Incoming Finals and Assignments

Week 11

Sunday

I was reviewing the previous tutorial problem sets, and I reversed one of them (assume - implement) and disproved the statement.

The original one is proof 5n^4 - 3n^2 + 1∈ O (6n^5 - 4n^3 + 2n)

Therefore, the reverse one is 6n^5 - 4n^3 + 2n ∈/ 5n^4 - 3n^2 + 1

To disproof the statement

∃ c' ∈ R, ∃ B ∈ N, ∀ n ∈ N, n >=  B => 6n^5 - 4n^3 + 2n =< c (5n^4 - 3n^2 + 1)

∀ c' ∈ R, ∀ B ∈ N, ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)

assume c' ∈ R
   assume B' ∈ N
       let n' = 6, then n ∈ N
       assume n >= B
          then 5n^4 - 3n^2 + 1 < 5n^4 + 1
                                            < 5n^4 + n^4
                                            =< 6n^4
                                            =< n^5
                                            =< 2n^5
                                            =< 6n^5 - 4n^5
                                            =< 6n^5 - 4n^3
                                            =< 6n^5 - 4n^3 + 2n
          then 6n^5 - 4n^3 + 2n > (5n^4 - 3n^2 + 1) 
       then ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)
   then ∀ B ∈ N, ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)
then ∀ c' ∈ R, ∀ B ∈ N, ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)          

So, there is less than a week left for all the lectures. I will sure work slight harder since it's almost the finals. Beside that, I still got the CSC165 assignment 3 due on next Wednesday, CSC148 project II due on Friday. Well, gotta continue get my hands working on them now. MARCH madness is not over yet!!!

Good luck everyone!

Thanks for reading!

Roland Qiyang Liu