How do we solve for n:
n / log_2(n) < c : c is a constant
[Original Equation: n^2 < c n log_2( n ) ]
(Sorting algorithms using the iterative method.)
Ughh... It feels like I had done these kinds of algebraic problems 5+ years ago... can't remember.
Solve for n/log2(n)=c first.
If you're using an iterative method, then you first find an equation with n on its own on the left and n with other stuff on the right. This is fairly easy:
n = c * log2(n)
Then you'll want to convert that log2(n) into something you can resolve easily on a calculator; if memory serves then it is this:
n = c * ln(n) / ln(2)
where ln is the natural logarithm, ie base e. You could use log10 instead if you wanted.
Then you choose an arbitrary value of n which is reasonably close to what you expect, and call that n0 - lets say n0 =2.
Then you do this:
n1 = c * ln(n0) / ln(2)
n2 = c * ln(n1) / ln(2)
n3 = c * ln(n2) / ln(3)
until you have the value accurate to the level of precision you want.
Say that n5 is accurate enough for you, you can then say that n=n5, correct to x sf.
Then you can substitute that value of n back into your original equation, and then find the range where n/log2(n) < c.
Please double-check everything I have said here to make sure that it is mathematically sound before using it. This is provided in the hope that it will be useful, but I disclaim all responsibility for any errors.
That said - HTH.