Asymptotics and logarithmic exponents

Question

I was tasked to order some functions in respect to the $\mathcal{O}$ notation, and have some problems with the following ones:

$n^{\ln(7)} = \mathcal{O}(n^{3})$: I can't find a more elegant proof for this other than using L'Hôpital's rule and a numerical approximation of $\ln(7)$:

$\lim_{n\to\infty} \frac{n^{\ln(7)}}{n^3} = \lim_{n\to\infty} \frac{n^{ \approx1.94591014906}}{n^3} = \lim_{n\to\infty} \frac{n^{\approx1.94591014906-3}}{3*2*1} = 0 $

(The last equal sign is valid, since the exponent is negative.) Is there a better way to do this ?

Furthermore I have no idea how to prove $8^{\log_2 n} = \mathcal{O}(5^n)$. I tried to apply the limit rule but to no avail. I think it is intuitively clear, since $\log_2 n$ grows slower than $n$, but I don't know how to argue that rigorously.

Thanks for any answers.

Answer 1

First, notice that $3 > \ln 7$. You can prove this formally using the Taylor expansion of $e^x$: $$ e^3 = \sum_{n=0}^\infty \frac{3^n}{n!} \geq 1 + 3 + \frac{9}{2} = 8.5 > 7. $$ Monotonicity of $\ln$ implies that indeed $3 > \ln 7$.

Armed with this knowledge, we deduce that for $n \geq 1$, $$ n^3 = n^{3 - \ln 7} n^{\ln 7} \geq n^{\ln 7}. $$ This shows that $n^{\ln 7} = O(n^3)$, using the definition of big O.

For your other question, try using $8^{\log_2 n} = n^3$ and induction. The idea is that as you increase $n$ by 1, the value of $n^3$ hardly increases, whereas $5^n$ increases by a factor of 5.

Finally, let me say that there is absolutely no reason to be so formal. It's enough to go through the formalities once, just to get convinced that they work, and afterwards you can just use well-known facts such as $n^k = o(c^n)$ for all $c > 1$.