I'm trying to order some numbers....For n very large, I need to order:
2^n, 4^n, 2^(n^2), n!, and n^n
I know 2^n<4^n<n^n, but I'm not sure where the 2^(n^2) and the n! come in...I want to say it goes 2^n<4^n<2^(n^2)<n!<n^n, but I'm not too sure.
Is this correct?
Logarithms will buy you the first three.
That is now rather easy to order.
The last to are easy enough by themselves. One is n factors of n and the other is n factors, but only the largest is as great as n.
Two logs will settle everything but the factorial. You do that.
All we've left is the factorial vs. the
No ideas? Search the brain.
Note: The log is useful because it is monotonic increasing. This is a useful transformation that does not change the order of anything.
Actually, I think I have it. 2^(n^2) is the biggest of all of them since 2^(n^2)=2^(n*n)=(2^n)^n, which is larger than n^n.
so essentially, we have and I think you have it. (But I am still a little worried about just how big 'n' is.)