@blue You recommended me Susan Colley's Vector Calculus. Do you have any recommendations of someone who treats both single and multi variable calculus (in 1 or more books)?
@robjohn I have a problem with the current analysis books. Books like Rudin treat theory but almost no applications or computations. Are there any good calculus books out there with both theory and applications, other than Spivak and Apostol?
@robjohn I was going to post an answer, but then I realized I couldn't' justify that $ \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x) \cos(ux)}{xu} \ du \ dx = \int_{1}^{\infty} \int_{0}^{\infty} \frac{\sin (2x) \cos(ux)}{xu} \ dx \ du $. Any suggestions?
@ThomasAndrews you're a year late to try brilliant.org. Back then, they weren't money-driven, this wasn't some social network, and there were really interesting weekly challenges.But it's all gone now...
@ThomasAndrews They are a relatively new organisation. Trying to get things in place. The emails aren't necessarily spam but rather promoting problems to you. You can turn them off. I agree though, at first they send way too many. They are not sending to spam though.
@BalarkaSen I am not joking :| ,,, you can use $x+(y+z)$ and $(x+y)+z$ to expand on $f(x+y+z)$ using that relation in two ways :) .. do that and tell me :D
This looks so symmetric that I must ask whether there's a category theoretical reason for this (both the theorems and the requirement for finiteness of the collections in c and d).
@BalarkaSen Yes and therefore a single equation will not be helpful. That is starting with a single equation. It is like giving an abbreviation and asking the person what does this tell you, or an acronym one is trying to interpret.
Also, is there any category theoretical link between the $A \subseteq B \subseteq C \Rightarrow [A:C]=[A:B][B:C]$ that appears both in group and field theory?
I've got Ahfors at the moment and don't like it that much.
@BalarkaSen I had a feeling you would ask me that. I have not read a single paper or book in its entirety in the oeis or other places, about the Riemann hypothesis.
@user2804303 If you feel that's enough for you, read up Titchmarsh "Theory of Riemann Zeta function". In case you think you need to understand complex analysis more, read up Titcmarsh. (This time "Theory of Functions" =P)
@user2804303 Well, I can't recommend anything unless I know how much you have studied.
@BalarkaSen When the proof of the Riemann hypothesis comes it will be either through bounds on arithmetic sequences that Riemann hypothesis predicts, or it will be proof that "If the Riemann zeta zeros are of this form then their real part must be one half".
@MatsGranvik It's hard to tell. AFAIK, the most popular argument these days is Iwaniec's, by inspecting trace formulas for L-functions of certain elliptic curves.
@DanielFischer If I could bother you for a minute, how could you show that $\int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (x) \cos (xu)}{xu} \ du \ dx = \int_{1}^{\infty} \int_{0}^{\infty} \frac{\sin (x) \cos (xu)}{xu} \ dx \ du$? Changing the order of integration is not immediately justified by Tonelli's or Fubini's theorem.
@Alyosha $|G : H| = |G : N||N : H|$ fact for groups (assume inverse galois for some base field with which you want to work with) is just a realization of Galois's theorem : if $K/E/F$ is a tower of galois extensions then there is a short exact sequence of galois groups $1 \to \mathbf{Gal}(K/E) \to \mathbf{Gal}(K/F) \to \mathbf{Gal}(E/F) \to 1$.
Note that $\mathbf{Gal}(A/B)$ has order precisely $[A : B]$
So once you know Lagrange's theorem for groups, your degree-of-field-extension theorem (which indeed mimics Lagrange's theorem) is no accident.
@TedShifrin A very faulty way to evaluate the integral : $f(x)$ be $\int \frac{dx}{x}$ without the arbitrary constant. $f(uv) = \int \frac{udv + vdu}{uv} = \int \frac{dv}{v} + \int \frac{du}{u} = f(u) + f(v)$. One can prove that $f(\exp(x)) = x$ by the limit definition of $e$ now.
@TedShifrin Try it. I'm pretty sure it works. It's the hint Otto Forster gives for exercise 18.3 in his Analysis I, and how we did it in the first semester. Of course you need to know the derivative of $x \mapsto a^x$.
Ah, @Daniel, I assign that exercise (which I put in Spivak) for all other $p$. Are you suggesting you get $p=-1$ by taking a limit, then? Believable. I couldn't see how a log could show up :)
You can read if you care, @Balarka. You were here fir the altercation with me.
@Rainbolt it seems there's no pattern for who gets the maximum value in a given trial. So choosing a winner according to the max value is inconsistent. I would balance things like $$\frac{0.1\times \max + \min + 2\times \text{other values}}{0.1+1+6}$$ It still rewards people with large maximum to a lesser extent, putting a larger emphasis on other values
I don't remember months ago. Have I tangled with him before? No, yesterday with his "Needy" comment to me. I still don't know if @N3 has read my answer. I guess this abbreviation doesn't ping him.
@Alizter No, certainly not. Since $\operatorname{Re}$ and $\operatorname{Im}$ are standard, pretty much everybody uses those, I wonder what devil rode Knuth to choose $\Re$ and $\Im$.
It's only by luck that they worked over $\Bbb Q$ and by luck again that one implies another. The resolutions of Galois and Abel-Ruffini are vastly different.
@TedShifrin I lecture geometers about some small algebras I know off the top of my head time-to-time =P
@RandomVariable Unless I'm being stupid, integrating by parts gives you $$\int_0^\infty \int_1^\infty \frac{\sin x \cos (ux)}{ux}\,du\,dx = \int_0^\infty \int_1^\infty \frac{\sin x\sin (ux)}{u^2x^2}\,du\,dx - \int_0^\infty \frac{\sin^2 x}{x^2}\,dx,$$ and you can change the order of integration in the remaining iterated integral. Now, seeing that that is the same as changing the order of integration in the original, well, it's not obvious to me ;)
@G.T.R Yes, it was purely by chance that Crowley and AverageLoser did so well in the first trial. They were paired with other "simple" submissions. I don't know how to parse your notation. Are you viewing this chat room in some sort of viewer? It was literally displayed to me as $$\frac{0.1\times \max + \min + 2\times \text{other values}}{0.1+1+6}$$
