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1:22 AM
Hello!! I am looking at an exercise about central limit theorem and I got stuck. The exercise is :
0
Q: Central limit theorem : manufacture of chocolates

Mary StarIn the manufacture of chocolates of a specific variety with a nominal weight of 20.4 grams, there may be fluctuations in the actual weight of a praline. We describe the weight of a praline of this type using a random variable X and assume that X has mean 21.37 and variance 0.16. One box contains ...

I got stuck at (b). Do we use the formula Iwrote there? Or can we nontuse there the central limit theorem?
 
i will crawl out on a limb here and suggest that this problem has never actually come up in the context of chocolate manufacturing.
i don't know what their tolerances are but i imagine that decimal points are not involved.
we did work for a biotech startup whose main thing required manufacturing lots of stuff with diameters varying less than 2 microns. pralines were not an input to that process.
hi ted. there was a weird ibis-like bird at the duck pond again today. and the northern shovelers were back.
 
1:41 AM
Thank goodness. Munchkin would have called out the marines.
 
she also got yelled at by a black crowned night heron.
 
Stomping feet too loudly?
 
getting too close. trying the routine that she does with the cat on a wild bird whom she does not know.
 
No, that will. Not. Do.
 
nero wolfe could not have said it better.
 
1:44 AM
To the orchid sanctuary.
 
fetch my beer, archie.
some of the dishes described in those books... i need fritz, here and now.
 
You have the cookbook, or not?
 
i don't have it. i should probably get it.
 
I got locally fished halibut today. Gotta go cook it!
 
sounds delicious. best of luck.
 
1:54 AM
I can will you mine if you want. I guess you’d have to come get it :)
Some of the recipes are absurd, but plenty are manageable.
 
i don't think i can feed chickens on blueberries from birth to... chicken time.
 
No. Nor muscovy duck.
 
we see tons of those at the duck pond.
 
Oh, well, maybe then.
Who will kill and prep?
 
Munchkin.
 
1:57 AM
That might turn her vegetarian.
 
i hope it doesn't.
although i have plucked a shot waterfowl for dinner and it did make me less hungry.
 
I’ve never gotten that intimate.
 
 
1 hour later…
Bob
3:21 AM
good evening
bye
 
spent most of the last two days dealing with computer hardware issues. pita
 
4:22 AM
Pita bread?
 
livvy flipped out at bedtime and clubbed me. my daughter said "why is there blood on your face?" because we live with a cat.
 
pain in the asterisk
 
math.stackexchange.com/questions/4358730/… interested to hear other views of this. a lot of my anti-visualization thing is an act i do for fun, but in some cases i think insisting on a visualization inhibits actual analysis.
venn diagrams are great examples of this, actually. they absolutely do not scale.
 
Hello!
 
hola!
 
4:36 AM
I would like to ask as I was confused: From Abbott's Understanding Analysis, about the construction of $\mathbb{R}$: Why is it that the upper bound of $A \subseteq \mathbb{R}$ is defined as $b \in \mathbb{R}$ instead of $b \in A$ such that $a \leq b$ for all $a \in A$?
 
the upper bound of a set need not lie in that set.
i don't know the particulars of the construction of $\mathbb{R}$. there are multiple constructions.
 
I mean, we still don't know what the properties of real numbers are, aside from $\mathbb{R}$ also being a field (assumed) and ordering.
Isn't it safer, probably to first assume that $b \in \mathbb{Q}$?
 
the canonical example would be $A = \{x \in \mathbb{Q}: \text{$x \geq 0$ and $x^2 < 2$}\}$.
definitely bounded above, definitely has a least upper bound in $\mathbb{R}$, definitely does not have a least upper bound in $A$.
 
@leslietownes What I am trying to ask is that, why say that the upper bound is a real number when we still don't know other properties aside from the inherited properties from the rationals?
 
simpler examples would be $A = \{1 - \frac{1}{n}: n \in \mathbb{N}\}$ which also has an upper bound that is not in the set. this example does not truly test the order completeness of the rational numbers.
soupless if you have a page in abbott i can look at it. in general, the least upper bound of a set $A$ will not lie in the set $A$. it might lie instead slightly outside of it.
 
