Mathematics - 2017-05-11 (page 1 of 3)

LegionMammal978

00:42

facepalm

Just realized that $\sin^2x+\cos^2x=1$ trivially follows from the definition of the unit circle

Akiva Weinberger

It's called the "Pythagorean identity" for a reason

arctic tern

04:38

@Danu Allegedly physicists write $e^{iX}$ (with an extraneous $i$) for $\exp(X)=\sum_{n=0}^\infty X^n/n!$. Is this true?

Leaky Nun

@arctictern where did you see that?

arctic tern

first paragraph of this answer

Balarka Sen

04:56

@arctictern O_o

PVAL-inactive

05:38

@Mike Do the sequences $z \mapsto nz$ (or $z \mapsto z^n$) converge to as maps $\Bbb CP^1 \to \Bbb CP^1$?

I imagine the answer is nothing

SL_2(\Bbb R) is connected but open.

Danu

@arctictern Yes, that is correct.

arctic tern

Danu

I am pretty sure that this derives from the following combination of facts: (i) physicists like to work with $\mathfrak{su}(n)$ (ii) physicists like to work with Hermitian (as opposed to skew-Hermitian) matrices.

Hermitian matrices represent observables in quantum mechanics, in case you didn't know.

If one wants to describe a particle with spin 1/2, for instance, $\mathfrak{su}(2)$ is useful.

arctic tern

for me hermitian matrices are a linearized version of projective geometry: every hermitian matrix is uniquely a linear combination of projection maps (onto pairwise orthogonal subspaces). the other kind, skew-hermitian matrices, are a linearized version of the oriented grassmanian manifold Gr(2,n): every skew-hermitian matrices is uniquely a convex linear combination of right-angle plane rotations (with respect to mutually orthogonal planes).

(this is in the real case of course) (convex meaning nonnegative scalars)

Danu

Right. So projective stuff is related to quantum mechanics (since in quantum mechanics the states of a system are described by rays in a Hilbert space)!

arctic tern

05:47

mmhmm. so Baez tells me.

Danu

Is it in the octonion paper? Or another one?

PVAL-inactive

@Mike According to a comment here math.stackexchange.com/questions/633567/… The complex automorphisms of a complex torus are always $GL_n(\Bbb C)$ with the automorphisms of the lattice. In particular in the standard lattice this will contain $GL_n(\Bbb Z)$ as a subgroup (I really mean n here not 2n) which will be an infinite countable group.

arctic tern

I think it's in the octonion paper, and in one of <this week's finds>

Danu

Right. Baez always seemed to be right on the border between math and physics.

Until he switched to really different stuff :D

BAYMAX

06:13

V= set of all solutions of the differential equations $y'' + by' + cy = 0$ where b,c are real numbers satifying $y(0) = y(1)$ ,then dimension of V is ?

like there are two solutions to the above homogeneous ODE and other possible Linear combination of them are also solutions to the ODE and hence dimension is 2 as the two solutions generate the other solutions and they span the whole set of solutions but then

when we apply initial conditions $y(0) = y(1)$ does that affect the dimension ?

Leaky Nun

I think that the initial conditions narrow down the solution set to one solution... so the dimension is ???

BAYMAX

1

Leaky Nun

I would say that a point has no dimension...

BAYMAX

but I thought of like this - the two solutions may cross each other

like x axis and y axis

still they are the basis of whole RxR plane

it is not a point

they are the solutions!

Leaky Nun

are there two solutions?

BAYMAX

06:19

It is a 2nd order ODE

arctic tern

If $y$ is a solution then so is $ay$, where $a$ is a scalar. (It still satisfies the initial conditions.)

Leaky Nun

Oh, then the dimension is $1$

arctic tern

@BAYMAX If you put enough initial conditions on a DE, it will only have one solution (or even no solutions).

