Mathematics - 2020-12-17 (page 1 of 2)

copper.hat

00:01

@John_Krampf there are many candidates here, but i think the worst prog. language that sees 'popular' use is Tcl.

John_Krampf

@copper.hat haha is TCL really still used that much though?

copper.hat

in my world unfortunately. (what is referred to as electronic design automation)

i will rack up 4 hours consulting today as a result of the language doing something that in py would be a few minutes.

i suppose that it good is some semi-evil sense.

John_Krampf

Hmmmm

copper.hat

i exaggerate slightly, of course, for emphasis.

i view it as the cobol of eda

and john ousterhout is such a nice fellow, i don't why he foisted that on the world.

later folks, may all your manifolds be smooth...

Lucas Henrique

normed spaces are really cool

now I want to understand how algebra works with topology

Thorgott

00:17

in lots of really interesting ways

user2103480

in mysterious ways

Alessandro Codenotti

If only algebraists and topologists also worked together

Insteas we get 57 "algebra brain" messages

Lukas Heger

@LucasHenrique a starting point would be learning some algebraic topology

Thorgott

whilst I regularly try and do topology in here

Alessandro Codenotti

@user2103480 I want to make clear that if I say I'm going to sleep and then you still see me posting I didn't lie, I failed.

Thorgott

00:22

I'm tragically misunderstood

user2103480

@AlessandroCodenotti topology brain

Black Panther

@John_Krampf it looks to me like x, y = place splits the two tuple into the variables x and y respectively, is this right? In C# that would assign the tuple place to x and also y.

user2103480

@AlessandroCodenotti lmao

Lucas Henrique

a topology brain is just a sphere

user2103480

@LucasHenrique smooth brain

that proves it @MikeMiller

and now, since we live in somewhat 4-dimensions including time, prove what the number of smooth structures is

Alessandro Codenotti

00:26

Topology brain

user2103480

ah, horned conway

Lukas Heger

lmao

Ted Shifrin

@Thorgott Verily.

robjohn

@TedShifrin: I posted the code for the Yellow Brick Road on Mathematica.

@AlessandroCodenotti I got to see a seminar by Conway while I was at Princeton.

user2103480

not to confuse with horned @Thorgott: chat.stackexchange.com/transcript/message/56454365#56454365

Ted Shifrin

00:28

He gave a bunch of lectures at UGA.

Alessandro Codenotti

@robjohn Nice, do you remember what was the topic?

Thorgott

horned Thorgott has a much less interesting complement

robjohn

@AlessandroCodenotti It was about Fractran.

Coding with rational numbers

Thorgott

@user2103480 p sure that's still open

Alessandro Codenotti

Ah nice, what a weird language

robjohn

00:29

indeed

Karim Mansour

that is cool

love_sodam

One exam left

Mike Miller

@Thorgott I do algebra I just keep quiet about it

user2103480

@Thorgott that's the joke

Lukas Heger

I'm pretty sure we know that there are continuum many smooth structures on R^4

Thorgott

00:35

yes, that's a result by Freedman/Donaldson, I believe

the question whether there are any exotic smooth structures on S^4, however, is the smooth Poincare conjecture and still open

Lukas Heger

yeah

I wasn't getting that we were talking about spheres

Mike Miller

00:49

@Thorgott Indeed. Existence of continuum-many is due to Taubes

Indeed the r-balls in a specific exotic R^4, for r large, are pairwise non diffeomorphic

Thorgott

ah, I see

Mike Miller

(Not every exotic R^4; if there are any exotic S^4s, then deleting infinity will give you an exotic R^4 so that all sufficiently large balls around the origin are diffeomorphic

Thorgott

Here are two cool theorems:
Theorem: Let $M,N$ be oriented closed connected manifolds of the same dimension and suppose there exists a continuous $f\colon M\rightarrow N$ of non-zero degree. Then the growth type of $\pi_1(N)$ is dominated by the growth type of $\pi_1(M)$.
Theorem: Let $X,Y$ be compact Riemann surfaces and assume there exists a holomorphic $f\colon X\rightarrow Y$ that is not constant. Then $g(X)\ge g(Y)$.
Note that a non-constant holomorphic map between compact Riemann surfaces automatically has positive degree and Riemann surfaces are automatically oriented (and connected …

Mike Miller

I assume you meant growth type there

Oh genus

Thorgott

genus in the latter case, yes

Mike Miller

00:58

Sure, you can intuit that a nonzero degree map means the domain is "larger" than the codomain in some essential sense

