Mathematics - 2022-06-24

Bald eagles are not actually bald; the name derives from an older meaning of the word, "white headed". The adult is mainly brown with a white head and tail. The sexes are identical in plumage, but females are about 25 percent larger than males. The yellow beak is large and hooked. The plumage of the immature is brown

leslie townes

that makes less sense. they're not white eagles. unless 'bald' for hair has the same origin, in which case i'd bet the name came through that.

Joe Shmo

well their identifying feature is the white feathers on their head

(which if anything should make it clear that they are anything but bald)

but, as it turns out, they are balde.

leslie townes

well when i wrote "they look bald" above, i didn't mean literally bald.

Devansh Bhardwaj

Middle English: probably from a base meaning ‘white patch’, whence the archaic sense ‘marked or streaked with white’. Compare with Welsh ceffyl bal, denoting a horse with a white mark on its face.

origin of bald

leslie townes

most birds have more than one identifying feature. in many places you can identify juvenile bald eagles by their size and behavior. they are huge and they hunt fish and they are not ospreys. that's enough to narrow it down.

Joe Shmo

2:40 AM

they are truly a majestic animal

gorgeous bird

also lethal

hide yo dogs

leslie townes

someone, i think it was ben franklin? at least apocryphally? wanted to make a turkey the national bird.

it might have made more sense, but looked less cool on military and state logos.

Joe Shmo

come on would you take seriously a storming marine with a turkey emblem?

leslie townes

i'd have to judge that on an emblem-by-emblem basis.

Joe Shmo

indeed

Koro

Suppose I’m alone in a wide open space. Some 6 to 10 eagles are flying above. Should I worry?

leslie townes

2:44 AM

i wouldn't think so.

Joe Shmo

you're dinner.

geocalc33

only if your a baby

Joe Shmo

6-10 of them at a time could actually take you

(do they work in teams?)

in fact my money is on the eagles if youve got nowhere to run

geocalc33

good question they might cooperate in hunting to some degree

this somehow reminds me of the guys who dressed up as a zebra and ran around and got attacked by lions

Joe Shmo

as one does

robjohn

4:02 AM

@JoeShmo If they try to work in Teams for Windows, they usually crash in Windows.

Devansh Bhardwaj

4:30 AM

Hi. I have a symmetric system of four nonlinear equations which I could find one solution for by setting three terms equal; but I cannot solve it analytically

May I share? It is needed to answer a question

Mad Spaces

Good morning dear Friends

I have a question regarding Multilinear algebra,

Devansh Bhardwaj

Eagles consider humans as predators in general; so it would be rare for predator to get predated. That said, eagles are variable in size, and larger ones may hunt down; though going violent can deter them

Mad Spaces

4:49 AM

We defined the wedge product (outer product) of a linear mapping
$ \varphi: V \rightarrow W :\, V,W \text{Vectorspaces} \\ \bigwedge^2 \varphi :
\bigwedge^2 V \rightarrow \bigwedge^2 W \\ v_1 \wedge v_2 \mapsto \varphi(v_1) \wedge \varphi(v_2) $
Furthermore, we have stated that for such Function, we obviously recieve the Functionsrepresentative matrix in relation to some arbitary bases in the selected vector spaces. Let these be for example
$B,C \\$
It is now said that the wedge product of such matrix, (which i do not really understand, is it the same as taking the wedge product of a funct…

I am trying to prove that statement, however, i am having difficulities since i do not have a grasp on the notation.

I believe the last statement should say less or equal than 1

robjohn

Ted Shifrin

@Mad Start with linear maps $\Bbb R^2\to\Bbb R^2$. Then do $\Bbb R^3$.

Mad Spaces

I have already showed (proved) that for a enodmorphism in any field of F^2, the defined operation is equivalent to multiplying the wedge product of two vectors with said determenant of a 2 x 2 matrix. ( so i am guessing thats what your suggesting, but before helping me with the proof, can you expand on the notation)

Ted Shifrin

OK, next case.

