I'm given the runtimes for input size $n=100$ of some polynomial-time (big-Oh) algorithms and an $\mathcal{O}(n \log(n))$ one. I want to calculate the runtimes for: $200$, $1000$ and $10000$.
For the polynomials it is trivial. The results I should get for the $n \log n$ case are: $0.06$, $1$ and $20$ seconds respectively, where $0.03s$ was given for $n=100$.
I have tried: (examples should have $0.06$ as solution)
(1) $\frac{f(2x)}{f(x)}$ similarly to how I've got the right solution to the polynomials, $\frac{2x \log(2x)}{x \log x} = \ldots= 2 \log((2x)^{1/\log x})$, not getting anywhere, and substituting $n=100$ gives $2.3010$ for any $\log$ base.
(2) calculating the "unit time of execution" $x$, from $100 x \log(100x)=0.03$ after approximating the inverse of $y=n \log n$ for $y=0.03$, $n=1.03$ for natural $\log$, substitution gives: $1.48$. $n=1.02058$ for binary $\log$, substitution gives: $2.10$
(3) calculating "unit time of execution" from the polynomial cases, which yield different for each, and result in negative runtimes predictions for the nlogn algorithm
(?) I also thought of searching for a function that is $\mathcal{O}(n \log n)$ and fits onto the points given by the solution, but thats over my head.
How to get to the solution ? Did any of my attempts made sense ?