Mathematics - 2022-03-10 (page 2 of 2)

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love_sodam

13:22

I have one simple question : What is the maximal possible value of $|Tr(M)|$ where $M\in SU(n)$? Here, $SU(n) = \{M\in GLn(\Bbb C)\mid MM^* = M^*M = I,\det M =1\}$.

Since $M$ is a unitary matrix anyway, there is $A\in U(n)$ such that $AMA^{-1} = D$ where $D$ is the diagonal matrix.

But the problematic spot is that $A$ does not necessarily contained in $SU(n)$.

Koro

@love_sodam 3?

love_sodam

Where did that number come from?

Koro

13:40

M is also an isometry.

and isometry’s eigenvalues have modulus 1.

This is what I thought when I looked at your question.

love_sodam

Ah yes eigenvalues have norm 1. Make sense. Thanks

Koro

:)

love_sodam

13:59

I think whether $A\in SU(n)$ or not is not important. Since trace is invariant under similar matrices, $Tr(A)$ is anyway $Tr(D)$ so maximum value is again $n$ this time.

Koro

I don’t yet know SU(n). What I said was keeping in view GL_n(C) only.

love_sodam

I wrote the definition at first.

Koro

14:19

Oh, okay. Fine then. I thought this was from 'Lie algebra'.

1 hour later…

robjohn

15:45

@Koro I saw modulus 1 and thought, "that's a simple ring"

Hey, @Ted!

Koro

16:05

^_^

unit 1991

16:15

Is open parallelepiped in $R^m$ open set? I am thinking taking $\delta$ to be minimum of $c_i-a_i,c_i-b_i$ where $c_i$-s are coordinates of our point

Saying open parallelepiped I mean $(x^1,..,x^m): a^i <x^i < b^i$

user193319

16:29

What does it mean for a group to act on a tree without inversion?

leslie townes

unit: if by that you mean that when c is given in the parallelipiped, you choose delta like that (except: b_i - c_i, not c_i - b_i), then whenever x satisfies |x - c| < delta then x will also be in that parallelipiped, then yes. you are blending superscripts and subscripts a bit but the idea is OK

robjohn

@leslietownes don't discriminate due to index location! Subscripts and superscripts feel marginalized already.

Voilet Flame

@leslietownes My proof should work if we require that $a_{n}$ are bounded away from, 0 right?

8 hours ago, by Voilet Flame

22 hours ago, by Voilet Flame

Given a dedekind cut $x$, If $x \gt 0$, I want to define the inverse of a cut $x^{-1}= \{ r \in \Bbb Q^{+}\mid \exists p \in \mathbb{Q}-X \ ( pr\lt1)\}$, it’s easy to check that it’s a cut, but how to prove that $xx^{-1}=1^{\ast}$ where $1^{\ast}$ is the usual dedekind cut cut for $1$, it’s easy to see that $xx^{-1}\subseteq 1^{\ast}$ but how to show the other direction?

8 hours ago, by Voilet Flame

15 hours ago, by Voilet Flame

I suppose it can be shown that there exists $a_{n}$ in $x$ such that $1/(a_{n}+1/n)$ is in $x^{-1}$ and then using the fact that $a_{n}/(a_{n}+1/n)$ eventually approaches $1$, but is there an easier proof especially one that avoided the Archimedean property and does my proof hold water?

@robjohn Does it work?

I think it does because the $a_{n}$ are bounded away from $0$, so Leslie Townes criticism does not hold.

unit 1991

16:46

@leslietownes thanks is there need to write that more rigorously or it's enough?

leslie townes

that's more a question for whoever you are writing this for. personally, i would include more detail if someone were looking at it later. but if this is self study i think you are done.

robjohn

@VoiletFlame no idea. I'd have to read up on Dedekind cuts.

Voilet Flame

@leslietownes Does my defence against your criticism work?

leslie townes

violet i am opting out of further dedekind cut stuff

Voilet Flame

Why

leslie townes

16:50

because it's spread across multiple hours of chat, because i linked to an answer on math.SE that solves this problem, and because i can

i looked in rudin to see what he did with this. he left it to the reader!!

if rudin leaves it to the reader, what chance do i have

Voilet Flame

its such a simple proof , it shouldn’t even take a minute for an expert like you,

I want to know if my proof is correct I am aware of the other proofs

robjohn

@VoiletFlame However, you do need to use \lt and \gt and \ast instead of < and > and * in MathJax in chat

They can be interpreted as HTML or Markdown

leslie townes

if you're interested in whether it holds in non archimedean fields why don't you try the proof in a non archimedean ordered field

Voilet Flame

But I first need to know if my argument is correct @leslietownes

leslie townes

testing it in a non archimedean field might help you evaluate that

Voilet Flame

16:56

How? My proof makes fundamental use of the Archimedean property(Twice in fact)

leslie townes

then i guess you don't expect it to hold in non archimedean fields. so why are you so concerned about avoiding the archimedean property?