What notation? That’s matrices with respect to bases …

Mad Spaces

An additional questions that arises, since it is not well written in my scriptum, is it a necessity that the vector spaces be of same dimension? IE $dim V = dim W = n $

Ted Shifrin

5:02 AM

No, they need not.

Do 3 to 2 and 2 to 3.

Mad Spaces

Are you suggesting induction as a method of proof?

Ted Shifrin

No. I’m suggesting you understand by working examples.

Mad Spaces

In the notation, we write $ \bigwedge^2 C$ i am not sure i understand this notation, we can take a wedge product of a vector space, $C$ is merly a basis (set bare of structure)

leslie townes

C is definitely a vector space. in fact, it is a number of them.

i'm late to this, i just wanted to make that clear.

Mad Spaces

Do you imply we take the "linear span" of C?

Ted Shifrin

5:06 AM

They mean the basis for the exterior power you get from an ordered basis for the space….

No it’s a basis

Take $\{e_i\wedge e_j: i<j\}$

Mad Spaces

Oh Ted i believe i understand what you mean!

I see i see...

Now the notation makes more sense.

We are building the representing matrix of $\bigwedge^2 \varphi $ in relation to the basis in the outer product space.

Ted Shifrin

Right.

Mad Spaces

What would be a good method to start a general proof of the statement mentioned? Brute forcing with indicies ?

Ted Shifrin

Understand examples and you’ll have it.

Mad Spaces

Alright, thank you.

As a last note, i tried to search this on the web before asking teh question, i wrote terms like " Wedge prouct of a linear function " and what not , seems i never found what i looked for, are there some english terminology for said topic?

Ted Shifrin

5:11 AM

Induced map on exterior algebra

leslie townes

exterior algebra is the biggie

oh ted is there

Mad Spaces

Thank you :=)
Have a good day, time to go to work :) or sleep tight i guess.

Ted Shifrin

LOL

leslie townes

ted: was reading online today about the closure of a restaurant on college ave (shen hua near where ozzies soda fountain used to be, probably after your time, started late 90s) and it heartbreakingly mentioned as an aside that another restaurant on college (that my wife and i used to live next to and ate at) had already closed

Devansh Bhardwaj

One solution I found was $a=b=c=\frac{3}{2}$

But could not solve analytically

Tried analytically to get $96\frac{a+b}{c^2}+64\frac{a+b}{c}+\frac{96}{c}=24$, and other such combinations

Mad Spaces

5:17 AM

Is this a topic concerning laplacian border conditions ( initial conditions )

Devansh Bhardwaj

@MadSpaces IDK just Lagrangian Multipliers to solve an ineq.

leslie townes

sometimes it can be helpful to reparametrize the constraint set so that a constrained optimization problem becomes an unconstrained optimization problem in a smaller number of variables. it might help to state the original problem if you are posting on SE.

generally the lagrange multiplier equations just are what they are, and sometimes can be quite ugly to solve.

Devansh Bhardwaj

@leslietownes Actually I was trying to answer a question

about inequalities

Mad Spaces

Yes i meant Lagrangian, not laplacian.

I dont evn think such thingv as laplace border conditions exist. anyways, good day must run to work now.

Devansh Bhardwaj

math.stackexchange.com/questions/4478525/…

This is the inequality

Raphael J.F. Berger

6:40 AM

Hi there, I posted the Q math.stackexchange.com/questions/4477140/… 2 days ago but haven't got any resonance as yet. Has someone some idea what I could improve with the question or on the topic itself?

one potato two potato

9:09 AM

Basic algebraic topology written in category theory language is hard to understand although I already know what that is.

Not Tfue

10:50 AM

anyone knows what that vertical line $x=0$ means?

Calvin Khor

evaluated at x=0 after differentiating

Not Tfue

anyone knows what that vertical line $x=0$ means?

