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12:05 AM
@PeterTamaroff Yo whatup
12:26 AM
@KarlKronenfeld Hey.
My computer froze.
@PeterTamaroff It must not have taken very long to thaw.
@KarlKronenfeld Heh, nah.
@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.
12:52 AM
@anon how did the e^(-jk(2*pi/To)) got distributed in the integrals?? theres no addition between them
please link back to the image or convo
@IvanMatala They wrote $t=t-\tau+\tau$.
my same response from last time applies
@anon Hey, how are you?
12:55 AM
oh, never mind, I see what you mean Iva
as Peter says, $e^{a+b}=e^ae^b$
or equivalently $e^a=e^{a-b}e^b$, or other forms
ohhh... i didnt see that :)
thank you anon and Peter
@anon I'm sick. =/
I can never remember the formula for the generalized binomial theorem
Not totally toasted, just feeling kinda tired, eyes sandy and stuff.
1:01 AM
I know much more complicated formulas than that
but for some reason
I have some mental block on the binomial theorem
@Bitrex $$(x_1+\cdots+x_m)^n=\sum_{|\alpha|=n}\binom{n}{\alpha} x^\alpha$$ =)
There you go.
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.
@KarlKronenfeld Oh. Sheaves. Reminds me of shawarma. Makes me hungry.
Just watched number 23
1:04 AM
@Alizter That's a nice movie.
@PeterTamaroff It's funny how all the math in it is just addition of digits. You do also see the odd meaningless integral on the wall
@Alizter Didn't pay attention to that.
@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 Do you know what a multinomial is?
@PeterTamaroff I DO! :D
1:09 AM
The multinomial formula is all about counting.
Sure, and please don't ask me to remember that one.
@Alizter Good.
@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?
Stirling numbers?
You have to choose from each $(x_1+\cdots+x_m)$ an $x_i$, right?
Think it for special cases, $m=3,4$ to get a feel of what happens.
Sorry, was AFK for a minute.
@PeterTamaroff I think it's the same as the formula for combinations with repetition
it'll take me forever to bang out the LaTeX
$\binom{n + k - 1}{k}$
1:23 AM
What does this sigma notation mean without the short cuts?
@Alizter It means you sum throughout all possible indices $k_i$ with the property their sum is $n$.
The indices are assumed to be non-negative.
@PeterTamaroff Can it be rewritten?
@Alizter "rewritten"?
What is the difference between the multinomial coefficient and the Stirling numbers of the second kind?
wrt counting partitions of sets
oh, multinomial is disjoint
1:28 AM
@PeterTamaroff Don't worry I just realised the complexity
Q: how many ways to make n by adding k non-negative integers (proof)?

palatokProblem : 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...

@Peter: Sorry you're sick. Students and faculty sick here ... Crazy for hot summer. Drink plenty of fluids and get some rest!
Coursera is giving a complex analysis course starting in October.
@TedShifrin Yes. I will rest. But there is so much time I can waste! =D
@PeterTamaroff I can expand this for 2 terms:
However when I get to 3 terms such as
I don't know how to rewrite it like I did above.
1:43 AM
@Alizter First, kill of an index. The continue like you did before.
That is, $$\sum_{k_1+k_2+k_3=n}=\sum_{k_3=0}^n\sum_{k_1+k_2=n-k_3}$$
@PeterTamaroff So expansion is basically recursive?
@Alizter Yes.
@PeterTamaroff Okay so I think I got it:
$$\large\sum_{k_1+k_2\dots k_m=n}=\sum_{k_m=0}^n\sum_{k_{m-1}=0}^{n-k_m}\sum_{k_{m-2}=0}^{n-k_{m-1}}\dots\sum_{k_1=0}^{n-k_2}$$
1:59 AM
@anon I know :C
@Alizter Nope. You need more subtractions.
Looks at time. Time says 3:00am. Thinks to self .... naaaa.
yes, the upper limits should go $n$, $n-k_m$, $n-k_m-k_{m-1}$, ...
