« first day (2761 days earlier)      last day (2263 days later) » 

6:00 PM
Does this not suffice
 
Main thing of interest for me is that the determinant in that case is $$\det M = 1+2\cos \theta_{ab}\cos \theta_{bc}\cos \theta_{ac}-\cos^2 \theta_{ab}-\cos^2 \theta_{bc}-\cos^2 \theta_{ac}$$
 
But in that case $\det M$ is the square of the volume.
 
if you let $u,v,w$ be these cosines then that's $\det M = f(u,v,w)=1+2uvw-u^2-v^2-w^2$
right
 
@TedShifrin Good Morning :)
 
oh, I forgot something important
 
6:02 PM
Good day :)
 
I'm assuming that $a,b,c$ are unit vectors
So $M$ in this case will be symmetric and have ones on the diagonal
 
@TedShifrin Are you busy atm ?
 
Right, I knew that, @Semiclassic.
 
Not too busy, @Tanuj.
 
6:02 PM
main thing is that the set $f(u,v,w)=0$, corresponding to when the volume vanishes i.e. the unit vectors become linearly dependent
 
@TedShifrin If you get time have a look here math.stackexchange.com/questions/2664316/…
It sadly failed to get the attention it deserved
 
but $f(u,v,w)=0$ is Cayley's cubic surface!
That's the business I had talked about a few months ago. The present problem is an outgrowth of that.
 
When you remove the unit vector thingy.
 
No, I actually still have unit vectors in the present problem. But now instead of having {a,b,c} and their pairwise dot products, I have {a,b}, {c,d} and the dot products between these sets
 
@TedShifrin Will I find you here mostly this time of the day ?
 
6:05 PM
So [a,b]^T [c,d] instead of [a,b,c]^T [a,b,c]
 
I'm unpredictable.
 
lol okay
 
I meant for the Cayley cubic, Semiclassic.
 
@Tanuj: First of all, how do you take derivatives of $x|x|^3$?
 
6:05 PM
You need that unit vector condition.
Oh, wait
Are we talking projective or affine?
 
@TedShifrin I don't know.I know derivative of |x| is x/|x|
 
Well, can you rewrite that without the absolute value?
 
@TedShifrin Ypu mean splitting it for when x<0 and x>0 ?
 
Right. Use a piecewise definition.
 
okay
 
6:07 PM
@TedShifrin If you want to have the projective surface, then I'll agree. But I'm happy with the affine surface.
 
Now I'm confused about vectors versus coordinates, but never mind.
 
for x<0 , it becomes -x^4 and for x>0 , it becomes x^4
 
user337082
Hello guys
 
user337082
0
Q: Find the cross-section when a pentachoron is cut along a hyperplane under the given conditions

user532368 Given a regular pentachoron ($ABCDE$), find the cross section when it is cut along a hyperplane which is equidistant from and parallel to line $AB$ and plane $CDE$. Here, a regular pentachoron means a $5-$ cell, i.e., a $4D$ figure with $5$ vertices A,B,C,D,E all at equal distances from o...

 
user337082
6:08 PM
Please check this
 
You want $\ge 0$, @Tanuj. OK, so what is $f'(x)$ for $x>0$, $x=0$, and $x<0$? Now can you compute $f''(0)$ by the definition of the derivative?
 
user337082
dudes, please
 
user337082
0
Q: Find the cross-section when a pentachoron is cut along a hyperplane under the given conditions

user532368 Given a regular pentachoron ($ABCDE$), find the cross section when it is cut along a hyperplane which is equidistant from and parallel to line $AB$ and plane $CDE$. Here, a regular pentachoron means a $5-$ cell, i.e., a $4D$ figure with $5$ vertices A,B,C,D,E all at equal distances from o...

 
user337082
anyone?
 
Spamming us is going to make us angry, @user532368.
 
user337082
6:10 PM
Oh okay
 
Especially when it's something technical and complicated.
 
