« first day (2050 days earlier)      last day (2978 days later) » 
00:00 - 22:0022:00 - 00:00

12:12 AM
@Ted: Of course, it depends who you are and what you're doing. Some of these people who are truly exceptional at finding a differential on the third page of a spectral sequence don't always think geometrically.
At the level of Hatcher I firmly agree.
Someone patient should write a careful answer for this guy that doesn't just say "read Hatcher", which his question runs the risk of.
 
@MikeMiller I'm not sure I could think of a terribly more useful answer than that.
Maybe take a basic algebraic topology course
is a better answer than throwing a textbook at someone.
 
One can still give a description of what's going on and why one would develop the machine of the fundamental group, in the same sense that someone asking a naive question about irrotational and conservative vector fields being the same deserves some honest discussion about topology without telling them to read a book.
On the other hand I'm not going to write it.
 
0
Q: If $f(z)$ analytic on $E(L)$, and $\lim_{z \to \infty}f(z)=A$, prove that $\frac{1}{2\pi i} \int_{L}\frac{f(\zeta)}{\zeta-z}dz = A$ or $-f(z)+A$

Jessy CatI am working on the following problem: Let $L$ be a closed, rectifiable simple curve, traversed counterclockwise. Let $f(z)$ be a differentiable function on a domain $G$ where $L \cup E(L)$ ($E(L)$ is the exterior of $L$, $I(L)$ is the interior of $L$) is contained in $G$. (That is, $f...

I needs some help please.
 
0
Q: Free Strict Monoidal Category is Noetherian?

Julian RachmanLet us define the free strict monoidal category $\Sigma(\textbf{Pos})$ on the category of posets $\textbf{Pos}$. How may we prove that $\Sigma(\textbf{Pos})$ is a noetherian category if we know that $\textbf{Pos}$ is noetherian?

 
Mine is more interesting ;P
 
12:20 AM
@MikeMiller If someone is asking a question like this,I don't think they likely need any motivation for why one would develop the machinery of the fundamental group.
 
r9m
@robjohn ever seen anything bizzare like $\displaystyle \prod\limits_{n=1}^{\infty} \cos^2 \left(\frac{1}{6^n}\right)$? :o
 
what is spectral flow symmetry?
can someone tell me what it is? I can't find in on wikipedia
 
@MikeM: My comments were directed at Karim, not at world experts. :)
 
12:51 AM
The pi day question is nice.
What are the possibilities for the number components of a link of round unknots of the same diameter with all 2 component sublinks equivalent and a pair of 3-component sublinks non-equivalent?
 
1:35 AM
So is no one able to answer this question? math.stackexchange.com/questions/1697861/…
 
I don't know why you think posting it thrice in two hours will get more responses. Surely anyone who saw one of those saw the others.
 
Did anyone take a look at my question? It's only had 7 views :(
And who says thrice?
I ate chicken & thrice soup for dinner tonight ;)
 
1:52 AM
XD
So my background in complex is not so good but it sure does look like you're doing some sort of "inverted" Cauchy's Integral Formula
 
It's called Cauchy's Integral Formula for Unbounded Domains.
 
sensible :P
 
Who does know a lot about complex?
I've gotta say, I've never been dissatisfied with anything from @robjohn
So, if you're listening @robjohn, please take a look! You're amazing! :)
 
I would like an explanation for the downvote.
 
good luck
 
2:16 AM
Any explanation at all?
 
the chance that the downvoter is in chat is minimal, and the chance that even then they would explain themselves is also not good
so no probably not
 
 
2 hours later…
4:01 AM
I'm in calculus and I keep getting hung up on remembering to apply formulas, especially when the math lets me go down one path until it becomes a dead end. Students generally take 1/2 - 1 hour per problem whereas it takes me up to 3 hours.
An example of what I'm talking about would be integrating $x^3/6 + x/2$. I didn't have the wisdom to magically auto-recognize that I needed to combine the terms there; I thought that would make a more difficult problem (of probably partial fraction decomposition), but I was wrong. I later got stuck trying to factor a nasty polynomial. Is there any specifi
 
Does $$\frac{||W\vec x||}{||\vec x||} = \frac{||W^TW\vec x||}{||W\vec x||}$$?
 