@anon This is my reaction to what the formula will look like.
2:04 AM
@Alizter $$\sum\limits_{{k_1} + {k_2} + {k_3} + {k_4} = n} {} = \sum\limits_{{k_1} = 0}^n {\sum\limits_{{k_3} = 0}^{n - {k_4}} {\sum\limits_{{k_2} = 0}^{n - {k_4} - {k_3}} {\sum\limits_{{k_1} = 0}^{n - {k_4} - {k_3} - {k_2}} {} } } } $$
No wonder the original notation was used. The end. Good night everybody. Including @anon, @PeterTamaroff and square dude @robjohn.
@anon Hey. Do you think you can tell me stuff about finite fields?
It seems they are always made out of roots of polynomials.
2:09 AM
that is the go-to way to construct ring extensions in general (well, quotienting by ideals of polynomial extensions, which is more general)
Today in my Linear Algebra class we got to the classic $K$ a finite field then $|K|=p^n$.
@anon Oh? For example, we were given that $\Bbb F_9=\Bbb Z_3[i]$.
@anon So, what are the basics?
If $L/K$ is an extension of finite fields, $L$ is obtained as $K[x]/(\pi(x))$ where $\pi(x)$ is any degree $[L:K]$ irreducible over $K$
(IOW finite fields are simple extensions)
${\Bbb F}_q$ is the splitting field of $x^q-x$, where $q=p^s$ is any prime power
@anon OK, that is new to me. What is $L/K$?
2:12 AM
in field theory L/K means "L is a field extension of K"
so it's not a quotient of any sort, it's metadata of sorts
@anon Oh. So $L$ is a field such that $K\subseteq L$.
And what is $\pi(x)$ "any degree"?
"Any irred. polynomial of degree..." maybe?
pi is any polynomial of degree [L:K]
irreducible is short for irreducible polynomial
@anon Right. Guessed so. And what is $[L:K]$?
2:14 AM
the degree of the extension, $[L:K]:=\dim_KL$
@anon As vector spaces?
@anon And what is the "splitting field"?
the smallest field containing all roots of a given polynomial
@anon Oh, OK.
2:16 AM
being a root of $x^q-x$ is essentially being either 0 or having order dividing q-1 in the group of units
if |F|=q then |F^x|=q-1 is cyclic and everything has order dividing q-1
"Splitting" because the polynomials will "split" in linear terms?
yes, into linear factors
@anon Right.
@anon $F^x$?
F^\times, group of units
@anon Oh. LOL.
@anon Yes.
2:20 AM
the group of all field automorphisms of Fq is cyclic and generated by x->x^p
${\Bbb F}_{p^r}$ is contained in ${\Bbb F}_{p^s}$ if and only if $r\mid s$ (not just $r\le s$)
this can be proved with (p^r-1)|(p^s-1)<=>r|s
anything in particular you wanted to know?
@anon Heh, dunno. I was curious about finite fields. They popped up in class, and I was wondering what they looked like.
the only concrete idea of what they look like I am aware of comes from doing the quotient-by-an-irreducible process I mentioned
@anon Right.
Like $\Bbb Z_3[x]/(x^2+1)$.
I think I read $x^2+x+1$ and $x^2+x+2$ in Wiki.
it doesn't matter which irreducible you quotient by, only its degree
since finite fields of a given order are unique up to isomorphism
@anon Ah. OK.
2:30 AM
x^2+x+1 is reducible over F3 though
it is (x-1)^2
as a fun exercise, you may try to find an isomorphism between ${\Bbb F}_3[x]/(x^2+1)$ and ${\Bbb F}_3[y]/(y^2+y+2)$
@anon Heh, OK.
2:42 AM
@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
2:57 AM
finite fields are not niche, they are a standard component of abstract algebra, number theory and field theory
so I would not think it too uncanny
@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
3:25 AM
@GustavoBandeira I lol'd
@KevinDriscoll =D
@KevinDriscoll Are you new here?