BAYMAX

yeah

Leaky Nun

@BAYMAX maybe you meant two linearly independent solutions, which is not present here...

please do not use ambiguous terms

BAYMAX

06:21

so if in the question I will not mention the initial condition will the dimension still be one ?

ok will take care @LeakyNun

arctic tern

no

without initial conditions, the solution space is spanned by exp(rt) and exp(st), where r and s are solutions to x^2+bx+c=0.

(well, assuming x^2+bx+c is separable, otherwise the answer needs amending)

BAYMAX

So if one specifies initial conditions then it is mandatory that the dimension will reduce ?

like it is intuitive

Leaky Nun

Can anyone figure out what this means?

BAYMAX

they are certain characters of a cartoon ?

every1 has a face

arctic tern

those are letters, not people

Leaky Nun

06:26

I tend to believe that it is graffiti

BAYMAX

oh

Maximum of {$x+y : (x,y) \in \bar{B}(0,1)$}

Leaky Nun

What is $\bar{B}$?

arctic tern

presumably closed ball of radius 1 centered at 0

BAYMAX

I know that maximum of $f(x,y)$ always occurs on the boundary of $B$

Closure of $B(0,1)$

oh

Leaky Nun

The maximum is the supremum which is $\sqrt2$

arctic tern

06:38

:37317958 yes there is a maximum

cuz the ball is closed

well, don't just give the answer :P

Leaky Nun

just draw a graph

arctic tern

yes, x+y is the length of the projection of (x,y) onto the line y=x.

BAYMAX

projection of a point onto a line

?

arctic tern

yes

(orthogonal projection)

that's for geometric intuition

you can just use calculus or whatever to prove it

polar coordinates -> maximize cos(theta)+sin(theta), derivatives, blah

BAYMAX

any other intuition to this lik any other approach

arctic tern

06:51

I gave you the most straightforward "student do what she's told" approach - use calculus. the most fundamental intuition is geometric, which is the orthogonal projection one.

BAYMAX

ok

arctic tern

any line x+y=c is perpendicular to y=x, and they intersect at (c/2,c/2). thus, x+y is the signed length of the orthogonal projection of the point (x,y) onto the line y=x. if you draw a closed circle, you want to find the point on the circle that is as far along in the direction of that axis as possible, which is (1/sqrt2,1/sqrt2) just by looking at the circle (and using pythagorean theorem or trig to get coordinates).

Mike Miller

@PVAL Good point with the torus example. It makes me want to restrict to simply connected a little.

BAYMAX

thanks@arctictern

Marine1

09:34

Hi there ! I had a very simple question in Lagrangian multiplier ...

When defining new $u_1$ and $u_2$ in Lagrangian relaxation :

$$u_1=\max(0,u_1-tL_1),u_2=\max(0,u_2-tL_2)$$

what is $t$ ? How much is it ?

Julius

11:06

Hi. I got a $K^2 \times K^2$ matrix $A$ that consists of $K^2$ block matrices of size $K\times K$, each of which is symmetric and with real entries from [0,1] (mostly zeroes). Can I say something about positive-defiteness of $A$ in such a case? I'm aware of lots of results when $A$ has only 4 blocks, but in this case $K$ could be large.

Vrouvrou

hello

Leaky Nun

@Vrouvrou bonjour

Vrouvrou

bonjour

M'vy

11:39

salut

Sha Vuklia

Guys, why is the expected value a time and not the size of the expected population? I mean, it isn’t the time that is decaying, but the population..? I don’t really understand the conversion to a probability density function. So I’m guessing if we’re given a nonnegative $f(x)$ (whatever meaning that its values may have), we somehow end up with the distribution of the $x$, instead of the distribution of the values of $f$.

So what really happens when you normalise a function? (except for the obvious, which is calculating this normalising constant)

Oh maybe I see it. So by normalising a function, you basically write it as a scalar multiple of a density function. Then whatever the argument is in the function (say $x$), these $x$'s will be the possible values of the random variable associated with the distribution.

