Mathematics - 2021-07-08

Koro

02:39

Suppose I say G=\{z\in \mathbb C: z^n=1 \text{for some $n\in \mathbb N} \} $. Then from this, I understand that G is a group of all roots of unity under complex multiplication.

Is that correct or should I think it as for fixed n? That is, is G a finite group containing only nth roots of unity?

That is I think that G={$z\in \mathbb C: z^n=1 \text {for some $n\in \mathbb N$}\}$ is an infinite group.

copper.hat

03:06

@hyper-neutrino thanks :-)

@Koro if $G_n = \{ z | z^n=1\}$ then $G=\cup_n G_n$. It is infinite.

my son passed his driving test today. this was his 11th day driving.

very happy :-)

Ted Shifrin

Your sentences are contradictory, @Koro. It did not say to fix $n$.

That’s nuts, copper. Most of us needed lots of practice.

copper.hat

03:22

@TedShifrin He still does, and he knows it. We will do lots more driving together.

Koro

@TedShifrin Hi Ted! So if only G={$z\in \mathbb C: z^n=1 \text {for some $n\in \mathbb N$}\}$ is given then I should interpret it as an infinite group (that is the group of all roots of unity under complex multiplication)? I believe that I am right (please correct me you think otherwise). I ask this question because I saw it in a test and they stated the reason as G is finite, which I disagree with.

@copper.hat Hi copper, I wish the group was given in this form $G_n$. It would have been more understandable and not caused any confusion.

I totally agree with you professor Ted-about not having fixed $n$. It should have been done outside curly brackets.

in the same test, $G$ infinite was marked wrong.

gaufler

Is there a ring with unity but no units other than unity ?

Koro

trivial ring where $0=1$? @gaufler

Ted Shifrin

Whoever wrote that stuff needs to go back to learn basic quantifiers. You’re right.

gaufler

Other than the trivial ring.

Koro

03:33

@TedShifrin professor Ted, is this about my $G$?

Ted Shifrin

Yes

Koro

Yes. Thank so much Professor Ted. :-) I don't care if it was marked wrong. I am happy that I was right :-)

leslie townes

you didn't ask, but i agree with ted

again.

Koro

Leslie and Ted agree :-)

again

Ted Shifrin

The math world should all agree — even the idiot who wrote wrong stuff.

leslie townes

03:35

i agree, it isn't a close question.

we might disagree about what to feed our cats.

Ted Shifrin

@gaufler what about $\Bbb Z_2[x]$?

Milk is actually not good for cats , it turns out,

leslie townes

oh, i believe it. that is 1000% cat.

they like eating it but they are lactose intolerant after they are no longer kittens.

Koro

my college was the first place where I saw cats and dogs lived in harmony

leslie townes

we serve our cat about 50% dry food because it is convenient and easily stored, and 50% wet food mixed with water so that she has to hydrate. she also visits her water dish but not very much.

she also eats whatever is left on my daughter's plate or our soup bowls.

Koro

I didn't know milk is not good for cats :'(

leslie townes

03:40

my cat loves to clean my daughter's cups of yogurt.

it's small enough a quantity that we aren't worried about it. my daughter is pretty good about eating yogurt.

gaufler

@TedShifrin It would work over any Z_p[x], where p is a prime. Right ?

*for any

Ted Shifrin

No @gaufler

leslie townes

characteristic 2 is important.

gaufler

@leslietownes Would you kindly explain, why it is important ?

leslie townes

anything in the prime subfield will be invertible. if i'm not mistaken. so if the characteristic is not 2 you will have more than 1 unit.

Ted Shifrin

03:44

I hate to agree, but of course.

leslie townes

i should point out that i'm not an algebraist and could be totally wrong about any of this.

Ted Shifrin

Shaddup.

leslie townes

math.stackexchange.com/questions/687267/… some useful thoughts here.

gaufler

@leslietownes Thank You!

Ted Shifrin

I wonder if we just solved an exam question.

