@AmithKK Owow, I'll watch that tonight. Seems really exciting or something. About what I would suggest to watch, I'm not sure if we share common interest. What kind of movie are you in the mood for right now?
@robjohn yup! because EH intersects FG and the opposite angles of that intersection are the angles of congruent triangles. cool
now i have a geometry question (it's geometry day in mathchat!): I have a circle with inscribed quadrilateral. i know two angles of the quadrilateral (70 and 80 degrees). I need to find one other angle of the quadrilateral.
@Jeff nor mine, but there was enough of a subculture for some of it that it has stayed around. I did buy CDs long ago with a lot of it though, and that helps.
I always buy my music on CDs. I rarely, if ever, download my music.
I did download the album with "Grandma got run over by a Reindeer" on it, because I could not get it any other way at one point.
i had a bunch of old CDs around somewhere, too. i used to like CDs, too. because i want to own what i buy (in a way) and not be at the mercy of renting from some company
@meg_1997 They shouldn't, I was confused for a moment as to why Wolfram doesn't show $-1 \leq x \leq 1$ but then I saw it shows $x=-1$ and $x=1$ separately (as integer solutions). That's why I linked to it.
@Eugene what would be the motivation to close? I don't think we close questions for being too hard. It seems that the OP is asking about a certain approach to the RH.
@Eugene It's unclear whether the OP was taken in by the joke exercise or is asking a serious question about the relative merits of a quasicrystal approach to the RH. Even if he was taken in, he may change the question to be the more serious question and it will end up more respectable.
@Gigili sorry to interrupt..But i have a last doubt :in the answer you gave is it $-1 \leq x \leq 1$ instead of $-1 \leq x \leq 0$ for the union of last two cases?
I reserve the label crank, beyond just ignorance or pedestrian familiarity with math, for those who claim expertise and understanding far beyond their actual state of such.
So e.g. if someone has no idea what they're talking about, and indeed is talking about silly nonsense, but has no qualms admitting this, I wouldn't call them a crank. I also think cranks tend to be stubborn and hard-headed in the face of community response.
Jose Garcia often pushes the line there. I do however remember there being one attempted proof at RH that was met warmly on MSE. Someone found an error and the author thanked that person.
@meg1997: I wonder what that user meant by those comments! But we will wait for him to respond. I checked it again but found nothing. The picture clearly shows that final answer.
I have to memorize the 20 basic integral forms, simpson's rule, trapezoidal rule, area of a surface of revolution formulas and arclength formulas, so much to have memorized
@Gigili oh well what's done is done i guess. i hate that he asked me a whole bunch of questions after answering and after all that showed no gratitude.
Jordan: you could factor and use partial fraction decomposition and then integrate with shifted logarithms, or you could complete the square and use an affine-linear substitution to integrate $1/(u^2+1)$
"I have to memorize the 20 basic integral forms, simpson's rule, trapezoidal rule, area of a surface of revolution formulas and arclength formulas, so much to have memorized"
@PeterTamaroff The book really grew on me. I did not like it at all in the beginning, but after going to some lectueres, and doing some exercises I enjoyed it much more. Although some of the problems can be a tad hard sometimes.
