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12:03 AM
@loch aha the answer comes one page after
$\Lambda_M$ is filtered iff $M$ is flat
i see
3 hours later…
2:52 AM
if $\mathfrak a$ is an ideal in $R$ then $\mathscr F(N) := \mathfrak aN$ is a functor R-Mod => R-Mod?
@LeakyNun yes
and there is a natural transformation $\mathfrak a \otimes_R - \implies \mathscr F(-)$
3:11 AM
3:43 AM
heyo chat
3:55 AM
Please someone answer this:
Q: Circle can always represented as $\{z:|z-c|=k|z-d|\}$

SilentI am reading Howie's Complex Analysis. There I see this remark: The observation that $c$ and $d$ are inverse points is the key to showing that every circle can be represented as $\{z:|z-c|=k|z-d|\}$. Suppose that $\Sigma$ is a circle with centre $a$ and radius $R$. Let $c = a + t$, where $0 < t

any online tool to do the symbolic calculation ?
I have two nonlinear equations in two unknowns
any suggestions?
2 hours later…
5:40 AM
user image
Oh no!!
5:51 AM
Devils rep
@Secret quack
1 hour later…
7:09 AM
Ugg, I'm struggling with understand the tricks used in holder duals, they basically substitute in a second variable and maximize over it. Then the dual becomes the auxiliary space the second variable lives in?
8:06 AM
Say we have two concentric circles, one of radius $r$ and one of radius $R$
Choose a point from each at random. What's the expected distance between them?
just integrate some expression
"bUt YoU cAN dO iT eLEMentARiLy ThO"
In other words, if we have two planets in circular orbits, what's their average distance
@LeakyNun I mean I dunno what the answer is so any technique is fine
I've just been too lazy for the integral so far
$\displaystyle \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta$
this looks familiar
is this right?
do I need to divide by 2pi or something
$\displaystyle \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta = \int_{0}^{2\pi} \sqrt{R^2 + r^2 - 2rR\cos\theta} \ \mathrm d\theta$ (hey I derived cosines theorem!)
I'd need some way of evaluating $\int_0^{2\pi}\sqrt{a+b\cos\theta}d\theta$
yeah no you can't evaluate this integral directly, you need to recognize that this is actually the area of the large circle
so it evaluates to $\pi R^2$
8:12 AM
That sounds wrong
I'm fixing the point on the inner circle and moving the point on the outer circle around
the line connecting the two points trace out the large circle
That would only work if sweeping around the outer point kept the angular velocity of it around the inner one constant
angular velocity is always constant
no matter around which point of reference
That's clearly not true
why not?
8:16 AM
If the inner point is right next to the circle
then the angular velocity would be really fast as the outer point goes right by it
and it would be slow as it's far away
but you need to multiply by... wait wtf
@MarioCarneiro hi!
lol hi
@AkivaWeinberger you need to multiply by the distance so it's ok
@MarioCarneiro I hope you recognized me lol
8:17 AM
yes I did
no need to stop on my account
This is like when you do a change of variables and you need to multiply by something to make the differentials work out
@LeakyNun Maybe you could multiply by the distance from the inner point to the tangent line of the outer point
in order to? @AkivaWeinberger
You're doing $\int d~d\theta$
yeah and I said the answer is $\pi R^2$
but the area of that triangle isn't $d~d\theta$
it's $h~d\theta$
where $h$ is the distance to the tangent
so $\int h~d\theta$ will equal $\pi R^2$ but $\int d~d\theta$ won't
8:25 AM
The answer has to be symmetric in $r$ and $R$
Let's do the special case where $r=R$
We have $\int_0^{2\pi}\sqrt{(1-\cos\theta)^2+\sin^2\theta}d\theta$
Well I guess this is $r=R=1$
are these on the circle or in the circle?
