@KarlKronenfeld I'm wondering about the following.
Suppose $T$ is an elementary linear transformation $T:\Bbb R^n\to\Bbb R^n$. I have proven that $$\int_{T(A)} 1=|\det T'|\int_A 1$$ But I am wondering how it should be properly done.
Here $A$ is an open rectangle.
For example, I think we can simply consider the case $n=2$ by Fubini's theorem.
@TedShifrin Hey. I should get a better anti-virus. Now I'm under the weather.
I don't know how to do that properly, unless it is proper to draw a picture and say "hey, it works!". I am also having too much fun with sheaves right now. I hope you aren't sick for too long, btw. I was felt like I had a bug for the last couple of days, which mainly tired me out.
@PeterTamaroff I can remember the form for when n is a positive integer. I usually block on the formula for the generalization to complex exponentials, viz $(x + y)^r = x^r + r x^{r-1}y + \frac{r(r-1)x^{r-2}y^2}{2!}...$
I'm just going to stare at it until I remember it forever.
@Bitrex Well, consider the expression $$(x_1+x_2+\cdots+x_m)^n$$
Let $\alpha$ be a partition of $n$ into $m$ parts, that is, a set of non negative integers such that $\alpha_1+\cdots+\alpha_m=n$.
Then note that in the expansion of the expression above, you'll get sums of the form $x_1^{\alpha_1}\cdots x_m^{\alpha_m}$ where $\alpha_1+\cdots+\alpha_m=n$.
The question is: in how many ways can you obtain such string?
Problem : How many ways are there to make $n$ by adding $k$ non-negative integers, where order matters. Suppose $n=4$ and $k=3$. There are 15 solutions using $0, 1, 2, 3, 4$:
$(0,0,4), (0,1,3), (0,2,2), (0, 3, 1), (0, 4, 0), (1, 0, 3), (1, 1, 2), (1, 2, 1), (1, 3, 0), (2, 0, 2), (2, 1, 1), (2, 2...
@PeterTamaroff I can expand this for 2 terms: $$\sum_{k_1+k_2=n}a_{k_1}b_{k_2}=\sum_{k_1=1}^{n-1}a_{k_1}b_{n-k_1}$$ However when I get to 3 terms such as $$\sum_{k_1+k_2+k_3=n}$$ I don't know how to rewrite it like I did above.
@PeterTamaroff Okay so I think I got it: $$\large\sum_{k_1+k_2\dots k_m=n}=\sum_{k_m=0}^n\sum_{k_{m-1}=0}^{n-k_m}\sum_{k_{m-2}=0}^{n-k_{m-1}}\dots\sum_{k_1=0}^{n-k_2}$$
@PeterTamaroff That's uncanny. Finite fields popped up in my reading today because I was looking at regularization techniques, in particular zeta functions and that led to the finite field case where zeta functions are constructable explicitly
@anon It is uncanny that I would read about this on the same day as Pedro though because my usual work is applied and has nothing to do with abstract algebra or analytic number theory
@BrianM.Scott: I think in the case of the question we were discussing the problem is that the "homework" tag is poorly named. What seemed quite clearly to apply is the "hints, not a solution" aspect of its specification.
@dfeuer I interpret the tag very literally: it’s for assigned homework, and especially for assigned homework that is to be graded. And whether I choose to leave a big hint, a small hint, a sketched solution, or a full solution depends on a lot more than the presence or absence of a homework tag.
@dfeuer Indeed. But that’s a separate matter from whether a problem is actually homework as I use the term.
@BrianM.Scott I appreciate your selection of which answers should be answered. ometimes people want only to answer hard questions. i guess the participation of people with higher qualification is a good thing.
@BrianM.Scott: since we never have any way, short of trust, to know that something is not assigned homework, does the homework tag really serve any genuine function? Would it not be better to tag based on what kinds of answers the OP desires, at which point we can make our own judgements as to whether we wish to provide such?
@BrianM.Scott I've read somewhere, I guess that it's in the Klein's elementary mathematics from the advanced standpoint that usually undergrad and grad students omit themselvses of communicating with the general audience.
@dfeuer I’m perfectly willing to trust. Yes, sometimes it’s misplaced, but I don’t really care. And as I say, I base my judgement on how to answer on many things. The only consideration that in itself ensures that I will offer only a hint is the OP’s explicit statement that that’s all that’s wanted.
@dfeuer I think that Gustavo’s saying that he’s read that undergrads and grad students don’t make any real effort to communicate with a general audience, in particular with a less mathematically sophisticated audience.
