问题
I have the formula a(n) = n * a(n-1) +1 ; a(0) = 0
How can i get the Omega, Theta or O Notation from this without the Master Theorem or did anyone have a good site to understand the explanation
回答1:
The Master theorem doesn't even apply, so not being able to use it isn't much of a restriction.
An approach which works here is to guess upper and lower bounds, and then prove these guesses by induction if the guesses are good.
a(0) = 0
a(1) = 1
a(2) = 3
a(3) = 10
a(4) = 41
A reasonable guess for a lower bound is that a(n) >= n! for n>1. This is true for n=1. Suppose it is true for n=k-1.
a(k) = ka(k-1)+1
>= k (k-1)! + 1
>= k!.
So, if it is true for n=k-1, then it is true for n=k, so a(n) >= n! for all positive integers n, and a(n) = Omega(n!).
A reasonable guess for an upper bound is at a(n) <= 2(n!). This is true for the first few values, but it turns out to be a little awkward to prove using induction. Sometimes with inductive proofs, it is better to prove something stronger. In this case, it's better to prove that a(n) < 2(n!), or that a(n)<=2(n!)-1. This is true for n=1. Suppose it is true for n=k-1 for some k-1>=1. Then
a(k) = k(a(k-1))+1
<= k(2(k-1)!-1)+1
= 2(k!) +1-k
<= 2(k-1)!-1.
So, for any n>=1, a(n) < 2(n!). Since we have a lower bound and an upper bound of the form c*(n!), a(n) = Theta(n!).
回答2:
You have already noticed that your formula is very close to the factorial n!. Now you can express this finding in a more formal way: you can prove, for example, that
n! < a(n) < 2*n! (for big enough n)
If this is true, then all of O, Θ and Ω are n!.
I believe you can prove the inequality above, or some variant of it, using induction (haven't tried it though).
回答3:
Hint:
Dividing a(n) by n!, the first terms are
a(1)/1! = 1/1! = 1
a(2)/2! = (2.1+1)/2! = 1 + 1/2!
a(3)/3! = (3.(2.1+1)+1)/3! = 1 + 1/2! + 1/3!
a(4)/4! = (4.(3.(2.1+1)+1)+1)/4!= 1 + 1/2! + 1/3! + 1/4!
...
This establishes the tight bracketing
n! <= a(n) < (e-1).n!
and a(n) is in Θ(n!).
来源：https://stackoverflow.com/questions/30015614/how-to-get-omegan