@CalvinKhor thanks

Calvin Khor

np

Not Tfue

I am an engineering student so is there any advice for someone who is changing their major to mathematics.

I have like 50 page experience in rudin lol.

And recently started real group theory.

i am going to second yr

robjohn

1:01 PM

@NotTfue so assuming the series converges, the whole expression is $a_1$

unit 1991

1:50 PM

Hi. Is there a undergraduate math class that has differential geometry as prerequisite?

Joe Shmo

3:09 PM

quack quack: youtube.com/watch?v=kotWv4MCxNI

user10478

4:46 PM

I am confused about order statistics. In this explanation (en.wikipedia.org/wiki/Order_statistic#Notation_and_examples), how is the transition made from observed values to random variables? When I see observed value notation ($x_{(1)}...x_{(n)}$), it makes sense to find the max, min, or any other function of the set of observations, but when I see random variable notation ($X_{(1)}...X_{(n)}$), this makes little sense.

Here it seems the goal is to define a meaning to the notion of order for the probability distributions associated with each random variable, just as one must manually define a meaning to the notion of order for objects of user created types in a programming language.

Ted Shifrin

5:42 PM

@unit1991 Not typically. Perhaps a rare second course in differential geometry. More typically, graduate level material comes next. Differential topology is sometimes taught undergrad, but it is a parallel course depending more on multivariable analysis.

user193319

6:11 PM

Let $X$ be a topological space (compact Hausdorff if necessary). Let $Homeo(X)$ denote the group of all homeomorphisms of $X$ equipped with the compact open topology, and let $A$ be some subspace of $X$. Is it true that $\overline{ \{f \in Homeo(X) : f(A) = A\}} = \{f \in Homeo(X) : f(\overline{A}) = \overline{A}\}$? It's clear that the set on the LHS is contained in the RHS, but what about the other inclusion?

I tried proving it but it isn't clear that it is true (in fact, I am currently trying to find a counterexample).

Alessandro Codenotti

Interesting question

Sounds plausible to me

user193319

Sounds plausible as in true, or plausible that there is a counterexample?

I would like for it to be true.

Alessandro Codenotti

The former

Ted Shifrin

I think I have a counterexample.

Alessandro Codenotti

Let's think about the case in which $X$ is compact metric so that the compact open topology is metrized by $d(f,g)=\sup_{x\in X}\{d(f(x),g(x))\}$ because it's nicer

Ted Shifrin

6:19 PM

I think you can have a homeo on $A$ whose extension to the closure is not bijective.

Oh, I don’t have it right.

user193319

@TedShifrin Maybe I'm misunderstanding you, but $f$ is assumed to be a homeomorphism on the whole space.

Well, unfortunately the space I'm dealing with is not metrizable...but maybe it'll be clear how to generalize it by looking at the metrizable case.

Ted Shifrin

Yeah, I messed up. But is the limit of homeos necessarily a homeo? That’s false.

Alessandro Codenotti

@TedShifrin Limit in which topology? But yes that's usually false

user193319

Probably uniform convergence. I know compact-open topology can coincide with the topology of uniform convergence, so I guess $x^n$ on $[0,1]$ would be an example, right?

Oh, wait...

No, that doesn't converge uniformly...it converges pointwise, but that's no relevant.

Ted Shifrin

I was thinking $f(z)=z/n$ on $\Bbb C$ or the Riemann sphere.

AMDG

6:48 PM

Hey, can anyone help me locate terminology for operating on graphs?

Specifically, the formal names for operations of converting a node into two, and making two or more nodes into one.

copper.hat

7:13 PM

split, coalesce?

Mad Spaces

7:34 PM

how is the formula for $sign(\sigma)$ as a product derived?
$ \Pi_{1\leq i < j \leq n} \frac{\sigma(j)-\sigma(i)}{j-i}$
For $ \sigma \in S_n$

Ted Shifrin

I've never seen such a formula. And it is clearly wrong, as typed. You can't have $i=j$.