I've been around for a few months, but not regularly @GustavoBandeira
Got it.
What are you doing?
@KevinDriscoll Are you there?
What language would you say is more influential por mathematics nowadays? korean or japanese?
I'm thinking of learning one of the two
any input?
3:43 AM
@Omnitic: I have no idea.
@Omnitic Akkadian.
@GustavoBandeira: Good one.
@BrianM.Scott Hello, Brian!
I guess I'lllearn Ithkuil.
3:49 AM
@GustavoBandeira Hullo; you’ve changed your appearance again!
@BrianM.Scott Do you remember me?
It's weird. You answer a lot of questions. I thought you forget people easily.
@GustavoBandeira I don’t remember exactly what avatar you were using before, but it was definitely different.
@BrianM.Scott Yes. But you remember the name.
@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.
@GustavoBandeira Absolutely. That’s probably because we interacted several times.
3:51 AM
@BrianM.Scott: someone facing a similar problem in an application would have come in with a realistic one.
@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?
@GustavoBandeira I’m currently reading things written in forníslenska — but not mathematics!
@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.
3:57 AM
@GustavoBandeira, I don't understand your last sentence.
@BrianM.Scott Brian!
@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 also don't understand my last sentence.
@BrianM.Scott I have quite a sillypantalones of a question.
@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.
4:00 AM
@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.
Exactly what Brian said.
@PeterTamaroff Go ahead; I just hope that it isn’t a breech(es) birth!
@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 $I\times f(I)$.
@dfeuer I have a bit of an advantage on that second point: I’ve been figuring out what students mean for a very long time now.
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.
4:02 AM
Note that for example $f:(0,\infty)\to (0,\infty)$, $x\mapsto x$'s graph is not closed in the whole $\Bbb R^2$.
@PeterTamaroff True.
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.
So a set $A$ is closed in $B$ if every sequence of points in $A$ with limit in $B$ is in $A$.
4:05 AM
@PeterTamaroff I think that you mean that whenever the limit of such a sequence is in $B$, then it’s actually in $A$; if so, yes.
@BrianM.Scott Yes, that is what I am saying.
@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.
4:11 AM
@dfeuer This is something that varies enormously from person to person. I’ve had quite good students who would not have done well with such a text.
Hi @BrianM.Scott how are you?
@skullpatrol Not too bad; and you?
@PeterTamaroff It looks okay to me.
@BrianM.Scott Fine thanks :-)
@BrianM.Scott That is enough for me =). How have you been Brian? Long time since I bothered you with my topology question!
@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
4:16 AM
@BrianM.Scott Ah! Well, guess it was worth the while.
@PeterTamaroff Yes, this is one of my closest friends. A former girlfriend, actually, who’s getting married next month.
@PeterTamaroff: I seem to remember there being a much more general theorem about graphs of functions being closed.
I'm trying to remember/prove such now.
@BrianM.Scott Oh. That is so movie like! =D
@PeterTamaroff I seem to remember the Hausdorff condition being a key element?
@dfeuer If $f:X\to Y$, you want $Y$ to be Hausdorff.
4:19 AM
@BrianM.Scott, yeah, that much is clear. I'm not looking for a hint though.
@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 $y\notin \partial A$."
I'm not looking for a hint; I'm trying to remember it myself.....
@dfeuer Sorry, I meant to ping Brian.
@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.
@BrianM.Scott Yeah, I know. I'm doing it for this
4:34 AM
@PeterTamaroff Ah, yes; yet another example of a text (or instructor) doing things in what I consider a rather unnatural way.
@BrianM.Scott Yep. Sucks, but I cannot go around telling the guy what or what not to do.
Maybe you can! =)
What in the world is the boundary of the boundary of A?
@KevinDriscoll =)
@PeterTamaroff Usually there’s not much point. But I am surprised at how often that sort of thing crops up here.