4:44 AM
@soupless here it is
 
the rationals just don't allow you to talk about least upper bounds. they don't exist there.
or rather, sometimes they do, and sometimes they don't. if they do it is something of an accident.
 
it is on page 15/slide 28.
 
ok. you could do definitions 1.3.1 and 1.3.2 for any ordered field or perhaps even for any ordered set. the only issue is that sets that are bounded above might not have least upper bounds.
they seem to be specifying to $\mathbb{R}$ because that is a context in which bounded sets do always have least upper bounds.
abbott is my mother's maiden name, i wonder if we're related.
 
Is there an alternative way to first know what the properties of real numbers are before defining upper bounds as elements from $\mathbb{R}$?
not that Dedekind thing yet
 
if we're not doing anything yet, it's not clear in any context that an ordered set does, or does not, contain least upper bounds of bounded sets.
you can define the notion in any ordered set, and then it is just a set-dependent question of whether the least upper bound exists or not.
the example of $\mathbb{Q}$ is helpful. the positive stuff that squares to something less than $2$ is bounded. the least upper bound would be something that squares to something that is exactly $2$. this takes some work, but if you assume it, it helps.
and $\mathbb{Q}$ just doesn't have such a thing.
 
4:59 AM
I found an alternative construction through Cauchy sequences.
 
those are the two usual constructions. dedekind completeness and cauchy completness.
one requires a notion of order and the other requires a notion of distance.
 
5:23 AM
Consider two intersecting circles. What is the locus of points with the same signed distance from each circle?
(Signed meaning the distance is negative for points inside the circle and positive for points outside it)
 
@AkivaWeinberger You can probably start with the midpoint of the centers?
 
@soupless Only if they're the same size
I know the answer, by the way, and it's really nice
 
6:11 AM
@AkivaWeinberger Internal tangents?
@AkivaWeinberger Internal tangents?
 
If a point is on a circle then it has zero distance from it
so the only way a point that's on one circle is on our locus is if it's on the other circle as well
 
 
2 hours later…
8:24 AM
@Koro Hey Koro. I copied this from Profs lectures. Why do you think it is wrong?
Having trouble to understand this seemingly simple proof of the Taylor series in one dimension.

Theorem: $f(b)=f(a)+\frac{b-a}{1!}*f'(a)+\frac{(b-a)^2}{2!}f''(a)+...+\frac{(b-a)^{n-1}}{(n-1)!}*f^{n-1}(a)+R_n(x_o)$

Proof let $g(x)= f(b) - [f(x)+\frac{b-x}{1!}*f'(x)+\frac{(b-x)^2}{2!}f''(x)+...+\frac{(b-x)^{n-1}}{(n-1)!}*f^{n-1}(x))]$
Then is $g'(x)= \frac{(b-x)^{n-1}}{(n-1)!}*f^n(x)$ and $g(b)=0$

So ... what did we prove here? Deriving it gives that but i am not sure how to understand this
Naturally for $f$ diff n times at interval [a,b] with x_o in that interval..
Nevermind... understood it while writing haha!
Me stoopid
 
 
3 hours later…
11:36 AM
@robjohn hello , what is H and G and M ?
 
Those are Latin letters. :D
In Algebra H is a half gorup, G a group, M monoid.
 
what it means i just want to prove this inequality
$\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$
 
Google inequality between geometric and arithemetic mean
 
i don't understand
 
11:56 AM
Google that. you will see something that helps you
Question. If we have a function $h: R \rightarrow R^n $ and $f: R^n \rightarrow R$ then the function $g:=f \circ h$ have derivative of $D(g) = D(f(h))*D(h)$ suppose the function $h : t \rightarrow t*x, x \in R^n$ then simply we have $dg/dt = \sum_{i=1}^n \partial f(h)/ \partial x_i * x$ right?
where as $x$ is a vector, but why i have in my notes the following written $dg/dt = \sum_{i=1}^n \partial f(h)/ \partial x_i * x_i$ Ok this is bad notation from my side where as the first x_i is a variable and the second x_i is a vectorcomponent nevertheless why are we taking the singular components of the vector? it doesnt add up!
 
@Vrouvrou did you try to google amgm?
 
12:17 PM
Okay sorry i think i reliazed why it is like that... it is because $D(f(h)$ is basically gradient. and then $x = D(h)$ so we have a multiplication of vectors.. we need to write it in that order.. never mind.
I always get the derivatives mixed up when it is in R^1 to R^n and R^n to R^1
confusing
@robjohn Hello Prof Rob i hope you are doing good!
 
