I just want review the stuff during this week's tutorial .
So the definition of O is /Exist c in positive real number , /Exists b in natural number , /For all n in Natural number , n>=b => f(n) <= O(g(n))
During tutorial we have to proof 5n^4 - 3n^2 + 1 <= c(6n^5 - 4n^3 + 2n) , which is pretty obvious if we think in mathematical way, but we need to prove it. lets put the structure aside for now and focus on the actual prove.
5n^4 - 3n^2 + 1 <= 5n^4 + 1 # when we remove negative term , we get bigger number
<= 5n^4 + 1 # we want remove the 1 , so multiply n^4 , thus n>= 1
<= 5n^4 + n^4
<= 6n^5 # we now multiply n again .
#now we start from the other side
c(6n^5 - 4n^3 + 2n) >= c(6n^5 - 4n^3) # we keep the negative term , so the number always smaller
>= c(6n^5 - 4n^5) # when the negative term gets bigger , the number itself gets smaller
>= c(2n^5) # simplify
>= 3(2n^5) # i pick c = 3.
>=6n^5
#we we have 5n^4 - 3n^2 + 1 <= 6n^5 <= 3(6n^5 - 4n^3 + 2n)
#implies 5n^4 - 3n^2 +1 <= 3(6n^5 - 4n^3 + 2n) , c = 3 and b = 1 (since n>=1)
I'm not sure if I done this correctly until the tutorial solution is posted , for whoever reading my slog please feel free to leave a comment to correct me.
also , http://csc165blog1.blogspot.ca/ this blog actually contains really helpful note .