@GustavoBandeira, as for hard vs. easy questions, I tend to find it most fun to answer questions at my own middling level, but I'm also quite happy to help someone out with one I find easy as long as they put some effort into 1. thinking about it themselves and 2. presenting it so that I can understand exactly what they're asking.
@BrianM.Scott Well, I am claiming that if $I$ is an interval and $f:I\to\Bbb R$ is continuous then $\operatorname{graph} f$ is closed in $I\times f(I)$.
Oh, and I tend to refrain from answering "easy" questions if I don't think I'm able to formulate an answer that will further the goal of helping the asker learn about the topic at hand.
Because outside the context of applications (which I'm unlikely to know the math to answer anyway), "easy" questions don't serve any purpose if we don't learn from them.
@BrianM.Scott So for my claim above I just simply want to take a sequence in $I$, which amounts to taking a sequence in the graph of $f$ and show that if it converges to some element of $\ell \in I$, then the associated sequence which by continuity is $(\ell,f(\ell))$ is in the graph of $f$. So I need sequences but in the subspace topology, yes?
@dfeuer A carefully explained, fully worked-out solution does in fact have a positive probability of helping the OP learn. Sometimes it’s even the best choice, though I agree that more often a carefully crafted hint is a better choice.
@BrianM.Scott The problem I often find that with my engineering students is that if I work out a solution for them, they attempt to 'copy' the solution on other problems without considering the differences between the problems. But I agree that, especially when the student doesn't even know where to begin, a fully worked solution can be very helpful
@KevinDriscoll I’m most likely to do that when either I can’t see any way to give a useful hint short of working the problem, or I can say something about how to approach problems of that kind.
@BrianM.Scott: my own experience suggests that fully worked solutions are of very little value, although I do think it's valuable for a textbook to offer proofs of theorems even when those proofs have a lot of "fallen from the sky" steps that required some genius to discover. For me, an ideal textbook would have the student prove most of the theorems as exercises, with good suggested outlines and hints.
@PeterTamaroff I’ve been up and down. A couple of weeks ago I spent a week with a friend $400$ miles away, and the drives out and back have left me with residual back pain. And I’m doing it again in about three weeks. urgh
@BrianM.Scott "We prove $\partial A$ is closed. If $x\notin \partial A$ then there exists a nbhd $N$ of $x$ disjoint from $A$ or $X\setminus A$. But if $y$ is any point in $N$ then $N$ is itself a nbdh of $y$ and $y\notin \partial A$."
@PeterTamaroff Yes, that works. Or you can prove that $\operatorname{bdry}A=\operatorname{cl}A\cap\operatorname{cl}(X\setminus A)$ and is therefore the intersection of two closed sets.
@PeterTamaroff If x is not in the boundary of A then for each neighborhood N of x, then either N is a proper subset of A or a proper subset of X mod A but then for each y in N, N is a neighborhood of y that contains only points in A or point in X mod A, therefore y is not in the boundary of A. The neighborhood of a point in the boundary of the boundary of A must contain at least one point in the boundary of A.
But we have already showed that if x is not in the boundary of A then no point in any neighborhood of x is in the boundary of A therefore x is not in the boundary of the boundary of A.
@PeterTamaroff Did you delete your comment on your answer?
Haha @AlecTeal I'll pass along the message. I dont spend any time on stackoverflow (your know there ARE other overflow sites? like MathOverflow!) but I have heard things can get out of hand
They're tards, there's no entry requirement. Here there's not but .... it's implicit that one can count before coming here, with SO, it's not.
So while I never really get to answer any questions because someone who's forgotten more than I yet know has got there first, I just ask the odd ones.... thanks guys
Like that Ted Shifrin guy linked earlier, never asked a question.... wow
@KevinDriscoll just found a question I can do, 7 answers. Hah!
I've often submitted an answer while someone else was submitting and didn't notice the duplication. Certainly if $f$ is integrable on $R$ and $S\subset R$ is a region over which integrability makes sense (e.g., subrectangle), then $f$ is integrable over $S$. I don't like the vagueness of "bounded sets," although I suppose the intersection of eligible sets should agin be eligible. We're not doing Lebesgue measurable or Jordan measurable ... Just bounded? Ugh.
Yes, @Peter, although in my course I stick to regions, which I define to be the closures of bounded open sets with the property that the boundary has volume (Jordan content?) $0$.
