NOV 15th.2014 CSC165 big omega & prove general statement about 2 functions

big omega:
∃C ∈ R+，∃B∈N，∀ n ∈ N,n >=B ⇒ f(n)>=c.g(n)

prove general statement about 2 functions:

I think this is just a little change from the 1function's style.  One more assume for the second function and leads to the conclusion of the connection between the 2 functions.Pay attention to the choice of picking B,C sometimes we use the max or min of the two numbers, of we might use the ceiling or the floor.

 ∀ f,g∈F, (f∈O(h))∧g∈O(h))⇒(f+g)∈O(h)

Prove:
Assume f,g∈F  #Just the old thing
    assume f∈O(h))∧g∈O(h) #Just the old thing
       ∃cf ∈ R+ ,∃Bf∈N,∀ n ∈ N, n>=Bf ⇒f(n)<=cf *h(n)# nearly the od thing ,just a little bit different on the c, which becomes cf(also B) now.
       ∃cg ∈ R+ ,∃Bg∈N,∀ n ∈ N, n>=Bg ⇒g(n)<=cg* h(n)#now the f(n) becomes g(n),which is for the second function.
       Pick C = Cf + Cg,Then C∈ R+
       Pick B = max(Bf,Bg)
       assume n ∈ N ∧ n >=B
            Then (f+g)(n) = f(n)+g(n) <=Cf*h(n)+Cg*h(n)
                                                      =(Cf+Cg)h(n)=Ch(n)#now we can find how to pick C,which is Cf + Cg
      ∀ n ∈ N, n>=B,(f+g)∈O(h)
       ∃cf,cg ∈ R ,∃Bf,Bg∈N,∀ n ∈ N, n>=B⇒(f+g)∈O(h)
∀ f,g∈F, (f∈O(h))∧g∈O(h))⇒(f+g)∈O(h)