How do you realize that $$(3log^2(n) + 55log(n^{10})+8log(n))*log(n) \neq \Omega(log^{10}(n))$$ ,where $log^x(n)$ means $(log(n))^x$

I know that by definition, if $f(n) = \Omega(g(n))$ then there exists a constant $c_1$ such that $$0 \leq c_1 *g(n) \leq f(n)$$ for all $ n > n_0$. In our case it yields $$c_1*log^{10}(n) \leq (3log^2(n) + 55log(n^{10})+8log(n))*log(n) $$


$$c_1 \leq \frac{3log^3(n)}{log^{10}(n)} + \frac{55log(n^{10})*log(n)}{log^{10}(n)} + \frac{8log^{2}(n)}{log^{10}(n)}$$

RHS: It's fairly easy to see that the denominator is larger than the nominator of both the first and third term, which means that $55log(n^{10})*log(n) > log^{10}(n) + c_1$ for some $n_0$ if the first equation holds.

I am just not able to see how to manipulate the equation to realize it...

  • $\begingroup$ you should mention that there exists $c_1 > 0$ and $n_0 > 1$ such your inequality holds for all $n > n_0$. $\endgroup$ – Ahmad Bazzi May 17 at 10:22
  • $\begingroup$ Good point, thx $\endgroup$ – Alex5207 May 17 at 10:24

I'll take it from where you paused. Assume your claim is true, that is the following is correct $$(3\log^2(n) + 55\log(n^{10})+8\log(n))\log(n) \neq \Omega(\log^{10}(n))$$ Then there exists $c_1 > 0 $ and $n_0 > 1$ such that for all $n > n_0$,we have $$c_1\log^{10}(n) \leq (3\log^2(n) + 55\log(n^{10})+8\log(n))\log(n) $$ Divide both sides by $\log(n)$ $$c_1\log^{9}(n) \leq (3\log^2(n) + 55\log(n^{10})+8\log(n)) $$ Use $\log a^b = b \log a$ $$c_1\log^{9}(n) \leq (3\log^2(n) + 550\log(n)+8\log(n)) $$ Divide again by $\log (n)$ $$c_1\log^{8}(n) \leq (3\log(n) + 558) \tag{1}$$ Your task now is to find me a $c_1 > 0$ and $n_0 > 1$, such that for all $n > n_0$ such that $(1)$ is true. Written differently, $$\exists n_0 > 1 \quad \mid \quad 0<c_1\leq (\frac{3}{\log^{7}(n) } + \frac{558}{\log^8(n)}), \ \forall n \geq n_0$$ The above is clearly not true. The larger $n$ is, the upper bound goes to zero. By the sandwich theorem, you get that $c_1$ has to be zero. By contradiction, the initial hypothesis is false.

  • $\begingroup$ The $loga^b = b*log a$ trick did it! Clear explanation - Thanks! $\endgroup$ – Alex5207 May 17 at 12:53
  • $\begingroup$ btw try using $\log$ not $log$ .. just add a backslash ;) and no need for an asterisk for multiplication .. good luck $\endgroup$ – Ahmad Bazzi May 17 at 12:54
  • $\begingroup$ Thanks - Got it :-) $\endgroup$ – Alex5207 May 17 at 12:55

Let us denote $y = \log n$, then our inequality transforms to $$c_1 \leq \frac{3}{y^7} + \frac{550}{y^8} + \frac{8}{y^8}$$

and we need to prove that for any $c_1 > 0$ this inequality is violated for arbitrary large $y$.

Let us take any $y > \max(1, \frac{550 + 8 + 3}{c_1})$. Then we have $\frac{3}{y^7} + \frac{550}{y^8} + \frac{8}{y^8} < \frac{550 + 8 + 3}{y} < \frac{550 + 8 + 3}{\frac{550 + 8 + 3}{c_1}} = c_1$


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