Koro

03:49

A very small problem please for finding index of a subgroup in a given group

I'll share shortly, the latex file got deleted by mistake

:'(

gaufler

@TedShifrin I am preparing for an entrance exam and I seem to have a lot of doubts. I suck at Abstract Algebra. I get stuck with problems and that follows this dilemma whether I should work out the problem or look at the solution manual. If I look at the solution manual I feel bad but If I won't then a lot of time is wasted thinking about it and I have a lot to cover.

Ted Shifrin

Ah, gotcha @gaufler. There is no shortcut, though. Looking at answers is like reading texts. If you don’t work lots of problems and struggle, you just don’t learn mathematics.

troll bot

04:05

man

Koro

I have included a short solution to the question of finding index of H in G. I would request if someone can review it and let me know of any mistakes there, if any

troll bot

a lot of americans who have an 'americacentric' world view

Koro

I want to find index of H in $\displaystyle G$ that is $\displaystyle [ G:H]$, where

$\displaystyle G=\Bigl\{\begin{bmatrix}
v & z\\
0 & 1
\end{bmatrix} :v\in F,\ z\in C\Bigr\}$, where $\displaystyle F=\left\{z\in C:\ z^{2020} =1\right\}$

$\displaystyle H=\Bigl\{\begin{bmatrix}
1 & z\\
0 & 1
\end{bmatrix} :\ z\in C\Bigr\}$

I simply take $\displaystyle g\in G$ such that

$\displaystyle g=\begin{bmatrix}
w & z'\\
0 & 1
\end{bmatrix}$ and here $\displaystyle < w >\ =F$

I observe that

$\displaystyle \begin{bmatrix}

troll bot

ay

you could use pastebin

but ok

Koro

With that I conclude that $[G:H]=2020$

Ted Shifrin

04:15

I thought you promised not to write <>

Koro

Professor Ted, on that editor when I tried \langle \rangle, it was showing some norm type of symbol

that's why i wrote that

Ted Shifrin

Huh?

Koro

It's showing correct now :'(

Ted Shifrin

Uh huh :)

shoteyes

I am convinced that this answer to a question about a simple logarithm inequality has an error. I have tried to convince the answerer of this, but it is possible that they have not been online to see it. Somehow, three upvoters don’t seem to agree or didn’t notice which leads me to think that my critique is somehow flawed. Are comments here incorrect?

I want to clarify that I am not intending for this to be a downvote brigade or anything. I actually think the answer is quite clever and deserves recognition, but it also deserves to be fixed.

Koro

04:22

sorry about that professor Ted, I may have typed something wrong there that showed something different than $\langle \rangle$ and in hurry I switched to <>.

Adil Mohammed

hii

i got a small doubt, if there are 2 elements in a set what is the number of symmetric relations. I used the formula 2^[n(n+1)/2] and got 2^2=4 as the no of symmetric relations. is it correct?

shoteyes

@AdilMohammed The formula is correct, but shouldn’t the exponent be $3$ if $n = 2?$

Adil Mohammed

but if elements of my set is (a,b) then arent (i) empty relation, (ii) {(a,a)} ,(iii) {(a,a),(b,b)}, (iv) {(a,b),(b,a)} and v) {(b,b)} all symmetric relations?

@shoteyes oh yes you are right, i made a mistake

sinclair

according to the definition of a probability space, it includes a probability function P that assigns a real value between 0 and 1 to an outcome in the sample space. But there is also the concept of probability distribution function PDF that assigns a real value between 0 and 1 to a random variable X. What is the difference between P and PDF?

Is PDF constructed since we cannot work with outcomes analytically?

robjohn

04:53

@sinclair By P, do you mean the cumulative distribution function?

If so, then in general, the PDF is the derivative of P (in the sense of distributions).

Ted Shifrin

05:06

Probably too fancy, @robjohn

Koro

Professor Ted, my question is answered now. My solution was correct :-)

robjohn

@TedShifrin and the PDF does not assign a value between 0 and 1, either.

shoteyes

05:19

Meant to say “Are my comments here incorrect?” a few messages back. Sanity check.

