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Questions about asymptotic notations and analysis

1
vote
Hint: use $\log(a+b) = \log a + \log(1+b/a)$ this; Write the $\log(n^n+n)$ as $\log(n^n+n) = \log(n^n) + \log(1+n/n^n)$ $ \log(n^n) + \log(1+n/n^n) \in \mathcal{O}(n \log n)$ note $n/n^n \rightarro …
answered Oct 11 '18 by kelalaka
1
vote
$T(n)= a T(n/b)+ f(n) $ A generalization that usually works; set $n= b^k$. With backward substitution method. $T(n)=4T(n/2)+n^2 log^2 n $ take $n = 2^k$ \begin{align} T(2^k) & = 4 T(2^{k-1}) …
answered Oct 11 '18 by kelalaka
2
votes
We can write the for loop as the sums; $$\sum_{i=1}^{n} \sum_{j = 1}^{i} 1 = \sum_{i=1}^{n}i = \frac{n(n+1)}{2} \in\mathcal{O}(n^2) \, .$$ Note: set the starting values from $i = 1$ and $j = 1$, and …
answered Dec 23 '18 by kelalaka
1
vote
$\log_{100}n \in \mathcal{O}(log(n))$ since base change is a constant. $\log n \in \mathcal{O}(n) $ $\log n \in \mathcal{O}(n^2) $ $\log n \in \mathcal{O}(n^3) $ $\log n \in \mathcal{O}(n^{3/2}) $ …
answered Oct 17 '18 by kelalaka
3
votes
Using David's definition; a function is super-exponential if it grows faster than any exponential function. More formally, this means that it is $\omega(c^n)$ for every constant $c$, i.e., if $\lim …
answered Nov 10 '18 by kelalaka
1
vote
Here a list from Wikipedia, The lower in the table the bigger complexity class; $$ \begin{array}{|l|l|} \hline Name & \text{Running Time} \\ \hline \text{Constant time} & \mathcal{O}(1) \\ \text{Inver …
answered Oct 6 '18 by kelalaka