2
$\begingroup$

Given two natural numbers $p$ and $i$, such that $0 < i \leqslant 2^p$, let $$ \Phi(p,i) := \frac{1}{2^p+1} + \frac{1}{(i+1)^2} - \frac{1}{2^p}\lg\left(\frac{2^p}{i}+1\right), $$ where $\lg x$ is the binary logarithm. With the help of a Computer Algebra System, it seems that

  • If $0 \leqslant p \leqslant 3$, then $\Phi(p,i) < 0$.

  • If $4 \leqslant p$, there exists $i_p$ such that $\Phi(p,i_p) = 0$ and $\Phi(p,i) > 0$ for $1 \leqslant i < i_p$, and $\Phi(p,i) < 0$ for $i_p < i \leqslant 2^p$.

How can I prove this?

Just in case, the partial derivative with respect to $i$ is: $$ \frac{\partial\Phi}{\partial i}(p,i) = \frac{1}{i(2^p+i)\ln 2} - \frac{2}{(i+1)^3}, $$ where $\ln x$ is the natural logarithm.

$\endgroup$
0
$\begingroup$

Peter Mueller at MathOverflow solved the problem: https://mathoverflow.net/questions/108816/root-and-sign-of-a-complicated-bivariate-function

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.