That's really hard to read on my monitor
 
@JoshuaLamusga: Not really, you just have to force yourself to try something else after a half hour or so
 
If you take the rectangular prism whose diagonal is $\vec x$ and transform it by $W$, you increase the area by a factor of $\det(W)$.
 
I'm also really bad at this (at a higher level, admittedly), so generally I have to work on a different problem
to get my brain unstuck
 
4:06 AM
That's better. I didn't know you could change the size of latex stuff. That said, I wouldn't know the answer to that question.

Well, it was worth a shot, Eric. It wouldn't be a problem if I could factor well and my tests weren't timed.
I just had one of those fantastic moments where I studied really well for 2 weeks only to score 53.5% on a test.
 
Ha finally!!! Solved the integral. Thanks @MikeMiller
 
I am unsure what thanks I deserve :)
 
Assuming $$f(x)\doteq\begin{cases}
1 &\text{if}\;x\geq 1\\
2 &\text{if}\;x\geq 2
\end{cases}$$
Is it well defined that $f(2)=1$ ?
 
4:23 AM
The point of mathematical notation is clarity. While you could make an argument that your expression isn't self-contradictory, I could make a better argument that your expression is dumb.
 
It's obviously dumb. It's completely synthetic.
More precisely what I'm after is whether or not the conditions of a piecewise function are to be evaluated strictly in their listed order, or if that's not well defined.
 
There is no arbiter of such things.
 
 
1 hour later…
5:41 AM
i flagged this question earlier today but i guess it's okay to be asking the pizza hut problems at this hour since the pizza hut page has thousands of answers already? math.stackexchange.com/questions/1697186/…
although it's not clear if anyone got the third one right since it asks for all of the possible numbers of rings
obviously 3 rings works but i haven't figured out any other possibilities yet
oh wait there is an update on the pizza hut page! OPTION C: YET TO BE SOLVED, No one has gotten this one exactly right yet! Hint: It helps to show your work!
 
You can probably assume from the fact that nobody has guessed it that it's not less than a hundred.
 
A and B are solved already though
(and the MSE question was about B)
 
Yes, I'm referring to C.
 
i assume it requires a list of some numbers. maybe they are all less than 100, if so i have a 2^-100 chance of free pizza by guessing
well no fair talking about it here either i guess
but i answered my own question when i checked the blog, B) is solved so the MSE question should stay open
i actually woke up this morning and started working on A, 10 minutes later i was still debugging my program which had 32 bit integers in it. then i reloaded and realized there were hundreds of correct answers for it already.
 
6:31 AM
Is 0 a real and imaginary number?
$0i = 0$
 
Hello
I am brand new here and was wondering if someone could help answering a question about conformal mappings in complex analysis :)
 
GGG
6:46 AM
Any tips on calculating $5^{2009}\mod{31}$ and $5^{2009^{1492}} \mod{503}$
Done the first one! The second one looks a lot trickier.
 
7:04 AM
@GGG: You only need to know $2009^{1492}$ mod 503. Figure out why, then use that.
 
@MikeMiller doesn't @GGG need to know $2009^{1492} \mod 502$, instead of $\mod 503$?
 
GGG
Lol! What sorcery is this? Is it because $503$ is a prime number?
 
Not really (I think).
I just have computational evidence for my point.
In particular, I know that $5^{502} \equiv 1 \pmod {503}$. Can you see how that helps to greatly reduce the computation required?
 
GGG
7:23 AM
Yeah, how did you get that?
 
I just made a quick Python program to compute $5^1 \mod {503}$, $5^2 \mod {503}$, etc., and then it found that $5^{502} \pmod {503} = 1$. So that's why I say I don't know why it is, I just know that it is.
Do you see how to compute your second number now?
 
GGG
@feralin No, in fact. I'm still puzzled about it.
@feralin with what you have given I can calculate the first one (the one I had already done) in like one step.
@MikeMiller I can't figure it out. Please reveal!
 
