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I have an inequality:

$ n \ge C\log^k n \times (A\log(n^r\log^l n) + B)$

I want to turn it into an inquality

$n > F(A,B,C,k,l,r)$

that implies the first inequality, while making it $F$ as tight as possible. How can I do that?

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    Sorry, no $s$. Just $A,B,C,k,l,r$.2010-10-12

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I don't think you will get a satisfying bound on n without iteration. You can expand the right side to $n \ge AC*log^{(k+r)}n+AC*log^kn*log log^ln+BClog^kn$. If n is very large, you can even ignore all but the first term as it will dominate. Using the fact that $log n$ varies slowly with $n$, start with some guess for $log n$, such as 1 or 2. Plug it in to the right hand side and calculate a new value of n. Then calculate $logn$. Continue to convergence, which should be quick.

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    If you ignore the later terms, Lambert's W function will provide the answer-see http://en.wikipedia.org/wiki/Lambert_W_function under Applications. Of course, you have to have a way to calculate W...2010-10-13