Or more complicated than

But I don't think there's a general philosophy/theory of degree-monotonic invariants

One really nice invariant which behaves well with positive degree maps is the simplicial volume. If M is an oriented closed manifold we say the simplicial volume of M is the infimum over the ell^1 norm of every real singular cycle representing the fundamental class of M

It is clear from the definition that if f: M -> N has degree k, then |M| >= k|N|

It follows that anything which has a map to itself of degree > 1 has zero simplicial volume

What I love is that you can use this property to calculate the simplicial volume of hyperbolic surfaces

You have a candidate with 4g simplices all with coefficient 1, coming from the 4g-gon representation, so that |S_g| <= 4g

Thorgott

so a sphere has zero simplical volume

Mike Miller

Yeah

Thorgott

how does one see this explicitly

Mike Miller

Try to unwind the proof to construct a sequence of real chains with ell^1 norm going to zero

Normally one doesn't see this explicitly

Anyway one also knows there is a covering map of degree d $f: \Sigma_{d(g-1)+1} \to \Sigma_g$ (take a connected sum of the torus with $d$ copies of $\Sigma_{g-1}$ placed at $d$ points rotationally symmetric around the torus; the deck transformations are rotation)

Therefore $$|\Sigma_g| \leq \frac 1d |\Sigma_{d(g-1) + 1} \leq \frac 1d 4(d(g-1) + 1) = 4(g-1) + \frac 4d,$$ so taking infima we get $|\Sigma_g| \leq 4(g-1)$ in fact for $g > 1$

This is in fact an equality. There's a beautiful argument going the other direction, wherein you take your fundamental class, lift it to the hyperbolic plane, and use it to compute an estimate on the volume of an ideal simplex in the hyperbolic plane

Thorgott

01:18

ah ok, it's crucial that you take real coefficients, so if we do S^1 for simplicity, the fundamental class can be represented by the simplex $t\mapsto e^{2\pi it}$, which norm $1$, but we can also represent it by $1/2$ the simplex $t\mapsto e^{4\pi it}$, which has norm $1/2$, etc.

that's of course precisely what you get by repeatedly applying the degree 2 squaring map

Mike Miller

yeah, a priori it seems fine if you do rational coefficients so idr why real is desirable to be honest

Anyway the above actually proves the full classification of when there is a map $f: \Sigma_g \to \Sigma_h$ of nonzero degree, or even of degree >= d

Thorgott

How does it give a full classification? It clearly gives an obstruction to the existence, but I don't see the converse

Mike Miller

You just construct the rest by hand

I think this proves that if you have a degree d map S_g -> S_h then g >= d(h-1)+1, but you can take the covering map above then just add on more tori and crush them to obtain a map for g > d(h-1)+1

Thorgott

01:34

oh right, I mixed up my inequalities

this is really cool

Mike Miller

I learned it from Balarka

Thorgott

is there a reference I can add to my ever-expanding reading list?

Mike Miller

I'll try to find one for you a bit later

I don't know more than this, I just like surfaces

Thorgott

niceg

also we obtain a vast generalization of the second theorem I quoted above as corollary of this, any continuous of non-zero degree must shrink genus

Mike Miller

yup, which is not too hard to prove by cohomological considerations

but d>1 I think is hard to do cohomologically

Thorgott

01:52

is there some geometric way of thinking about the simplical volume or is it "just" a useful invariant?

I have a hard time picturing in which sense spheres could have zero volume

Mike Miller

for hyperbolic manifolds there's a proof showing that it's a c(dim M)-multiple multiple of vol(M)

Ted Shifrin

I haven't been following, but do only null-homologous spheres have zero volume?

Thorgott

what's a null-homologous sphere?

the top homology of a sphere isn't zero

Mike Miller

Simplicial volume is an invariant of manifolds, not of submanifolds. But I'm interested in what you're remembering.

Ted Shifrin

I don't know the notion. I thought it might be a relative notion.

Sort of like incompressible surfaces.

Mike Miller

02:04

This is something that gives zero for spheres and tori, but a multiple of the volume on hyperbolic things. Who knows for most everything else.

Ted Shifrin

Weird.

Mike Miller

@Thorgott The manifold atlas page is good and has references: map.mpim-bonn.mpg.de/Simplicial_volume

Thorgott

oh lol, that's written by the same author as the GGT book I'm reading

very nice

Mike Miller

I guess this notion corresponds to a seminorm on H_*(M;R) and the simplicial volume is the norm of the fundamental class.