12:30 PM
$$\frac{\mathrm{d}g}{\mathrm{d}t}=\sum_{i=1}^n\frac{\partial g}{\partial x_i}\frac{\mathrm{d}x_i}{\mathrm{d}t}$$
Then you can apply the chain rule to that
 
Yes Prof i have figured that out. One just needs to be a bit careful with the indicies and the dimensions and it will work out quite fine.
Dimensional Analysis is very useful! In physics as also in mathematics (IE comparision of dimensions or units and figuring it out how th eresult should be like)
 
 
2 hours later…
VLC
2:41 PM
Does anybody know a nice proof of the combinatorical theroem : $\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{r!}$
 
That's usually the definition for the binomial coefficient.
What is your definition of n choose r?
 
VLC
@anakhro My definition is: $\binom{N}{r}$ is all possible subsets of $N$ of cardinality $r$ , and then $\binom{n}{r} = |\binom{N}{r}|$
So I need to get from that definition $\binom{N}{r} = \{A: A\subset N, |A| = r\}$, to that formula
 
So basically unordered subsets of size r from a set of N elements?
And you count the number of them total.
 
VLC
Yes that is correct
 
Great, so if you have to start making choices for one of these subsets, how many different choices do you have for a first element of one of these subsets?
 
VLC
2:50 PM
I have $n$ choices
 
How about for your second element?
 
VLC
$n-1$ choices
 
Third?
 
VLC
$n-2$
 
... rth?
 
VLC
2:50 PM
$n-r+1$
 
Great, so you see how they get the numerator, right?
 
VLC
Yes I understand that part
 
Great, so now we have the problem where we implicitly have an order here.
 
VLC
Ok so we have $r$ elements that shouldnt be in order right ?
 
Indeed.
So how many different orders of these r elements are there?
 
VLC
2:53 PM
I see it's $r!$
 
Great, do you see how it works now?
 
VLC
I understand it now
thanks
 
:)
 
VLC
3:15 PM
We have $R_A = \{ (a_1,...,a_r) \in v(N,r): \{a_1,...,a_r\}\}$, where $|v(N,r)| = n(n-1)...(n-r+1)$. How do we know that $R_A \cap R_B \neq \{\}$
So that it is not equal to null set
 
What is A and B?
 
VLC
$A \in \binom{N}{r}$
where that is a subset of cardinality r of N
Oh and $B$ is also in the set $\binom{N}{r}$
We somehow got to that if that is true then $A = B$ and is not a null set, but I don't get it how we got there
 
And what is v(N,r)?
Or what does it have to do with A and B, because so far your definition of R_A and R_B look identical, I don't know what R_A has to do with A.
 
VLC
v(N,r) is all the subsets of r that have cardinality r
subsets of N*
Well yes, if we have (a_1,..a_r) =A and (a_1,...,a_m) = B, and both are in that set R_A and R_B, why are A =B
Well, because we can have n elements and let's say an element that is in B is not necessarily in A is it?
Ok so basically my question is: We take two subsets of an r-tuple from $N$ without repetition.
and if order doesnt matter we call that set A
and the other one B
Why are A = B
 
I don't quite understand. If N=3 and r=2, then A could be {1,2} and B could be {2,3}. A is not B in this case.
 
VLC
3:30 PM
Yes exactly
That is what I don't understand too
Looks like I messed it up in my notes
 
Perhaps, though I don't quite understand what your notation meant earlier.
 
VLC
So the notation $R_A$ was a set of all orders of a set that has cardinality of r. So in other words $|R_A| = r!$
And $\bigcup_{A \in \binom{N}{r}}R_A = \text{All subsets of N with cardinality r}$
 
Do you mean that R_A is the set consisting of ordered r-tuples with the same (unordered) entries as A?
 
VLC
Yes exactly
So basically A is a subset of R_A
 
Okay, so maybe you are showing that if R_A and R_B are not disjoint, then A=B?
 
VLC
3:36 PM
And R_A is a subset of all possible subsets like $R_A$ that cover up all the set N
Yes that would be it
 
Which makes intuitive sense to you, right?
 
