Mathematics - 2023-02-15 (page 1 of 2)

Sine of the Time

00:00

@Obliv this step is wrong

Ted Shifrin

No, they won’t, and certainly the economic interpretation of L.M. Is important for you!

Obliv

so it should be $\frac{-1}{e^y} = \frac{-1 + C}{2x^2}$?

Ted Shifrin

It’s not that hard. I required it for all the econ folks in the class.

Sine of the Time

@Obliv the RHS is still wrong

Ted Shifrin

@copper.hat I had the Saab 96, not 99.

Obliv

00:04

It should be $\frac{-1}{e^y} = \frac{2x^2C -1}{2x^2}$ then?

D.C. the III

@TedShifrin I did the economic one. It was the other two theoretical ones I was curious about

Sine of the Time

@Obliv yep

Obliv

so do I cross multiply then take the logs then? @Sineofthetime

Sine of the Time

@Obliv yes

Obliv

I have $\log{e^y(2xC-1)}=\log(-2x^2)$

I can take the y* out of the log and divide both sides by the lhs?

Sine of the Time

00:07

@Obliv you must isolate exp(y) from the rest

Obliv

like $y = \frac{\log(-2x^2)}{\log(2x^2C-1)}$

Oh

nvm

$y = \log(-2x^2) - \log(2x^2C-1)$

shintuku

Ted, did your suggestion for induction use something else than the fact a polynomial has a form $f(z) = (z-w)g(z)$, with $g$ being an $n-1$ degree polynomial? I'm trying to use this specific fact to get the full form $w_0(z-w_n)\cdots(z-w_1)$ for an arbitrary $n$

Ted Shifrin

@D.C.theIII Oh, so you did 32a as part of 33? Save the bordered hessian 32b for later.

@shin No, you assume the result for degree $n-1$, so you get the full factorization.

D.C. the III

@TedShifrin Yes. It was more so 32b and 34 I was asking about.

Ted Shifrin

34 is a chain rule computation. Save 32b until after you do implicit fn theorem.

I was never taught 32a. It was so obvious, too.

Obliv

00:11

sorry to bug you @SineOfTheTime but I don't know what to do now to get it in the form of the choices above'

shintuku

the induction step is getting from $w_0(z-w_n) \cdots (z-w_1)$ to $w_0(z-w_{n+1}) \cdots (z-w_1)$ using said fact?

Obliv

I could just verify those solutions I guess, instead of trying to take the integral

Sine of the Time

I'll be back in a few minutes

@Obliv the answer is B

Obliv

did you use verification of solution or took the integral ?

Ted Shifrin

@Obliv That will give you zero experience with what you need to learn.

Obliv

00:16

I should have been trying to verify, but I thought it would somehow be faster to take integral..

Sine of the Time

@Obliv this is right, now do a little trick: $\log(-2x^2)-\log(2x^2C-1)=-(\log(2x^2C-1)-\log(-2x^2))=-\log(\frac{1}{2x^2}+C)$

Obliv

I think you meant to factor out a negative so it's plus

Sine of the Time

It should be $-C$ but since the constant is arbitrary you can write +c

@Obliv there are different ways to do it

Obliv

Oh okay I see

Sine of the Time

the problem of having choices is that you've to rewrite your solution to befit the answer

I don't know if it is good to study ODE

Obliv

00:21

yeah, I guess not but I like the feedback system of Khan Academy. It served me well to get a 5 on my AP Calc exam in high school

shintuku

or, Ted, does the inductive step instead go from $(z-w_1)g(z)$ to $(z-w_2)(z-w_1)g(z)$?

Sine of the Time

@Obliv you can also use wolfram alpha to check your solution, but keep in mind that they can be written in different ways so the solution is not univocal

wolframalpha.com/…

Obliv

should I practice variation of parameters for linear 1st order ODE?

given this linear ODE $y' + 2x -3y + 5 = 0$ I can't easily identify P(x)

oh wait

should be in the form $y' - 3y + 5 = -2x$ so it's nonhomogeneous

then I set the rhs to 0 and do the variation of parameters I think

$y = y_c + y_p$ to get the particular solution after

Sine of the Time

@Obliv yep, but I think it's better to put the 5 in the rhs

Thorgott

my favorite impractical proof of the FTA is that a field extension $L/\mathbb{C}$ with $[L\colon\mathbb{C}]=n$ makes $\mathbb{P}(L)\cong\mathbb{CP}^{n-1}$ into an abelian Lie group, whose classification then forces $n=1$, hence $\mathbb{C}$ is algebraically closed

Sine of the Time

00:33

hi @Thorgott. I can't understand it by now, I don't have the tools

Thorgott

you don't have to, it's mostly a joke

Sine of the Time

I'm in my second year :(

Obliv

2nd year on this earth, as an AI?