Mad Spaces

Now its correct

Good evening Ted!

Ted Shifrin

Maybe. Convince me that it's always $\pm 1$.

Mad Spaces

I want to convince myself. LOL!

Alessandro Codenotti

One can convince himself of many wrong things, but it's harder to trick Ted

Ted Shifrin

7:37 PM

If it was obvious that $\text{sign}(\sigma\tau) = \text{sign}(\sigma)\text{sign}(\tau)$, that would be helpful.

Mad Spaces

Vorzeichen (Permutation)

Das Vorzeichen, auch Signum, Signatur oder Parität genannt, ist in der Kombinatorik eine wichtige Kennzahl von Permutationen. Das Signum einer Permutation kann die Werte + 1 {\displaystyle +1} oder − 1 {\displaystyle -1} annehmen, wobei man im ersten Fall von einer geraden und im zweiten Fall von einer ungeraden Permutation spricht. Es gibt mehrere Möglichkeiten, gerade und ungerade Permutationen zu charakterisieren. So ist eine Permutation genau dann gerade, wenn die Anzahl der Fehlstände in der…

I know its german, but the english version does not have the Formula

Obviously, because we germans are mathematically superior

Ted Shifrin

Yeah, right.

Mad Spaces

Scroll to the part where it says "Darstellung als Produkt"

Ted i found your proof, and in doing so, i found my proof

Ted Shifrin

I find it less convincing that their example has only $n=3$, so that all the terms are $\pm 1$ to start with.

Mad Spaces

Since you can go the other direction ))

de.wikiversity.org/wiki/Permutation/…

Also German, but its obvious

Ted Shifrin

7:39 PM

Anyhow, your wiki link proves the product formula. From that it should follow immediately by checking a transposition.

Mad Spaces

Well, thats done!
Nice

Ted Shifrin

It's curious that in my 50+ years of being a mathematician, I've never encountered that formula. I wonder if it's useful for some reason.

Mad Spaces

Well Ted, now you did. Thanks to me. surrender old man, its time for the young and dumb

Ted Shifrin

There's lots of useless stuff out there. One can't be bothered to know all of it.

Ethakka appam with Chai

7:55 PM

Hallo leute

wie gehts euch

Mad Spaces

Hallo Kamerad

Ethakka appam with Chai

Ja wazzup

Mad Spaces

your name is GREAt dude!

kuddos!

Ethakka appam with Chai

Ich kann dich nicht verstehen. Ist es wirklich so dass die titel nicht in die Englischen artikel gibt?

Wirklich? Kommst du aus Kerala?

Mad Spaces

Your german is not bad.

But Flawed.

Ethakka appam with Chai

7:59 PM

hahahahahaha :D

Bitte korriegieren mich

Mad Spaces

I am not sure what you meant to be honest

Ethakka appam with Chai

I can't understand you, is it really so that the title isn't there in the English article?

second message: Please correct me

Mad Spaces

but what do you mean with title?

Ethakka appam with Chai

I should have said formular

Mad Spaces

anyways it is "der Titel" und "in dem englischen Artikel"

Ethakka appam with Chai

8:02 PM

Noting that down

Mad Spaces

more colloqually you would say "aufm englischen"

Ethakka appam with Chai

The noun gender is the hardest part of this language xD

Mad Spaces

Yes yes, it is not easy.

Gotta run, gladly talk later. See you.

Ethakka appam with Chai

are you a native speaker?

Okie see u

robjohn

8:29 PM

@TedShifrin it is curious

Ted Shifrin

8:58 PM

@robjohn I gather you've encountered. Have you used it for something interesting?

Julius

Hello. Suppose that $a$ and $b$ are $1\times d$ real vectors and $U_1,U_2$ are $d\times d$ orthonormal matrices (potentially $U_1=U_2$). I am looking for operations $\star$ such that $(aU_1)\star(bU_1) = c(a\star b)U_2$ for a constant $c$. So far I only see that $\star$ can be the addition operator. Am I right that that is all?