Sha Vuklia

12:12

In any case, I've formulated my question here: math.stackexchange.com/questions/2276252/…

SteamyRoot

Might we worth noting that, in the case of the exponential, you're integrating from $0$ to $\infty$, not $-\infty$ to $\infty$

Sha Vuklia

technically, you integrate from $-\infty$ to $\infty$

it's just zero on $-\infty$ to $0$

SteamyRoot

Indeed, but writing $N(t) = $N_0 \exp^{- \lambda t}$ doesn't state that (although, to any experienced user, it's of course clear from the context)

Sha Vuklia

yea okay, fair enough

SteamyRoot

And, in regards to your question itself: isn't it just the distribution of the lifetimes?

As in, I'm assuming this is some exponential decay thingy from $N(t)$

Sha Vuklia

12:20

well it must be, but the only reason I know that, is because I read that the mean is considered as the mean life time

so I know it "in hindsight"

I would kind of like to interpret it on a little more abstract/general level, so that I can recognise it on my own

SteamyRoot

Well, in a very vague way: $N(t)$ is the number of particles. You normalise everything, so it's unitless, but then you integrade w.r.t time; so the unit is time

Sha Vuklia

okay, that already helps a bit

so whatever we were working first, it will get normalised, so it loses its dimension, and that's why we end up working with the input technically

I think your answer "distribution of lifetimes" is indeed what I was looking for actually

Oh yea, because is if you consider some time $t$, then $N(t)$ will give you the amount of particles that are still present at that time, so it gives you the amount that have a lifetime that is $\geq t$

How would you interpret it if we're considering the "decay" of an amplitude? So say we have $C(t)=C_0e^{-\gamma t}$. Like, what is the mean life time of the amplitude then? :P

(this is where my question originally came from actually)

Fawad

13:05

Hello why math.se is not opening? Anyone else facing same?

Sha Vuklia

it works for me

Leaky Nun

and for me also

Fawad

Ok, working now

VenkiPhy6

13:42

The distance of the plane 3x-4y+12z=3 is 1/2 . Am I wrong?

from the origin

Leaky Nun

14:06

@VenkiPhy6 why is it 1/2?

Secret

14:29

One of my maths friends said there is no general formula that can compute all possible carry overs for a base n system. If that's the case, I hope I can at least find a second best option that can model the "dynamics" of carry overs in base n.

This is because one reason I found number theory so confusing is because most of the weirdness (such as n parasitic numbers, last digit of a given natural number, divisibility etc.) all seemed to have something to do with the carry overs

For an illustation on what I am trying to say , consider the following:

Akiva Weinberger

You might find this paper interesting, Secret

(I shared it before, I don't know if you saw it)

user1091558

Hey all, I have to shorten a formula. my formula looks like this z = xyyyy. Y is either 0 or 1 but each Y represents different values. Is there a way to shorten that formula something like z = xy^4?

Akiva Weinberger

(but it might be related to what you want)

Secret

Probably not read before, it looks very unfamilar on opening the pdf. Reading it now cause today my chemistry calculations are screwing up and they cannot be fixed until tomorrow, thus I had some free time with the maths today

Akiva Weinberger

@user1091558 Welcome to the chat! Do you have LaTeX on? (There's a link on the top-right in the room description)

user1091558

14:35

No, let me turn it on

Akiva Weinberger

To answer your question: If they're all different values, maybe you'd want to write it as $z = x y_1 y_2 y_3 y_4$ (with subscripts) to differentiate all of them

VenkiPhy6

@LeakyNun because of formula here(mathinsight.org/distance_point_plane). I am asking because a multiple choice question that I am solving doesn't have 1/2 as the option...

Waiting

Awesome @ShaVuklia, how are you doing?

@Hippalectryon hey! How is it going?

Leaky Nun

@VenkiPhy6 can you show me how you got 1/2?

Hippalectryon

@Waiting Great :D what about you ?

user1091558

14:39

Akiva, Thanks. But is there a way to shorten that formula like $z = xy(subscript)4$ and tell my readers explicitly that first y represents a, second y represents b etc ?