Is it not possible to reply to oneself in chat?

hyper-neutrino

when you reply to someone else, note how it prepends :####### to your message

that is the message ID

you can just take your message's ID and put it in manually

alternatively, this userscript lets you reply to yourself with the reply button at the right side of each message

shoteyes

@hyper-neutrino Much appreciated.

leslie townes

the Church is against replying to yourself. with the right amount of willpower, you don't need to do that.

shoteyes i think you are right.

with things where the result is right but maybe the details aren't all there it's a question of how much is missing vs. expected to be filled in by the reader.

but i think the gap you identify is, as we would say in my profession, material

but i'm also dumb, so i should point that out.

shoteyes

05:40

Yes, I thought it was strange to use the equality condition of AM-GM to assert some absolute bound in the inequality. But, I was a bit apprehensive to correct a high-reputation user; I thought I was going crazy!

leslie townes

i am one of the world's greatest mathematicians, and i have only 6000 reputation. that should tell you all you need to know.

i am also known for being very humble.

the fact that nobody is disagreeing with this means they all agree with it.

Koro

@leslietownes I also that think you are very humble :-)

leslie townes

yes. a lot of people are saying this. i've been getting a lot of praise for this aspect of my personality.

sinclair

06:07

Is the event space the powerset of the sample space iff the sample space is finitely countable?

abenthy

Hi, I'm a little bit confused. I have found that the parametrization $(r,\phi)\mapsto\begin{pmatrix}2 \cos(\phi) + r \cdot \cos(\phi/2) \\ 2 \sin(\phi) + r \cdot \cos(\phi/2) \\ r \cdot \sin(\phi/2) \end{pmatrix}$ for $r \in [-1/2,1/2]$, $\phi \in [0,2\pi]$ gives me a Möbius strip. But all parametrizations I can find online seem more complex. Can my parametrization be derived from the "standard" Möbius strip parametrization? I don't see how. Or is there something wrong with it?

robjohn

08:20

@leslietownes it's an abomination against nature.

@robjohn that's for sure!

Mad Spaces

09:04

Hey, if you want to write a complex number as a vector, do you write for example for $ a +ib $ that it is equal to $ \left (
\begin{array}{c}
a \\
ib \\
\end{array}\right ) $ or do you just use b instead ib?

robjohn

just $b$

Mad Spaces

the i is implied?

robjohn

you are plotting in $\mathbb{R}^2$ where does $ib$ go in $\mathbb{R}^2$?

Mad Spaces

I am asking because i am having trouble understanding the definition of the complex space definition of the eucledian metric which we defined to be $ d(a,b) = \sqrt{\sum (a_i-b_i)* \bar{(a_i-b_i)}} $

Sorryy wait let me correct that

and i am trying to construct easy example so i understand what the complex conjugate in this case means, for in R for example the complex conjugate of $\bar {(a-b)} $ is just = $ (a-b ) $ resuting in the easy equation from the trianlge hypothense

But i am not sure how this translates into complex numbers? excuse my dumbness.

robjohn

don't use $i$ as an index if you're using $i^2=-1$ Try \overline{(a_i-b_i)} which gives $\overline{(a_i-b_i)}$

Mad Spaces

09:11

Oh yes that must be confusing

then $ d(a,b) = \sqrt { \sum{ (a_j-b_j )* \overline{(a_j-b_j)}}} $

robjohn

Also, use \cdot or nothing instead of * for multiplication

but I get what you're saying

Mad Spaces

Yes sorry i am bad with latex.

eventually one should get $ \sqrt {\sum{(a_j-b_j)^2}} $ Even for complex numbers, but what would the components transate into , or basically said, can you write me an example of a b complex and what the complex conjugate transate to ?

Also, my l key is kind of not typing haha

robjohn

@MadSpaces No. You want to get $\sqrt{\sum\limits_j|a_j-b_j|^2}=\sqrt{\sum\limits_j(a_j-b_j)\overline{(a_j-b_j)}}$

The absolutes are different than the parentheses

Mad Spaces

Well yes makes sense.

