Can you explain that in plain language? I still don't quite understand the difference. Is it something like $f+g$ is the name of the new function, whose variable is $x$, and the new function is created by the operation of adding two functions $f(x)+g(x)$?
@Eulb Hi Jeff. I had gone to bed. So what's the use or advantage of the different notations? Surely it's not for the sole purpose of confusing students?
@BalarkaSen, if differentiable function $f$ is $\Bbb R\to\Bbb R$, then we can talk about whether derivative continuous or not. But can we do something similar with total derivatives? I am not sure because, now derivative at each point itself is a function.
What is meant by learning math historically (not learning math history only, but learning math with a historical development perspective)? I've seen some sources say that to learn a math topic X, you need to look at the historical development of the topic X and go over the famous questions by yo...
Claim: If $X$ is a topological space and $(A,B)$ a separation, then $X$ intersects both sets in the intersection. Proof: Suppose that $X \cap A = \emptyset$. Then $X = X-(X \cap A) = (X-X) \cup (X-A) = X-A$ and therefore $A \subseteq A \cup B = X = X-A$, which is absurd. Hence, $X \cap A \neq \emptyset$. The same work shows that $X \cap B \neq \emptyset$...How does that sound?
> Do all the rings here have identity? Yes: the reason is that a great deal of the logic connecting properties here uses the assumption that the ring has identity.
no, the right reason is that the other rings don't exist
Hmm...you're right...Perhaps the result should be rephrased in terms of subspaces of $X$, say $Y \subseteq$, so that $A$ and $B$ are not necessarily subsets of $Y$. I think the same proof works in that case, right?
I mostly think about singular cohomology of topological spaces. The intuition I have in mind is it extracts out the cochain complex $C^\bullet(X)$ from a given topological space $X$ where elements of $C^k(X)$ ("$k$-cochains on $X$") are functions on the "$k$-skeleton of $X$"
If $X$ is a simplicial complex, then this "$k$-skeleton of $X$" is nothing but the union of $k$-simplices of $X$.
Think of piecewise linear functions on $X$ if you like
@BalarkaSen Recently I learned how an algebraic result about group rings had some reflection in group cohomology, and I wished I understood more. mathoverflow.net/q/297043/19965
Excuse me, I come for a reference request. I finished my first linear algebra course and I would like to brush up on linear maps and projections because I feel I didn't really understand them
I think one intuition for many stuff in derived functor cohomology is to measure how far away a left/right exact functor is from being exact
e.g. often in AG people you want to understand global sections, and cohomology inevitably comes up because the global section functor is not right exact
@loch Btw, I think I saw you register at the DaRT site :) any plans to submit a ring suggestion? Your help, especially with commutative algebra would be appreciated.
i'm at a little bit of a disadvantage because I don't know a lot of AG. I don't know how to wield the weapons of homological algebra... I mainly engage problems with fisticuffs.
Not that I don't appreciate examples given in that vein... I'll probably eventually catch onto them!
@BalarkaSen, if differentiable function $f$ is $\Bbb R\to\Bbb R$, then we can talk about whether derivative continuous or not. But can we do something similar with total derivatives? I am not sure because, now derivative at each point itself is a function.
Let's say you have a smooth function $\mathbb{R}^n \to \mathbb{R}^m$
Then the derivative of the function at a point is a linear map $\mathbb{R}^n \to \mathbb{R}^m$
So the map that takes in a point $p$ and spits out $Df(p)$ is a function from $\mathbb{R}^n$ to $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m) \cong \mathbb{R}^{nm}$
That's the analog of thinking about $f'$ as a function $\mathbb{R}\to\mathbb{R}$. So this does allow you to talk about a function's derivative as itself being a continuous/differentiable function
@JakeS: I'll also add some self-horn-tooting and suggest my Linear Algebra book with M. Adams. (I'm assuming you can find it in a college library, rather than buying it.) We do lots more than most books with geometric stuff, and there are several discussions of linear maps and projections.
@Silent: To add to Demonark's reply, you can also then think about differentiating the function $g(p) = Df(p)$ as a map $g\colon \Bbb R^n\to \mathscr L(\Bbb R^n,\Bbb R^m)$. This will be the second derivative, and the deep result is that when you view $Dg(p)$ as a linear map $\Bbb R^n\to \mathscr L(\Bbb R^n,\Bbb R^m)$, you can associate this to a bilinear map $\Bbb R^n\times\Bbb R^n\to\Bbb R^m$, and, as such, it is symmetric.
pretty well, Demonark, and how's your summer research going?
