Function g of the smallest order; f(x) is big-O of g(x).

Question

Find a simple function g of the smallest order so that f(x) is big-O of g(x). In your estimation you can use theorems learned in lecture about the big-O estimates of sums/products of functions.

f(x) = ((x^2.3) + xln(x^5))(1.1^(x+1) + 1.2^x) + (x^2 + 1.2^x)(x^3 + 0.92x)

I'm not understanding big-O order (or big-Theta/big-Omega) at all, and the lecture notes aren't helping. If anyone could break this down for me in somewhat Layman's, that would be amazing.

Answer 1

As x → ∞, f(x)=O(g(x)). if f(x)/g(x) is bounded. In particular, this is so when f(x)/g(x) → a finite limit. This is the meaning of big O notation which is a measurement of order of growth. e.g. x^a and a^x (a >1) both → ∞ as x → ∞, but x^a/a^x → 0. So the power function is of lower order than that of exponential functions. Similarly, any logarithms are of lower order compared with power functions. In the question, the function f can be written as

f(x)=x³1.2^x+ lower order terms(l.o.t.). Let g(x)=x³1.2^x. Then as x → ∞, f(x)/g(x) → 1. Therefore f(x)=O(g(x)).