big O, you only take the parts of the term that are most relevant, right? O(n^2 + 3n + 9) == O(n^2). The n^2 has a lot more influence to the result of the term than the 2^(log n)-part. http://www.wolframalpha.com/input/?i=Plot%5B%7Bn%5E2,+2%5ELog%5Bn%5D%7D,+%7Bn,+0,+50%7D%5D
O(n^2 * 2^(log n)) == O(n^2)
big o
Your reasoning is incorrect. The rule that you can discard irrelevant terms only applies to additive terms, but in your case the parts are multiplied.If the logarithm in your case has base 2, 2^ld(n) is actually n, so the total complexity is n^3, which most definitely is not n^2.