Big O simplification of factorial

Question

To calculate BigO and constant for the following function : I have don't know how to simplify it further. Any advise ? Thx

 T(n) =  (n!n+n^3)(n^2+7logn)      
      <= (n!n+n^3)(n^2 +7n)   (since n>= log n)
      <= (n!n+n^3)(n^2 +7n)    (n^3 >= 7n if n > 3) 
      <= (n!n+n^3)(n^2 + n^3)   
      <= (n!n+n^3)(n!n + n^3)  (n!n >= n^2)    
         :

hi Simon, have you tried your luck on [Computer Science Stackexchange](https://cs.stackexchange.com/)? — Bagus Tesa, Feb 19 '20 at 07:13

Patrick87 · Accepted Answer · 2020-02-19T18:25:16.890

1

The highest-order term can be found by expanding:

 T(n) =  (n!n+n^3)(n^2+7logn) 
      =  n!n^3 + 6n!nlogn + n^5 + 7n^3logn

At this point we can simply compare terms and see which one is biggest:

n!n^3 vs 6n!nlogn
since n^2 > 7logn for n >= 4, n!n^3 > 7n!nlogn for n >= 1

n!n^3 vs n^5
since n! > n^2 for n >= 4, n!n^3 > n^5 for n >= 4

n!n^3 vs 7n^3logn
since n! > 7logn for n >= 4, n!n^3 > 7n^3logn for n >= 4

Based on all this, for n >= 4,

 T(n) =  (n!n+n^3)(n^2+7logn) 
      =  n!n^3 + 6n!nlogn + n^5 + 7n^3logn
     <=  n!n^3 + n!n^3 + n!n^3 + n!n^3
      = 4n!n^3
      = O(n!n^3)

If you want to find another expression that bounds n!n^3 that's well and good, but I'd recommend another question for that.

edited Feb 19 '20 at 18:25

answered Feb 19 '20 at 14:30

Patrick87

25,592
3
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how to get 6n!logn and 6n^3logn ? Thx – Simon Leung Feb 19 '20 at 15:35
I think your `6`'s are supposed to be `7`s and `6n!logn` is supposed to be `7*n!*n*logn`. – walnut Feb 19 '20 at 16:33
@SimonLeung I edited to fix some typos that you and/or walnut pointed out. Please see if this makes more sense. – Patrick87 Feb 19 '20 at 18:25

Big O simplification of factorial

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