VLC
Actually it doesn't because like you said, we could have {1,2,3,4} = N and then for r = 2 we pick orders of R_A = {(1,2),(2,1)} and R_B = {(2,4),(4,2)}
so A not eqaul to B
 
But R_A and R_B are disjoint.
So of course, A \neq B
 
VLC
Oh wait i see....the n-tuple is like one element, so if one is in intersection with another it means that A = B
I get it
 
This is all hypothetical, assuming I understand what you are trying to show in your notes. I cannot verify what your notes should have been.
 
VLC
3:40 PM
So if $R_B$ would have B ={2,1} then $R_B = R_A$
It is ok, I think I got it right now
 
 
1 hour later…
4:58 PM
meth is boring
 
meth is a lot of things, but boring is not one of them
 
5:26 PM
Are there any analogs to $\int \frac{1}{x} dx = \ln(x)$ for the exponential?
I want to see if there's another way to speed up convergence for approximating the exponential by using the sum I found for division.
Multiplication is still relatively expensive, let alone using multiplication and exponentiation by squaring to do exponentiation for arguments that have been reduced for a given $e^x$ approximation.
Taking a simple approximation that first computes $(e^{2^{-64}x})^{2^{64}}$ for 64-bit precision would be expensive even with a fixed procedure that uses an addition-subtraction-chain.
 
SNR
6:01 PM
Hi dear gentleman ,
10
Q: Solve $\int _{x=0}^{\infty }\int _{t=-\infty }^{\infty }\exp \left(\frac{-a t^2+i b t}{3 t^2+1}+i t x\right)\frac{x}{3 t^2+1}\mathrm{d}t\mathrm{d}x$

granular bastardHow to solve this double integral? $$f(a,b)=\int _{x=0}^{\infty }\int _{t=-\infty }^{\infty }\exp \left(\frac{-a t^2+i b t}{3 t^2+1}+i t x\right)\frac{x}{3 t^2+1}\mathrm{d}t\mathrm{d}x$$ $$\text{with }a>0,b\in \mathbb{R},i^2=-1$$ Known special solution for ${\bf b=0}$ $$f(a,0)=\frac{\pi}{\sqrt{3}...

Is it allowed here to change the order of the integrands?
That is to say to integrate first x and vice-versa
 
ooh, this looks tough. i'm worried about the x-integral.
robjohn loves these.
 
SNR
Certainly, but is it allowed to change the order of the integrands?
 
@robjohn pinging you as you might like this.
 
6:21 PM
@robjohn i don't understand this i never see it
i know that $2ab\leq a^2+b^24$
 
I assume he means arithmetic-geometric mean
 
but i don't know how to use it
@AMDG yes
 
Just look up AGM and you'll find the formuler
 
that's AMGM and not AMDG
you get different results for the latter
 
kek
 
6:23 PM
:)
 
i don't really understand
i'm not familliar with this
 
smth about elliptic integrals idk
 
is there a direct methods to prove this inequality please
 
Cring. wolfram alpha can't figure out what I'm trying to say for computing the limit of an integral.
I want to know $\lim_{t\to u} \int_t^u \frac{1}{u^2 + 1} du$.
 
6:54 PM
@Vrouvrou use that and then also the reciprocal of the inequality you get by substituting $a\mapsto\frac1a$ and $b\mapsto\frac1b$
@SNR the integral in $x$ for each fixed $t$ diverges, so no, I don’t think changing order is going to help.
 
@robjohn i don't understand how to do at all
 
ah.....just got a fresh dose of computer chips injected.....ready for any 5G programmingg coming my way
 
Clearly some sort of complex domain, but I don't know what.
 
7:13 PM
> The special contour $\mathscr{L}$, which is used in the definition of the Meijer $G$ function and its numerous particular cases.
 
Nice, more functions I haven't learned about yet.
Thank you Xander
 
EM4
welcome to mathematics :) learning never ends :D .
 
I like how Weisstein included a joke as part of the page on contour integrals.
> Renteln and Dundes (2005) give the following (bad) mathematical joke about contour integrals:

Q: What's the value of a contour integral around Western Europe? A: Zero, because all the Poles are in Eastern Europe.
> As a result of a truly amazing property of holomorphic functions, a closed contour integral can be computed simply by summing the values of the complex residues inside the contour.
Hey wait a sec
So you're saying I can compute indefinite integrals by applying a (convenient and easily manipulable) transform that makes a function holomorphic and summing the residues?
 