Ted Shifrin

So what makes it be $0$?

Thorgott

02:09

seminorm meaning that norm 0 doesn't force 0? then ye

Mike Miller

It's the infimum of $\{|c|_1 | c = \sum_{i=1}^n a_i\sigma_i, c \text{ represents the fundamental class }\}$

where here $|c|_1 = \sum |a_i|$

Thorgott

@TedShifrin If $\gamma$ is the loop once around the circle, it represents the fundamental class. So does $\frac{1}{2}\gamma^2$, $\frac{1}{4}\gamma^4$, etc. But these have norms $1,\frac{1}{2},\frac{1}{4}$, etc., hence going to $0$

Ted Shifrin

Oh, there's no notion of primitivity.

So why does that fail in the hyperbolic case?

Hard to see why Riemannian curvature would be relephant.

Mike Miller

You can think of the above as the existence of the simplex [0, n] in R, an arbitrarily large simplex. In hyperbolic space there's a largest simplex. You can't keep scaling it up like you do under the covering maps here.

I think the formula is $|M|_1 = \text{vol}(M)/s_n$, where $s_n$ is the volume of that largest simplex in hyperbolic space.

Maybe a factor of 2?

Ted Shifrin

Ah, I see. But why can't I still do Thor's trick of just taking multiples of the fundamental class?

Thorgott

02:16

there won't be a covering self-map with degree >1 (in absolute value), I guess

Ted Shifrin

But the $\sigma_i$ can be any chains?

Mike Miller

That's where he's using a self-covering. The general fact is that if you have a self-map of degree > 1, you have zero simplicial volume. Keep pushing forward your triangulation under that map and scaling down by 1/d, and your 1-norm keeps shrinking by a factor of d, goes to zero in the limit.

Ted Shifrin

Yeah, I get that, but the defn must have some constraint.

Mike Miller

Nope

Ted Shifrin

Then why can't I just multiply all the $\sigma_i$ by $n$?

Thorgott

02:20

that doesn't represent the fundamental class anymore, no?

Mike Miller

Ah, yeah, specifically you want it to represent the integer fundamental class.

Ted Shifrin

I rescale the coeffs as you did.

Nah, you're cheating somehow.

Thorgott

yes, but you also scale integrating over it by the same factor

Ted Shifrin

Huh?

I'm looking at Mike's defn.

Mike Miller

I'm confused. If you have $c = \sum a_i \sigma_i$ and $[c] = [M]$ and you rewrite $nc = \sum na_i \sigma_i$, then $|nc|_1 = n |c|_1$ but also $[nc] = n[M] \neq [M]$.

Thorgott's specifically using that you can write $n[S^1]$ as a single simplex.

Can't wind a hyperbolic manifold around itself so as to write $n[M]$ with not so many simplices, and then scale that down.

Ted Shifrin

02:23

No, I multiply by $n$ and divide the $a_i$ by $n$.

Mike Miller

So that doesn't change volume?

Ted Shifrin

Just as in Thor's example.

Thorgott

but $\sum n\frac{a_i}{n}=\sum a_i$?

Mike Miller

In Thor's example there is a sequence $\sigma_n$, the map with $\sigma_n(z) = z^n$. There is always a single $a_1 = \frac 1n$.

Ted Shifrin

So we need simplicial chains that are reduced.

Mike Miller

02:25

What does that mean?

Ted Shifrin

No, @Thor. I am putting the $n$ on the $\sigma$.

Mike Miller

That's not a thing.

Thorgott

fundamental class is the homology class you integrate over to get usual integration of forms on the manifold, scaling a representative of the class with a coefficient scales the integration, so doesn't give you the fundamental class anymore; that's what I meant with my integration comment

Mike Miller

A singular chain is an element of the free vector space over the set of maps $\sigma: \Delta^n \to M$.

Here we're taking the element with weights $a_i$ on the maps $\sigma_i$.

Multiplying by $n$ means multiplying the scalar of the basis element by $n$. It doesn't make sense to multiply the basis element by $n$, just as it doesn't make sense to take, say, $7e_2$ and call it one of $e_1, e_2, e_3$ in Euclidean space.

Ted Shifrin

Why not?

Mike Miller

02:27

Is $7e_2$ one of $e_1, e_2, $ or $e_3$?

Ted Shifrin

This is what I was getting at with my reduced question.

Mike Miller

You have not explained what that word means, and it is not standard.

Ted Shifrin

I thought the sigmas were arbitrary integral chains.