you're right, the answer doesn't match up
on the circle
when r=R=1 the answer turns out to be 8
8:29 AM
= 8
which doesn't add up
yeah no
should be divided by 2 pi
Oh that's true @MarioCarneiro
but we get $4/\pi$
which is still not the area of the circle
anyway I've confirmed (numerically) that the answer does depend on r also so
it's def not the area of the circle
8:30 AM
Oh, it couldn't be the area of the circle anyway 'cause the largest distance is $2$
the pi r^2 expression doesn't have the right units
it's definitely true when r=0 right
Oh well I think he would've updated his conjecture to $(\pi R^2)/(2\pi)$
still not the right units
this is very interesting
8:31 AM
it's a distance, if you double r and R the answer should double
Hm when $r=0$ you get $R$ 'cause it's constant
but what happened to my area interpretation
but whatever, your conjecture was that it doesn't depend on $r$, only $R$
which means our example with $r=R=1$ still means it's wrong ($4/\pi\ne1$)
wolframalpha refused to compute the integral
so let's call it a day
it's an elliptic integral
8:33 AM
(You use $1-\cos\theta=2\sin\frac\theta2$ to finish that example)
@MarioCarneiro Yeah that's what I got
which I guess means it depends on the perimeter of some ellipse?
21 mins ago, by Leaky Nun
$\displaystyle \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta = \int_{0}^{2\pi} \sqrt{R^2 + r^2 - 2rR\cos\theta} \ \mathrm d\theta$ (hey I derived cosines theorem!)
there must be some 3B1B way of somehow turning this into an actual ellipse or something
something like that, although the modern study of elliptic integrals has diverged somewhat from actual ellipses
Right so extra credit would be to find the average distance from Earth to Mercury and to Venus
Hm. Fixing $R$, varying $r$, what does the graph of this look like
I hope it's strictly increasing
Wolfram Alpha lets you plug in "distance from Earth to sun" and "distance from Mercury to sun", doesn't it
@AkivaWeinberger it's... actually linear??
r = 0 to 1, R = 1
8:37 AM
that looks like matlab
also didn't divide by 2pi
also it actually wobbles afterwards
$$\frac{r+1}{\pi } E\left(\frac{4 r}{(r+1)^2}\right)-\frac{r-1}{\pi}E\left(-\frac{4 r}{(r-1)^2}\right)$$
You want sin(t)
not sin(x)
oops, changed, it no longer wobbles
8:40 AM
Really close to $\sqrt{x^2+1}$
looks about right. It's too flat to be $\sqrt{x^2+1}$ though
the second derivative of the function is discontinuous at 1
So the reason I was thinking about this was 'cause I saw this
I didn't read the whole thing but it sounded wrong
Oh and now it won't load so I can't look at it again
If it loads for one of you can you read it and tell me what it says
Venus is not Earth’s closest neighbor
Calculations and simulations confirm that on average, Mercury is the nearest planet to Earth—and to every other planet in the solar system.
$\sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} = \sqrt{r^2+R^2 - 2rR\cos\theta} = \sqrt{r^2+R^2-2rR(2\cos^2(0.5\theta)-1)} \\ = \sqrt{(r+R)^2 - 4rR\cos^2(0.5\theta)} = \sqrt{(R-r)^2 \cos^2(0.5\theta) + (R+r)^2\sin^2(0.5\theta)}$
8:50 AM
Wait a moment
Wait did your graph actually confirm that
'cause it's increasing in $r$
(Units in AU so $R=1$)
and $r_{\rm Mercury}<r_{\rm Venus}$
@LeakyNun The half-angle thing only works for $r=R$ I think
I said you could evaluate it using the half-angle identity
but unless $r=R$ you still end up with an integral with no closed form
but now $\displaystyle \frac1{2\pi} \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta = \frac1\pi \int_0^{\pi} \sqrt{(R-r)^2 \cos^2\alpha + (R+r)^2 \sin^2\alpha} \ \mathrm d\alpha$
so it's half the perimeter of an ellipse with major axis R+r and minor axis R-r
Oh is that the formula for the perimeter of an ellipse
which we all know is (approximately) $2\pi\sqrt{\dfrac{(R+r)^2+(R-r)^2}2}$
so the whole integral becomes $\approx \sqrt{r^2+R^2}$
confirming your approximation of $\sqrt{x^2+1}$
(it actually shocked me to find out that there is no closed form for the perimeter of ellipse: I only found out just now)
8:57 AM
For $(r,R)=(1,1)$ we had $4/\pi$
Semimajor axis $R+r$ and semiminor axis $R-r$
so for $(1,1)$ we have a line
of length 4 (so "circumference" 8)
and then 8/2π
okay, so now @LeakyNun come up with a 3B1B style explanation that relates this value to the perimeter of an ellipse
lol no
I'm still annoyed that the article won't load for me anymore
Punishment for my skepticism
Did they mention it being related to the perimeter of an ellipse?