Its after all positive distance.

robjohn

That is why the conjugates

$|z|^2=z\bar{z}$

$z^2\ne|z|^2$

Mad Spaces

09:18

so for example let $ a,b \in \mathbb{C} $ where as $a = (a_1, a_2 ) , b = (b_1, b_2) $ what is then in this case $\overline {(a_1 - b_1) } = ? $ is it simply $(a_1 - b_1) $

robjohn

No. If you're separating the real and complex parts, then it is $(a_1-b_1)^2+(a_2-b_2)^2$

because those are reals

you need to know when your variables are real and when they are complex

$a=(a_1,a_2)=a_1+ia_2\in\mathbb{C}$ while $a_1\in\mathbb{R}$ and $a_2\in\mathbb{R}$

Mad Spaces

yes but then you would have that $(a_1 - b_1 ) $ is real, which means the complex conjugate of it, is itself, right?

robjohn

yes.

Mad Spaces

I am not sure why you answered "no" then to my previous statement

robjohn

$|a|^2=a\bar{a}=(a_1,a_2)(a_1,-a_2)=(a_1+ia_2)(a_1-ia_2)=a_1^2+a_2^2$

@MadSpaces because I thought your $a_1$ and $a_2$ were complex since you were using conjugates

Mad Spaces

09:24

Oh alright.

But why then not write in teh definition just that $ d (a,b)= \sqrt { \sum_j^n {(a_j-b_j)^2}} $

Is it not the same.

robjohn

are your $a_j$ and $b_j$ real or complex?

and what is $n$?

Mad Spaces

Well they are real, since even if a and b are complex, their components are real.

n is basically index of the nth vector a or b

robjohn

if you are talking $a_1+ia_2$ then the sum is only over 2

Mad Spaces

True

robjohn

$n$ was confusing

Mad Spaces

09:28

My apologies.

robjohn

@MadSpaces In $\mathbb{R}^n$, that is the distance formula

Mad Spaces

But wouldnt it be the same also in the complex space?

robjohn

That is also the distance formula in $\mathbb{C}$ if $a_j$ and $b_j$ are the components of the complex numbers (and therefore, real)

Mad Spaces

Yes!

They are.

which makes me onfused, why my professor wrote it like this $d(a,b) = \sqrt { \sum{ (a_j-b_j )* \overline{(a_j-b_j)}}} $ instead of the simple $ d (a,b)= \sqrt { \sum_j^n {(a_j-b_j)^2}} $

robjohn

because then you are looking at $\mathbb{C}$ as $\mathbb{R}^2$

@MadSpaces If he is looking at $a_j$ and $b_j$ as complex numbers in $\mathbb{C}^n$ then that is correct

Mad Spaces

09:30

No, they are components of a complex number $ a, b $

Thus, they are real.

robjohn

then he is unnecessarily making things complicated, since the conjugate of a real number is itself

Mad Spaces

Well, i just wasted forty minutes on this, which led me to believe that it might be something different. But i guess now i know what is up. Thank you robjohn for clarifying.

robjohn

np

yw

we

epsilon-emperor

10:10

mathb.in/59705

Let $\mu$ be the Lebesgue measure on $(0,1)$, and let $\lambda$ be the counting measure on the $\sigma$-algebra of all Lebesgue measurable sets in $(0,1)$. Prove that $\lambda$ has no Lebesgue decomposition relative to $\mu$, and although $\mu\ll \lambda$ and $\mu$ is bounded, there is no $h\in L^1(\mu)$ such that $d\mu = h\, d\lambda$.

Notation:

1. $\mu\ll\lambda$ ($\mu$ is absolutely continuous w.r.t $\mu$): $\lambda(E) = 0$ implies $\mu(E) = 0$ for all $E\in\mathfrak M$.
2. $\mu\ \bot\ \lambda$ (mutually singular): There are disjoint sets $A$ and $B$ such that $\mu$ is concentrated on …

Could I get some hints for how to go about this?

Mad Spaces

10:26

On the subject of proofs and such, have there been tries to construct a super computer program that finds or denies proofs of certain statements based on logical statements and such? that would be def reviolotionarry right?

porridgemathematics

11:13

there is the following definition of 'simply connected open set' i've come across in my notes: $\Omega \subset \mathbb{C}$ (open) is simply connected if its complement in $\hat{\mathbb{C}}$(the riemann sphere) is connected, im a bit confused because I thought simply connected sets need to be path connected

but it seems like with this definition its enough for the path components of $\Omega$ to be simply connected in the usual sense

then even more confusingly, the riemann mapping theorem is stated after this definition of simply connected, which according to the notes states 'a proper subregion (open subset) of $\mathbb{C}$ is conformally equivalent to $\mathbb{D}$ iff it is simply connected', so obviously now proper simple connected subregions must be path connected

should I just take this to mean they have forgotten to add that a simply connected subregion needs to be path connected?