@TedShifrin main thing is that the 1 which comes out of that determinant is from the ones on the diagonal
 
And we're already busy with other people.
 
user337082
Oh okay, sorry.
 
@Balarka Why is the straight line homotopy an isotopy?
 
6:10 PM
heya @MikeM
 
and that requires the unit vector condition since the diagonal elements are the squared lengths
anyways
 
guys, in a degenerate stable node, where does the solution go to for $t\to -\infty$? (I'll post an example)
 
@TedShifrin Do i compute f''(0) piecewise too ?
 
is there an (informal) justification for this (informal) rule with higher differentials that $\mathrm d(\mathrm dt) = 0$?
 
@TedShifrin, Am i right about this: conditionally convergent series has infinitely many positive terms and infinitely many negative terms
 
6:11 PM
You do $\lim\limits_{x\to 0} \dfrac{f'(x)-f'(0)}x$ using what you've computed already, @Tanuj.
@Leaky: If you mean exterior derivative, it's skew-symmetry of differential forms. If you mean an engineer's notion of "differential" it's not true.
 
@TedShifrin okay let me check
 
@ShaVuklia I don't think you can say more than "they start infinitely far away"
 
@TedShifrin because according to wiki, $\mathrm d(f \ \mathrm dt) = (\mathrm df)(\mathrm dt)$, implying that the second term generated by the product rule ($f \ \mathrm d^2t$) vanishes
 
since, as that picture shows, there's different directions along which the solution can enter in from
 
@Semiclassical but how do we justify the "turn"?
oh hm
 
6:13 PM
No, @Leaky, that's crap. You need wedge product in there.
 
so whatever direction it enters from, we just know how it will tend to the origin
 
They're using differential forms, and then, yes, $d(d\omega)=0$ all the time.
 
and that should give a good enough phase portrait
 
@TedShifrin this usage in wiki seems like the engineer's use to me
 
@ShaVuklia note that, if you have a degenerate unstable node, then this picture just flows outwards
 
6:14 PM
@TedShifrin Looks like you're too busy , I'm sleepy too , maybe hit you up in the morning :)
 
No, not with that rule. Read a real math book (or watch my lectures) instead of using wiki.
 
@Semiclassical but they drew a particular direction where the solution are coming from
 
@MikeMiller Oh. Flubberduck.
 
they could have also chosen the reverse direction for the entering of the solution
isn't there more to say about that?
 
@Tanuj: You should figure out that there's no problem with the second derivative of that function at $0$. You have to do similar things with the more complicated part. You have to find $f'(x)$, $f'(0)$, and use the definition of the derivative.
 
6:15 PM
I need an isotopy through isometries don't I
not just a random ass isotopy
 
sure. I'm just saying that replacing stable with unstable just reverses the directions of the arrows in that picture. There's more that can be said
 
okay , I'll try it one more time @TedShifrin
 
oh, that's now what I meant by direction @Semi
I means rather the "angle" from which the solution came into the picture
not the direction of the arrow
 
right
 
@Tanuj: Sleep probably helps.
 
6:16 PM
my point just being that you can equally well phrase this as "what happens to solutions near a degenerate unstable node as $t\to+\infty$"
no need to talk about $t\to -\infty$
 
they seem to following $v$
 
Well...
 
ooooh
it has to be like that
it can't be any different
I think I see it
 
First, suppose your solution starts on the line $v$. Does it ever leave that line?
 
no it doesn't
 
6:17 PM
Right.
 
but I see it now
it can't come in from the other 'direction' of $v$
 
Hang on.
 
because $v$ plays an important role too
 
what if you start that solution a little bit off that line?
 
let me think
 
6:18 PM
And hi @Sha
 
hi @Ted
 
Try looking at the other flow lines as well.
 
well @Semi
it will still go to "$-v$"
because of the $t$ in front of the $v$
 
Right.
 