I know.

Thorgott

no, the sigmas are the simplices

Mike Miller

OK, I understand the confusion now.

Ted Shifrin

02:28

Aha.

Thorgott

and then we're taking formal linear combinations with real coefficients

which are the $a_i$

Ted Shifrin

OK, I get it now.

Mike Miller

My bad. I should have been more explicit in the definition.

Ted Shifrin

I was lacking a specificity in the defn.

I get it.

Simplex = reduced :)

Mike Miller

Got it.

Ted Shifrin

02:29

I meant it by analogy with scheme stuff.

Thorgott

glad we're on the same page now

Mike Miller

Anyway, Thor's example wasn't formal, since $\gamma^4$ is a different element than $2 \gamma^2$. That's all what we were saying.

Ted Shifrin

Neat idea.

Mike Miller

It's still kind of magic that you get something interesting for hyperbolic things, and that it tells you everything you need to know about positive-degree maps between surfaces.

Ted Shifrin

Nah, it makes sense.

Thorgott

02:31

sounds like the average Gromov idea

John_Krampf

03:50

@BlackPanther correct

Black Panther

04:20

@John_Krampf thanks. I've been trying to implement your logic and solution in a C# program but I am getting an infinite loop when execution enters the while loop (i.e. while(pointsToCheck,Count > 0). I've compared my C# program to your Python solution, and I don't think my implementation is faulty. I suspect the problem is that pointsToCheck increases much faster than pointsChecked, and despite pointsChecked increasing albeit slowly, pointsToCheck never decreases towards zero.

@John_Krampf Please can you look at my implementation of your solution in C#, it is on pastebin

copper.hat

04:34

@BlackPanther why don't you print pointsToCheck.Count in the loop to see what is going on.

John_Krampf

@BlackPanther one thing I notice on line 78, when you convert to string in order to add the digits, is that for negative numbers you will have a "-" in front, right? so this would throw an error?

@BlackPanther In line 40 does your code recognize (x, y) as a point instance?

@copper.hat @BlackPanther this is a good idea

DarkRunner

I'm confused about an application of Gauss-Jordan Elimination

If anyone can help that would be great

John_Krampf

@DarkRunner "Associated with Math.SE; for both general discussion & math questions alike. Just ask; don't ask to ask."

DarkRunner

@John_Krampf OK Got it

The problem states "Write a set of simultaneous linear equations to describe the network and then show how to use the Gauss-Jordan elimination process to solve the system"

My guess is we have four variables (x1,x2,x3,x4), thus a square matrix of order 4, and we seek to find the vector x such that Ax=b, where b is [150,100,50,200].

However, I have no idea what the coefficient matrix A should be

Also, are the components of my b vector in the right order?

If anyone can help, that would be great

copper.hat

04:53

@DarkRunner looks like 4 equations in 5 variables?

Black Panther

05:08

@copper.hat Thanks for the suggestion. The real code is full of print tests. Here is some of what printing pointsToCheck.Count in the loop shows:

Infinite loop?1
Inside the conditional statement 6407
pointsChecked: 3086
Infinite loop?2
Inside the conditional statement 6408
pointsChecked: 3086
Infinite loop?3
pointsChecked: 3086
Infinite loop?4
Inside the conditional statement 6409
pointsChecked: 3086
pointsToCheck.Count: 1938
Infinite loop?1
Inside the conditional statement 6410
pointsChecked: 3087
Infinite loop?2
Inside the conditional statement 6411
pointsChecked: 3087
Infinite loop?3
pointsChecked: 3087
Infinite loop?4
Inside the conditional statement 6412

@John_Krampf Yes that was happening, so I had to use lines:

int x = Math.Abs(point.Item1);
                int y = Math.Abs(point.Item2);

copper.hat

@BlackPanther i'm not exactly sure what you are trying to do, so my ability to assist is limited...

Black Panther

On line 75 and 76 respectively, to make all x coordinate and y coordinate numbers positive because trying to sum their digits.

copper.hat

@BlackPanther sorry, i mean i don't know what problem you are solving...

Black Panther

@John_Krampf You mean this code pointsChecked.Add((x, y));? (x, y) creates an instance of a Tuple class which has the values that x and y respectively reference.

