Prove that an upper bound is incorrect

Question

Probably a simple question that I cant figure out from data structure course: I need to disprove the following statement: $$ 8n^3 + 12n + 3\log^3n \ge n^4 $$

Now I know that from some value $n_0\in\mathbb{N}$ this statement is incorrect but how do I write it/ show it? Any help please. Thank you

Answer 1

How about just calculate the expression for $n=1000$ (just a quite big number in comparaison to the coefficients).

The left side will then become $8000012081 \approx 8\cdot 10^9$ while the right side will become $10^{12}$, thus the right side is larger than the left side, which contradicts the statement.

Answer 2

Let $f(n)=8n^3+12n+3\log^3n$, we have $$\lim_{n\to \infty} \frac{f(n)}{n^4} =0.$$ then by definition of limit, there is some $n_0\in \mathbb{N}$ such that for all $n>n_0$ we have $$\left|\frac{f(n)}{n^4}\right|<1.$$ note that $f(n)\geq 0$, then the previous inequality is just $f(n)<n^4$ for $n>n_0$.

Answer 3

Either you can attempt to guess a value for $n$ where it fails, and then just calculate the value of the two sides and conclude that the inequality is false. That's probably the way I'd go on this.

The more rigorous way is to solve the equation: $$ 8n^3+12n+3\log^3 n= n^4 $$ and use that to find a value where the inequality fails.

Answer 4

I guess $n$ shall be an integer. Simplify check the values $n=1,2,\dots$ and find that it is incorrect or $n\ge 9$. If $n$ is real, you find numerical solution $n_0$ of the corresponding equation as $n_0\approx 8.227685157$