I put quotation marks, because I don't mean asymptotic approach
but it has to be somewhere there
 
6:19 PM
What I'd say is that the velocity will be asymptotic to $-v$
 
o okay
 
As you say, it's not clear exactly what line it'll be asymptotic to as it flows out
But that line will point in the $-v$ direction
 
I know shamefully little hyperbolic geometry
 
@BalarkaSen I know special relativity, but I don't know if that counts
 
That's Lorentzian geometry isn't it
Not even Riemannian
 
6:21 PM
@Balarka: If knowledge were about shame, we'd all be in bad shape.
Correct.
 
True I suppose :)
 
Pseudo-Riemannian, yeah
 
well, special relativity is Lorentzian.
 
Lorentzian is a special case of pseudo-Riemannian, no?
 
@Semiclassical yea, thanks!
 
6:24 PM
Sure, Semiclassic.
 
Like Euclidean is a special case of Riemannian.
 
right
flat in both cases
 
Sure.
 
general relativity, on the other hand...noooope
 
6:25 PM
Did you figure out the proof of your lemma?
 
no, but that's because I got caught up in conversation
 
I'm surprised you ever get anything done. I only spend time here because I don't have to :P
 
latex test:
$[a\quad b\quad c]$
$\bigl a\atop b \bigr$
$\left[a \atop b\right]$
there we go
 
I always make a little matrix to do that.
 
usually I would to
but I want to have that be itself an element of a matrix
and I'm not sure tex likes nesting that
 
6:32 PM
Ah, I never put the bars for block matrices.
 
hrm
aww
$$[a^T \quad b^T\quad (c\times d)^T] \begin{bmatrix} c \\ d \\ a\times b \end{bmatrix} = \begin{bmatrix} [a^T \quad b^T]\left[c\atop d\right] & 0 \\ 0 & (c\times d)\cdot (a\times b)\end{bmatrix}$$
there we go.
I don't like how it looks, but the idea is right
 
You seem to have rows and columns interchanged, unless you're using vectors as rows.
 
hrm, you're right.
 
But, yes, now you need to know the determinants of the two square matrices on the left.
 
6:35 PM
Right.
so volumes of parallelipipeds (sp?) formed by $a,b,c\times d$ and $c,d,a\times b$
 
Right.
So think through the lemma: What is the determinant of $[a\ b \ v]$ for general $v$?
 
I want to say it'll just be $(a\times b)\cdot v$.
 
Righto.
 
@AkivaWeinberger I summon thou
 
oh, I see
 
6:39 PM
thee? thoust? theemersty?
 
That would be thee, Balarka :P
 
Nailed it
 
let's say I know that for every proper ideal $P$ there's a maximal ideal $M \supseteq P$, and that for every ideal $I$ contained in a prime ideal $Q$, there is a prime ideal $J$ that is minimal with respect to the property $I \subseteq J \subseteq Q$. I wonder if I can prove the following without Zorn's lemma: for every element $f$ of the ring and ideal $I$, if $f^n \notin I$ for all $n$, then there is a prime ideal containing $I$ not containing $f$.
 
so that's $[(a\times b)\cdot (c\times d)]^2 = \det([a,b]^T [c,d])[(a\times b)\cdot(c\times d)]$
 
There you go, @Semiclassic.
 
6:40 PM
yeah
cute
Of course, one can just prove $(a\times b)\cdot (c\times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)$ directly
 
Yeah, that's the natural inner product on $\bigwedge^2$.
 
in index notation that's $\epsilon_{ijk} a_j b_k \epsilon_{ilm} c_l d_m$
 
Pfeh.
 
and that product of levi-civita symbols is just $\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$
which gives the right-hand side as it should
 
I guess nobody cares about algebra :P
 
6:45 PM
Calling @Mathein nobody isn't very nice, Leaky.
 
the usual proof involves looking at $\Sigma$, the set of ideals that avoid every power of $f$, and then using Zorn's lemma
@TedShifrin :P
 
"I guess nobody cares about abstract algebra" ftfy
 
but I wonder if I can do away with Zorn's lemma because I've already used it twice to prove like two facts
 
dont scorn and zorn
 
Balarka, DogAteMy appeared.
 