oh maybe im misunderstanding them and they've made no mistake, perhaps their term 'subregion', and 'region' means open + path connected

yeah that must be it, sorry for the wall of text :)

orange is the new f

12:38

How do I show that $\langle Cx, x\rangle \geq \lambda_{min} \|x\|^2$ where $\lambda_{min}$ is the minimum eigenvalue of the positive definite Hermitian matrix $C$

Martin Sleziak

12:51

@orangeisthenewf Cannot we use that $C=UDU^*$ for some unitary matrix $U$?

And $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$.

This should basically reduce the problem to the case when we're working with a diagonal matrix.

epsilon-emperor

14:06

Rayleigh quotient

In mathematics, the Rayleigh quotient () for a given complex Hermitian matrix M and nonzero vector x is defined as: R ( M , x ) = x ∗ M x x ∗ x . {\displaystyle R(M,x)={x^{*}Mx...

Semiclassical

14:35

Another way to state the problem: Show that $C-\lambda_{min} I$ is positive semidefinite.

love_sodam

0

Q: Definition of a map $\delta^q :\operatorname{Ext}(C_q,B)\to\operatorname{Ext}(C_{q+1},B)$

I'm currently reading Spanier AT Chapter 5 section 5. In page 242, Spanier defined a map $$\delta^q :\operatorname{Ext}(C_q,B)\to\operatorname{Ext}(C_{q+1},B)$$ where $C=\{C_q\}$ is a graded module and $\operatorname{Ext}(C,B) = \{\operatorname{Ext}(C_q,B)\}$ and $C$ is a chain complex. How the m...

Paras Khosla

14:56

i.sstatic.net/Wtdpa.jpg Could someone verify if the expression for $S_{n}$ is correct. $S_{n}$ denotes the sum of product of integers from 1 to n taken 4 at a time.

Thorgott

@porridgemathematics just confirming, yes, region in complex analysis always means open + connected

the equivalence of the definition is still non-trivial, of course

Nazmul Hasan Shipon

15:28

Can anyone please tell me, is it legitimate to integrate this math in the way he did it?

epsilon-emperor

15:39

@NazmulHasanShipon Looks weird.

kunal Ch.

16:14

There are $n$ songs segregated into $3$ play lists. Assume that each play list has at least one song. The
number of ways of choosing three songs consisting of one song from each play list is:

EM4

16:49

@NazmulHasanShipon I seen his video and looks weird to me. Imagine someone gives you integrate x/dx or (dx)^(dx) it makes no sense and its weird.

Semiclassical

yeah, $x^{dx}$ is nonsense

full stop

at best you could say that you're really interested in $f(t)=\int (x^t-1)\,dx$ for small $t$

EM4

yep.

the craziest video my friends showed me was the "horseshoe method" of integration.

biggest troll video I have seen LOL.

Semiclassical

whence $f(t) = \frac{1}{t+1}x^{t+1}-x+C\approx t x (\ln x-1)$ for small $t$.

leslie townes

EM4 do you have a link?

oh i think i found it.

EM4

of the horseshoe method @leslietownes ?

leslie townes

16:58

subject of 11 upvotes and a successful round of voting to close on main.

that video has more than a million views, haha

what an enormous waste of everybody's time

EM4

HAHHA.

he says he will have to take extreme integration class.

leslie townes

this is quite funny. i like deadpan comedy. ideally, it would be impossible to tell if someone is joking or not. this video comes very close to that

EM4

@leslietownes you want to study extreme integration - math 1052? HAHHA.

copper.hat

differentiation should be called disintegration.

3

leslie townes

i have a phd but i never made it to math 1052. i think the highest i got was, 260? 270.

Semiclassical

17:06

i feel like that entirely depends on how the uni structures things

leslie townes

ecclesiastes speaks of there being a time for every thing, including "a time to cast away stones, and a time to gather stones together." because calculus is latin for pebble, this is clearly a reference to differential and integral calculus.

i checked and i only got to math 260.