9:01 AM
it loaded for me, eventually... it looks like they got the right answer, with the elliptic integral and everything
@AkivaWeinberger net neutrality starter pack
try this
Fun fact, the minimum-energy trajectory from Earth to Mars starts when they're furthest apart from each other
Actually I think that's wrong
Never mind
oh yeah what the actual, it's $\sqrt{r^2+R^2}$ so it increases with $r$ so yeah actually mercury should be the closest to all other planets lol
but that's not very useful is it
I'll just go like oh mercury is r=0.3 and venus is r=0.7 and earth is r=1 so like "morally" venus is closer to earth
like that's not what people actually mean when they say venus is closest to earth right
they mean the orbits are the closest
"orbits are closest" = what?
@AkivaWeinberger does my link work for you?
9:06 AM
I guess they have small Hausdorff distance
the distance being exactly R-r
@LeakyNun It does
(though the images aren't there)
PCM = point-circle method
9:08 AM
@AkivaWeinberger Earth at launch time is furthest way from Mars at arrival time
would probably be the more accurate way of saying it
@LeakyNun Doesn't show for me
@MarioCarneiro What's the Hausdorff distance
That works
@MarioCarneiro That's not $R+r$ (for concentric circles), is it?
it's how you can measure "distance between sets" in R^2 or other metric spaces
@AkivaWeinberger it's R-r
it's $\max_{a\in A}\min_{b\in B}d(a,b)$
9:10 AM
How's the Hausdorff distance defined
@MarioCarneiro Ahh
Wait why is that symmetric in $A$ and $B$
in other words "you are always somewhat close to a point on the orbit of venus"
oh, it's actually the max of that and its symmetric counterpart
You need to restrict to compact subsets if you want to get a real metric though @AkivaWeinberger
@MarioCarneiro anyway I'm glad to see you here lol
9:12 AM
There we go
Thay actually does look like $R+r$
no it's R-r
the inf is R-r at any point, so it's taking the sup of a constant
Inf isn't constant in x
It's like a two player game. You want to maximize $d(a,b)$, I want to minimize it. I go first and choose $a$, you go second and choose $b$
it is, it's always R-r
you always choose the co-radial point
@LeakyNun $d(x,y)$, where $y$ is fixed and $x$ varies, depends on $x$
@LeakyNun The player who wants to maximize it goes second
because sup is on the outside
9:14 AM
$\forall x \in X, \inf_{y \in Y} d(x,y) = R-r$
player 1 (maximizer) chooses x, player 2 chooses y
player 2 chooses the co-radial point, achieving $R-r$
@LeakyNun Ohh
Oh OK you're right
I get it now
So the game goes outside-in then
Oh crap I'm back to 98 tabs
I thought I was recovering
how to count the number of tabs
On the bottom
I have about 3000 tabs open in FF
9:19 AM
Guess what I'm doing on Friday
I have like 80 tabs
Guess what I'm incredibly scared of doing on Friday
marathon in israel?
good luck!
9:21 AM
(Not the best translation but I guess they don't want to pay a translator)
Gonna tell you right now
I'm not gonna win any prizes
it isn't about the prizes
Will probably take me like 90min or something
It is about finishing without collapsing
@TobiasKildetoft I just learnt yesterday that a surjective endomorphism on a f.g. module is in fact bijective
and I'm like wat
I still haven't learned modules
What are they for
9:24 AM
@LeakyNun With no other assumptions at all?
they're like vector spaces but your scalars don't form a field, just a ring
they come up like everywhere
@TobiasKildetoft yes
@AkivaWeinberger if k is a field and G is a group then k[G]-modules are k-representations of G (so study of modules subsume (modular) representation theory)
Not sure I ever came across that before.
