@Eugene I believe there are two sites: computer science - cs.SE (which is in beta) and theoretical computer science - cstheory.SE. One of them is for research-level question, one is more-or-less on MSE level. I believe the lower level is at cs.SE.
@Eugene I just wanted to make sure that you know that there are two sites, because of your comment here. I am not saying that some of those sites would be better than MSE.
Note that the inverse function theorem, as the sketch I gave highlights, is sort of a special case of the implicit function theorem. The implicit function theorem allows you to solve for one set of variables in terms of another set of variables, when both sets of variables are all related to each other in a complicated implicit equation.
@Eugene Perhaps you should explain in your question that you want something more/something different than Cormen et al. Otherwise you probably receive it as an answer.
I went through most of middle school not realizing my last name is a sexual innuendo. I even made an email address out of the nickname I was given, totally oblivious.
Thanks! Ok, so I asked it here math.stackexchange.com/questions/158774/…. a) and b) I already have answered. I believe c) and d) should be easily extracted from b), but I'm missing something. Martin is trying to help me at his comments but I still don't get it
Sorry, I have no idea about difficulty! For me everything is hard, but I thought it was simple because Martin made it sound like so. It's just that I thought that having b) answered should make c) easy?
@PeterTamaroff \[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} \] seems to be easiest way. Sometimes the spacing is not so good and you have to muck around with array, though.
@Gigili It's a comment that is certainly critical of you and attacks your actions. It's not on the level that a spam/offensive flag would be valid. For general disagreement I'd recommend using the ignore user feature
[For the record I declared "Not Sure" (with the description "no strong opinion") on the flag. I'm surprised four people starred the comment given the unfair accusation against Gigili in it, and I'm not really comfortable with the comparison.]
@PeterTamaroff There is no fuzz, Ragib simply made a comparison (or a contrast rather) between my suspension and her incorrect help to Jordan for differential equations. Gigili and I are amiable.
@N3buchadnezzar Yes, musical chairs was fun. I did not like the very first round of the manga, it was so predictable and the characters were 0-dimensional. (Not like they've improved much, but the games got more interesting.)
Could someone explain me why I can't evaluate this integral: $$\int_0^1{\frac{e^{2x}+2e^x+1}{e^{2x}+e^x+1}}\,dx$$ Doing: $$\int{\frac{N(x)}{D(x)}}\,dx = \int{Q(x)*D(x)}\,dx + \int{R(x)}\,dx$$?
@unNaturhal Then you should get the same as what you started with, if you put on a common divisor right? Try that! The only problem is that e^x/(D(x)) is harder to evaluate than your original integral.
yup. whoever it is has been going through my questions and answers and downvoting without explanation. about 3-4 a day so it doesn't get reversed due to serial downvoting.
@leo There is not much a meta post can achieve, the community can't do anything about serial downvoting. This is better handled by the automatic vote fraud script, or in extreme cases by moderators or SE employees directly.
@unNaturhal Do you realize you've just written $$\int \frac{N}{D}=\int\frac{QD+R}{D}=\int Q+\int R~?$$ In other words, you have $\frac{QD+R}{D} = Q+R$. You sure you aren't forgetting a little something something?
@PeterTamaroff No, the Dirichlet inverse of a function means that $f*f^{-1}=\mathrm{Id}$, meaning that $\sum_{d|n}f(n/d)f^{-1}(d)=\color{Red}\delta_n$. (This is not the compositional functional inverse.)