@Ted it's going quite well. Right now I'm reading some algebraic number theory, after which I'll get to the main paper I'm reading, on Kronecker-Weber for $\mathbb{Q}(i)$
Hmm, I'm not on as much, and when I am on I usually just tab in for a second to see what's up, maybe say hi to people, then tab out. So with that noted, I haven't seen much go on
Nope, and they did say that in all likelihood at this point the guy doesn't even have the laptop anymore. So I got my dad's old chromebook that I'll be using for the time being
I think what "far from exact" and "close to exact" means for functors kind of depends on context --- for example something similar to what @Daminark - but for modules, is that if you have two $A$-modules $M,N$, then $Ext^1_A(M,N)$ is in bijection with short exact sequences of the form $$0\rightarrow N \rightarrow P \rightarrow M \rightarrow 0$$ up to isomorphism.
So here how far away $Hom_A(-,N)$ is away from being right exact is the same thing as saying how many ways can you have an exact sequence of the above form (if Ext is trivial, then $P$ is isomorphic to $M\oplus N$.
Question: Is there a word for a 1-dimensional slice? I have a solid with a line going through it. The word slice (to me) conjures up a 2-dimensional (planar) section, is there an analogous word for slice for the 1-dimensional case?
Nice! But yeah I haven't made much progress, Mathein told me some stuff on DC but in terms of working through things carefully and all I basically just made it through a section and a half. Hopefully I can get through a few more, and ideally reach ramification and the like soon
@Semiclassic: i've never thought about scaling the polar variable $\theta$. If anything, I'm inclined to say we should scale with $1/2$ rather than $2$ here. I'm in the middle of writing my fourth paragraph or so. :)
@Semiclassical I've encountered logic textbooks using $\to\phi\psi$ rather than $\phi\to\psi$ because clearly using prefix notation is waaaaaay less of a pain than having to write brackets
So I programmed in Racket for a little bit and you used Polish notation, but you still used parentheses (the joke, or maybe not even joke, is that racket is supposed to say "bracket")
@Semiclassic: I don't think the fourth power is relevant. I said in my answer that perhaps the difference in degrees of homogeneity of the two sides should lead to an $\alpha$, but I don't believe it now.
I really like it when an author introduces important examples (and sometimes even important theorems!) as guided exercises and lets the reader work through them (Of course that's risky because people might just skip the exercises :/ )
@Alessandro: Jeanne Clelland's new book on geometry done with moving frames is exactly that. @EricSilva loves it. I was a reviewer for it and made various suggestions, but that style is better for beginning grad students than for typical (US) undergrads.
As you've heard me say, Demonark, problem sets/exercises are what makes a course/textbook great. I mean, the text/lectures are significant, too, but ...
(The path I wanted to lead you down was the same proof written differently: whatever the distance is from x in U to some n points, there is a minimum distance, say $d$. If you pick $\epsilon$ small enough so that $\epsilon < d$ and $B(x, \epsilon) \subset U$, then...)
$T\vDash A$ if $A$ is true in the theory $T$, $\mathfrak A\vDash A$ if $A$ is true in the structure $\mathfrak A$ but I don't know if that's what you're looking for
You can let $x= k t$ to find
$$ I(k) \equiv\int \limits_0^\infty \frac{-k \ln(x)}{k^2+x^2} \, \mathrm{d} x = \int \limits_0^\infty \frac{-\ln(t)}{1+t^2} \, \mathrm{d} t + \int \limits_0^\infty \frac{-\ln(k)}{1+t^2} \, \mathrm{d} t \, .$$
Your substitution $t \to \frac{1}{t}$ shows that the first ...
I guess I mean that the only thing I can really compare x with in that function is k
e.g. if I had k=100, then I'd naturally be most interested in the case when x=100
anyways, once you have it in your head that x and k are more or less on the same footing in that denominator, that to me does suggest the substitution x=kt
since in that case one has k/(x^2+k^2) = 1/k*1/(t^2+1)
so the k-dependence has been factored out of that rational function
this comes at the cost of dx = k dt and log(x) = log(k)+log(t)
What I hadn't accounted for was being able to show that the log(t) term vanished
Let $(u_n)_n \in \mathbb Q^{\mathbb N}$, with $\forall \alpha>0, |u_{n+1}-u_{n}|=o(\alpha^n)$. Is it true that the limit of $(u_n)_n$ is a transcendental real ?