7:51 AM
Why hello there / Good morning
 
@GGG well we know that $5^{502} \equiv 1 \pmod {503}$, so then $5^n \equiv 5^{n \mod {502}} \pmod {503}$. Then if we want to find the value $5^{2009^{1492}} \mod {503}$ all we need to to is find $2009^{1492} \mod {502}$
So what is that value?
It turns out that $2009^2 \equiv 1 \pmod {502}$.
So then $2009^{1492} \equiv 2009^{1492 \mod 2} \equiv 2009^0 \equiv 1 \pmod {502}$, so then $2009^{1492} \equiv 1 \pmod {502}$.
Therefore $5^{2009^{1492}} \equiv 5^{2009^{1492} \mod {502}} \equiv 5^1 \equiv 5 \pmod {503}$.
 
GGG
@feralin That's neat! I would never think of that!
Thanks.
 
@GGG no problem! Glad to help :)
 
 
4 hours later…
11:41 AM
Hey @MikeMiller!
 
11:51 AM
@JC574 you know some topology right ?
just want to make sure my understanding is correct for something
 
i'll give it a shot
if it's easy topology i might be able to help
go ahead :)
@L33ter
i'm going to eat soon so ask quick :P
 
12:09 PM
alright
@JC574 when we prove the snake lemma we will have a map between $H_n(X,A)$ into $H_{n - 1}(A)$
 
ya
 
where $H_n(X,A)$ is the relative homology
how come it is enough to show that it preserve kernels and not boundary.
The reason for this is the following I think.
The elements of the boundary of $H_n(X,A)$ is of the form $\partial(\Beta) + \gamma$ where $\beta \in C_{n + 1}(X)$ and $\gamma \in C_n(A)$
So, if it preserve elements of kernel it will directly preserve elements of the boundary
right?
 
alright
 
1:14 PM
I get it now after thinking about it the reason isn't as I stated above
but because of the fact that if $\alpha$ is a cycle and we send it somewhere else, which is a cycle and define the map as the equivalence class, so that map would make sense.
 
1:36 PM
@L33ter There ought to be a simpler reason. The snake lemma is completely unrelated to topology and is basically just a statement about abelian categories
 
Hey Tobias
 
@JC574 Hey
 
I'm trying to prove something using some homological algebra
not sure whether it's going to work, or whether what i'm doing is pointless and the statement in alg geometry follows obviously from something
can I get your opinion?
 
Hey @MikeMiller
 
@JC574 Sure
 
1:39 PM
Let $X$ be a complex algebraic curve
hang on my chattex aint working
OK
the end game here is to show $ H_{\text{dR}}^i(X) = H^i(X, \mathcal{O}_X) \oplus H^i(X, \Omega^1_X)$
 
Trying to prove that $M=\Bbb CP^2\#\Bbb CP^2$ is not the boundary of an orientable 5-manifold, @MikeMiller. So I computed the signature of $M$ but unfortunately I got that it's zero (is that correct?). Tried playing around with Euler-characteristic then but didn't get me anywhere, and now I'm out of ideas.
 
so take a Cartan-Eilenberg resolution $I^{\bullet,\bullet}$ of the complex $0 \to \mathcal{O}_X \to \Omega_X^1 \to 0$
and eventually we'll consider $H^i \text{Tot}_\oplus(\Gamma(X,I^{\bullet,\bullet}))$
it seems this total complex will pretty much be a direct sum of injective resolutions of the two sheaves in the statement we're proving
 
@JC574 I don't know what $\Omega_X$ is
 
sorry um
$\Omega^1_{X/\mathbb{C}} $ the sheaf of germs of $1$-forms?
I'm pretty new to this stuff
but I think you can define it in several ways
 
This seems way too geometric for me
 
1:48 PM
too geometric for you? or too geometric a proof?
 
i haven't been using any of the geometry yet though, good point
seems like it would be necessary
but basically my question is -
the complex we start with has just two nonzero terms
and the map between those is injective
or a monomorphism or whatever
is there anything we can say about the horizontal maps in the Cartan-Eilenberg resolution of that complex from this info?
 
cartan eilenberg
sorry
E
 
not familiar with that resolution
 
1:55 PM
basically it's a double complex such that the columns are injective resolutions of each term of the complex we start with
i want the horizontal maps to be injective or something
 
