This is a question based on Exercise 7.6.5 of Humphreys', "Linear Algebraic Groups". For a solution to that particular problem, see:
A closed subset of an algebraic group which contains $e$ and is closed under taking products is a subgroup of $G$
The Question:
Is a closed, nonempty subset $H$ of...
Quanto , @robjohn , and some other users are so good at integrals....not trying to be look like a fan...they inspired me to get good in this integrals game.
@TedShifrin Sure, but the point was that it worked out as an application of some other tool that had just been taught. So instead of being a rabbit-pull, it flowed in the the narrative of the class.
Like, I agree that there is no reason to teach specific integrals just to teach them, but they are useful if they serve to demonstrate some technique or idea that might come along later, or that students are trying to master.
@TedShifrin I didn't claim it was useful for your students. But I have found it to be useful for some of my students in the past (at UCR---almost certainly not here).
And I now have two feeders on the back porch (they were going through one feeder in three or four days---I'm hoping that with two, I'll be able to go a week without refilling them).
@leslietownes Vodka and simple syrup sounds pretty boring. Are you adding some bitters, at least?
Also, am I being too snarky?:
If $a,b,c\in \mathbb{N}$, it seems that the maximum of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is $3$ (if $a=b=c=1$). I am fairly certain that there are rational numbers larger than this, so it seems that the answer to your question is "No". — Xander Henderson ♦48 secs ago
if $U$ is an open set, if ${U_i}$ is an open covering of $U$ and if $s$ in $\mathcal{F}(U)$ is an element such that $s|_{(U_i)}=0$ for all $i$, then $s=0$. In the definition of a sheaf, what does this $s$ restricted at $U_i$ means? whether $s$ is in $\mathcal{F}(U_i)$ or not?
@SoumikMukherjee I don't know what $\mathcal{F}(U)$ is (functions on $U$?), but if $s : U \to X$ and $V \subseteq U$, then $s|_{V} : V \to X$ is defined by the property that $s|_{V}(t) = s(t)$ for all $t\in V$. Note that $s(t)$ makes sense for all $t\in V$, since $V \subseteq U$ by assumption.
I have never really done much with sheaves, but my recollection is that there are two basic ideas: (1) you can always restrict to "smaller" open sets in a nice way (I think that the general phrasing is that if two things restrict to the same things on every set in a cover, then they must have been the same thing in the first place), and (2) you can "glue things together" in a reasonable way (I forget quite how this works---something about things agreeing on intersections...?).
In any event, as @TedShifrin says, "restriction" is pretty fundamental to the whole field.
@SoumikMukherjee Don't start by working abstractly. Pick an example, and work through that example to understand the definition. What is your goto example of a presheaf?
I am confused by the notation, say we have a presheaf of sets, now $\mathcal{F}(U)$ is a set, we pick an element $s$ in $\mathcal{F}(U)$. Now what does $s|_{U_i}$ means? My confusion is that $s$ is an element on the image set, not a function, then what does this restriction means?
@SoumikMukherjee Again, pick a concrete example of a presheaf. in that example, what are the elements of $\mathcal{F}(U)$?
My goto example is holomorhpic functions (which is actually a fairly trivial example, thanks to the identity theorem): the elements of $\mathcal{F}(U)$ are functions $s : U \to \mathbb{C}$ which are holomorphic.
Then $s|_{U_i}$ is simply the "normal" restriction of $s$ to the domain $U_i$.
@SoumikMukherjee The elements of an arbitrary $\mathcal{F}(U)$ in an arbitrary presheaf are just some generalization of functions. They have simply had everything abstracted away, except for the specific properties required to be a presheaf.
So if you think you understand sheaves of functions, pick a second goto example which does not consists of functions, and try to work through that.
@SoumikMukherjee Again, this is really not my area of expertise, but I am fairly certain that the elements of $\mathcal{F}(U)$ could be morphisms in any kind of appropriate category. Not all morphisms are functions.
Depending on what you are actually trying to do, however, this likely doesn't matter.
And it is probably just fine to treat all of these objects as though they were functions, until and unless you end up in a situation where they aren't.
@SoumikMukherjee $s \restriction U_i$ is the image of the element $s \in \mathcal{F}(U)$ along a restriction map $\mathcal{F}(U) \rightarrow \mathcal{F}(U_i)$
presheaves have restriction maps as per the definition of presheave
$s$ itself is an element of $\mathcal{F}(U)$ which is always a set, but can of course have more structure, e.g. you can have a presheaf of abelian groups
in which case $s$ is an element of an abelian group
and the restriction maps then respect the additional structure
which are local inverses to the espace etale projection, i.e. they are locally elements of the presheaf, and they take values in the stalks of the espace etale
yeah this is all a bunch of abstract nonsense lol
but it is helpful at least to me to think of everything post sheafifying
maybe someone who has taken it more recently than me has a suggestion. if "completely mathematical" means what i think it means, there's some tension between that requirement, and wanting a "calculus" book (at least in the USA, "calculus" describes a calculational subject that is often not oriented toward any kind of mathematical rigor)
if you think of them as a product being sold, the markets are different. most physics students would not want a book that goes into proofs and theory, at least early on.
my general impression is that all of these books are kind of the same. i learned from a book by james stewart, around 1000 pages long, called "calculus: early transcendentals", which is a popular choice in the US.
there's nothing mathematically wrong on it, and some of it is just as rigorous as a more formal book. but that isn't really what it's for. the 1000 pages are mostly exercises in calculation, or examples of calculation.
