@robjohn I used that there are infinite parallelograms with same area with a fixed base and a parallel line to the base...will that be applicable or is that wrong?
@robjohn Sorry, I get it now. Both the diagonals can't be smaller than 4, but if one of them is bigger than 4 then the other one can be smaller than 4. Right?
Oh no! Don't you know the formula that the area is half the product of diagonals into the sine of angle between them @hawk Its verrry easy to prove, do you want it?
@Hawk Draw the diagonals of the convex quad. Its divided into 4 triangles. Sum their areas using $2A=ab\sin C$. Done! Its quite a useful formula, you should remember.
@DanielFischer It's been bothering me for some time and I can't seem to get over it. Every time I see boldface it just annoys me. I can understand that some undergrad new and low rep users might use "??" to put emphasis on their question but when it is used in comments by established users... it really annoys me. And that's an understatement. Does it not bother you?
@robjohn something must be done about this!! I just hate people, who hide themselves and unnecessarily(without playing around) attack others or try to harm others
@MattN. Since that comment was originally on the (now deleted) pretty senseless answer, I would cut Asaf some slack for expressing extreme incredulity by a double question mark.
@MattN. Something like this math.stackexchange.com/questions/728173/… irritates me. Fine answer, but did the colors and the different font really enchance the answer?
@robjohn Yes, I know that...and I always liked your answers...others who are my favourite answerers are DanielFischer, Andre Nicolas, Gerry Myerson, Don Antonio...and some other people...whom I cannot recollect now...
@DanielFischer By Riemann mapping, shouldn't we only need to find what class of functions has a unique solution to the Dirichlet problem for the unit died?
Could please somebody explain question? What are the similarities and differences between a function existing at a point and a limit of a function approaching a point? Really can't get it.
@DanielFischer What I'm actually interested in is harmonic measure. I've gotten this far: if a domain is bounded by smooth arcs, then we can solve the Dirichlet problem for the indicator function of the union of some of those arcs.
@KirillZhukov Define $f(x) = 1$ on $\mathbb R \setminus \{0\}$. Then this function is not defined at $0$ (it does not exist there) but yet the limit of $f$ for $x \to 0$ is $1$.
I tried to come up with a more interesting example.
Without knowing $\zeta(s)$'s Laurent series, is there a way of directly working out $$\int_{|z-1|=R}\frac{\zeta(z)}{(z-1)^n}dz$$ for $n \in \mathbb{Z}$?
@meer2kat While I do agree in my own infallibility, there's a bit of a difference between "people should love everything I say" and "someone went through various things I posted and decided to downvote a few without having taken the time to look through them".
@Mike it sounds like a revenge thing, but the script did not see it as such. I think the downvotes I have gotten recently are similar, but they are spaced much further apart. The script cannot tell if the answers are good or not.
@Chris'ssis That was my first approach (except using $\{\dots\}$ without changing to $\lfloor\dots\rfloor$). When I saw that the sum simplified back to the symmetric integral, I just used that.
Quick question - If I have a probability, say throwing a regular dice until I have tossed two values that are the same, e.g. I can throw 1, 2, 3, 1 then I'm done, what is the lower bounds for this (in term of dice tosses)? It must be two right?
@meer2kat Are you sure? It seems only logical with that, but we have a homework with hash collisions, and one of the questions are what is the lower bounds for a hash collision i.e. x1 != x2 but h(x1) = h(x2)
and the hash function is a "random oracle model", that is like it's with the birthday problem
@robjohn Quite nice. My answer to the question is the way Naslund did it.
Anyone with an expertise in Galois theory here? I have a question.
Is there a condition for transformation of polynomials with variable coefficients over $\Bbb Q$, i.e., any particular restriction for the Galois group?
Of course, the variable coefficients must be independent in a sense to avoid the decomposition of galois groups which would make the solubility function field much much smaller.