@copper.hat This is the problem I am trying to solve:

There is a robot which can move around on a grid. The robot is placed at point (0,0). From (x, y) the robot can move to (x+1,
y), (x-1, y), (x, y+1), and (x, y-1). Some points are dangerous and contain EMP Mines. To know which points are safe, we check
whether the sum digits of abs(x) plus the sum of the digits of abs(y) are less than or equal to 23. For example, the point (59,
75) is not safe because 5 + 9 + 7 + 5 = 26, which is greater than 23. The point (-51, -7) is safe because 5 + 1 + 7 = 13, which is

@John_Krampf solved it, but in Python. I'm trying to get it to work in C#

copper.hat

@BlackPanther what do you mean by area? is it guaranteed that the area must be bounded?

Black Panther

05:25

I understood area to be the number of 1x1 squares on the grid.

I'm not the one who wrote the question/puzzle so I'm not sure.

sorry

According to @John_Krampf , his Python code on pastebin works.

copper.hat

@BlackPanther i need to convince myself that the area is necessarily bounded :-)

Black Panther

:)

copper.hat

it may be 'obvious' but it takes me awhile. i need to check in some code and do some admin and will take a look then assuming the wine does not interfere :-)

Black Panther

@John_Krampf do the following lines do anything special that I'm missing?

place = places_to_check.pop()
    x, y = place

@copper.hat Thanks :D. The wine might be a stimulant :)

@John_Krampf inspecting this while loop:

while places_to_check:
    place = places_to_check.pop()
    x, y = place
    if safe_condition(x, y):
        verified_safe_and_reachable.add((x, y))
        for xp, yp in [(x+1, y), (x-1, y), (x, y+1), (x, y-1)]:
            if (xp, yp) not in places_checked:
                places_to_check.add((xp, yp))
    places_checked.add((x, y))

Since places_to_check.add((xp, yp)) is executed inside a for loop which has four rounds of execution, where as places_checked.add((x, y)) is outside the for loop places_to_check will increases at a faster rate than places_checked.

Black Panther

05:54

You know I thing it is the following thing that Python does with a set that is the reason why your Python solution to the problem works, but my similar C# solution to the same problem results in the while loop causing an infinite loop:

> In python all elements of a set are unique, so if I wind up adding the same point twice then the code will automatically make it so the point is only checked once

John_Krampf

06:13

@copper.hat It is easy to prove to yourself that no point with a coordinate greater than 9950 could ever be reached.

@BlackPanther that should work

yes that is correct, but if I understand your lines 43-44
IEnumerable<(int, int)> uniquePoints = pointsToCheck.Distinct<(int, int)>();
pointsToCheck = ToStack<(int, int)>(uniquePoints);
That should take care of it?

copper.hat

06:35

@John_Krampf still struggling with the 9950 part.

Black Panther

@John_Krampf Yes, I tried to replicate some properties of a stack, i.e. the unique elements of a stack

@John_Krampf What is the advantage of:

> In python all elements of a set are unique, so if I wind up adding the same point twice then the code will automatically make it so the point is only checked once

in your code?

@John_Krampf I copied and pasted your code into an online Python ide, but it didn't run. Is the code you pasted on pastebin missing anything that is needed for it to run?

Georg Friedrich Bernhard Riema

Does someone know where can I get works of famous/great mathematicians?

copper.hat

@GeorgFriedrichBernhardRiema that is a bit vague

John_Krampf

Premise 1, if a point where the first co-ordinate X is reachable, then there must be points reachable that have the first co-ordinate be X' for every X' between 0 and X.
Premise 2, the sum of the digits of 9951 is greater than 23.
Suppose for the sake of contradiction there is a point with a co-ordinate greater than 9950 ( meaning it is 9951 or greater). By Premise 1 this would imply by there is a point that is reachable with a coordinate of 9951, which contradicts premise 2. Thus there are no points reachable that have a coordinate greater than 9950 so the region of reachable points is bou…

copper.hat

@DarkRunner i get 592597

John_Krampf

06:45

@GeorgFriedrichBernhardRiema Princeton Excyclopedia of Mathematics is good articles of mathematics written by great contemporary mathematicians. If you're looking for older mathematicians the Great Books of the Western World series has volumes from ancient greece up to the 19th and maybe early 20th century. For 20th century great Soviet mathematicians surveying topics you can try Mathematics: Its Content Meaning and Methods

Georg Friedrich Bernhard Riema

Thanks

This was what I needed

John_Krampf

@GeorgFriedrichBernhardRiema You're welcome, glad to help. One last thing if you like Princeton Encyclopedia of Mathematics there is also Princeton Encyclopedia of Applied Mathematics and it's sufficiently mathematical to be of interest to mathematicians.

copper.hat

@John_Krampf nice & simple. basically an 'intermediate value theorem'.