Z-modules are abelian groups (so study of modules subsume study of abelian groups)
Anyway, I will need to step out of the room shortly, as I will be having a phone meeting negotiating my starting salary and stuff like that.
if M is a manifold then (locally free) $C^\infty(M)$-modules are vector bundles
@AkivaWeinberger structure theorem for finitely generated abelian groups say that they are all $\Bbb Z^r \oplus \bigoplus_{i=1}^t \Bbb Z/p_i^{n_i} \Bbb Z$
the generalization is structure theorem for finitely generated modules over principal ideal domains (Z being a PID)
they're all $R^r \oplus \bigoplus_{i=1}^t R/p_i^{n_i} R$
and this provides a "straightforward" proof of Jordan Normal Form theorem
because a $k$-vector space with a linear transformation to itself is just a $k[X]$-module, and $k[X]$ is a PID
9:38 AM
> I just learnt yesterday that a surjective endomorphism on a f.g. module is in fact bijective
That sounds like a variation on the pigeonhole principle
If you restrict to vector spaces, then it's saying that the basis is finite, so any surjection from the basis to itself is a bijection
sure, but what is shocking is that it holds for modules
(that's handwaving but should give the intuition)
because one expects modules to be very ill-behaved
submodule of f.g. module isn't f.g. etc
9:42 AM
I think of modules as vector spaces...
quotients don't have sections etc
no division is like, okay
i.e. submodules don't have complements
I usually think of that stuff as a miracle / quirk of AC that this happens at all in vector spaces
constructive vector spaces :P
how about C being a C[X]-module
i.e. a "smaller" thing somehow being a module over the "bigger" ring
9:45 AM
are you interpreting X as some random number?
yeah sure
or is the multiplication some crazy nonconstructive thing
no, $p \cdot m$ is just "evaluate $p$ at 7 and then multiply by $m$"
yeah okay, so this is true in general... if you have a ring hom $R \to S$ then $S$-modules are $R$-modules, and the hom need not be injective or surjective, so "bigger" doesn't matter
right, so how do you translate this to vector spaces :P
9:49 AM
why wouldn't this still hold?
I mean, you can't have a non-trivial vector space that's "smaller" than the field
oh, this is just the thing about field homs being injective
yeah actually I don't know how I should visualize modules
as vector spaces...
like year 1 vector spaces
just ignore the fact that there is no division
and do linear algebra anyway
9:58 AM
all the structure is there, but some of the theorems require more assumptions
every module is flat and projective and injective and torsion-free and free
those are all words for "modules aren't vector-space-like enough, I need this property of them for my argument to go through"
I guess category theory does the same thing... when an argument needs an assumption, give it a name
I see
Say we have a great circle that passes very close to the poles but doesn't touch them
I guess a manifestation is in the Künneth formula @MarioCarneiro
10:01 AM
and we take the inverse image of that in the exponential map
What does the result look like
where if your base ring is a field then you get to omit a term from the short exact sequence
so Tor is kinda like the "measure for how much your ring is away from being a field"
also if every module over your ring is flat and your ring is local then your ring is a field
@AkivaWeinberger type mismatch
are you interpreting your sphere as $\Bbb CP^1$
The exponential map from Riemannian geometry
which is defined for all manifolds
(well, with a basepoint, but I was assuming the basepoint was the north pole)
For $v\in\Bbb R^n$, $\exp_p(v)=\gamma(1)$ where $\gamma$ is the geodesic with $\gamma(0)=p$ and $\gamma'(0)=v$
is it a flow along some vector field?
