Timeline for Root and sign of a complicated bivariate function
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Oct 5, 2012 at 18:32 | vote | accept | Christian Rinderknecht | ||
| Oct 5, 2012 at 13:01 | answer | added | Peter Mueller | timeline score: 0 | |
| Oct 5, 2012 at 11:54 | comment | added | Christian Rinderknecht | The context in which this question arises is noble enough (analysis of algorithms), but giving too many details here would further discourage knowledgeable people to try and help, I think. $\partial\Phi/\partial i = 0 \Leftrightarrow i^3 + (3-2\ln 2)i^2 + (3 - 2^{p+1}\ln 2)i + 1 = 0$. Solving another cubic equation, we find that if $p=0,1,2$, then there is only one real root and, if $p \geqslant 3$, there are three real roots. The exact expression of these roots is so unwieldy it is unhelpful. How can I show that $\partial\Phi/\partial i < 0$ if $0 < i \leqslant 2^p$ and $p \geqslant 4$? | |
| Oct 5, 2012 at 9:33 | comment | added | Peter Mueller | I doubt that this is research level. To start with, I would analyse the cubic polynomial in $i$ which arises from $\frac{\partial\Phi}{\partial i}(p,i)=0$. | |
| Oct 4, 2012 at 13:09 | history | asked | Christian Rinderknecht | CC BY-SA 3.0 |