6:46 PM
just use it and continue reading porn (aka algebra)
 
Scorn Zorn sounds like a math metal band
I have no idea what the lyrics for such a band would be like, but there you go
 
Probably involves a lot of "Satan"
@Ted He did but I wonder if it's just his avatar playing tricks
 
no
they'd just replace Satan would Cartan
(I know the two don't actually sound quite the same but oh well)
 
anything rhyme / with anything / if you chime / it / a little bit
 
I'm outta here for now. You all misbehave without me.
 
6:49 PM
later
 
Cya
 
7:18 PM
guys, how again does it follow from the group property that $y_0=\phi(t_0-t_1;x_0)$? The group property says $\phi(t+t_1;x_0)=\phi(t;\phi(t_1;x_0))$.
 
Well, take $t=t_0-t_1$ in that last statement. Then $\phi(t_0;x_0)=\phi(t_0-t_1;\phi(t_1;x_0))$
 
English is not my first language, but the word "series" doesn't exist, does it?
 
it does
 
"serie"
 
@Semiclassical let me see
 
7:25 PM
My auto-correct agrees with me
 
in English, we don't say "serie" no
so series can be either plural or singular
 
That's what I thought
Kind of confusing
 
hm I don't see why that equality holds
I'm also confused by $t=t_0-t_1$. So we don't consider $t$ as a variable?
You say "take $t=t_0-t_1$", but you don't seem to do anything with that?
 
I mean that, if it's valid for all $t$, then it's valid in the particular case $t=t_0-t_1$
 
7:35 PM
if what is valid for all $t$?
 
But it should work for other values as well. When $t=0$, for instance, it states that $\phi(t_1;x_0)=\phi(0;\phi(t_1;x_0))$
@ShaVuklia the group property
 
oooh
like that
 
I'm not entirely sure how that notation is functioning, but if it's going to be valid for all times then those cases are certainly true as well.
That second instance, I guess, just says that if you know the position at one time and then wait 0 seconds then the position isn't changed.
One thing I'd suggest if you want to get a handle on this is to take the special case where $t_0=0$, $t_1=1$
Then the group property states that $\phi(t+1;x_0)=\phi(t;\phi(1;x_0))$
Stuff like that
 
ah okay
I'm a bit tired, so I'll have a look at it later (thanks for the help tho)
 
7:52 PM
Hello chat
 
@AkivaWeinberger, sorry for asking this very late but, why is no subsequential limit of the partial sums of (25) has limit less than $\alpha$ or greater than $\beta$ here?
I tried doing this way, but it seemed that then every subsequential limit of the partial sums of (25) would be $\alpha$ or $\beta$ and nothing between them, which seemed unnatural.
@robjohn, Why is this on hold? What part of the question is unclear? Will it get deleted? What part of the question is unclear?
 
8:14 PM
hello
someone know this :
Let $F\subset E$ a vector subspace, such that $\overline{F}\neq E.$ There exists a functional $f\in E^*$ such that $$\langle f,x\rangle=0, ~ x\in F,$$ but $f\not\equiv 0.$
i need a reference
please
 
yo I would like to check something out with you guys
 
@Vrouvrou take $v \in E \setminus \overline{F}$, define a functional on $v \oplus \overline{E}$ that takes $v$ to $1$ and is $0$ on $\overline{E}$, then extend by Hahn-Banach
 
so if we have two holomorphic functions such that $(f \circ g^{-1})(z) = e^{i\theta}z$ then $f = g$ because of analytic continuition
right ?
oh yeah
that is good problem @Vrouvrou
actually there is shorter solution than @MatheinBoulomenos but it uses a corollary that depend on Hahn Banach
 
I think the proof of the corollary of Hahn-Banach is similar to my solution
maybe more general
 
yeah
what do you think of my reasoning above @MatheinBoulomenos ?
 