Voilet Flame

Ok, I’m not. Is my proof correct? I think that’s the question that I should focus first

leslie townes

a lot of proofs about R really do use all of the properties of R. there's that theorem, there's only one dedekind complete ordered field

Voilet Flame

Ah! Is my proof correct?

leslie townes

as i said, i'm opting out of going into that

good luck

Voilet Flame

17:00

Ok, it’s correct thanks.

Alessandro Codenotti

@user193319 No element of the group swaps the two endpoints of an edge

Ted Shifrin

@robjohn I have never in my life used \lt or \gt shrug

amWhy

17:18

Yay, @robjohn! I like the outfit! I came to alert to the one week count-down! You beat me to it! :)

robjohn

17:34

@TedShifrin I had a problem last week with a < being mistaken for the beginning of an HTML tag

or perhaps it was a > closing a tag

leslie townes

<marquee> that sounds excellent! </marquee>

<blink> or does it? </blink>

robjohn

@leslietownes The < and > are more of a problem in the posts rather than the chat

I don't know if any of those tags are supported even on the main site.

<i>shrug</i>

leslie townes

this violates my first amendment rights. lawsuit forthcoming.

robjohn

yeah, they don't interfere in chat

amWhy

Sorry if interrupting, no immediate response needed. But do you know, @robjohn, how I can change the coloring of my keyboard gravatar to shades of green?

robjohn

17:38

@leslietownes we haven't restricted your right to say it, we are just laughing at you for saying it ;-)

amWhy

@robjohn Hah!

robjohn

@amWhy You have to edit the image and use the new image for your avatar

amWhy

@robjohn I figured that, and ensure "my regular" is saved on my computer. Would that editing the image through, e.g., photoshop, or paint, or something like that?

leslie townes

you'll all look very foolish when i put the entire system on trial

Xander Henderson

Because it only took 10 seconds with GIMP (with 8 of those seconds being the time it takes for the software to load):

user image

leslie townes

17:42

my avatar is already irish, and there's some green in it.

amWhy

@XanderHenderson So cool! Thanks!

Xander Henderson

Were I a St Patrick's Day celebrating kind of person, I think that I would opt for orange, rather than green. I feel like I come down more on the side of the Protestants than those Papist blowhards.

On the other hand, Purim is next week. Maybe I should give my avatar a tiara?

Purim is a great holiday. I think that people use St Patrick's Day as an excuse for drinking, whereas you are more or less commanded to drink on Purim, until you can no longer remember Haman's name. Why tolerate public drunkenness when you can actively encourage it?!

I have a precalculus lecture to give in 10 minutes. Spring break is next week, and I finished up a chapter on Tuesday. I don't want to start a new chapter, so I have decided to riff for 70 minutes on matrices (the chapter we just finished introduces vectors; matrices were supposed to have been introduced in the prereq class)

I am hoping to show them how a matrix acts on a vector (by scaling and rotating and whatnot), I want to show them how the angle sum formulæ "just pop out" of this interpretation, and I want them to see the determinant as an area. None of this is vital to the class, and I don't plan to test them on it---I just want them to see something interesting before spring break.

Are there any other easy results which might be interesting?

Again, my plan is total improvisation today, because f*** it, spring break is next week and I don't care.

leslie townes

ooh, thanks for reminding me. gonna put an out of office message saying SPRING BREAK!!! on my office email.

Xander Henderson

I post an agenda at the beginning of class, noting what I plan to cover. Today's agenda: "whatever pops into my head". Because holy s***, I just cannot be arsed today.

leslie townes

run down the top 10 viral videos of the week

Xander Henderson

17:57

@leslietownes I'm not "with it" or "hip" enough to know what those videos might be.

:/

Alright... time to "teach".

leslie townes

it's mostly ted shifrin doing interpretive dances to illustrate differential forms and complex geometry

and my 20-minute supercut of funny car accidents

PM 2Ring

18:22

Last week, I posted a song from an all-girl Ukrainian grunge rock group, The Sixsters. Here's their latest song. They probably won't be posting more songs for a little while: their drummer's house got trashed. Fortunately, they got out in time, so nobody was hurt.

The Sixsters - Правда (official audio 2022)

robjohn

18:46

@amWhy here is another. Might have better resolution, not that it matters that much.

