Is it in general true that $O((x-a)^n) = O(x^n)$? I have the feeling that the answer is no because I can only make the following be true if $x\geq a$:
$f(x) = O((x-a)^n) => |f(x)|\leq K|(x-a)^n|\leq K|x^n|$
If someone can confirm or say otherwise, that would be great. Thanks!
It doesnt have to be the same constant. You have to show that there is some constant $C > 0 $ such that $|(x-a)^n|\leq C|x^n|$ for all sufficiently large $x$. To do this, write down the binomial expansion for $(x-a)^n$ and estimate every term in terms of a constant times $x^n$.