10:09 AM
You can also think of it as $\gamma(\|v\|)$ where $\gamma$ is the normalized (unit-speed) geodesic with $\gamma'(0)=v/\|v\|$
oh, so it's just the polar angle for each point on the circle
Basically that map projection
Notice that everything of distance $\pi$ away from the origin maps to the south pole
Also it's defined for the whole plane, not just the disc of radius $\pi$, but beyond that it just starts repeating itself
I can get an exact figure, but you get something like a rounded half-circle
The reason it's called the "exponential map" comes from Lie theory
(Fun fact: if you take a radius and circle centered on the origin and apply $\exp_p$ to them, the images are still perpendicular)
Tissot ellipses
They're all squished radially
This is true for all manifolds
@MarioCarneiro do you come here often?
10:14 AM
(Well, if it has constant negative curvature, they'd all by stretched radially)
@AkivaWeinberger have you heard of Morse theory?
Heard of it, don't know much about it
Inverse exponential map of the hyperbolic plane
Another pic of the exponential map
Man, negotiating salary is hard. Especially when my experience is in something completely different.
I'm merely an undergraduate so I don't know how to respond
I can't life
I am now having to learn how to live in the real world, rather than the nice and sheltered world of academia. It is scary out there.
I mean, they actually expect you to do work once in a while.
10:21 AM
that's when you go to grad school
@MarioCarneiro I did that already
Also the vid it refers to
Then I stayed in academia for another 4 years
10:21 AM
keep pumping that well
@AkivaWeinberger cool
@TobiasKildetoft can I think of a local ring as a "field with some noise"?
@LeakyNun I suppose
can we do galois theory with local rings?
no idea. We can try
@LeakyNun So a quick check, what goes wrong here. I take the polynomial ring in infinitely many variables, I kill the first variable and identify the result with the original ring by shifting all indices by one. Why is this not a surjective map that is not a bijection?
so $f: k[X_1, \cdots] \to k[X_1, \cdots]$ with $f(X_1) = 0$ and $f(X_{n+1}) = X_n$ etc
10:32 AM
I think this isn't $k[X_1, \cdots]$-linear
might be the issue
$f(X_2 X_2) = X_1^2$ but $X_2 f(X_2) = X_1 X_2$
possibly we are killing more than intuition tells us when we kill off a variable out of infinitely many
A $k[X_1, \cdots]$-module is a $k$-vector space with countably infinitely many endomorphisms
and a linear map is a linear map that commutes with each endomorphism
10:36 AM
Ahh, so it is the shifting of the variables that is the issue here
which provides an isomorphism of rings, but not of modules, since we would have to twist the action on the target, which would make it no longer an endomorphism.
@TobiasKildetoft if I set $\mathcal O_\Bbb Q = \Bbb Z_{(p)}$ then what would happen to my number fields?
@LeakyNun I have no idea what that means
like for $L/K$ fin. gal. you first set what $\mathcal O_K$ is
and then define $\mathcal O_L$ to be the integral closure of $\mathcal O_K$ inside $L$
11:01 AM
Is a subgroup of a Hopfian group itself Hopfian?
Obviously there are no finite counterexamples (I initially foolishly tried searching through finite groups for a counterexample).
@user193319 no
Shoot. Do you have a counterexample in mind?
it's in the link
$F_2$ contains a copy of $F_\infty$, the former is Hopfian, the latter is not
Shoot...I am trying to prove that all finitely generated free groups are Hopfian and I was hoping to reduce it to the $F_2$ case.
2 hours later…
1:07 PM
20k rep!
1:48 PM
I have an interesting integral equation problem, I've been sent here from Phys.SE
So I'm looking at the following integral equation
$$\rho(\lambda) + \int_{-\infty}^\infty d\lambda' \rho(\lambda') \mathcal{C}(\lambda-\lambda') = (2\pi)^{-1}$$
Numerical integration does not exactly work with "infinity". One may choose a finite symmetric interval of integration and take the limit afterwards (which is the traditional way of doing these integrals), but Mathematica is not equipped to solve this integral completely symbolically
Although I conjecture that it is theoretically possible
Regardless, I wish to find a constant $\lambda_F$ defined implicitly by
$$N = \int_{-\lambda_F}^{\lambda_F} \rho(\lambda) d\lambda$$
I don't see how I can solve the integral equation subject to a constraint of effectively two unknowns...
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