there are some more complicated exercises in each section, but again, most instructors would skip them. and it is not a book that i would use to learn theory.
but because it's a popular book, it should be easy to find, both new, used, and, in, uh, electronic form
it's hard to say. "phd in mathematics" could cover a lot of different skill sets. it's maybe less like engineering where a specific degree is pretty closely tied to a specific set of job openings.
i wouldn't think so. i don't work in the area but my general understanding is that a lot of software jobs are more about the skill set than what you have a degree in.
yes. the titles are a little confusing. he has books called "calculus" that are available in several "versions" (at least two). the "early transcendentals" version puts things like trig functions, logarithms, exponentials, etc. earlier in the book than the other version, and maybe includes more and more difficult problems.
i think the other versions are intended for calculus classes for people who might not go into those subjects but still have to take calculus for some reason.
it's pretty common for US universities to have at least two 'flavors' of calculus, with engineering students taking one, and, say, biology students taking another.
i taught at a place that had three flavors of calculus, i forget exactly how it broke down. engineers and math majors did not take the same one and there was a third one.
i mean, whatever definitions you pick, it turns out that a limit either exists or it doesn't, whether or not you are choosing think of it as a pair of "one-sided" limits.
the one-sided limits come up a lot in textbook exercises where functions are "piecewise defined," i.e. f(x) is given by one formula on one interval and then another formula on an adjacent interval, etc.
because if you have nice formulas for the pieces, the "sided" view is the one that maybe most expressly draws your attention to how those pieces fit together.
(and from a calculational point of view, allows you to substitute individual formulas in evaluating the limit, instead of dealing with two-sidedness)
I'm doubting a simple concept. Consider the function $$f(x) = \left\{\begin{array}{l}x^2 \text{ if } x > 0 \\ 0 \ \ \text{ if } x \leq 0\end{array}\right.$$ If I use the definition of the derivative, i.e. $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, to calculate the derivative at $x<0$, then what does $f(x+h)$ simplify to?
@Jakobian Hmm, but a priori, I just want to simplify the fraction $\frac{f(x+h)-f(x)}{h}$ (THEN use the limit operation). I don't know yet that we are interested in small $h$, so that's why I'm doubting that $x+h$ isn't smaller than zero.
@sunny well, I'm not exactly taking the limit with that comment. One got to realize that the limit is only determined by values as $h$ is small i.e. $h\in U$ for some open $U$
If want to bound a function $f(x)$ which is $f(x)=g(x)-h(x)$ where $|g(x)|=g(x)$ and $|h(x)|=h(x)$, then is it true that we first must have bounds on the functions $g(x)$ and $h(x)$ in order to be able to bound $f(x)$?
@Jakobian ok, then simplifying the fraction as I suggested and as you did above probably doesn't make much sense, since the limit of $(x+h)^2/h$ doesn't really exist, or?
@sunny it does make sense, I guess, but it won't matter in the end, because the only thing that matters is how the expression behaves as $h$ is close to $0$
look at the $\varepsilon-\delta$ definition of a limit for example
This is too long to be a comment so I'm making it CW.
The following was written in GAP (because that's the only programming thing I know):
Squares:=[];
Values:=[];
m:=0;
t:=0;
while not m>79601 do
t:=t+1;
m:=t^2;
AddSet(Squares ,m);
od;
for n in [1.. 199] do
a:=n^2+(n+1)^2;
if ...
What could I have done better, @XanderHenderson? That's what I want to know. I don't want to repeat the same mistake. I see that there's a quick way to use GAP (in the comments) but that's no reason to downvote.
@Shaun Like I said, I have no idea why your answer was downvoted. Only the downvoter can explain.
As to the question, there's no motivation, nor any indication of what a good answer will consist of (maybe the asker doesn't want code, but wants some more abstract solution---I have no idea), nor any source for the problem. it is, essentially, a homework problem whose solution is being outsourced to Math SE.
Basic question. If two functions have the same derivative at a point, are they then equal on some small neighborhood around that point? It holds the other way around I guess; if they are equal on a neighborhood around that point, then surely they have the same derivative at each point in that neighborhood.
Suppose we have a CT irreducible Markov process on a countable space S such that $\lim P(X_t=x)=b(x)$ and $\lim\frac1t\int_0^t1_{x}(X_s)\,ds=a(x)$. Then prove that $a=b$
What I have: (1) If x is recurrent then $\lim\frac1t\int_0^t1_x(X_s)\,ds=\frac{1_{\{T_x<\infty\}}}{q_x m_x}$ and (2) If the process is irreducible and positive recurrent then unique stationary distribution exists and limiting distribution=stationary distribution = $\pi(x)=\frac1{q_x m_x}$
there's a fact that lim t->0 sin(t)/t = 1. you can potentially use that fact here to write that limit as a product of other, potentially easier to evaluate limits. i would be careful to stick to what theorems limits actually say and provide, over "just replacing" things with other things.
@sunny "Whenever"? I would be very careful about that...
If I were you, I would multiply upstairs and downstairs by $(x+h)x$, then use the fact that the limit of a product is the product of the limits (assuming that all the limits exist).
Hi, so I'm try to write $\{\emptyset\}$ but the preview doesn't show the braces and just displays the empty set. Is that something that only happens on the preview? Or will the post look like that as well?
i ran into some inconsistencies with the previewer once (i think also about the use of curly braces) and it drove me up a wall until someone (perhaps PM 2Ring) told me that