Sha Vuklia

@Waiting Haha, I'm good:P how are you?

Waiting

@Hippalectryon Glad for you! Good in some sense, and less good in other sense. Overall, I'm doing some research and preparing to publish something. :D

@ShaVuklia Cool!:P Still at uni, learning some new interesting stuff? Me? Not that bad, there is room for improvements. :P

Hippalectryon

@Waiting Yay :-) btw what happened to the other articles you wanted to have published ?

Waiting

@Hippalectryon It's going to be published these days. It's just a matter of days.

VenkiPhy6

@LeakyNun In the formula in the link I gave I used the following: (x1,y1,z1)=(0,0,0); A=3; B=-4; C=12 and D=3 and thats how I got 1/2...

Hippalectryon

14:42

:D @Waiting

Sha Vuklia

@Waiting yea, still doings physics

Leaky Nun

@VenkiPhy6 firstly D=-3, and I don't see how that formula gives you 1/2.

Waiting

@Hippalectryon Perhaps I shouldn't say that about my work, but it's damn awesome. :D

Hippalectryon

@Waiting Don't forget to send me a link to the articles when it's out :-)

Waiting

@ShaVuklia Nice! Looks like you have a great time there. :P

Sha Vuklia

14:43

btw, some Jason is looking for you:P @Waiting just letting ya know

VenkiPhy6

@LeakyNun Okay I'm sorry I just found out what I am doing wrong... I was not doing 12^2 at all! Sorry again for wasting your time. Thanks, anyways

Waiting

@Hippalectryon OK OK OK

@ShaVuklia Who is Jason?

Let me check.

Sha Vuklia

uhm i donno, he changes his user name i think

Waiting

@ShaVuklia I got it. It's OK.

Sha Vuklia

good:d

Balarka Sen

14:59

Hi @PaulPlummer

Paul Plummer

@BalarkaSen Hey

Secret

(modulo the cohomology section which I only partially understood) It seemed to be more or less a rigorous version of my sketchy attempts on that carry over problem this afternoon (where I tried to write a natural number as a finite string, and then noticing how carry over occurs whenever the units sum larger than the base), as they formalise whether a carryover occurs with the map $z(b_1,b_2)$.

So yes it is quite relevant. Now to figure out how to generalise $z$ so that it can handle carry overs that occurs when we multiply two digits together, if we can do something like that, then solvin…

I also notice from reading the paper that the phenomenon of carry over is "nonlocal" (i.e. it depends on all digits of smaller powers of base $b$ before it). I wonder if something similar can be found in arbitrary summations, that might help me to better understand how to decompose infinite series...

Secret

Now a completely unrelated question:

$$\epsilon_0=\sup(\{0,1,\omega,\omega^{\omega},\omega^{\omega^{\omega}},\cdots \})$$

When we build those $\omega$ power towers, are we acting the exponentiation from the left or the right side?

That is, to go from $\omega$ to $\omega^{\omega}$ to $\omega^{\omega^{\omega}}$ do we put our new $\omega$ beneath the old one or above the old one each step?

Sha Vuklia

Guys, this is probably very stupid, but given that $\gamma>\omega$, I don't see immediately how $\gamma-\sqrt{\gamma^2-\omega^2}<\omega$. Maybe I could show that $\sqrt{\gamma^2-\omega^2}>\gamma-\omega$. However, $\gamma-\omega=\sqrt{\gamma^2+\omega^2-2\gamma\omega}$, so that's not helping either.

I don't see it. I also can't do something like this; $\gamma-\gamma\sqrt{1-\omega^2/\gamma^2}<\omega$

Martin Sleziak

15:20

@Secret I am not sure I understand the question. Are you asking whether the notation $a^{b^c}$ means $(a^b)^c$ or $a^{(b^c)}$?

If that's the case: What is the order when doing $x^{y^z}$ and why? and Can anyone explain why $a^{b^c} = a^{(b^c)} \neq (a^b)^c = a^{(bc)}$.