@JC574 Hmm, for the case of such a short sequence it seems to be clear, as you have each of them "surrounded" by three injective maps
Hmm, not quite actually
something not in the image of the first map might be in the kernel of a horizontal map
 
What kind of meaning does "$x_1, x_2$ are strict relative minima of $f$" have? Google is not helpful, and (because of the word strict) this appears to be stronger than simply $x_1, x_2$ being local minima of $f$.
 
yeah @TobiasKildetoft hmm
the columns are exact when we adjoin it to the original complex though
does that help
(that's how resolutions work right?)
@s.harp maybe it means $<$ rather than $\le$ ?
hmmm
 
@iwriteonbananas: No, it is not zero.
 
no @TobiasKildetoft that doesn't seem to help with what you're saying
Hey @MikeMiller :D
 
2:10 PM
@JC574 that part is definitely true, and it explains the word strict. Relative is still a bit vague though, from the text it looks like I am looking at simply $2$ strict local minma
 
 
how are you?
 
hi
is the cross product only defined for two $3$D vectors?
 
@MikeMiller any idea on my problem?
I guess if we tried to compute hypercohomology with a suitable Cech resolution it'd come out in the computation
but that's not nice
 
2:29 PM
@MikeMiller Oh yeah, my computation was wrong...the signature is 2.
So $\Bbb CP^2\# \Bbb CP^2$ is not the boundary of an oriented 5-fold. How about non-oriented?
Btw. dumb question, but what's an example of an orientable manifold w/ odd Euler-characteristic?
 
@JC574: I don't
@iwriteonbananas: You can figure out both of those questions. as inspiration for the first, think about how you bound CP2 # CP2bar
 
ah i think i have an alternate way
at least to get some kind of decomposition
 
What's CP2bar?
 
to what extent is an algebraic curve a kahler manifold?
over complex numbers
 
does erf(1) have a nice closed for approximation?
as opposed to 0.842700792...
 
2:43 PM
yeah sorted out my proof - didn't need all that homological algebra, I forgot $\mathcal{O}_X = \Omega^0_X$
 
3:02 PM
@Lembik $\operatorname{erf}(1)$ is the nicest closed form...
 
@robjohn thanks
 
@Lembik not much help, I know. Otherwise, there is an integral, but that is probably where you started.
 
@robjohn I really want to find an easy way to prove an upper bound
I should have just asked that :)
 
@Lembik you can look at this answer. That gives some methods of computation that might help.
 
@robjohn ok thanks
 
3:08 PM
$$\begin{align} \mathrm{erf}(1) &=\frac2{\sqrt\pi}\int_0^1e^{-t^2}\,\mathrm{d}t\\ &=\sqrt{\frac2\pi}\int_0^{\sqrt2}e^{-t^2/2}\,\mathrm{d}t \end{align}$$
 
Is there a nice intuitive explanation of the Lie derivative $\mathcal{L}_X$?
 
You're taking the directional derivative in the direction of the vector field.
You can also think of it as saying that you're taking the derivative on each of the flowlines.
 
r9m
@RandomVariable you there?
 
@MikeMiller I'm really struggling with all these calculation rules (doing Symplectic Geometry), is there some big picture that I am missing?
 
I don't understand the question.
 
3:22 PM
@r9m Yes.
 
Well, for example
Wait, let me just write out the question that I am struggling with
 
I'm guessing it's a Moser trick type question?
 
$M$ manifold, $\alpha$ 2-form on $M$ and $X, Y, Z$ vector fields: Is the following always true: $$\mathcal{L}_X(\alpha(Y,Z)) = (\mathcal{L}_X \alpha)(Y,Z) + \alpha( \mathcal{L}_X Y, Z) + \alpha(Y, \mathcal{L}_X Z)$$
 
r9m
@RandomVariable I was looking at the second integral Cody put here .. :O I opened the thing inside the integral into a series but I am having a hard time justifying interchange of sum and integral .. maybe you can help?
 
So I know things like $\mathcal{L}_X(Y) = [X,Y]$ and so on
But I feel like I'm missing most of the "intuition" and am just doing some juggling with symbols
 
3:26 PM
I don't remember, that's true up to signs or something.
Using the Leibniz rule in introductory calculus is symbol juggling; same with the chain rule. Why should this be any different?
 
