« first day (2769 days earlier)      last day (2549 days later) » 

18:00
@nbro Well you need to define it as a measurable function $(X,\mathcal{A},\mathbb{P})\to(\Bbb R,\mathcal{B}(\Bbb R))$ where the first is a probability space and the second uses the Borel $\sigma$-algebra if you want to be precise, but those are concepts most students won't be familiar with when taking a first course in probability
Hello
please what happen if the intersection in this lemma is empty
$\varphi^{-1}([c-\varepsilone,c+\varepsilone])\cap S_{2\delta}=\emptyset $
what happen is this case ?
18:46
hello, someone here ?
2
Q: Sum involving binomial coefficients.

user326254 Prove that $${^{404}\mathrm C_4}-{^4\mathrm C_1}\cdot{^{303}\mathrm C_4}+{^4\mathrm C_2}\cdot{^{202}\mathrm C_4}-{^4\mathrm C_3}\cdot{^{101}\mathrm C_4} =(101)^4$$ I tried writing $101=102-1$, but couldn't move forward. Sorry for the inconvenience, I am new here and couldn't type it. '.' Me...

Can this be done using only algebra (not a combinatoral proof)
Sorry, if this is too basic a question. But I'm confused:
https://math.stackexchange.com/questions/2676401/can-we-think-of-k-as-a-linear-operator?noredirect=1&lq=1
19:11
@MoreAnonymous Just wait for an answer, Tony S.F. is being a bit rude so ignore him
@MoreAnonymous I don't really understand what is going on there. You have two specific matrices, $A_0$ and $A_1$ (though they depend on a choice of a small enough $s$ to make everything converge). Defining $\widehat{K}A_0 = A_1$ doesn't say anything about what $\widehat{K}$ should do to any other matrix, so it makes no sense to ask if it is linear.
Did
Did
20:03
@ÍgjøgnumMeg "so ignore him" Really an excellent piece of advice, to "ignore" mathematical objections made at one's own post, simply by inventing a convenient "rudeness".
@Did Perhaps I should've been more tactful with the wording, but I don't appreciate the un-pedagogical "tone" some people take when commenting on this website so my usual reaction is to take onboard the mathematical content and ignore the wording :)
Just my interpretation
Did
Did
20:27
@ÍgjøgnumMeg Sorry but you explicitely recommended to ignore the maths in TSF's comment, for (dubious) considerations of rudeness. Doing so, you aligned with a strong, but most unfortunate, tendency of the site. In the end, TSF did make a mathematical point, right? If the point is sound, it should be addressed, for the OP's own sake (if it is not, one may explain why it is not and pass to something else).
@Vrouvrou yep
hello @GrantGarrison
0
Q: Module Structure Induced by Quotient Ring

user193319 Proposition 7.7.15: Let $\frak{a}$ be a two-sided ideal in the ring $A$. Then an $A$-module $M$ admits an $A/\frak{a}$-module structure compatible with the given $A$-module structure if and only if $\frak{a} \subseteq \mbox{Ann}(M)$. Such an $A/\frak{a}$-module structure is unique if it exist...

20:45
@GrantGarrison do you know "deformation lemma"?
20:57
@Vrouvrou I can't math. LOL. Is this Calc BC?
Calculus
 