Akiva Weinberger

@user1091558 $z=x\prod\limits_{i=1}^4y_i$ would work, I think. You can learn about summation and product notation here.

Secret

@MartinSleziak well it's kinda related. cause the power tower can be built either by raising the index by omega on the right for each step, (which corresponds to the left associative case $((a^b)^{{}^c})$) or by attaching a new base omega beneath the power (which will correspond to the right associative case $(a^{(b^c)})$). so the question is that for the definition of $\epsilon_0$ how is the $\omega$ power tower built, from the left or from the right?

Astyx

Hi chat

Sha Vuklia

hi @Astyx

Akiva Weinberger

@Secret On the left. If you did it on the right, you'd just end up with $((\omega^\omega)^\omega)^\omega\dotsb$

which would equal $\omega^{\omega^\omega}$.

Secret

15:27

Ah I see

Sha Vuklia

never mind, I showed it!

Martin Sleziak

$x-y<\sqrt{x^2-y^2}$ $\Leftrightarrow$ $(x-y)^2<(x-y)(x+y)$ $\Leftrightarrow$ $x-y<x+y$

Sha Vuklia

oh yea I did it like this: $$
\sqrt{\gamma^2-\omega^2}>\gamma-\omega\iff \gamma^2-\omega^2>\gamma^2+\omega^2-2\gamma\omega\iff\omega^2<\gamma\omega
$$

Martin Sleziak

I see.

Astyx

What's up ?

Sha Vuklia

15:33

not much, how about you?

Astyx

Finished my written exams 30 minutes ago

Waiting

@ShaVuklia Take care! I have to leave now. One thing: follow your intuition (in general, you can't be wrong). ;)

Sha Vuklia

@Waiting alright, see ya!

Astyx

It feels so relieving

Sha Vuklia

oh sorry I got interrupted irl :P

i wanted to say: congratz!

Astyx

15:47

Haha don't worry

Sha Vuklia

I got interrupted for that question about graphs that i asked here yesterday or the day before yesterday. We think that we have a solution now:P

but it's too easy, so I'll ask my teacher if it's correct

so you finished everything now?

Astyx

Care to tell me ?

Sha Vuklia

or your written?

do you remember the question? it was: if we have a 8-regular graph, show that is a spanning subgraph that is 4-regular.

Akiva Weinberger

@ShaVuklia $\gamma-\sqrt{\gamma^2-\omega^2}<\gamma-\sqrt{\gamma^2-(\gamma)^2}=\gamma<\omega‌$

Note that, to maximize $\gamma-\sqrt{\gamma^2-\omega^2}$ (keeping $\gamma$ fixed), you want to minimize $\sqrt{\gamma^2-\omega^2}$ which means you want to maximize $\omega^2$.

Sha Vuklia

oh cool, thanks @Akiva!

Akiva Weinberger

15:49

And if $\omega<\gamma$, then to maximize it you'd substitute in $\gamma$.

Astyx

Only my written, I'll get the results mid June and then I'll know if I have oral exams (and I really hope I do, the contrary meaning I failed everything)

Sha Vuklia

ohh gee:P that's a long time

but it's normal i guess

anyhow, about the question:

we thought of making an Euler-cycle

and then removing each other edge

then we are left with a graph that has degree 4

Astyx

Why is that ?