No.
An integral along a contour is not indefinite.
 
Ah
 
7:24 PM
would anyone care to look at the topology exercise I'm solving, I don't really know how to progress
 
@Jakobian Don't ask to ask. Just ask.
 
well, alright
I have a map $f:X\to Y$ between compact metric spaces which is irreducible. I'm trying to prove that for every non-empty open $U\subseteq X$ there is open non-empty $U^\#\subseteq Y$ such that $f^{-1}(U^\#)\subseteq U$ is dense in $U$.
Irreducible means that it's a continuous surjective map such that for every closed proper $A\subseteq X$ we have $f(A)\neq Y$
I've tried setting $U^\# = Y\setminus f(U^c)$ which is a non-empty open subset such that $f^{-1}(U^\#)\subseteq U$ but I have trouble showing if this is dense in $U$.
Taking a non-empty open $V\subseteq U$, and assuming that density fails for this open set, I have that $f(V)\subseteq f(U^c)$. I don't really see any contradictions from this.
 
If a function is holomorphic, it has no singularities and hence no residues. Badly written.
 
@TedShifrin Holomorphic needn't mean entire. Though I agree that meromorphic would be better.
 
Holomorphic in a domain (of which the contour is a subset) certainly implies no residues. I never mentioned "entire."
 
7:35 PM
But the domain can be specified so that the poles are excluded.
 
I'm just saying the quote is bad ... and I actually know one of the authors of that. He was a student of mine when he was an undergraduate.
@Jakobian Since $f$ needn't be injective, I don't see how you control the preimage. Why do you say $f^{-1}(U^\sharp) \subset U$?
 
If $x\in f^{-1}(U^\#)$ then $f(x)\notin f(U^c)$ so necessarily $x\in U$.
If it were $x\in U^c$ then $f(x)\in f(U^c)$ which is impossible.
 
@TedShifrin I agreed with you that it could be better written, but holomorphic doesn't necessarily mean "no singularities". Depending on how, precisely, an author is defining the term, I suppose.
 
Holomorphic certainly means no singularities in the domain under discussion.
 
The function $z \mapsto 1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$, and has a singularity at zero.
I don't see a contradiction here to what was written in the definition.
 
7:41 PM
I'm not going to continue this.
Let me try to think about Jakobian's thing.
 
7:58 PM
I have another idea to take $U^\# = \bigcup_{V\subseteq U, V\neq\emptyset} Y\setminus f(V^c)$ for which we also have $f^{-1}(U^\#)\subseteq U$.
 
I don't recall, is the Hodge star C^\infty linear, or just R-linear?
 
don't mind what I wrote, this is just $Y\setminus f(U^c)$ again
 
8:20 PM
So if $f^{-1}(U^\#)\cap V = \emptyset$, then $f^{-1}(U^\#)\subseteq V^c$ so that $U^\#\subseteq f(V^c)$ but this means that $Y = f(V^c)$, in particular $V$ is empty.
I think the above is correct
yep, pretty sure this solves the problem of density
 
8:39 PM
@Vrouvrou here it is in detail:
You want to show
$$
\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}\tag1
$$
You know that if $a\ne b$,
$$
2ab\lt a^2+b^2\tag2
$$
Take reciprocals
$$
\frac1{2ab}\gt\frac1{a^2+b^2}\tag3
$$
Multiply by $2a^2b^2$
$$
ab\gt\frac{2a^2b^2}{a^2+b^2}\tag4
$$
Substitute $a\mapsto\sqrt{a}$ and $b\mapsto\sqrt{b}$
$$
\sqrt{ab}\gt\frac{2ab}{a+b}\tag5
$$
Divide $(2)$ by $2$ and take square roots
$$
\sqrt{ab}\lt\sqrt{\frac{a^2+b^2}2}\tag6
$$
Putting $(5)$ and $(6)$ together gives
$$
\frac{2ab}{a+b}\lt\sqrt{ab}\lt\sqrt{\frac{a^2+b^2}2}\tag7
 
thank you very much
i try this
 
Good afternoon, @Ted
 
Hi, @robjohn
 
can you tell me if it is right ?
 
If what is right?
 
8:43 PM
$\left(\frac{2ab}{a+b}\right)^2\leq \frac{a^2+b^2}{2}$
then
 
@TedShifrin how does a Riemannian metric induce a metric on 1-forms?
 
$8 a^2 b^2\leq (a^2+b^2)(a+b)^2=(a^2+b^2)+2ab(a^2+b^2)\leq 2(a^2+b^2)^2$
then we get $4a^2b^2\leq (a^2+b^2)^2$
 
A metric on a vector bundle induces metrics on the dual bundle, tensor bundles, etc.
 
which is true, is my solution is ggood @robjohn
 
@Vrouvrou For future reference, you can always use Lagrange multipliers to prove these sorts of inequalities. Just state it as a minimization subject to a constraint.
 
8:46 PM
@Vrouvrou what solution?
@Ted: OMG, really?
 
growls at sarcasm
 
@TedShifrin Does this mean it's just the dual metric tensor, [g^{ij}]?
 
@Vrouvrou I thought you were trying to show $\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$
 
yes
i squar the two sides
and i use that $2ab\leq a^2+b^24
 
@anakhro That's the inverse matrix, of course. But if you have an orthonormal basis, then the dual basis is declared to be orthonormal. You can work out what that means in terms of matrices. Mumble mumble inverse transpose ...
 
8:53 PM
I am just trying to work out the expression of $g(\alpha,\alpha)$ for a general 1-form, $\alpha$.
Since I have not seen it written anywhere...
 
@Vrouvrou you are using the inequality to prove the inequality
You cannot start with $\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$
 
i try to do equivalence with a true inequality
 
You have to have $\iff$ between all steps if you are doing that
It is often easier if you can start with a true statement and derive what you want
 
ok
thank you very much
 
@anakhro Just expand $\alpha$ in terms of an orthonormal coframe.
 
9:06 PM
@Vrouvrou: it is not immediately obvious that you have that:
You have
$$
\left(\frac{2ab}{a+b}\right)^2\leq \frac{a^2+b^2}{2}\iff8a^2b^2\le\left(a^2+b^2\right)(a+b)^2=\left(a^2+b^2\right)^2+2ab\left(a^2+b^2\right)
$$
The questionable step is
$$
8a^2b^2\le\left(a^2+b^2\right)^2+2ab\left(a^2+b^2\right)\iff8a^2b^2\le2\left(a^2+b^2\right)^2
$$
If you can show that last biconditional, you are good, but then you should start at the end and go backwards.
starting with a true statement and deriving what you want is always more convincing
 
jay
Hi
If I know that $F_n$ $\Gamma$ converge to $F$, is it true that $(1+\frac{1}{n})F_n$ $\Gamma$ converges to $F$?
 
You should make this clearer. It should be $\Gamma$-convergence. I thought you were multiplying $F_n$ by $\Gamma$.
At any rate, I've lived almost to the age of 70 without ever hearing of $\Gamma$-convergence.
 
jay
9:22 PM
I see your point!
No worries :)
 
So this is convergence of functionals (i.e., functions of functions).
 
jay
yes
usually used in calc of variations
 
Except that post has regular functions and input $x$ rather than $u(x)$. Totally confusing.
 
jay
it is described as if $F_n$ $\Gamma-$converges to $F$ then minimisers of $F_n$ converge to minimisers of $F$
yeah in that post $x$ will be an element of some topological space
That is the "Fundamental Theorem of Gamma convergence"
 
It's clear why sums won't work, but I don't see why scalar multiplication would be a problem. If $c_n\to c$ (in the usual sense), doesn't $c_n F_n$ $\Gamma$-converge to $F$ if $F_n$ $\Gamma$-converges to $F$?
 
jay
9:28 PM
should that of been $cF$ in the last line?
 
Yes, I got lost with all the $\Gamma$ typing.
 
jay
thats a good point though I was thinking in terms of sum property but maybe I look again just thinking of multiplication.
I know $\Gamma$ is not fun to write ey
We need Elon Musk and his Neuro link
 
The issue with general sums is that the optimizers won't be related to one another at all. But if you take $F$ and $cF$, they have the same optimizer for any $c$.
 
jay
yesyes I see
 
jay
9:59 PM
but @TedShifrin the $\liminf$ of a product is not the product of the $\liminf$ 's
 
When one factor is constant?
 
jay
its not constant though $c_n$ depends on $n$?
 
