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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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I am unclear on what you want... Are you looking for a constant, possibly in terms of A, B, C, k, l, and r, such that the first inequality will hold? That is, you are assuming/guessing/know that the inequality is true in a set which includes an interval of the form $(a,\infty)$, and you want to find the smallest such $a$? –  Arturo Magidin Oct 12 '10 at 19:43
    
Are you asking for a "simpler" function F greater than the given one? If so, simpler in what way, to comprehend, calculate, or what? Also, in your notation above, does $\log^2{x} = \log{\log x}\ $ or $(\log x)^2$ –  Bill Dubuque Oct 12 '10 at 19:54
    
$\log^2 x = (\log x)^2$ Note that in the first inequality, there is dependence on $n$ on both sides of the inequality. The constants $A,B,C,k,l,r,s$ are fixed. I am looking for a way to find an inequality of the form $n \ge F(A,B,C,k,l,r,s)$ such that $F$ does not depend on $n$, the dependence on $n$ is only on the left handside. The new inequality should entail the first one. –  normvector Oct 12 '10 at 21:09
    
(Imagine solving the first inequality with respect to $n$. Then we would get $n >= ...$ like I want. But it is probably impossible to solve it directly for $n$ without using very complicated functions of the constants $A,B,C,..$ etc. So instead I am looking for an inequality on $n$ that would just entail the first inequality. –  normvector Oct 12 '10 at 21:10
    
Sorry, no $s$. Just $A,B,C,k,l,r$. –  normvector Oct 12 '10 at 21:12

2 Answers 2

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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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Thank you. It is unfortunate I cannot translate it into an analytic solution. –  normvector Oct 13 '10 at 17:03
    
If you ignore the later terms, Lambert's W function will provide the answer-see en.wikipedia.org/wiki/Lambert_W_function under Applications. Of course, you have to have a way to calculate W... –  Ross Millikan Oct 13 '10 at 22:25

Let $D=C\cdot(A\cdot(r+l)+B)$ and $c_k=(2/(k+1))^{k+1}$.

If $D\geqslant3$, a sufficient condition is that $n\geqslant3$ and $$ n\geqslant c_k\cdot(\log(D))^{k+1}\cdot D. $$ To prove this, use $\log(n)^l\leqslant n^l$ and assume at the onset that $n\geqslant3$, then rewrite everything in terms of $z=n^{1/(k+1)}$ and use simple inequalities like $x\geqslant2\log(x)$ for every positive $x$.

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