Semiclassical

at UMN the courses start at 1000

leslie townes

at berkeley introductory undergraduate subjects were 0-99, less introductory subjects were 100-199, and graduate was 200 to, well, i'd say 299, but i don't know if they ever had anything higher than 260.

copper.hat

seminar courses had higher numbers if i recall

leslie townes

oh that's true, maybe even stuff like 298.

math 1052 was not offered.

Semiclassical

17:13

the breakdown for UMN is here: policy.umn.edu/education/coursenumbering-appa

with it roughly being nxxx with n being the expected year in which you'd take the course

copper.hat

i think meaningful names would be better, but wouldn't fit in n characters on a punch card.

leslie townes

in law school none of the classes had numbers, they just had names.

Semiclassical

(so first-year courses are 1xxx, second year is 2xxx, etc)

though that's really more in terms of "years into a major"

leslie townes

at berkeley there was no correlation within a subdivision. some of the more remedial classes had high numbers when the standard calculus sequence was 1A and 1B.

i don't think i know of a single school where the introductory class in a subject is called subject 101. that might just be from the movies.

copper.hat

104a was the closest...

leslie townes

17:20

that was a funny joke, making math 104 one of the most difficult classes in the major

supposedly, anyway. one of my wife's friends could not pass the complex analysis class 185. she just didn't get it. which is weird because if you took it from a postdoc they'd usually teach it like a calculus class and just do tons of contour integrals.

copper.hat

i took it in sort of a remedial fashion (i really needed functional analysis). it was the third time i was exposed to real analysis but the first time that it actually 'took'.

leslie townes

she took it something like 3 times and then switched majors. it was really weird.

it was very common to have phd students from math adjacent disciplines in 104.

copper.hat

i took complex analysis to fill some prereqs. from jack silver. a bit silly an ee taking complex analysis

leslie townes

you'd also have that, someone showing up for like three weeks and realizing "what am i doing here"

copper.hat

well, it gave me time to work on stuff :-)

leslie townes

17:23

185 was another common one where you'd see folks from other departments.

i think some engineers do need to know it, maybe for fluid mechanics or something where conformal mapping is an issue. EE not so much.

copper.hat

a fair bit in 'modern' control theory, h infinity design etc.

leslie townes

i do like h infinity.

it annoyed me that the engineers had gotten to it before the mathematicians. a lot of stuff anyway.

i had a funny time once where i had a paper submitted and the editor said this is a known result, citing an engineering paper. i pointed out that the engineering paper did not contain anything resembling a proof. i asked him to point out where he saw a proof. publication followed. weird philosophical issues about whether a result can be known if nobody's put a proof in writing.

a friend of mine who worked in low dimensional topology said this was characteristic of his field, there was just tons of stuff that everybody 'knew' but nobody had bothered to write down.

orange is the new f

@epsilon-emperor @MartinSleziak thanks much appreciated my lin algebra is sad

AMDG

18:01

Implementing a division algorithm that is faster than hardware is pain.

Worthwhile, but pain. What's more annoying is that I can't seem to find a way to use SIMD to implement it because it needs to count leading or trailing zeroes.

Balarka Sen

hi @Semiclassical

Semiclassical

o/

Balarka Sen

its been a while!

Semiclassical

yeah

been on other orbits

orange is the new f

Anyone know the major differences between the linear ODE $\dot{x}=-Ax$ when $A$ is positive definite Hermitian and when $A$ is non-defective

Balarka Sen

18:06

how's life?

safe i hope

Semiclassical

yeah. sorta in a holding pattern at the moment

Balarka Sen

as in?

Semiclassical

as in not really finding a next stage

Balarka Sen

ah ok

orange is the new f

let $A$ be a complex valued matrix, why dont people talk about symmetric $A$ i.e $A_{i,j}=A_{j,i}$ ? Hermitian matrices seem to appear more?