John_Krampf

@copper.hat yep exactly. thanks I appreciate you said it's nice and simple, the highest compliment!

Georg Friedrich Bernhard Riema

I was wondering what would happen if the RH was false

havoc

copper.hat

07:01

@DarkRunner my version pastebin.com/FeR55t52

John_Krampf

@copper.hat Nice

copper.hat

@John_Krampf thx! not that it matters but with your 'connected' observation one could reduce the search space by a factor of 4.

Akash Karnatak

Can I get hints on, how to apply tremaux algorithm on this maze

Red mark is the starting point and green, end

This maze is open at some points, so consecutive intersections are confusing me

nathdwek

09:07

Hey guys, me again

Looking for a pair of fresh eyes to look at some quadratic programming/linear algebra issue I am having

Astyx

10:46

Hi Edward

I have a question if you have a minute:

Edward Evans

Hey @Astyx

Depends on the question but sure

Astyx

We know that if we have an extension $k\subset L$ it's possible to find subextensions that correspond to ramification/residual degree/decomposition of primes

Edward Evans

sure

Astyx

But is it possible to have subextensions that swap the order things are done? Say the lower extension is totally ramified, then we decompose, then we deal with the residual degree

Edward Evans

lemme think

So you have $L/k$ an extension, and inbetween you have these unique intermediate extensions that correspond to inertia, residual degree, decomposition

and you're asking if there are situations in which these guys are ordered differently?

Astyx

10:56

yes

Edward Evans

I don't think so, just because of how the intermediate extensions are defined

I need to give reasoning for this think though lol

Astyx

(nontrivial, arguably if f=1 you can do the totally residual extension anywhere)

@EdwardEvans My reaction as well lol

Edward Evans

Let's call $\mathfrak{P} \subset \mathcal{O}_L$ a prime lying over $\mathfrak{p} \subset \mathcal{O}_k$

You have the decomposition group $D_\mathfrak{P}$ and inertia group $T_\mathfrak{P}$. Do you know that $D_\mathfrak{P}/T_{\mathfrak{P}} \cong \operatorname{Gal}(\Bbb F_{\mathfrak{P}}/\Bbb F_{\mathfrak{p}})$ ?

Astyx

yes, that's how it was defined for me (the inertia group is the kernel of $D\to Gal$)

Edward Evans

uhhh

Nah I don't think you can because of the picture you have, all of the ramification occurs in the extension $L/K^{T_\mathfrak{P}}$, and all of the residual degree comes from $K^{T_\mathfrak{P}}/K^{D_\mathfrak{P}}$, and all of the splitting in $K^{D_\mathfrak{P}}/K$

That's not such a satisfactory answer I know, but these guys control the factorisation of your prime. There's no ramification occurring anywhere in $K^{T_\mathfrak{P}}/K$ by definition

Astyx

11:10

But you could have an extension that is not "aligned" with this specific tower

(I mean, you'd have to)

I gotta go eat, brb

Edward Evans

11:24

ergh I can't answer your question hahaha

Astyx

Haha no worries I doubt it's a trivial question anyways

Thanks for thinking about it!

DakkVader

11:46

Hey! How do i go about if i want to get (d/dE)(fft(E(t))

Or rather, i have (d/dE(t))(sum(E(t)E(t-tau)/(abs(E(t)E(t-tau))*scalar - E(t)E(t-tau))))

It looks messy, i know ...

the delay tau is done with ffts

Edward Evans

You need to TeX it, else nobody will read it

DakkVader

Can i do it here in teh chat ?

\sigma

Edward Evans

you need dollar signs

DakkVader

$\sigma$

is it tex for you ?

Edward Evans

tinyurl.com/cfqcvpc and you need to put this link on your bookmark bar

sorry there's a link on that link that says "start chatjax", you need that on your bookmark bar

DakkVader

11:50

ah ok ok i think i've got it now

hang on and i'll try to write my equation

this'll be messy, hope i avoid typos haha

Georg Friedrich Bernhard Riema

Riemann got a new mask

DakkVader

$\frac{d}{dE\left(t \right)}\left(\Sigma\|\frac{E\left(t \right)E\left(t-\tau \right)}{\|E\left(t \right)E\left(t -\tau\right)\|}\cdot \text{scalar} - E\left(t -\tau\right)E\left(t \right)\|\right)$ (just testing)

There it is!

