Is this generalization of Big-Theta notation correct?

Question

Say you have an algorithm that completes in a polynomial number of steps for the input of size n, like, for example, P(n)=2n^2+4n+3. The asymptotic tight bound for this algorithm Θ(n^2).

Is it true to say that the Big-Theta notation for any algorithm is n to the power of the degree of the polynomial P(n), or are there any cases where that is not true?

This doesn't account for Theta(log(n)), or Theta(2^n), or Theta(n!), etc — Zim-Zam O'Pootertoot, Apr 29 '13 at 15:54
Yes, that is true for polynomials. (Of course, not all functions are polynomials.) Proving it from the definitions is a good exercise. — Nemo, Apr 29 '13 at 15:54

score 1 · Accepted Answer · answered Apr 29 '13 at 17:30

The complexity of polynomial time algorithms are bounded by O(n^k), where 0 < k ≤ ∞. It doesn't mean that all the algorithms are having a polynomial time complexity. There are many algorithms with sub polynomial complexity. Examples include O(k) (constant complexity), O(^k√n) (k^th root of n, where 1 ≤ k ≤ ∞), O(log n), O(log log n), etc. There are also algorithms which are having super polynomial time complexity. Examples for such complexities are O(kⁿ), where 1 < k ≤ ∞, O(n!), etc.

Is this generalization of Big-Theta notation correct?

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