Semiclassical

18:19

Hermitian matrices have real spectrum and have a lot of nice properties under the spectral theorem, e.g., diagonalizable. Complex symmetric matrices generally have complex eigenvalues and don't enjoy such nice properties

orange is the new f

right so share a property with real symmetric matrices of always having real eigenvalues is the main point

Semiclassical

I'd put it the other way around. Real symmetric matrices are a special case of complex hermitian matrices.

The latter has real spectrum, therefore so does the former

orange is the new f

that way would make more sense :)

copper.hat

For any physically based system real matrices feature more frequently, hence the prevalence.

Semiclassical

that too. in QM, an observable is essentially defined as a Hermitian operator

(modulo subtleties about self-adjointness)

orange is the new f

18:37

And is there anything special to be said when an eigenvalue has geometric multaplicity equal to its algebraic multiplicity ?

whats the difference between stable and asymptotically stable? It is not mentioned on the wiki en.wikipedia.org/wiki/Equilibrium_point

copper.hat

it would be diagonalisable...

you might want to give some background here...

orange is the new f

@copper.hat sorry yeah I just found that must be super useful

I'm just making notes its been ages since I did any math like this.

Semiclassical

@orangeisthenewf maybe try the page on Lyapunov stability instead

orange is the new f

w.r.t to the stability question for a linear ODE like $\dot{x}(t)=-Ax(t)$, for say a real matrix $A$, then a stable point is of course at zero. In classifying the stable points someone describes them as stable or asympotically stable

@Semiclassical ok thanks !

"Lyapunov stability of an equilibrium means that solutions starting "close enough" to the equilibrium (within a distance {\displaystyle \delta }\delta from it) remain "close enough" forever (within a distance {\displaystyle \epsilon }\epsilon from it). Note that this must be true for any {\displaystyle \epsilon }\epsilon that one may want to choose.
Asymptotic stability means that solutions that start close enough not only remain close enough but also eventually converge to the equilibrium."

thanks ! :)

I am a person

18:55

This shouldn't be possible right?

right...

856 answers...

orange is the new f

link the question

please

I am a person

It's part of a different stack exchange but

474

Q: "Hello, World!"

So... uh... this is a bit embarrassing. But we don't have a plain "Hello, World!" challenge yet (despite having 35 variants tagged with hello-world, and counting). While this is not the most interesting code golf in the common languages, finding the shortest solution in certain esolangs can be a ...

code-golf string kolmogorov-complexity hello-world

orange is the new f

thanks

I am a person

no problem

hyper-neutrino

19:11

3396 answers

8439 if you count the deleted ones

leslie townes

20:03

aren't there a number of languages where hello world is the response to an empty input?

copper.hat

if i recall correctly, the early MATLAB response to the command why was RTFM.

AMDG

Brilliant

leslie townes

i miss stuff like that. although weirdly i think the world has gotten a lot better about documentation with online resources and not expecting everyone to be a hobbyist who has time to debug everything.

my copy of ms-dos 3.3 came with a manual for GWBASIC that was like 300 pages long. it could have been 30 pages.

or one page. "this is a variant of basic, which is one of the dumbest languages ever and it's right there in the name. type exit to leave the interpreter and find something better to do with your life."

when i was in high school someone told me and a friend that you couldn't make a TSR in gwbasic or any language that ran only via an interpreter. we wrote something in assembler and shoehorned it into basic via POKE commands. a recent version of windows found it in an old folder and identified it as a virus.

copper.hat

i think basic was a great way of learning stuff. when something is taught, one only gets the 'correct' way of doing stuff and many learning opportunities are lost. when you can experiment yourself, you learn more.

leslie townes

i'm glad i learned on basic and not on scheme

copper.hat

20:18

i love scheme, but that is after i learned why it is nice

leslie townes

i'm pretty basic myself. i used to do a version of what people would call s---posting on a forum for computer programming challenges. i would post one-page screenshots of a program that implemented things in GWBASIC.

here's some new technical challenge? ok, here's 20 lines of GWBASIC.

it was juvenile and yet also funny. people enjoyed it.

lisp might be the only good language

Ted Shifrin

juvenile + leslie ... imagine that!

leslie townes

it's really my life's story. i'm glad i have been able to share it with you. 40 is the new 12.

hyper-neutrino

@leslietownes probably - Stuck is an esoteric language that did that (though not really by design, it was a placeholder that ended up allowing a zero-byte answer to Hello World)