Sha Vuklia

oh shit

DAMN:P

it's not

Astyx

Lol

Sha Vuklia

15:51

we already tried this actually

but this friend came over and told me another person told him this solution

scheiße

nm

Astyx

Sorry :/

Sha Vuklia

:P

Astyx

I'm not sure I understand the question really, you mean there is a subset of edges $E'\subset E$ such that $(V, E')$ is 4 regular ? @ShaVuklia

Akiva Weinberger

@ShaVuklia Wait, ignore what I said

That was nonsense

Sha Vuklia

@Akiva lol I already didn't get it entirely to be honest :P

at least

the last part

actually

i donno. my brain is mush

@Astyx yes!

apparently this answer I found on stack

is really the way to go

i contacted my teacher

so yea

Akiva Weinberger

15:55

(Sorry)

Sha Vuklia

(nw)

Astyx

I'll give it some thoughts once I get some paper under my hand @ShaVuklia

Sha Vuklia

Sure! Though it's really a matter of me understanding some specific steps in the proof. When I start studying for my exam on graph theory (which is in 1.5 months), I'll probably get back at it

user1091558

Avika, Thanks. Thats what I was after :-)

Astyx

Will you start studying in 1.5 month ? Or is the exam in 1.5 month ? Or both ?

Sha Vuklia

16:04

lol the exam :P

i hope I don't start when the exam is XD

quick question guys:

One way to show that $e^{i\omega t}$ never vanishes, is to consider it as the polar form of some complex number with modulus 1. However, is there also a trigonometric approach? So we have
$$
\cos\omega t+i\sin\omega t=0\iff i\tan\omega t=-\frac{\cos\omega t}{\cos\omega t}.
$$
However, now that I come to think of it... why are we even allowed to divide by $\cos\omega t$?

oh wait

I have to consider cases of course

i completely forgot about that

Secret

you can divide by anything as long it is not 0 (in usual context)

Sha Vuklia

yea, i forgot that i could simply consider cases

Secret

I however forgot where the complex roots of cos x were

Sha Vuklia

complex roots?

Secret

well, if you are considering $e^{i\omega t}$ then naturally you need to consider the trig functions on the complex plane

Leaky Nun

16:11

I don't think cos x has any non-real roots

Sha Vuklia

of course it has

Leaky Nun

for example?

I said non-real

Sha Vuklia

sorry

i misread

too many negations dude :P

Leaky Nun

sorry

Sha Vuklia

haha

Tim The Enchanter

16:14

I don't think it should, cause we could rewrite it as $e^{2i \theta}=-1$, and a non-real number in the argument would make the modulus of the LHS not one.

Secret

ah ok

Sha Vuklia

@Tim you referring to what I said?

because maybe I should have specified that we're only dealing with real $\omega$ and $t$

Tim The Enchanter

@ShaVuklia Referring to there being no complex roots of $cos(x)$. Not strictly to what you said.

Sha Vuklia

oh ok

Astyx

16:40

You get a quadratic in an exponential term using euler

You only need to show the roots have modulus 1

The quadratic is $X^2-1=(X-1)(X+1)$ so that's settled

Nevermind that's exactly what Tim said

Santosh Linkha

Hello ... I need help interpreting an algebra question

Let $G$ be a finite abelian group. If $G$ has at most $m$ elements of order dividing $m$ for each divisor of $m$ of $(G:1)$, show that $G$ is cyclic.

what does $(G:1)$ means?

Astyx

Probably the cardinal of G

Santosh Linkha

hmm ... this is problem from this book jmilne.org/math/CourseNotes/FT.pdf

page no 25

thanks ... i'll take that into account and think about it.

Sha Vuklia

17:02

Uh, I don't follow what @Astyx is saying actually

and I don't understand why Tim takes $e^{i2\theta}=-1$

i'm kinda curious tho

Astyx

Do you know Euler form for cos ?

Sha Vuklia

ohh

yea give me 1 sec

i need to write it out:P

Astyx

${e^{ix}+e^{-ix}\over2}=0$

Sha Vuklia

haha yea that

Astyx

Then mutliply by $2e^{ix}$

Qed

Sha Vuklia

17:05

ohh

Tim The Enchanter

Just keep in mind that $e^{ix}$ is never zero.