Use the definitions @KevinDriscoll, heh.
@KevinDriscoll Try it for the unit open disk.
4:39 AM
@PeterTamaroff I am having trouble imagining a point which is part of the boundary of the boundary but is not part of the boundary
@KevinDriscoll Well, $\partial\partial\partial A=\partial A$ if that is of any reassurance. =P
@KevinDriscoll But, again, consider $D=B(0,1)$.
@PeterTamaroff What is capital X in your comment?
@KevinDriscoll That'd be the ambient space.
The closure operator varies if the ambient space varies.
The closure of a set in $X$ may (and usually does) differ from the closure of a set in $Y\subseteq X$.
5:01 AM
@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?
5:27 AM
@BrianM.Scott do you know Ted Shifrin?
6:15 AM
Hey guys
Good evening/morning
I just wanted to say thank you.
The maths SE site is so much better than overflow.
You guys are mature and there's actual maths, not trivial/stupid questions, thanks for being so grown up.
Carry on!
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!
@AlecTeal I am also out of my league here most of the time so I'm in much the same position.
6:38 AM
@skullpatrol Stalking him?
@skullpatrol YES!
What Overflow? Stack overflow or Math overflow?
@skullpatrol I'm sorry, but your question is too vague, broad, vietnamese with cheese and WTF?, please STFU.
@skullpatrol don't worry my friend.
It's not stalking if you love that person, and it's okay if they don't know they love you yet.
@AlecTeal XD
Imprints upon @GustavoBandeira
1 hour later…
7:53 AM
@GustavoBandeira @AlecTeal Come on guys, we need more Profs in this room, no?
8:12 AM
Oh man
I'm so proud of my answer
stackoverflow.com/a/18544093/2112028 it's not to anything math related, but look at that, perfect.
8:29 AM
I'm trying to beautifully compute $$\lim_{n\to\infty}\sum_{k=0}^{n} \frac{1}{\displaystyle \binom{n}{k}^2} $$
@Chris'ssis ask it as a question
Not here.
@AlecTeal does it seem to you that I ask a question?
This is a chat room, post questions in the question section :)
@Chris'ssis go ahead and post it
Go on :)
8:44 AM
@AlecTeal keep these things to you.
@AlecTeal ahhhhhhhhhhhhh $$\lim_{n\to\infty}\sum_{k=0}^{n} \frac{1}{\displaystyle \binom{n}{k}^{2013}}$$?
@AlecTeal nice, isn't it? You know what is the big problem with it?
/me can't read LaTeX properly yet.
What is the relationship between mathematical thinking and moral virtue?
@AlecTeal I'll go to shopping and plan to solve it mentally on my way ... (I'm a bit worried about my plans)
8:57 AM
@AlecTeal Just some imagination training, nothing special.
brb (a bit later)
@PSTikZ Nice question.
@skullpatrol :-)
@AlecTeal don't you have the LaTex support for chat?
9:10 AM
@AlecTeal Here you go pal :-)
9:22 AM
Yeah I don't know how to use that @skullpatrol
I'm not sure how a boomark can have so much power
Oh wait that was easy @skullpatrol
What's the difference, chat and mathjax?
9:50 AM
@AlecTeal Here are the details.
@PSTikZ I guess, if to view math as a form of disipline; you could connect it with virtue...
...what do you think?
@skullpatrol Yes. agree...
10:31 AM
@skullpatrol when you typed above "nice question" I thought for a second that you referred to my question. :-)
2 hours later…
12:25 PM
How is my answer?
12:38 PM
+1 :-)
@Alizter I can sense a lot of paint there
why not using TikZ =
the two circles for example don't have the same center
shall I show you what i mean ?
12:57 PM
hi guys
hi guy
@Alizter Something like that, but it is a vector graphic even though the upload is really bad for the quality
@Alizter here the quality isn't hurt that much
1 hour later…
2:11 PM
no good questions to solve .