I haven't come across too many that do that, really. most just do nothing or error

copper.hat

i think the focus should be on problem solving not the langauge

leslie townes

20:21

i met a coworker hired during the pandemic the other day and he said "you're even weirder in person." compliment accepted.

hyper-neutrino

@copper.hat for a challenge as trivial as Hello World, problem-solving only really happens with the weird languages :P

but yes, the merit of golfing languages has been debated on our site since they first started to exist

leslie townes

i love code golf although i do not participate in it.

some of the code golf quines made me laugh out loud.

copper.hat

i like novel solutions, things that are tricks often have a way of becoming useful or a seed for another idea

hyper-neutrino

I've definitely tried to promote upvoting novel solutions and unique ideas or clever approaches over just "this code is short because the language offered three built-ins that completed the task". Unfortunately, it doesn't always work as well as we'd like, but we do have several discussions and past/current events to try to balance out the voting issue

we even have a community ad dedicated to just that, lol.

leslie townes

somehow this reminds me of a cute problem. you have a medium like a CD-ROM that cannot be rewritten. you can poke out holes that represent bits but not un-poke them once burned in. it holds N bits. show that you can implement a scheme where you store 2N/3 bits on it, and then rewrite it with another 2N/3 bits.

very simple but very cute.

hyper-neutrino

20:35

ooh, that's very interesting. i'll give it some thought

are all 4N/3 bits known ahead of time or do you have to figure out a way to write the first 2N/3, and then rewrite with an arbitrary set of another 2N/3 bits later?

(not sure it matters, but I don't know the solution yet)

leslie townes

the bits are arbitrary.

hyper-neutrino

okay

leslie townes

the way i analyzed it was to draw a graph of all 3-bit strings, with edges between them if you could get from one to the other if you could get there by poking one bit out. there's enough room to encode two-bit strings twice in that graph and still be able to get from any string to any other.

i never implemented it but it would be a fun thing to do if i still had a computer with a cd-rom drive.

hyper-neutrino

oh that's a nice solution

leslie townes

i forget where that problem came from. i'm searching my notes now.

epsilon-emperor

21:09

leslie, i commented on one of your answers from ten years ago lol

found a really short way out instead

leslie townes

the golden oldies, we call them.

that makes sense if f is real valued. does it work if f is complex valued?

i remember a commenter dinging me on this.

epsilon-emperor

@leslietownes it does work, i've edited my comment

just use continuity of $|\cdot|$ then

$|\cdot|$ composed with $f$ should be continuous, and you can contradict that with the same argument

copper.hat

:-)

epsilon-emperor

:)

leslie townes

does this make me the epsilon emperor?

copper.hat

21:12

of course, i have no idea what you are discussing

epsilon-emperor

@leslietownes no lol my argument works right XD

leslie townes

copper reminds me of my cousins.

i'm still thinking it through but i don't see a problem yet. it would have been nice to get this comment ten years ago when i still had brain cells that did math.

epsilon-emperor

ten years ago i was a kid who did not even know trigonometry

damn, i didn't even know quadratic equations then

leslie townes

although maybe i would have resented it. i was immature in my younger days. still am, but also then.

copper.hat

i knew more when i was younger

epsilon-emperor

21:15

i learnt all the hardcore math only in undergrad

nothing proof based before that

nowadays high school kids are learning category theory tho

leslie townes

i was at peak knowledge in about 2006. i knew everything. now, the long decline into senescence and dotage.

i'm hitting all my triple word scores today.

epsilon-emperor

@leslietownes i learnt two new words today, thanks! hehe

leslie townes

i don't mean to stereotype but i think there's something irish about this. my dad and grandfather also loved words. the weirder the better.

but then you have people like copper who just say the crudest things they can think of with no poetry whatsoever.

copper.hat

22:16

@leslietownes i take my inspiration from the best. my favourite: twitter.com/irishlittimes/status/1319932030461878272?lang=en

sorry for repeat.

leslie townes

23:23

as always, behan brings the good stuff.

i'd forgotten how short-lived he was. he only barely beat dylan thomas. essentially the same cause of death.

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