1 hour later…
22:26
How is $$\int_0 ^1 \left[x^{n-1}(1+x)^n dx\right]$$
= $$\int _1^2 x^n (x-1)^{n -1}dx$$
I want to know how they changed limits.
I'll be really grateful to any kind of help.
22:38
@Abcd: They set $u=x+1$, so $du=dx$ and $u$ goes from $0+1=1$ to $1+1=2$.
@TedShifrin Why didn't they change the variable x?
Change it to $u$ and then let $u=x$ when you're all done. The dummy variable doesn't much matter.
Okay, thanks.
what's up @TedShifrin
Heya @Antonios
22:40
so, a bit of a disaster scene on that midterm... (the algebra course)
well, I'm not 100% sure, but there are 40 students. The prof showed me the marks for the first 6 exams she had scored.
Guess what the highest mark /50 was
Hard for me to do anything intelligent, not having seen the test or her grading style. Shall I guess 20?
Oh f***.
22:42
sorry, 14%
i.e. 7/50.
LOL
So, uh. I'm not sure what to say. I've never seen this before.
Unless she "randomly" picked the worst students in the class, this had better be a serious wake-up call for her.
Do you have a copy of the exam to send me?
I thought the exam was fairly reasonable, to be honest.
Yeah, maybe it'd be good for you to tell me what you think.
I'd also like to grade one of the exams and see how far off my grading is from hers. With my almost 40 years of experience teaching US students, I mean. :)
Should be in the inbox.
Yeah, that would be interesting to know.
I want more concrete stuff. And the wording in #1 will flip people out, even though the problem is meant to be very easy.
Other than mentioning a cyclic group, there's not a concrete group on the test.
Blah.
Not a check that a concrete subgroup of some group is normal. Not a straightforward application of the Fundamental Homomorphism Theorem.
What moon does this person live on?
3
22:45
yeah I'm not sure. It seemed to reasonable to me, but the other prof I asked (my advisor) thought it was a bit abstract so to say.
Every question is reasonable, but it's more abstract than I would give ...
Nothing for a C student to grab onto.
Yeah, well, she's adamant about no curve, so there may be no C students.
She may have only F students like this.
Did I send you my exams?
(OF course, group theory was the second semester in my course.)
22:47
Actually about 16 iirc.
Don't think I saw any groups problems in there
No, no, you're thinking of the calculus/multivariable math.
I meant algebra.
yeah, I don't think I saw algebra ones.
OK, I'll send. OF course, I did rings and modular arithmetic stuff all first semester.
Did they do (repeated) homework problems at the level of those exam questions?
22:48
someone drew me in xkcd
Both of you, Leaky.
@TedShifrin One of the reasons I found the exam questions a little more reasonable than you might have is: two of the more nontrivial problems were from inclass notes, and the homework exercises were a lot harder than these exam problems.
@Antonios If you don't mind, could I have a look?
yeah that should be fine
where shall I send the exam?
(won't bother deleting that message)
(it's public anyway)
received
@TobiasKildetoft I have a sequence and a method to go from sequence $1$ to sequence $2$. It follows properties such as $hat K (S_1 + S_2) = \hat K S_1 + \hat K S_2$
hence it seems like a linear operator to me
would it be rude for me to say that I don't find it hard?
problem 4 has some resemblance to "hausdorff iff diagonal image is closed"
22:56
I didn't particularly either @LeakyNun, but it's more a question of your background vs. the background of the students who took it.
is there any relation? @TedShifrin
@LeakyNun I don't think I see any relation other than the fact that these are criteria using the diagonal mapping
X is hausdorff iff {(x,x) | x in X} subset X x X is closed in the product topology
G is abelian iff {(g,g) | g in G} subset G x G is normal in the group product
@Antonios: I'm having gmail issues trying to send you my exams. Hence my quiet absence.
Can you google drive it?
23:00
Every time I take a trip and log into my gmail account with my phone/iPad, Google goes nuts and blocks my accounts. Drives me insane.
sup chat
(e,h) (g,g) (e',h') = (n,n), oh... lol
hi Eric
23:01
how's it going
well for problem 3 it's even a normal subgroup...
Can someone explain how this inequality holds?
I was afk after I posted this ... Can anyone help?
https://math.stackexchange.com/questions/2676401/can-we-think-of-k-as-a-linear-operator?noredirect=1&lq=1
Actually I think I see it now
guys?
(sorry if its too basic)
23:17
any luck @TedShifrin? not a huge deal if its too much effort
I need to get this working, dammit. Gmail frustrates me so much.
So they've IP blocked you?
sorry if this is stupid question, but if we have $\frac { 10*{ log }_{ 7 }10 }{ log_{ 7 }abc...j } =10!$, can we simplify to $\frac { 10*10 }{ abc...j } =10!$
No ... I have two different email accounts. I can't send uga email through the uga (Microsoft) server, so I have a dummy gmail account for sending with my uga address. It used to work fine. But it's messed up.
No, @DarkRunner.
@Antonios am I not supposed to share the exam paper?
I won't send it to anyone if you say so
23:20
best not to @LeakyNun
ok then
I don't think it ought to matter too much, but I didn't write it, so I can't authorize that
@TedShifrin oh
To get rid of log base 7, @DarkRunner, you must take 7 to the expression you have.
@TedShifrin can you help me out with this?math.stackexchange.com/questions/2676401/…
23:24
@TedShifrin like "let y represent log base 7"?
No, @DarkRunner. $7^{\log_7 x} = x$.
But you'll end up with $7^{10}$ and $7^{10!}$, which are horrendous. So I have no idea what you're wanting to do.
that's great
thanks @TedShifrin