2:32 PM
@MarianoSuárez-Alvarez could you give me a quick proof of this please?
By the way: I do not ignore your «hi»s on purpose or anything :-)
np pal :-)
It's just that one of my computers logs into this thing and I am never sitting in front of it :-)
what definition of $\frac ab$ are you using?
2:35 PM
So what exactly would showing that $\frac{a}{1}$ is $a$ entail?
by your definition, $\frac a1$ is $a\cdot\frac11$, and we want to see this is $a$.
one way to do this would be to show that $\frac11$ is $1$, for you already know that $x\cdot1=x$ for all $x$
can you show that $\frac11=1$?
I'm lost in the ones :(
When I asked what you meant by $\frac ab$ you answered $a\cdot\frac1b$, and that can only mean $a\cdot b^{-1}$.
(Otherwise the definition is circular, precisely in the case of $\frac11$)
Now $\frac 11$ is, according to that $1\cdot 1^{-1}$.
What is $1^{-1}$? Is the unique real number which when multuplied by $1$ gives you $1$.
Since $1$ has that property, we see that $1^{-1}=1$
so that $\frac11=1\cdot 1^{-1}=1\cdot1=1$.
@MarianoSuárez-Alvarez Thank you, using the -1 exponent really helped a lot :-)
2:50 PM
it is not an exponent in this context
I am using $(\mathord-)^{-1}$ to denote the multiplicative inverse
well, it sort of looks like it is in the exponent's place... no?
@MarianoSuárez-Alvarez what I meant was using a different symbol for the reciprocal really helped
22 mins ago, by skullpatrol
I'm lost in the ones :(
Thanks again.
3:25 PM
@MarianoSuárez-Alvarez Heya.
3:37 PM
Hi @TedShifrin how are you?
Heya @skull. Fine, thanks, and you? @Peter: You feeling better?
hi @PeterTamaroff @MarianoSuárez-Alvarez @skullpatrol @TedShifrin
Greetings, @what's ....
@TedShifrin Fine thanks :-)
Someone could come help me grade homeworks ...
3:39 PM
how are you ?
@TedShifrin Yeah. Doing good today. It is quite hot.
@TedShifrin ...for grad students?
Isn't it supposed to be winter there, @Peter? :) Here, sadly, it is still in the 90s (F) ...
Yeah, @skull
3:41 PM
@TedShifrin Heh, we're actually hopping into spring.
Hopping, eh? Rabbit-like ...
What's the hard math problem of the day?
@TedShifrin I am not looking at any right now. I was wondering if you can help Mika here. I don't want to tell her/him things that are not true.
Peter Halburt must delete his answer
$$ \lim_{n \to \infty} \sum_{k=0}^n \frac{1}{\dbinom{n}{k}^2} $$
3:51 PM
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.
good question :-) $$ (x - a)(x-b)(x-c) \cdots \cdots (x-z) = ????? $$
@what'sup Look up "Vieta's Formulas".
@PeterTamaroff focus
@what'sup Eh?
@what'sup That is an interesting one. I think I have an article where it is evaluated.
@PeterTamaroff how don't i know vieta's formulas ??
3:54 PM
@what'sup If you know them, why are you asking?
Ugh , $$ (x-a)(x-b)(x-c)....(x-x)(x-y)(x - z) = 0 $$ :-) @PeterTamaroff
@what'sup =)
@Peter, you saw my ugh above?
@TedShifrin Aw, yiss.
3:57 PM
@PeterTamaroff did you see user1892's answer to your question?
help him then =)
@TedShifrin With multiple Riemann integrals, we usually integrate over Jordan measurable sets, yes?
@skullpatrol Link?
@PeterTamaroff Here
@skullpatrol I don't know what a "parse tree" is, nevertheless I see no reason to use $\bar a$ instead of $a^{-1}$
@PeterTamaroff conceptually simpler
3:59 PM
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?) $0$.
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