0
$\begingroup$

Consider this function $f(x)=-x\log\log(bx)+x\log\log\log(bx)+ax$, where $a$ and $b$ are positive constants, and the logarithms are 2-based.

Is it possible to find the maximum value (or even with approximation) of $f(x)$, in terms of $a$ and $b$, for $x>\frac{16}{b}$?

Note, it can be shown that $f(x)$ is concave for $x>\frac{16}{b}$.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Why would you want to do that? $\endgroup$
    – Will Jagy
    Commented Feb 6, 2012 at 8:37

1 Answer 1

Reset to default
1
$\begingroup$

If you call that maximum value $M(a,b)$, you can first of all get rid of $b$ since $ M(a,b)= M (a,1) / b $. And $ M(a,1) = g ^ \star (a) $, the Legendre transform of the convex function $g(x):= x\log \log x -x\log\log\log x$ for $x \ge 16$, and $g=\infty$ for $x < 16 $. Then, you may at least estimate $g ^ \star $ integrating $(g^\star)'=(g')^{-1}$. Note that you may also consider the somehow more natural definition of $g$ with finiteness domain $]2,\infty[$, which seems still convex.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for you suggestion. However, I think to find the inverse mapping $(g^')^{-1}$ is even harder. $\endgroup$
    – Allen
    Commented Feb 6, 2012 at 17:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .