Mathematics - 2022-01-17

Mary Star

01:22

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

In 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?

leslie townes

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.

Ted Shifrin

01:41

Thank goodness. Munchkin would have called out the marines.

leslie townes

she also got yelled at by a black crowned night heron.

Ted Shifrin

Stomping feet too loudly?

leslie townes

getting too close. trying the routine that she does with the cat on a wild bird whom she does not know.

Ted Shifrin

No, that will. Not. Do.

leslie townes

nero wolfe could not have said it better.

Ted Shifrin

01:44

To the orchid sanctuary.

leslie townes

fetch my beer, archie.

some of the dishes described in those books... i need fritz, here and now.

Ted Shifrin

You have the cookbook, or not?

leslie townes

i don't have it. i should probably get it.

Ted Shifrin

I got locally fished halibut today. Gotta go cook it!

leslie townes

sounds delicious. best of luck.

Ted Shifrin

01:54

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.

leslie townes

i don't think i can feed chickens on blueberries from birth to... chicken time.

Ted Shifrin

No. Nor muscovy duck.

leslie townes

we see tons of those at the duck pond.

Ted Shifrin

Oh, well, maybe then.

Who will kill and prep?

leslie townes

Munchkin.

Ted Shifrin

01:57

That might turn her vegetarian.

leslie townes

i hope it doesn't.

although i have plucked a shot waterfowl for dinner and it did make me less hungry.

Ted Shifrin

I’ve never gotten that intimate.

Bob

03:21

good evening

bye

copper.hat

spent most of the last two days dealing with computer hardware issues. pita

Ted Shifrin

04:22

Pita bread?

leslie townes

livvy flipped out at bedtime and clubbed me. my daughter said "why is there blood on your face?" because we live with a cat.

copper.hat

pain in the asterisk

leslie townes

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.

soupless

Hello!

leslie townes

hola!

soupless

04:36

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 $a \in A$?

leslie townes

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.

soupless

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.

math.ucdavis.edu/~babson/MAT127B/abbott-second-edition.pdf

Isn't it safer, probably to first assume that $b \in \mathbb{Q}$?

leslie townes

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 $A$.

soupless

@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?

leslie townes

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.

soupless

04:44

@soupless here it is

leslie townes

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.

soupless

it is on page 15/slide 28.

leslie townes

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.

soupless

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

leslie townes

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.

soupless

04:59

I found an alternative construction through Cauchy sequences.

leslie townes

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.

Akiva Weinberger

05:23

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)

soupless

@AkivaWeinberger You can probably start with the midpoint of the centers?

Akiva Weinberger

@soupless Only if they're the same size

I know the answer, by the way, and it's really nice

soupless

06:11

@AkivaWeinberger Internal tangents?

Akiva Weinberger

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

Mad Spaces

08:24

@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

Vrouvrou

11:36

@robjohn hello , what is H and G and M ?

Mad Spaces

Those are Latin letters. :D
In Algebra H is a half gorup, G a group, M monoid.

Vrouvrou

what it means i just want to prove this inequality

$\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$

Mad Spaces

Google inequality between geometric and arithemetic mean

Vrouvrou

i don't understand

Mad Spaces

11:56

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!

robjohn

@Vrouvrou did you try to google amgm?

Mad Spaces

12:17

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!

robjohn

12:30

$$\frac{\mathrm{d}g}{\mathrm{d}t}=\sum_{i=1}^n\frac{\partial g}{\partial x_i}\frac{\mathrm{d}x_i}{\mathrm{d}t}$$

Then you can apply the chain rule to that

Mad Spaces

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)

VLC

14:41

Does anybody know a nice proof of the combinatorical theroem : $\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{r!}$

anakhro

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

anakhro

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

anakhro

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

14:50

I have $n$ choices

anakhro

How about for your second element?

VLC

$n-1$ choices

anakhro

Third?

VLC

$n-2$

anakhro

... rth?

VLC

14:50

$n-r+1$

anakhro

Great, so you see how they get the numerator, right?

VLC

Yes I understand that part

anakhro

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 ?

anakhro

Indeed.

So how many different orders of these r elements are there?

VLC

14:53

I see it's $r!$

anakhro

Great, do you see how it works now?

VLC

I understand it now

thanks

anakhro

:)

VLC

15:15

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

anakhro

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

anakhro

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

anakhro

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

15:30

Yes exactly

That is what I don't understand too

Looks like I messed it up in my notes

anakhro

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 $\bigcup_{A \in \binom{N}{r}}R_A = \text{All subsets of N with cardinality r}$

anakhro

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

anakhro

Okay, so maybe you are showing that if R_A and R_B are not disjoint, then A=B?

VLC

15:36

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

anakhro

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

anakhro

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

anakhro

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

15:40

So if $R_B$ would have B ={2,1} then $R_B = R_A$

It is ok, I think I got it right now

MethNoob

16:58

meth is boring

leslie townes

meth is a lot of things, but boring is not one of them

AMDG

17:26

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

18:01

Hi dear gentleman ,

10

Q: Solve $\int _{x=0}^{\infty }\int _{t=-\infty }^{\infty }\exp \left(\frac{-a t^2+i b t}{3 t^2+1}+i t x\right)\frac{x}{3 t^2+1}\mathrm{d}t\mathrm{d}x$

How 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}...

integration complex-integration residue-calculus

Is it allowed here to change the order of the integrands?

That is to say to integrate first x and vice-versa

leslie townes

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?

leslie townes

@robjohn pinging you as you might like this.

Vrouvrou

18:21

@robjohn i don't understand this i never see it

i know that $2ab\leq a^2+b^24$

AMDG

I assume he means arithmetic-geometric mean

Vrouvrou

but i don't know how to use it

@AMDG yes

AMDG

Just look up AGM and you'll find the formuler

leslie townes

that's AMGM and not AMDG

you get different results for the latter

AMDG

kek

leslie townes

18:23

:)

Vrouvrou

i don't really understand

i'm not familliar with this

AMDG

smth about elliptic integrals idk

Vrouvrou

is there a direct methods to prove this inequality please

AMDG

Cring. wolfram alpha can't figure out what I'm trying to say for computing the limit of an integral.

I want to know $\lim_{t\to u} \int_t^u \frac{1}{u^2 + 1} du$.

robjohn

18:54

@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.

Vrouvrou

@robjohn i don't understand how to do at all

dc3rd

ah.....just got a fresh dose of computer chips injected.....ready for any 5G programmingg coming my way

AMDG

What is L here? functions.wolfram.com/ElementaryFunctions/Exp/07/01/0001

Clearly some sort of complex domain, but I don't know what.

Xander Henderson

19:13

@AMDG functions.wolfram.com/Notations/3

> The special contour $\mathscr{L}$, which is used in the definition of the Meijer $G$ function and its numerous particular cases.

AMDG

Nice, more functions I haven't learned about yet.

Thank you Xander

EM4

welcome to mathematics :) learning never ends :D .

AMDG

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?

Xander Henderson

No.

An integral along a contour is not indefinite.

AMDG

Ah

Jakobian

19:24

would anyone care to look at the topology exercise I'm solving, I don't really know how to progress

Xander Henderson

@Jakobian Don't ask to ask. Just ask.

Jakobian

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.

Ted Shifrin

If a function is holomorphic, it has no singularities and hence no residues. Badly written.

Xander Henderson

@TedShifrin Holomorphic needn't mean entire. Though I agree that meromorphic would be better.

Ted Shifrin

Holomorphic in a domain (of which the contour is a subset) certainly implies no residues. I never mentioned "entire."

Xander Henderson

19:35

But the domain can be specified so that the poles are excluded.

Ted Shifrin

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 $f^{-1}(U^\sharp) \subset U$?

Jakobian

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.

Xander Henderson

@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.

Ted Shifrin

Holomorphic certainly means no singularities in the domain under discussion.

Xander Henderson

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.

Ted Shifrin

19:41

I'm not going to continue this.

Let me try to think about Jakobian's thing.

Jakobian

19:58

I have another idea to take $U^\# = \bigcup_{V\subseteq U, V\neq\emptyset} Y\setminus f(V^c)$ for which we also have $f^{-1}(U^\#)\subseteq U$.

anakhro

I don't recall, is the Hodge star C^\infty linear, or just R-linear?

Jakobian

don't mind what I wrote, this is just $Y\setminus f(U^c)$ again

Jakobian

20:20

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

robjohn

20:39

@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

Vrouvrou

thank you very much

i try this

robjohn

Good afternoon, @Ted

Ted Shifrin

Hi, @robjohn

Vrouvrou

can you tell me if it is right ?

robjohn

If what is right?

Vrouvrou

20:43

$\left(\frac{2ab}{a+b}\right)^2\leq \frac{a^2+b^2}{2}$

then

anakhro

@TedShifrin how does a Riemannian metric induce a metric on 1-forms?

Vrouvrou

$8 a^2 b^2\leq (a^2+b^2)(a+b)^2=(a^2+b^2)+2ab(a^2+b^2)\leq 2(a^2+b^2)^2$

then we get $4a^2b^2\leq (a^2+b^2)^2$

Ted Shifrin

A metric on a vector bundle induces metrics on the dual bundle, tensor bundles, etc.

Vrouvrou

which is true, is my solution is ggood @robjohn

Ted Shifrin

@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.

robjohn

20:46

@Vrouvrou what solution?

@Ted: OMG, really?

Ted Shifrin

growls at sarcasm

anakhro

@TedShifrin Does this mean it's just the dual metric tensor, [g^{ij}]?

robjohn

@Vrouvrou I thought you were trying to show $\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$

Vrouvrou

yes

i squar the two sides

and i use that $2ab\leq a^2+b^24

Ted Shifrin

@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 ...

anakhro

20:53

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...

robjohn

@Vrouvrou you are using the inequality to prove the inequality

You cannot start with $\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$

Vrouvrou

i try to do equivalence with a true inequality

robjohn

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

Vrouvrou

ok

thank you very much

Ted Shifrin

@anakhro Just expand $\alpha$ in terms of an orthonormal coframe.

robjohn

21:06

@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 $F$?

I guess not by math.stackexchange.com/questions/3034353/…

Ted Shifrin

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

21:22

I see your point!

No worries :)

Ted Shifrin

So this is convergence of functionals (i.e., functions of functions).

jay

yes

usually used in calc of variations

Ted Shifrin

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"

Ted Shifrin

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

21:28

should that of been $cF$ in the last line?

Ted Shifrin

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

Ted Shifrin

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

21:59

but @TedShifrin the $\liminf$ of a product is not the product of the $\liminf$ 's

Ted Shifrin

When one factor is constant?

jay

its not constant though $c_n$ depends on $n$?

Ted Shifrin

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 $F$

Ted Shifrin

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.

anakhro

22:05

@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

Ted Shifrin

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.

anakhro

@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.

Ted Shifrin

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.

Amin Idelhaj

Hey everyone!

anakhro

22:10

Thanks Ted!

Ted Shifrin

In general, @anakhro, you also want to get induced connections on these associated bundles, too.

hi Demonark.

anakhro

THAT'S WHO AMIN IS?

jay

@TedShifrin I agree, and double agree about the unboundedness ! Cheers

anakhro

profile picture checks out

Ted Shifrin

This is who Demonark has always been?

Amin Idelhaj

22:12

Hahaha, so you knew of Daminark and of Amin separately? That's interesting

But yeah how've you guys been?

anakhro

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.sstatic.net/ohgtn.png

It never took off.

Ted Shifrin

I was never as handsome as Gauss.

Amin Idelhaj

Amazing

But yeah any fun math you guys have been up to?

anakhro

math? fun? what do you think this is? a math chat?

Let's chat cooking

Amin Idelhaj

Hahaha

anakhro

22:28

Have you been doing any math, DAMINARK?

IT EVEN HAS AMIN IN IT

Amin Idelhaj

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?

anakhro

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.

Amin Idelhaj

22:56

Automorphic forms pretty much, possibly bounding geodesic periods

anakhro

Nice! Is that mostly on the number theory side of things for you>?

Amin Idelhaj

It's kind of a mix of number theory, Lie theory, analysis

Astyx

23:25

hey amin

Ted Shifrin

Salut, Astyx

Astyx

Salut Ted

leslie townes

23:47

i think i have that shirt. in the ted meme.

ted's, not gauss's.

Astyx

I was gonna ask

Ted Shifrin

Pity.

leslie townes

i have a bid on the gauss shirt on ebay, nobody outbid me. auction closes in a few hours.

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$Mathematics$ Mathematics