Because I get in a mess and fail ¯\_(ツ)_/¯
 
Well, don't.
 
Ah, thanks.
 
@r9m I don't know if I can be of any help. I usually ask Daniel Fischer when I can't justify changing the order of limiting operations.
 
Sure.
 
r9m
3:31 PM
@RandomVariable 'kay .. I'll scratch my head some more .. but I was just curious, is there a way to attack the problem with residue theorems?
 
@Krijn: In practice you don't need to memorize whatever hundred rules you're deriving right now. The most important one is Cartan's Magic formula which is easy to remember. The rest, meh, if you need them again find them.
 
@r9m Probably not.
 
@MikeMiller $\mathcal{L}_X \alpha = i_X d\alpha + d(i_X \alpha)$, right?
 
r9m
@RandomVariable 'kay ,, thanks!
 
Yup.
I guess that's effectively what you just said but eh
 
3:38 PM
Yeah but I just can't make sense of $\mathcal{L}_X \alpha(Y,Z)$
 
morning
 
My course on differential topology has not prepared me for this :@
 
i'll say, whenever i see those kinds of 'symbol manipulation' issues, it makes me me sympathetic to things like Penrose notation
especially since, when one does physics calculations in that vein, things tend to be represented in index notation and therefore chock-full of dummy indices
 
r9m
@RandomVariable Also can you help me with evaluating $\displaystyle \int_0^{\pi/2} \frac{\sin^2 ax}{\sin^2 x}\,dx$? :-) ($a$ is a positive real number)
has this^ one been discussed in I&S before?
 
e.g. leibniz rule for covariant derivatives in penrose notation is just "circling multiple elements = circling single elements in parallel"
 
3:44 PM
@r9m I'm almost positive that sos440 did that one somewhere.
 
r9m
@RandomVariable oh! In I&S? or is it there in his blog? .. need to look!! Thanks man!! :D
 
@r9m If I recall correctly, it's in terms of the digamma function.
 
@Semiclassical: Nothing makes me sympathetic to that notation.
 
rolls eyes
 
r9m
@RandomVariable I wouldn't be surprised if it was .. thanks!
 
3:50 PM
@Krijn: $\alpha$ is a 1-form, and presumably you know the invariant formula for d of a 1-form
 
@MikeMiller It's a 2-form
 
Fine
 
I guess so
$d\alpha (X,Y,Z) = \mathcal{L}_X(\alpha(,Y,Z)) - \mathcal{L}_Y(\alpha(X,Z)) + \mathcal{L}_Z(\alpha(X,Y))$ and then some terms
Let's see
$-\alpha [X,Y], Z) + \alpha( [X,Z] , Y) - \alpha( [Y,Z],X)$
 
@r9m I am writing this up...
 
r9m
@robjohn okay!! lemme know when you do! .. I will post it in the main so you can answer there sir! :D Thanks!!
 
4:01 PM
Of couese you can write those Lie brackets as Lie derivatives.
 
Yeah, I feel like I'm almost done
The only obstacle left is $( di_X \alpha ) (Y,Z)$
But let me try the invariant formula on that one.
Oh, wauw, thank you @MikeMiller
I barely understand the symbols, though, but it worked!
 
hi
i have a question
is the cross product defined only for 3D vectors
 
4:17 PM
Mathematics is best done by the method of reasoning. When that fails one has to resort to the method of repeating and iterating. And when that fails one has to resort to the method of prolonged staring. And when all else fails one has to resort to the method of shouting.
 
4:31 PM
@r9m I found it. But you need to first integrate by parts. sos440.tistory.com/207
It's solution 3.
 
r9m
@RandomVariable Cool!! Thanks! :D
 
4:51 PM
@MikeMiller Could you help me with another question? I think I'm oversimplifying things and need a sanity check
For which $k$ is this true: Let $x_1, \ldots, x_k$ and $y_1, \ldots, y_k$ be any pair of ordered sets consisting of $k$ distinct points of $M$. There exists a symplectomorphism $\phi$ of $(M, \omega)$ such that $\phi(x_i) = y_i$ for all $i$
I claim that this is not true when $k>1$ because, take $y_1 = x_2$ and $y_2 = x_1$ then we would have $\omega(x_1, x_2) = \omega( \phi(x_1), \phi(x_2) ) = \omega(x_2, x_1)$ but $\omega$ is anti-symmetric
Ahhhh nevermind
 
It's true for all $k$. Goes to show the flexibility of symplectic geometry, and why some people prefer to say symplectic topology.
 