You're taking liminf over $x$ or over $u$.
 
jay
your using the notation from math.stackexchange.com/questions/3034353/…
which is written badly
lets go from this definition
en.wikipedia.org/wiki/%CE%93-convergence over first countable spaces
I am asking what can I say about $(1+\frac{1}{n})F_n$
if I know $F_n$ $\Gamma-$converges to $F$
 
So, if $u_n$ optimizes $F_n$, it also optimizes $c_n F_n$ with optimal value $c_n F_n(u_n)$. Maybe it's better to think about $c_n F_n - c F = (c_n-c)F_n + c(F_n-F)$. I see. So if the optimal $F_n(u_n)$ is unbounded as $n\to\infty$, we have a problem. But if it stays bounded, I think I'm right.
 
10:05 PM
@TedShifrin Just to confirm that I have the right idea: basically the metric g creates an isomorphism $f\colon TM\to T^*M$ and so for covectors $\alpha_p,\beta_p\in T^*_pM$ we define $g_p(\alpha_p,\beta_p) = g_p(f^{-1}(\alpha_p),f^{-1}(\beta_p))$.
 
jay
@TedShifrin whats weird about $\Gamma$ convergence is that even the constant functional $F$ may not $\Gamma$ converge to itself
 
I don't like thinking of it that way, @anakhro, but it's fine ... You need $\tilde g_p$ or something. I think it's worth understanding how to induce a metric on associated bundles in general. The easiest way is to work with orthonormal frames.
Well, if $F_n\to F$ (in the $\Gamma$ sense), then certainly $cF_n \to cF$ is fine.
 
@TedShifrin Is this something I could expect to come up with myself, or is there a good reference on this? I am not super informed when it comes to Riemannian geometry.
 
I think it's pretty much in every book, @anakhro. Certainly the more modern geometric analysis books, like Jost and the French guys' ... I'm sure it's in Spivak somewhere and probably even in Warner.
@jay I think you should work with the specific $F_n$ and $F$ you have. This stuff is not going to work in generality; that much is clear.
 
Hey everyone!
 
10:10 PM
Thanks Ted!
 
In general, @anakhro, you also want to get induced connections on these associated bundles, too.
hi Demonark.
 
THAT'S WHO AMIN IS?
 
jay
@TedShifrin I agree, and double agree about the unboundedness ! Cheers
 
profile picture checks out
 
This is who Demonark has always been?
 
10:12 PM
Hahaha, so you knew of Daminark and of Amin separately? That's interesting
But yeah how've you guys been?
 
Ted has been living his golden years.
Helping everyone do homework.
I only ever come around to bug Ted, as well as to listen to leslie say anything.
@AminIdelhaj I made a meme of Ted: i.stack.imgur.com/ohgtn.png
It never took off.
 
I was never as handsome as Gauss.
 
Amazing
But yeah any fun math you guys have been up to?
 
math? fun? what do you think this is? a math chat?
Let's chat cooking
 
Hahaha
 
10:28 PM
Have you been doing any math, DAMINARK?
IT EVEN HAS AMIN IN IT
 
Yup :PPPPP
Amin in the dark is how I came up with it
I've been taking it a bit slow during the winter break, just before that... I was slowly reading about quantum unique ergodicity
And I had given a presentation about mapping class groups which required learning about mapping class groups :P
And that was a lot of fun
This was in dynamics, and a friend of mine wants to do his masters thesis with the prof for that class, probably in translation surfaces and billiard stuff. I very much intend to learn some of the stuff with him
Also we're reading through some scheme theory together which is coo. And trying to learn some more rep theory
Not making it very far in any endeavor tbh but it's fun :P
You?
 
What sort of stuff are heading towards thesis-wise?

I find it hard to say I have been learning any math, but apparently I have been doing basic Riemann geometry computations while trying to follow along a few papers in geometric analysis.
 
10:56 PM
Automorphic forms pretty much, possibly bounding geodesic periods
 
Nice! Is that mostly on the number theory side of things for you>?
 
It's kind of a mix of number theory, Lie theory, analysis
 
11:25 PM
hey amin
 
Salut, Astyx
 
Salut Ted
 
11:47 PM
i think i have that shirt. in the ted meme.
ted's, not gauss's.
 
I was gonna ask
 
Pity.
 
i have a bid on the gauss shirt on ebay, nobody outbid me. auction closes in a few hours.
 

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