Merge pull request gzc#179 from DevDoggo/master

Added English Answers to a few Questions in 3.1

C03-Growth-of-Functions/3.1.md

@@ -4,21 +4,40 @@ Let f(n) and g(n) be asymptotically nonnegative functions. Using the basic defin
 notation, prove that max(f(n), g(n)) = Θ(f(n) + g(n)).
 
 ### `Answer`
-我们需要证明 c1(f(n) + g(n)) <= max(f(n), g(n)) <= c2(f(n) + g(n))
-
+我们需要证明 c1(f(n) + g(n)) <= max(f(n), g(n)) <= c2(f(n) + g(n))
+
 因为f和g都是非负函数，只需要令c1 = 0.5, c2 = 1即可.
 
 
+
+English:
+
+Use the definition of Θ-notation and set it up with the given values:
+
+0 <= c1(f(n) + g(n)) <= max(f(n), g(n)) <= c2(f(n) + g(n))
+
+---If we can Solve All inequalities, we can then use that as proof for the definition of Θ-notation.
+
+c2 = 1: This holds because the Max must be lower than the sum of the two functions.
+
+c1 = 1/2: This holds because the Max can at its lowest be equal to 1/2 of the Sum of the two functions. 
+
+0 <= c1(f(n) + g(n)) is trivial.
+
+Thus, since all inequalities hold, we've proven that: max(f(n), g(n)) = Θ(f(n) + g(n)).
+
+
+
 ### Exercises 3.1-2
 ***
 Show that for any real constants a and b, where b > 0,
 
 ![](http://latex.codecogs.com/gif.latex?\(n+a\)^b=\\Theta\(n^b\))
 
 ### `Answer`
-需要证明 c1n^b <= (n+a)^b <= c2n^b
-
-在 n > 2|a| 时, 令 c1 = 0.5, c2 = 2 即可.
+需要证明 c1n^b <= (n+a)^b <= c2n^b
+
+在 n > 2|a| 时, 令 c1 = 0.5, c2 = 2 即可.
 
 
 ### Exercises 3.1-3
@@ -37,6 +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