8:23 PM
is $\theta$ fixed?
or is it the argument of $z$?
 
fixed
 
I don't think this true. Take $g= \operatorname{id}$ and $f(z)=e^{i\theta}z$
 
hm
okay wait
let us say we fix the origin
suppose that $f \circ g^{-1}$ fix the origin
 
$e^{i\theta}z$ always fixes the origin
 
$e^{i\theta} z$ fixes the origin... It's literally a rotation
 
8:26 PM
Is it a corollary of Hahn-Banach theorem ? @MatheinBoulomenos
 
right
I don't understand the following statement then
how do we get $F = G$
I mean we kinda get like a rotated version
but not equality
 
@Vrouvrou yes
 
@MatheinBoulomenos i don't find it in Brezis's book, where i can find it please
 
@Adeek it's in the text. $H'(0)>0$, but $H'(0)=e^{i\theta}$
 
ohhh sorry
I am blind sometime
sometime I skip through word while reading lol
 
8:30 PM
...
 
@BalarkaSen what can I do I am blinddddd
 
@Vrouvrou I only have German functional analysis books, not sure if that's helpful for you.
 
@Vrouvrou Big Papa Rudin is your PAPA :D
you will find it there
 
??
what is this book?
 
don't mind me I am kinda buzzed
 
8:32 PM
Oh wait, I think one has been translated. It's in "Linear Functional analysis" from Werner Alt, chapter 4
It's theorem 4.3 in the German edition
 
@BalarkaSen wazzup
 
not sure if the numbering is the same in the English edition
 
@MatheinBoulomenos this book? springer.com/us/book/9781848000049
 
it's in the chapter about linear functionals
 
@MatheinBoulomenos wzzup
 
8:34 PM
but I'm sure this is in every text on functional analysis
 
Why must they have $e^{i\theta} = 1$?
 
@orbit-stabilizer because if you take the derivative
 
@orbit $1$ is the only number on the unit circle which is real and positive
 
it is bigger than zero
the only way that can happen is if $e^{i\theta} = 1$
 
Hmm. We have $H'(z) = e^{i\theta}$, right?
 
8:35 PM
because $H^{\prime}(z) = e^{i\theta}$
 
@orbit-stabilizer Understand that $F'(z_0) > 0$ means first and foremost that $F'(z_0)$ is real.
 
yeah
 
There is no order on $\Bbb C$. Only on $\Bbb R \subset \Bbb C$
 
@BalarkaSen, oh okay. then my question is, how do we know $H'(z) > 0$?
 
it is by
 
8:36 PM
Hypothesis
 
Gotcha
Makes sense then, thanks
 
@Adeek What you just wrote made 0 sense. Thank you for deleting that.
Please think before typing.
 
@BalarkaSen ...
 
Don't drink and drive :(
 
hey @MatheinBoulomenos do you know about the automorphisms of the disc
i.e holomorphic automorphism
 
8:42 PM
yeah, you can classify them by Schwarz' (or Schwartz'?) lemma
 
They're also precisely the isometries of the Poincare disk I think
 
Hey there! I'm taking a Linear Algebra practice test and am stuck on a question. I wanted to know what I'm doing wrong without looking at the exact solution. The question is Let A = [column1(5, -1), column2(-1, 5)]. Compute a formula for A^k where k is a positive integer. Your answer should be a single matrix.
 
the orientation preserving ones :)
 
yeah
 
So I did A^2 and got [col1(26, -10), col2(-10,26)], but figured repeated multiplication was not the way to go, so I used the eigenvalues and eigenvectors to diagonalize A to [col1(4, 0), col2(0, 6)] with the change of basis matrix [col1(1, 1), col2(-1, 1)]. Then I inverted the change of basis matrix to [col1(1/2, -1/2), col2(1/2, 1/2)].
I multiplied [col1(4^k, 0), col2(0, 6^k)] on the left by this inverse to get the putitive solution [col1((4^k)/2, -(4^k)/2), col2((6^k)/2, (6^k)/2)], but it does not satisfy the above result I got when initially computing A^2. Am I on the right track with a computation error, or am I using an invalid approach?
 
