done with question 1

HW1/hw1.tex

@@ -65,11 +65,11 @@
     \item $ n^{1.3logn} = \Omega (n^{1.5}) \; ~ \because$ asymptotically $n^{1.3logn} > n^{1.5}$
     \item $ 3^{n} = \Omega (n \cdot 2^{n}) \; ~ \because$ asymptotically $3^{n} > n \cdot 2^{n}$
     \item $ (logn)^{100} = O (n^{0.1}) \; ~ \because $ power of $ logn$ will only weaken the infinity of $logn$ and $n^{0.1}>(logn)^{100}$
-    \item $ n = \Omega ((logn)^{log(logn)}) \; ~ \because $ infinity of $(logn)^{log(logn)} > $ infinity of $n$
-    \item $ 2^{n} = \Omega (n!) \; ~ \because $ infinity of exponential function is greater than the infinity of factorial function  
+    \item $ n = O ((logn)^{log(logn)}) \; ~ \because $ infinity of $(logn)^{log(logn)} > $ infinity of $n$
+    \item $ 2^{n} = O (n!) \; ~ \because $ infinity of factorial function is greater than the infinity of exponential function  
     \item $ log(e^{n}) = O (n \cdot logn) \; ~ \because log(e^{n}) = n $ and we know that $n < nlogn$
     \item $ n + logn = \Theta (n + (logn)^2) \; ~ \because $ dominating term is $n$ and powers of $logn$ will not affect infinities
-    \item $ 5n + \sqrt{n} = \Omega (logn + n) \; ~ \because $ comparable functions are $\sqrt{n}$ and $logn$; $\sqrt{n} > logn$
+    \item $ 5n + \sqrt{n} = \Theta (logn + n) \; ~ \because $ both $logn$ and $\sqrt{n}$ grows slower than $n$. So, $n$ term will dominate
 \end{enumerate}