Ted Shifrin

I don't think so. Certainly $a\star b = ma+nb$ works for any real numbers $m,n$, doesn't it? Provided $U_2=U_1$ and $c=1$.

Julius

Ah yes, sure. And apart from that? It doesn't seem like any nonlinear operations are possible.

Ted Shifrin

Well, there aren't many nonlinear operations you can do on vectors in the first place.

Julius

For instance, for my purposes the Hadamard (elementwise) product would be great, but it doesn't seem to satisfy this property.

Ted Shifrin

9:09 PM

The Hadamard product is very basis-dependent. It is not, as we like to say, intrinsic.

But I have no idea how to prove any of this.

Julius

I see. Thanks.

Joe Shmo

suppose I have a root finding algo

can anyone think of anything that will give me the first root?

Ted Shifrin

Huh?

Joe Shmo

suppose $f \in C^0[0, 1]$ has multiple roots

Balarka Sen

@MadSpaces lmao

Ted Shifrin

9:13 PM

What does "first root" even mean?

Joe Shmo

first root $= inf\{f = 0\}$

Ted Shifrin

Ah, smallest. Inf = min. It's a closed set.

Joe Shmo

yes

youll also have finitely many 0s

Ted Shifrin

I will?

Joe Shmo

this is a practical question, not a math question

hence first

Ted Shifrin

9:17 PM

So, look at the output and take the smallest number.

Joe Shmo

the output converges on an arbitrary root

not all of them

(false position)

Ted Shifrin

No clue.

I suppose you could, with a "real-world function," divide by some power of $x-c$ and do it again. But who knows what power will give you a continuous quotient.

Joe Shmo

o I like that

what is c?

Ted Shifrin

(By continuous quotient I meant, of course, to remove the removable singularity .... once you know it's removable.)

$c$ is the root you found.

Joe Shmo

nah too slow

Ted Shifrin

9:22 PM

I revert to my "no clue."

Joe Shmo

9:44 PM

I mean the other way of doing this is just rerunning the algo on $[0, x_0)$ $x_0$ being the root you found, and repeating until you dont have any more

Mad Spaces

ny general rule of thumb to when functions commute under Composition? Biliniarity, or something idk?

Joe Shmo

you would generally not expect them to commute

leslie townes

the domain and hypotheses matter. carl cowen had a good paper on commuting analytic maps of the disc to itself called "commuting analytic functions." it gives a taste of what is out there.

lots of papers cite that paper.

Mad Spaces

Okay thank you this is then out o the scope of this proof and thus for me momentarily not relavant

leslie townes

operator theorists love learning when things commute with other things. sometimes it ain't easy.

Ted Shifrin

9:50 PM

I prefer skew-commutativity.

Mad Spaces

Teddy bear help me plS

leslie townes

we should enjoy it while it lasts, most states are going to outlaw things commuting or skew-commuting with other things.

Joe Shmo

I never found commuting enjoyable to begin with

skew-commuting sounds intriguing, however

Mad Spaces

I am trying to prove the universal property of the Tensor product for higher dimensions.
We had it defined for 2 Dim. i am trying to do it iteratively for n dimensions. So i was thinking of for example combining sets for example V x V x... x V n times to be V^n-1 x V and then i can do the properity, and then i thought maybe something like
V^n-2 x (V^n-1 ^V^n)
an then iteratively use the properity, i was then trying to think of a way of reaching the final function that is supposed to satisfiy it

with v^n-1 i mean like the vector space that is contrsucted by X ing the V, notationabuser begone

Ted Shifrin

@leslietownes And you haven’t started on associating!

Mad … are you misusing dimension?