Sha Vuklia

lol yea that was the entire point no? xD

Astyx

(I hope my latex is okay, i have no poit)

Sha Vuklia

yea it's fine

Astyx

Showing e^anything isn't zero is easy

Sha Vuklia

17:06

lol wasn't that the entire point??:P

Astyx

You just need to shiw there's an inverse

Sha Vuklia

like, what were we doing then?

oh

Astyx

Discussing cos

Sha Vuklia

no i donno

Tim The Enchanter

Yeah I'm just pointing it out to be pedantic

Sha Vuklia

17:06

oh yea

pedantic is good :P in math

because then i usually learn some tricks/details

Astyx

You show that exp is a morphism from (R,+) to (R*, *)

Actually take C and C* instead of R

And that's done thanks to Cauchy

Funnily (or kinda) this is how you show the exponential of a matrix is invertible

Sha Vuklia

17:29

Guys, say we're given $Ae^{i\theta}$. I would like to write this number in its complex form $a+bi$. I know that $a^2+b^2=A^2$. Now I also know that $\tan\theta=b/a$. So I guess I could write
$$
a^2+a^2\tan(\theta)^2=A^2\implies a=\sqrt{\frac{A^2}{1+\tan(\theta)^2}},
$$
and then obviously $b=\sqrt{A^2-a^2}$, where we substitute our $a$. Is this the only possibility? Or is there a cleaner expression possible?

Daminark

Hello there chat!

Sha Vuklia

and hi @Daminark

Daminark

How's it going?

Sha Vuklia

well i'm not dead, so i think good?

you?

Astyx

Hi daminark

@sha can't yiu just use e^it = cos t + isin t ?

Sha Vuklia

17:31

lol I thought you were talking to a chinese person for a sec :P

but ok i'll try i guess

Daminark

Almost dead, but not yet. And it's scav so ('-')/ woo

Sha Vuklia

oh wait, yea of course

Astyx

Sorry, i struggle too much with my autocorrect when typing mathjax on my phone

Daminark

As this message demonstrates :P

Sha Vuklia

what's scav? @Dam

Astyx

17:33

Especially since it's in french

Sha Vuklia

haha I feel for ya @Astyx

Daminark

It's a big scavenger hunt we have each year

It's the main feature of our school's existence

Sha Vuklia

lol I think I'd hate such an activity XD

but I hope you have some fun then:P

Daminark

Well it's not really a scavenger hunt, it's more about building stuff and cracking codes than searching for items lmao

Sha Vuklia

hahah oh K :P

Daminark

17:34

But yeah, in part for that reason I think I shall say adieu

Eric

Hi chat

Daminark

Hibye @Eric

Sha Vuklia

hi eric

Eric

Ew @Daminark I hate scav

Sha Vuklia

lol :P

Daminark

17:35

gasps

N3buchadnezzar

Hey all

Daminark

That's like, the reason why this place is great. That and HvZ :P

Eric

No one showers during it

Sha Vuklia

hi @N

Eric

And in my dorm they obstruct my path to the laundry room

Daminark

17:36

And there's a reason for that, because there are higher priorities at play

Eric

Lol @Daminark I also dislike hvz

Nah sorry that's gross

Daminark

@Eric I need to train your tastes more :P
Anyway I will actually get back to it now, I'm a page captain so I need to be on top of my shit, so see you!
Hibye @N3buchadnez

Sha Vuklia

haha g'luck

Astyx

Hi newly arrived people

And bye Dami

N3buchadnezzar

I know it is a bit fraudulent to ask about mathematics in chat :p However I have a question more along the lines of formulating a proposition in a clear matter, rather than the mathematics behind it.

If I have two inequalities $a < C b$ and $d < D e$, how can I say that the two inequalities are equivalent? Meaning that one holds if the other holds and vice versa.

Astyx

17:41

You mean how can you prove it ?