Wait what?!
How can that possibly be true for all $k$
 
Do you believe it if it said diffeomorphisms instead?
 
Not really, no
 
Oh, then that's more of a personal problem.
 
4:59 PM
Should it be that trivial?
 
Well, in the diffeomorphism case, you need to assume the manifold is dimension greater than one, and you always need to assume connected.
 
Ah yes, that was part of the question
Connectedness
 
But it's a standard exercise. Do you believe it for $k=1$ and "compactly supported diffeomorphism"?
 
Yeah
Okay I see why it would be true for diffeomorphism actually
 
Then you believe it for all $k$.
 
5:03 PM
Yes
 
Indeed, you only need to prove it for $k=1$, compactly supported diffeomorphism, $\Bbb R^n$.
 
It's just twisting and bending your manifold in some sense
 
It's an entirely local question. So the symplectic statement is just that there are an abundance of symplectomorphisms.
 
I'm baffled.
But what I don't understand is
We can take $x_1$ and $x_2$ such that $\omega(x_1, x_2) \neq 0$
But if we then pick $y_1 = x_2$ and $y_2 = x_1$ we would get that $\omega(x_1, x_2) = \omega(x_2, x_1)$
So what am I doing wrong here
 
What you just said makes literally no sense.
How often do you plug points into 2-forms?
 
5:15 PM
...
I was thinking of a symplectic form, shit
Like, on a vector space
That was silly
 
 
1 hour later…
6:31 PM
Hello!!!
Could I ask someone how the Lebesgue theorem is applied at an integral?
@robjohn @DanielFischer Are you familiar with this theorem?
Why does the following hold?
 
we have a be nice policy now
oh my
 
$\int_{\mathbb{R}^n} \phi(x, \epsilon) \frac{1}{(4 \pi \epsilon)^{\frac{n}{2}}} e^{-\frac{|x|^2}{4 \epsilon}} dx \to \phi(0, \epsilon) \frac{1}{\pi^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-|y|^2} dy$ @robjohn
 
@Evinda dominated convergence theorem?
 
@ForeverMozart Could you explain to me how to apply it? Because I haven't got taught it...
 
hello
 
6:46 PM
hi.. has anyone got any ideas for math.stackexchange.com/questions/1699050/… ?
I feel it must be standard for someone who knows
the question is to find an upper bound for $\sum_{x\in\mathbb{Z}^n} e^{-x^Tx}$
 
in the case of $n=1$, it can be computed in terms of the 3rd Jacobi theta function as $\vartheta_3(0,e^{-1})$
which i'm sure is very helpful. (not)
 
I don't know, is a an expression like $\vartheta_3(0,e^{-1})$ more complicated than $\pi$?
 
actually, it looks for $n>1$ like the expression factorizes i.e. $\sum_{x\in\mathbb{Z}^n} e^{-x^Tx}=\left(\sum_{k\in\mathbb{Z}} e^{-k^2}\right)^n$
so all one should need is an upper bound on the $n=1$ case.
 
 
1 hour later…
8:02 PM
For a manifold $M$ and a function $f \in C^\infty(M)$ we get a vector field $X_f$.
How does this work exactly?
 
hi
anyone got any ideas about math.stackexchange.com/questions/1699050/… ?
 
8:22 PM
@TobiasKildetoft still here ?
 
@Krijn: It doesn't. Look again at your assumptions - it's not just a manifold.
 
Symplectic manifold*
 
Better. Do you know how to define the gradient on a smooth Riemannian manifold?
 
Roughly.
If I ever pass this exam, I'm gonna throw a party
 
I'm not asking for roughly. How do you do it?
 