Mad Spaces

10:03 PM

No no.. with n i meant the times i am taking the cross or the tensor product

Because as said, we defined it on two vector spaces, every bilinar map has only one linear map such that yadda yadda? i am trying to show this for n vector spaces

Ted Shifrin

Same vector space?

Mad Spaces

yes

Ted Shifrin

Your use of language is very sloppy, mr superior.

Mad Spaces

Sigh.. i will write it CLEARLY...

Ted Shifrin

It must be induction ….

Joe Shmo

10:07 PM

thank you, o benevolent kind sir

Sha Vuklia

Ted, can I share sth random?

Ted Shifrin

Only if you answer Mad’s question.

Mad Spaces

$\text{Let $V,W$ be vector spaces, then the universal properity is} \forall \varphi \in Bilin(V \times W , M) \exists! \mu \in Hom(V \otimes W,M) \varphi = \mu \circ \tau $ where as $\tau$ is the tensor product

Sha Vuklia

yea my brain is just (algebraic) topology atm

Ted Shifrin

Yes, so you need to use it for different $W$, namely the $(n-1)$-fold tensor product of $V$.

Mad Spaces

10:10 PM

Exactly thats what i meant

But then , you get only a bilinear mapping, between some vector of the n-1 and $V$
But we want to be multilinear, so i thought iteratively and some composition mumbo jumbo would do the trick

Sha Vuklia

Tu has a nice section in his book DiffGeo on tensor products, but I suppose that's cheating

Mad Spaces

My experiance with these subjects, it is more time efficient to prove it alone then to look up books and get a grasp of what the hell the author is talking about and whats his notation.

Ted Shifrin

Inductive hypothesis … Start with a multilinear, then get a bilinear thing with the tensor product and $V$. In other words, think about a multilinear map and fix the nth input.

leslie townes

why not just stop at 2? let's not be greedy. think of icarus and daedalus.

Ted Shifrin

2 is higher than I can count. And no more associating.

Sha Vuklia

10:14 PM

maybe in the short term, but having books as your "extended brain" in the long run defo works efficiently for me

(though I'm ofc pro proving things yourself)

leslie townes

i can't find tensor products anywhere in the constitution.

Balarka Sen

The constitution itself is sufficient to make you tensor.

Joe Shmo

I can confirm that my reading of the US constitution does not protect tensor products either.

tensor products, beware.

leslie townes

i'm going to hassle my state representatives to ban n-fold tensor products higher than n=2.

Joe Shmo

that sounds like a good idea

and the national anthem clearly indicates that this is the land of the free and home of the brave.

it doesn't say anything anywhere about tensors.

so I'm with you. it should all be outlawed

and can somebody bring back the Ohio law that writes into law that $\pi=3.14$? in practice you only need 2 decimal places anyway

Ted Shifrin

10:19 PM

Nah, 22/7 in Alabama.

Joe Shmo

Ah, The Indiana Pi Bill !

Indiana Pi Bill

The Indiana Pi Bill is the popular name for bill #246 of the 1897 sitting of the Indiana General Assembly, one of the most notorious attempts to establish mathematical truth by legislative fiat. Despite its name, the main result claimed by the bill is a method to square the circle, although it does imply various incorrect values of the mathematical constant π, the ratio of the circumference of a circle to its diameter. The bill, written by a physician who was an amateur mathematician, never became law due to the intervention of Professor C. A. Waldo of Purdue University, who happened to be present...

think about how many conjectures we could settle this way !

The constitution doesn't say anything about pi

peek-a-boo

my second grade teacher told me $\pi=3.14$ and my third grade teacher told me $\pi=22/7$, and I was so confused that whole year because $22/7\neq 3.14$

leslie townes

i don't have memories of pi until high school

you must have had good teachers, to lie to you about pi so early on

peek-a-boo

:)

Mad Spaces

Time to crawl up in bed and cry myself to sleep

Good night!

Joe Shmo

10:32 PM

sweet dreams! come back to visit us soon!

Ted Shifrin