N3buchadnezzar

No, just state it

Secret

$a < Cb \Longleftrightarrow d < De$

Astyx

Isn't "$a <b$ iff $c <d$" good enough ? (Sorry my internet isn't very good)

N3buchadnezzar

I guess iff could be used, was just a bit unsure

Thanks =)

Astyx

Np

Julius

17:48

Hi. $A$ is a symmetric, real, indefinite matrix with entries from [0,1] and nonzero diagonal terms, that I'd like to make positive definite (without affecting the diagonal). I tried "smoothing" off-diagonal terms: decreasing "highest" entries and increasing "lowest" ones with some matrix $B$. It happens that $B$ also, at least sometimes, is indefinite, but I still get the desired result. So indefinite + indefinite = positive-definite. Any intuition or results on how this happens?

Astyx

Is B still symmetric ?

And what do you mean by indefinite ?

Julius

Neither positive definite, negative definite, positive-semidefinite, nor negative-semidefinite. And yes, $B$ is still symmetric. Both $A$ and $B$ have lots of zeroes, the entries of $B$ sum to 0

Astyx

Since they are symmetric you can diagonalise them. Suppose the basis of the eigenvectors of A is also a basis of eigenvectors of B

Ola

Yellow.

Astyx

Then the eigenvalue of the sum are the sum of each couple of eigenvalue from A and B

You want these to be all positive, which does not require the eigenvalues of B to be positive

Does this help ?

To see it clearly suppose both matrices are diagonal

Secret

17:58

Last night dream involve a very bizarre triple index notation for expressing a string of symbols in some compact form:
$$\mathbf{S}^i\vcenter{\tiny{j}}_{_k}=S^i T\vcenter{\tiny{j}}U_k=(((S)S)\cdots S)TT\cdots T(U \cdots(U(U)))$$

So for example $(T^2S)\vcenter{\tiny{4}}$ expands to $(((T)T)S)(((T)T)S)(((T)T)S)(((T)T)S)$

Well, I guess it will only be useful if some given nonassociative algebra have some nice rules despite nonassociativity

This dream also inspired an idea: I wonder whether a nonassociative derivative make sense...?

Astyx

Non associative ?

Secret

Well, I guess something like either left associative or right associative, or a mix of them but never (ab)c=a(bc) for any a,b,c

Astyx

But what do you mean by derivative then ?

Secret

We can potentially define a differential operator $d$ such that the product rule is obeyed i.e. $d(ab)=a(db)+(da)b$ (that's the minimal requirement for an element to be a differential operator). But on top of that, we also need to worry about where to place brackets when you have more than one $d$ in the expression

Julius

@Astyx, so $A=CDC{-1}$, $B=EFE^{-1}=EGG^{-1}FGG^{-1}E^{-1}=C(G^{-1}FG)C^{-1}$ and then the eigenvalues of the sum are the diagonal of $D+G^{-1}FG$, right? If so, this shows how this can happen. What about this "smoothing" of non-diagonal terms? It seems to be quite hard to relate it to the eigenvectors basis

Secret

18:10

So one possible way to construct such thing may be to generalise a differential algebra from a ring to some structure such that the multiplicative group becomes a loop (for example)

user193319

Let $D$ some countably infinite set in the plane $\mathbb{R}^2$. Is $D$ closed? If not, please don't provide a counterexample; I would like to find my own.

Secret

The usual topology of $\Bbb{R}^2$?

user193319

@Secret Yes.

Astyx

@Julius not sure what you mean by smoothing here

@user193319 no

Julius

I mean that $B$ is such that it decreases relatively high nondiagonal elements of $A$ and slightly increases the zero ones. In this way reducing the variance of nondiagonal entries

user193319

18:15

@Astyx Thanks! I'll have to think about it, and if I can't resolve the problem--I will definitely be back.

Alessandro Codenotti

@user193319 Can you find a counterexample in $\Bbb R$?

Astyx

It also feels like if it were, proving it wouls be rather intuitive

@Julius and I meant codiagonalisable matrices, whereas you did something more general

What you said is not true : the eigenvalues of the sum are not the diagonal of the matrice you wrotr.

What you said is not true : the eigenvalues of the sum are not the diagonal of the matrice you wrote.

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$Mathematics$ Mathematics