8:44 PM
do you guys know how this was discovered @BalarkaSen @MatheinBoulomenos ?
I am interested to know how people came up with the idea
 
No, I don't know much about the history of complex analysis
 
I guess it comes out from the proof, but I am interested to know its history
 
How can I rewrite $\prod_{j=1}^k(a_j-b_j)$ in a general form without parenthese?
i.e. multiply this out
I want write this in $\sum .....$
 
I see @MatheinBoulomenos I think complex analysis history seems pretty interesting from what I was reading in Stein
@MatheinBoulomenos btw I checked your blog the other day
it is nice
good job
 
8:52 PM
i don't find it @MatheinBoulomenos
 
the statement is a little different, but it implies directly what you ask for
It's Theorem 4.3 in my edition
 
in the version i have theorem 4.3 is Projection theorem
 
it is before or after ?
 
It's the statement if $Y \subset X$ is a closed subspace and $x_0 \notin Y$, then there is a $x' \in X'$ with $x'=0$ on $Y$ and $\|x'\|=1$ and $x'(x_0)=\operatorname{dist}(x_0,Y)$
It's the first thing after Hahn-Banach
 
8:58 PM
@BalarkaSen can I visit you one day ??
 
@MatheinBoulomenos this:
 
yes
this implies what you want
 
how please ?
 
@Vrouvrou try think about it it is more helpful
 
I don't really exist as a material being so no you may wish to but you can not, for you're constrained by the laws of quantum physics
@Semiclassical What is mysticism? Baby don't hurt me
No more
 
9:00 PM
I will go sober up
 
take $E=X$, $Y=\overline{F}$ and $x_0$ any element in $E \setminus \overline{F}$
 
I am becoming obnixous
brb
 
you can ignore the thing with $x'(x_0)$, $x'$ is non-zero because $\|x'\| =1$
and $x'$ is $0$ on $\overline{E}$, so also on $E$
 
why <x,x'>=0 ?
 
9:07 PM
This just says that $f'$ is $0$ on $F$
I confused $E$ and $F$ above, soory
$\langle x,x' \rangle$ is just $x'(x)$
x' is $0$ on $\overline{F}$
@orbit-stabilizer indeed
I once went to a concert where the cellist played all the Cello suites
it was really long
but great
 
I want to learn the Cello. It sounds amazing
 
I know a bit of Cello, but I haven't played in some time
 
$Y=F$ $X=E$ $f'=x',$ $<x',x_0>\neq 0$ here !!!
@MatheinBoulomenos
 
@Silent Hang on, I'll look.
 
@MatheinBoulomenos are you here ?
 
9:23 PM
oh man
I love classical music
I would like at some point to learn how to play the violin
do you know how to play the violin @MatheinBoulomenos
 
@Adeek no, only cello
@Vrouvrou but we only require <x',x> is 0 for x in F
x_0 is not in F
 
9:42 PM
I only listen to real music
(It starts slow but shit gets real quickly, especially where Takyon kicks in)
I also love JonTron's version of Firework way more than Katy Perry's lmao
 
)))
 
@Silent I posted a comment regarding this.
 
9:58 PM
OH goddamn how have I not seen this before
This is legendary artwork
 
@Silent additional context might get it reopened. Otherwise, a post to Requests for Reopen & Undeletion Votes, etc. might get it reopened.
 

« first day (2761 days earlier)      last day (2263 days later) »