Prove $10^3n^4+10^{-3}2^n=\mathcal O(2^n)$

Question

Prove that $10^3n^4+10^{-3}2^n=\mathcal O(2^n)$

I started this proof by trying to use induction, although as I put in $n=1$, although this gives: (when $n=1$) $1000.0002<2$ This is clearly untrue and I am not sure what method I can use to try and prove this statement.

Answer 1

If $\alpha,\beta\in\mathbb R$ then $$\alpha n^4=o(2^n)\ \text{since}\ \lim_{n\to\infty}\frac{n^4}{2^n}=0$$ and it's clear that $$\beta 2^n=O(2^n)$$ and finaly we have $$o(2^n)+O(2^n)=O(2^n)$$

Answer 2

I think you should better look at the definition of the big-O notation in order to fully understand what it means and therefore prove what you want.

Indeed, $f(x)=\mathcal O(g(x))$ means that there exists a positive real number $C$ and a real number $x_{0}$ so that : $| f(x) | < C | g(x) |$ for all $x>x_{0}$.

If you can find the constant $C$ and the $x_{0}$, your job is done!