8:34 PM
Pff, let's see
 
At least sketch the construction for me
 
My Riemannian Geometry is even worse than my Symplectic Geometry
$\frac{\partial f}{\partial x_k} \frac{\partial}{\partial x_i}$
 
We're on a manifold. That doesn't make any sense.
 
No, wait, I'm missing something
 
I don't want to see coordinates. I do not like or understand them.
 
8:37 PM
$g( \nabla f, X) = \partial_X f$
 
How would you change that if instead of a metric, you had a symplectic form?
 
Very naïvely, change $g$ to $\omega$?
 
Yes.
You have a nondegenerate bilinear form on TM in both cases. In one case it's stmmetric and in the other it's anti-symmetric.
But given a nondegenerate bilinear form, you can do exactly what you just did and change 1-forms to vector fields and vice versa.
 
So why is it $\omega(\nabla f, X)$ and not $\omega(X, \nabla f)$
 
After all, that's precisely what nondegeneracy means: that it defines an isomorphism $V \to V^*$, given by $v \mapsto g(v,\cdot)$.
 
8:41 PM
Ahhhh
 
Who cares? They differ by a sign.
 
It slowly starts to make sense
Very slowly :/
 
matters in Hamiltonian mechanics if you want to get a flow that makes physical sense :)
e.g. positive momentum shouldn't lead to things moving to the left
 
user189740
Given $f(4n) = [n]$ for all $4n \in 4\mathbb{Z}$

What do you think $n$ stands for exactly?
 
8:57 PM
hey @Semiclassical are you familiar with elliptic curves?
 
somewhat. though not with any degree of rigour.
 
maybe you can help me with something super simple
why here is the y intercept given by $v = y_1 - \lambda*x_1$
@Semiclassical ?
 
ah, this kind of thing. yeah, i don't know this kind've stuff off the top of my head.
though, doesn't that amount to $y_1=\lambda x_1+v$?
in which case $v$ is certainly the $y$-intercept of the line $y=\lambda x+v$
 
I see
so we can also equally choose also $y_2$
and $x_2$ right ?
 
sure. for that matter, you can just swap the labels on $P_1$ and $P_2$.
 
9:04 PM
I see
yeah
 
Hello!!
We have that $P$ is a $p$-Sylow subgroup of $G$ and that $P\leq N_G(P)$.
Is $P$ also a $p$-Sylow subgroup of $N_G(P)$ ?
 
9:18 PM
Hi: I have this: How many numbers less than 600 can be formed using only digits between 3 and 7?
I thought of it would be 2*3*3 = 18 choices, but my professor marked it as wrong, saying it would be 105.
How/why?
 
9:32 PM
Hello @robjohn
I have a question... I want to show that :

If $u \in $\mathcal{D}'(\mathbb{R}^n)$ and $\rho \in C_{C}^{\infty}(\mathbb{R}^n)$, then $(\rho \ast u)(x)=\langle u(y), \rho(x-y) \rangle \in C^{\infty}(\mathbb{R}^n)$.

So far I have thought the following:

$\langle \rho \ast u, \phi \rangle= \langle u(y), \langle \rho(x), \phi(x+y) \rangle \rangle$.

Is it right so far? How could we continue? @robjohn
 
@Link Why would you say 2*3*3?
 
Link, $105=7*5*3$: we must choose one of the $3$ numbers between 3 and 5 for the hundreds place. The $5$ then looks like choosing one of the 5 numbers between 3 and 7 for the tens place. But I'm not sure where the $7$ comes from...
 
I would have said 3*5*5 though :/
 
^
Mary, yes by Lagrange: a $p$-Sylow group is a group of maximum order $p^k$, and since $N_G(P)\leq G$, we know that the prime factorization of $N_G(P)$ has no more than $k$ factors of $p$.
 
r9m
9:59 PM
@RandomVariable correct me if I am wrong, for this we need to consider $f(z) = \frac{\exp \left(\frac{i}{z\pm ia}\right)}{z}$ and the contribution from the indentations at the essential singularities $z = \pm ia$ are $2\pi i e^{\pm 1/a}$ right?
 
00:00 - 22:0022:00 - 00:00

« first day (2050 days earlier)      last day (2978 days later) »