The delayed copies of E(t) i do with fft $\text{ifft}\left(e^{i\omega\tau}\text{fft}E\left(t\right)\right)$

I'm not sure hope i shoudl calculate this derivative

I want to derivate with regards to E and not to t, the fft's i don't know how to tackle

basically $E\left(t\right)E\left(t-\tau\right)$ is a guess i'm doing. The scalar fits it to my measured data, and then i compare how close i was (by taking the difference). I want to use this in a gradient descent way but i'm not sure how to do the derivative

vesii

13:10

Hi guys, I have small question on Catalan numbers. I know that $C_0=1$ and $C_{n+1}=\sum_{i=0}^{n}C_i\cdot C_{n-1}$. If I want to calculate $C_1$ then I get: $C_1=\sum_{i=0}^{0}C_i\cdot C_{0-1}$. What $C_{-1}$ should be?

NoName

13:22

Do you mean $C_{n-i}$

Manolis Lyviakis

14:17

Can i have some examples where it helps to know the kind of discontinuity of a function then extent the function to a continuous one and work with that function and get a good result? A student of mine when i was helping with calculus 1 asked why does it help to know that the discontinuity is removable ? I asked cause then u can extend to a continuous function and get results. but the information would be for the extention of the function not the orginal discontinuous one.

DakkVader

15:15

pi = 3 = e

höhöhöhöhö

robjohn

$\lfloor\pi\rfloor=3=\lceil e\rceil$

3

NoName

Quick mafs

robjohn

@ManolisLyviakis That's true, but you can lose sleep over it, or not.

DakkVader

The golkden equation on which all of my research is based

Manolis Lyviakis

well it must be helpfull to know what kind of discontinuity your function has. Somewhere maybe to physics or somewhere.

Georg Friedrich Bernhard Riema

15:25

Got these new glasses

John_Krampf

@GeorgFriedrichBernhardRiema swag

robjohn

@GeorgFriedrichBernhardRiema cool! did you get the red-tipped white cane that goes with them, or did you leave it with the guy you got the glasses from?

Georg Friedrich Bernhard Riema

Forgot them @robjohn

I look CoOl

@DakkVader I can smell engineering

Thorgott

16:18

Riemann's theorem states that if a holomorphic function has a removable singularity, the continuous extension of the function is also holomorphic (and some more stuff)

granted, singularities in complex analysis aren't quite the same thing as discontinuities in real analysis

Manolis Lyviakis

link?

Thorgott

16:34

Removable singularity

In complex analysis, a removable singularity of a holomorphic function is a point at which the function is undefined, but it is possible to redefine the function at that point in such a way that the resulting function is regular in a neighbourhood of that point. For instance, the (unnormalized) sinc function sinc ( z ) = sin ⁡ z z {\displaystyle {\text{sinc}}(z)={\frac {\sin z}{z}}}...

Manolis Lyviakis

thanks!

Astyx

18:02

Sanity check: the annihilator of B/I is I

Tobias Kildetoft

@Astyx You mean as a $B$-module?

Astyx

Yes

Thorgott

if $B$ is a ring, then ye

Tobias Kildetoft

then yes, assuming $B$ has a unit

Astyx

Thanks

sonicboom

18:11

Does anyone here have some probability theory knowledge? I'm wondering if we have a sequence of random variables $X_n$ such that $X_n = n \mu + O_p(n^{1/2})$, can we say that the constant $\mu = O_p(1)$ and thus get $X_n = nO_p(1) + O_p(n^{1/2}) = O_p(n) + O_p(n^{1/2}) = O_p(n)$?

Essentially can we treat the deterministic constant $\mu$ as a degenerate/constant random variable that is $O_p(1)$?

Thorgott

whats $O_p$

sonicboom

It is like Big O but it is for convergence of random variables (en.wikipedia.org/wiki/Big_O_in_probability_notation)

Thorgott

I mean, a constant sequence is obviously $O_p(1)$ then, no?

for any $\varepsilon>0$, $\mathbb{P}(|\mu|>|\mu|)=0<\varepsilon$

but saying that $X_n$ is $O_p(n)$ seems much weaker than saying than $X_n=n\mu+O_p(n^{1/2})$

sonicboom

Yes its weaker but I actually have a fraction $(n\mu_1 + O_p(n^{1/2}))/(n\mu_2 + O_p(n^{1/2}))$ and this would allow me to say $(n\mu_1 + O_p(n^{1/2}))/(n\mu_2 + O_p(n^{1/2})) = O_p(n)/O_p(n) = O_p(1)$..I think

