What Are the Big-O Notations for n^(n-1) and (n-1)^n?

[peterlam]

Hi!

For the following functions, what are their big-O notation?

1. n^(n-1)
2. (n-1)^n

Should their big-O notations be the same as the original functions? i.e.

1. O(n^(n-1)) = n^(n-1)?
2. O((n-1)^n) = (n-1)^n?

Please help!
Many thanks!

 

[jbriggs444]

If you look at the definition of big-O notation, it is an asymptotic inequality. Roughly speaking, f(x) = O(g(x)) whenever g grows as fast or faster than f.

In this case, g(x) = nⁿ grows faster than both. So it would be both convenient and correct to say that both are O(nⁿ).

Without more context, it is impossible to say whether this is "good enough" for your purposes.

 

[peterlam]

Thanks! I can understand that O(n^(n-1)) = n^n. But can I say O(n^(n-1)) = n^(n-1)? I am trying to find a tighter asymptotic upper bound.

Similar to the second case.

Thanks!

 

[jbriggs444]

Thanks! I can understand that O(n^(n-1)) = n^n. But can I say O(n^(n-1)) = n^(n-1)? I am trying to find a tighter asymptotic upper bound.

I have a quibble with the notation you use above. It is correct to say that n^n-1 = O(nⁿ). This is not a real equality. It's just a notation. One might loosely read "=O(g)" as "is of order g". It is incorrect to say that O(n^n-1) = nⁿ. The latter notation suggests that O() is a function which returns a single function as its value.

I can't think of any upper bounds that are both simpler than n^n-1 and tighter than nⁿ.