# Determining the order of a function with big Oh

#### RedKMan

I was wondering if anyone could double check my answers below please.

Give the order of the following functions,

$$\displaystyle 1. Ta(n) = 20^2 + (n + 4)^3$$
$$\displaystyle 2. Tb(n) = (6n + 4)^2 + 3nlog2(n)$$
$$\displaystyle 3. Tc(n) = (7n + 1)^2log10(n)$$

I got the following orders:-

$$\displaystyle 1. \theta(n^2)$$
$$\displaystyle 2. \theta(n log n)$$
$$\displaystyle 3. \theta(log n)$$

Item 3 is the most effecient for very large values of n.

1) the expression of the third order since it contains $$\displaystyle n^3$$ which dominates the whole expression, hence you get $$\displaystyle \theta(n^3)$$
2) you have a part which is of the second order: $$\displaystyle n^2$$ and another one of order $$\displaystyle nlog_{2}(n)$$ which is lower than $$\displaystyle n^2$$ hence the total order is $$\displaystyle \theta(n^2)$$
3) Here, you have multiplication of a polynomial of the second order with $$\displaystyle log_{10}(n)$$ hence the order is $$\displaystyle \theta(n^2log_{10}(n))$$
Clearly, the most efficient is the second one since $$\displaystyle n^2 < n^2log_{10}(n) < n^3$$