@MoreAnonymous: There's way too much there for me to read. I'm in the middle of fighting with Google right now.
Thanks anyways and best if luck ... But remember there's only one u and a google of them :P
And a googleplex of you ? :)
23:29
sigh .. and I thought I was punny :P
Perhaps you were :)
Anyone here familiar with Fourier analysis?
Anyone here able to help me with this problem: imgur.com/jvL4Ilt ? I was able to do (a) - (c), but I'm struggling on (d) and (e)
it is about functional/fourier analysis
23:36
Hi @Ted
Hi demonic @Alessandro
Hi @Semiclassic
@Antonios: I think it sent.
I'm thinking back to our conversation yesterday
ah ha!
yep.
i'll look in just a moment
Yes, Semiclassic? Was I completely unhelpful, as usual?
How exactly did we justify that one could choose the vectors of the 2-planes so that two of them would coincide?
Geometrically, I can see the idea---two generic 2-planes have a common vector
23:39
Well, you needn't choose them that way, but two planes through the origin always have a line in common (unless the planes are identical).
(Think the cross product of the normals, again.)
Sure, but the vectors are the starting point, not the 2-planes they generate.
I know. But I was suggesting analyzing by fixing the two 2-planes and varying the vectors, then varying the two $2$-planes. Of course, you can always rotate to make one of the $2$-planes the $xy$-plane, for example.
I mean, it's $U^T V=(a,b)^T (c,d)$
Hmm.
Yeah, I've been doing that
parts d and e above
Spamming doesn't help, @Drew. This looks technical enough you should be asking your professor for help.
23:42
e.g. $U=\begin{pmatrix} 1 & \cos\alpha \\ 0 & \sin\alpha \\ 0 & 0\end{pmatrix}$
Right, @Semiclassic. I still think the relative positions of the 2-planes gives some bounds and then varying bases allows a different variation.
My main concern about 'fixing' the two 2-planes and varying the vectors is that that amounts to replacing $U\mapsto UM$ where $M$ is 2-by-2
Yes, indeed, but not orthogonal.
But then it's not clear to me what $U^T V$ will have to do with $(UM)^T V$
i mean, obviously that's just $M^T (U^T V)$
Yeah, so we need to understand what is true of $M$.
23:45
but I mean that's it's not clear then that $(UM)^T V$ is actually in the image
I honestly don't see that your preoccupation with matrix formulas helps with this, but so be it.
Well, I mean
Huh?
You're confusing image of a fixed linear map with image of your entire mapping?
We're only allowing specific types of $M$'s here.
If I have $\frac { 1 }{ 10! } (mod\quad 7)$, will I get undefined?
my issue is that, given $f(U,V)=U^T V$
23:47
Yes, @DarkRunner, because you have a multiple of 7 in the denominator.
I don't see how I know that $M^T U^T V$ is itself going to be of the form $f(U',V')$ with $U',V'$ having unit columns.
@Semiclassic: But $UM=U'$ is a different $U$. I don't get it.
@TedShifrin is there any way I could avoid this from happening? Any theorem, or...?
It's because I'm very close to solving a problem and now I have this
Does $UM$ have unit columns? If not, then I"m not sure why it's at all relevant.
LOL, no, @DarkRunner, unless you cancel out the 7 with a 7 in the numerator somewhere.
23:49
MAYBE THERE IS HOPE
@Semiclassic: You ignored what I said ages ago. This is why I don't find this fruitful. But you do need to figure out what condition on $M$ will result in unit columns. I said that ages ago.
I guess $M = \begin{bmatrix} \cos\alpha & \cos\beta \\ \sin\alpha & \sin\beta\end{bmatrix}$ for $\alpha\ne\beta$.
No, that's not right.
@TedShifrin So there is no way to simplify $\frac { 10^{ 10 } }{ 10! } (mod\quad 7)$?
@DarkRunner: That won't be an integer, so what do you mean by mod 7?
It's not at all obvious to me that there's going to be a nice characterization of such matrices $M$.
And absent that, I don't see a good justification for why one can assume that the two 2-planes have a common unit vector WLOG
It's the product of matrices like I said — well, one with the inverse of the other.
23:54
I don't understand the parenthetical.
@Semiclassic: Of course the two planes have a common unit vector.
@TedShifrin I've been trying to solve
Bleh, yeah.
The bases needn't have a vector in common.
for the last hour
23:54
Right. That's my issue.
I can see why that case is very nice, but I don't see how it's generic.
I conceded that point in my original remarks yesterday, @Semiclassic. But I was suggesting studying how the function changes when you change the basis vectors. I guess that amounts to using $M=X^{-1}Y$ where $X,Y$ are as above.
I never said it was generic.
I suggested using it as a jumping-off spot and then seeing how the function varies when we vary the bases, which amounts to using what you just made me write down.
That is sort of interesting, though. I wonder if one can characterize maps that change one unit basis to another.
@DarkRunner: This has nothing to do with what you were writing.
One point of interest with that special case:
What is $10^{10}\pmod 7$?
23:58
it's equiv to 10^4 (mod 7), which is 4
Why is it that?
Why not say it's $3^{10} \pmod 7$?
since 10^6 is congruent to 1, by ruler's totient theorem
Ok, sure
If $a=c$, then $U^T V=(a,b)^T (c,d)=\begin{pmatrix} 1 & a\cdot d\\ a\cdot b & b\cdot d\end{pmatrix}$
OK, now how is $10^{10^2}$ related to $10^{10}$?
@TedShifrin that first test is really good.
23:59
First semester or groups, @Antonios?
@TedShifrin simply 10^10 squared, so the remainder doesn't change; No, my mistake, my mistake, it's not the same
In which case the three nontrivial matrix elements are exactly the same as the three you'd get in the 3-by-3 $U^T U$ case I had originally

« first day (2769 days earlier)      last day (2549 days later) »