Ted Shifrin

18:59

@robjohn Well, I don't usually expect my LaTeX formulas to end up in html :P

robjohn

19:41

They can if they end up on the main site here.

Ted Shifrin

Hmmm .... so not chat stuff, but main site questions/answers. I doubt I'm that popular :D

M17

20:22

Is it permissible to talk about things other than mathematics here? Because the last messages in the room are about other things

If (a + b)^2 we know that it has a general rule for its square and if the exponent is 3 then there is a general rule, my question is what is the highest exponent with a general rule in (a+b)^n?

Xander Henderson

@M17 Yes.

From the room description:

> Associated with Math.SE; for both general discussion & math questions alike.

@M17 I don't understand your question. Are you asking about the Binomial Theorem?

Binomial theorem

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial (x + y)n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. For example, for n = 4, The coefficient a in the term of axbyc is known as the binomial coefficient (...

M17

20:40

(a+b)^2=a^2+2ab+b^2

(a+b)^n What I mean is the highest number n has a general rule that we follow

@XanderHenderson

Vrouvrou

please i need help

Lukas Heger

there’s a rule for all n @M17

Vrouvrou

i have to find the series of $\sqrt{\ln(1+x)}$ at $\infty$

Xander Henderson

@M17 I still don't understand your question. Did you read the link I provided?

Vrouvrou

i put $u=\lfrac1x$ then $x=\frac1u$

M17

20:43

Sorry, I was typing my message while sending you the link

Xander Henderson

@Vrouvrou I am dubious about your statement here. What kind of series expansion? Doesn't the fact that this function diverges at infinity cause some problems around infinity?

(or, after you make the change of variables, you have a problem at $u=0$, as $1/u$ is undefined).

Vrouvrou

the exact question is the series of $\sqrt{\ln(1+x)}-\sqrt{\ln(x)}$

Xander Henderson

That doesn't look like an "exact question", as I see no question mark. Can you reproduce the entire text of the question?

M17

(a+b)^2=a^2+2ab+b^2

Vrouvrou

Guives the taylor series (développement limité in french) of $\sqrt{\ln(1+x)}-\sqrt{\ln(x)}$ at $\infty$

M17

20:48

I want to know how this was solved mathematically as well, what are the steps because I know the rule, but I know how this was achieved

@XanderHenderson

(a+b)^2=a^2+2ab+b^2

Xander Henderson

@M17 I still don't understand your question. How what was done? The steps to what?

The fact that $(a+b)^2 = a^2 + 2ab + b^2$ follows from the distributivity of multiplication over addition, and the fact that both multiplication and addition are commutative in the reals.

Vrouvrou

have you an idea @XanderHenderson

M17

I meant how to deal with the exponent with the expression inside the parenthesis

@XanderHenderson, yes

Xander Henderson

@Vrouvrou Not right off the top of my head, no, but I would not try to develop series expansions of each term separately---rather, I suspect that if you do some clever algebra first, you'll get something which behaves nicely.

Vrouvrou

like what ?

Xander Henderson

20:53

@Vrouvrou Like I said, I have no idea off the top of my head.

Vrouvrou

but why Algebra ?

Xander Henderson

@Vrouvrou Because my gut tells me that if you work out a Taylor series expansion at $a$, you will see some telescoping or other cute algebraic tricks with kill off a bunch of terms. After working out the general series, I suspect that you will be able to take $a$ to $\infty$ (or, really, to take $a \to 0$ and think about the expansion around $1/a$).

But that is just a gut reaction. I genuinely have no idea off the top of my head how to tackle that problem, and don't have the energy to think about it right now.

M17

My second question, is there a natural number that has only three prime factors? We do not count the same number and one among the factors

Xander Henderson

@M17 There are many numbers with exactly three prime factors. Choose your three favorite prime numbers, say $2$, $3$, and $47$. Their product is a natural number with exactly three prime factors.

M17

2×3×47

You mean?

What mean "What is their product?"

@XanderHenderson

M17

21:11

As long as general discussion is allowed here as well,
Why in sports when any external event happened was always separated from that, why is sports not separated now?

2 hours later…

anak

23:08

@Jakobian It's concerning pedagogy and accessibility.

robjohn

@Vrouvrou $\sim\frac1{2x\sqrt{\log(x)}}$

mick

23:28

0

Q: Show that If $x^3 x = x^2 x^2 $ then $x$ is power associative?

Consider a unital commutative algebra $A$. Let $x$ be an element of $A$. If $x^3 x = x^2 x^2 $ then $x$ is power associative. How to prove that ?

abstract-algebra commutative-algebra

rob john where is your mean square ???

amWhy

23:55

@robjohn Nice! Thank you!

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