I am comparing two functions for $x \ge 1$:
$$f(x) = \ln(\lfloor\frac{x}{9}\rfloor!) – \ln(\lfloor\frac{x}{10}\rfloor!) – \ln(\lfloor\frac{x}{90}\rfloor!)$$
$$g(x) = (2.07766)\sqrt{\frac{x}{9}} + (0.057712778)x$$
Comparing their values over a finite range suggests that $f(x)$ grows slightly faster than $g(x)$.
Since $f(x)$ is not continuous, I am not able to use its derivative to establish the rate of growth. How can I verify that as $x$ increases, $f(x)$ increases faster than $g(x)$?
Thanks,
-Larry
Edit: It looks like I copied the problem wrong. It should have read $(0.057712778)x$ instead of $(0.103883)x$. The correct term was $(1.03883)*\frac{x}{18} = (0.057712778)x$.
Best Answer
The problem is a bit messy since $f(x)$ is a step function. Initially, we can remove the steps with the restriction $x=90y$, under which conditions $f(y)=\ln (10y)! - \ln (9y)! - \ln (y)!$. We can now use Stirling's approximation to get $f(y)=10y\ln (10y)-10y -9y\ln (9y) +9y -y\ln y + y + O(\ln (10y)) = 10y \ln 10 + 10y \ln y -9y \ln 9 - 9y \ln y - y \ln y + O(\ln (10y)) = y(10 \ln 10 - 9 \ln 9) + O(\ln (10y))\approx 3.250829733 y + O(\ln (10y))$.
On the other hand, $g(y)=(2.07766)\sqrt{10}\sqrt{y}+(0.103883)(90y)=9.34947y+O(\sqrt{y})$.
Based on this, for at least those $x$ that are multiples of $90$, we can say that $g(x)$ grows faster than $f(x)$ (eventually), although they are both close to linear growth.