Added english answer to 3.1-4

C03-Growth-of-Functions/3.1.md

@@ -56,6 +56,15 @@ Is ![](http://latex.codecogs.com/gif.latex?2^{n+1}=O\(2^n\)?)Is![](http://latex.
 
 (2)不成立，假设存在常数c使得2^(2*n)<=c*2^n，则有2*n<=lg c+n，即n<=lg c，并不存在一个常数c使得这个不等式对n成立。
 
+English:
+for f(n) = O(g(n)), the definition of O-notation is:
+0 <= f(n) <= c(g(n)) for all n > n0.
+
+(1) 2^n+1 = 2*2^n <= c*2^n. If c = 2, the inequality holds and we prove 2^n+1 = O(2^n) to be True! Answer: Yes.
+
+(2) 2^2n = (2^n)^2. There is no possible constant c that can make 2^n into (2^n)^2. Thus, 2^2n =/= O(2^n). Answer: No.
+
+
 ### Exercises 3.1-5
 ***
 Prove Theorem 3.1. *For any two functions f(n) and g(n), we have f(n) = Θ(g(n)) if and only if f(n) = O(g(n)) and