Mike Miller

18:30

I don't want to think about the probabilistic notion but that already fails badly for the usual big-O notation

n/1 is not O(1) but n and 1 are both O(n)

f in O(g(n)) means, roughly, that f is eventually smaller than (a fixed multiple of) g. Then if f in O(g), then flipping everything upside down says "1/f is eventually larger than a fixed multiple of 1/g"

The opposite of what you want

Thorgott

yeah, the estimate is definitely too weak, but I believe the original fraction may still be $O_p(1)$

it's definitely true in the deterministic scenario, cba to think about the details in the probabilistic case

Mike Miller

how is it definitely true in the deterministic scenario

$n^{1/2}/1$ is not in $O(1)$

Thorgott

ok, guess I implicitly assumed $\mu_2\neq0$

Mike Miller

ah sure

that's probably true huh

Thorgott

yeah, you write $\frac{n\mu_1+O(n^{1/2})}{n\mu_2+O(n^{1/2})}=\frac{\mu_1+O(n^{1/2})/n}{\mu_2+O(n^{1/2})/n}\rightarrow\frac{\mu_1}{\mu_2}$

Mike Miller

18:36

exactly

Thorgott

of course, this fails if $\mu_2=0$

probably same idea works in the probabilistic case

Mike Miller

yeah i'd be surprised if there was any issue

just seems like an extra epsilon to book-keep

user2103480

@sonicboom this means that (X_n - n \mu)/n^{1/2} goes to zero in probability, which is a clearer formulation imo. Also, to make this clearer, why don't you just give those things names to get more used to this? Say X_1, X_2 are in o(...) and you consider the quotient (n \mu_1 + X_1)/(n \mu_2 + X_2). Divide by square root of n above and below

Thorgott

is it the same as saying $(X_n-n\mu)/n^{1/2}$ going to $0$ in probability?

I was trying to figure out such a formulation, but the quantifiers didn't seem to match up

user2103480

Or by n, that makes it probably easier. Then denominator goes to mu_2 in probability, and numerator to \mu_1, and thus we get convergence of the joint vector (numerator,denominator) and this implies by the continuous mapping theorem that this whole thing goes in probability to mu_1/mu_2

If I don't overlook something obvious

@Thorgott yeah this should be like that per definition

Thorgott

18:50

to say $X_n$ is $O_p(a_n)$ means that for every $\varepsilon>0$ I can find $M,N>0$ such that $\mathbb{P}(|X_n/a_n|>M)<\varepsilon$ for all $n>N$. But to say $X_n/a_n$ goes to in probability is to say for all $M,\varepsilon>0$ I can find $N>0$ such that $\mathbb{P}(|X_n/a_n|>M)<\varepsilon$ for all $n>N$.

the former is weaker, I think

user2103480

Are we talking about small or big O?

ah shoot I looked at small O

Then I don't care to actually calculate through the stuff to see if it works analogously

sonicboom

@MikeMiller If we consider the expression 1 + n, how can 1 be O(n)? That statement seems to undermine all of asymptotic theory. If we consider the asymptotic behaviour of n+1 for n to \infty we have n+1 \to n precisely because n = O(n) and 1 = O(1).

user2103480

but for the small o, the heuristic @sonicboom mentions does work

Thorgott

1 is O(n) because P(1/n>M)=0 for all n and any M>1

Mike Miller

Do you know what O(n) means?

user2103480

18:55

@sonicboom I think you should start working with definitions and stop working with heuristics. Saying "Z = O(f)" is incredibly misleading if you are not used to the notion

sonicboom

Did you mean Op notation?

Thorgott

writing things as equal O of something is honestly one of the worst notational abuses in all of math if you ask me

sonicboom

@Thorgott That statement you wrote is for Op not O

user2103480

@Thorgott agreed

but never forget

mathematicians are not to blame

computer scientists propagate and uphold that shitty convention

Thorgott

my argument is for O_p, but that doesn't matter

Mike Miller

18:58

@sonicboom O and Op are the same basic idea, one in the context of probability. If you understand one you can understand the other. I think worrying about whether or not Thorgott wrote O or Op here kind of misses the point.

sonicboom

Oh sorry, yes of course 1 is O(n)! I don't know how I messed up there, I've been using O and o notation for years

Thorgott

for deterministic sequences, O_p is the same as O

Mike Miller

OK, I am sorry to pick on you. It's jut really important to understand the difference :)

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