Sine of the Time

@Obliv I don't get it, my english is not so good

Obliv

oh sorry was joking that you were an artificial intelligence

my english is not so good either and it's my native language

Sine of the Time

00:37

@Obliv oh I see

@Obliv where are you from?

if you don't mind sharing

Obliv

originally from turkey, but came to u.s. at age 4 so i'm basically native english speaking

but i can speak turkish, english, and karachay

how about u

Sine of the Time

I'm from Italy but origin from Syria :)

But I was born in Italy

Obliv

Tough times for our people right now :[

Sine of the Time

@Obliv yes, heartbreaking images

Obliv

I want to visit Italy

Sine of the Time

00:39

It's a good place to visit

lots of attractions

Obliv

although seeing Rome would be cool, I kind of like the idea of visiting the countryside

I don't know haha

Sine of the Time

@Obliv sure, but there are breathtaking landscapes in the north

@Obliv take a look at the pictures of "Garda lake"

Obliv

01:01

@SineOfTheTime Thank you for adding another place to my list of places to visit! I can't even comprehend that beauty lol, i'm living in such a s*hole comparatively.

ペガサス Seiya

$\angle DAE$ is annoying

Obliv

@ペガサスSeiya angle DBC should be 30, making BCD 120?

ペガサス Seiya

@Obliv yes, that much is obvious. Its an isosceles triangle

Here's what I've done. Don't know if it's the correct answer

Obliv

If you made a parallel line from B that matched CD would that help @ペガサスSeiya

landing somewhere on ED

ペガサス Seiya

@Obliv don't think so, that construction doesn't lead anywhere

Obliv

01:11

yeah idk I forgot a lot of geometry, i was hoping it would give you the angle of B in that quadrilateral since it's opposite of D

like opposite angles of a parallelogram type deal

how's your injury feeling btw

ペガサス Seiya

@Obliv that angle B would just be 60. But, again, I don't know where you'd go from there.

@Obliv much better

Obliv

oh yeah ED and BC would have to be parallel nvm

That's good @ペガサスSeiya

ペガサス Seiya

@Obliv the solution I sent seems to be the only 'synthetic' way to do it and even then I don't know if I got the right answer

Obliv

I like how you said "off to study more calc 3" and return with a geometry problem :P

your enjoyment of geometry is truly something else

ペガサス Seiya

Yeah lol, though I reviewed all that I had to. It isn't that difficult and I had reviewed it yesterday as well

Obliv

01:17

yeah the only problem I had with calc 3 were surfaces and double/triple integrals and spherical coordinates.. so yeah a lot i guess lol

ok brb need to srsly study

ペガサス Seiya

I'm still kinda in a state of disbelief that there's no shorter way to compute to Gaussian integral than the polar coordinates trick, as Ted confirmed

Obliv

diff eq test tomorrow D:

oh no clue, but whatever Ted says is probably true.

He's like a giant and I'm an ant

ペガサス Seiya

Dang, didn't know you were that tall @TedShifrin. And here I thought I was the tallest guy here.

Ted Shifrin

01:41

As I said, there are other ways, but they are more sophisticated.

ペガサス Seiya

01:54

Wonder if I'll learn those ways too

Ted Shifrin

02:11

First you have to prove that you can “differentiate under the integral sign.”

leslie townes

seiya: have you looked at keith conrad's list of examples. kconrad.math.uconn.edu/blurbs/analysis/gaussianintegral.pdf

the concept of a 'shortest' way to prove anything is kind of unstable. the 'length' of any proof depends on whatever you need to be able to assume/justify/understand to follow the proof. which itself may not have a well defined length.

D.C. the III

Showing the normal density integrates to 1.....fun times

@leslietownes THis looks like something I should keep for future stats courses and beyond

leslie townes

conrad's notes attribute the polar coords method to poisson and include a link to a funny article that sets out a basis for a contention that poisson's trick is truly just a trick and not a more general 'method.'

Ted Shifrin

02:27

Well, there is a very closely aligned “trick” to compute integrals of polynomials over the unit sphere.

leslie townes

well,its an article in a education oriented journal that defines one abstraction of poisson's method and shows that this abstraction only applies to functions not too different from the one poisson was integrating. not something to be taken too seriously.

Ted Shifrin

I am not either maligning or being defensive. He has a point.

Munkres used to say that a method is a trick that you use more than three times.

leslie townes

the article implicitly makes the point that it's hard to denigrate a single-purpose trick, if its single purpose evaluates an integral that is both ubiquitous and tricky.

the fact that it's basically 'the' proof you see in calc books says something.

Ted Shifrin

Indeed.

Not just in calc books, but in every calc book (that does multivariable).

D.C. the III

When you refer to Munkres why do I have the inclination to believe that this was something he shared while you were interacting with him in person....

Ted Shifrin

02:39

Well, he was lecturing our 80-person point-set course.

This was also the course in which he pissed off a bunch of students by saying the first day that anyone hoping to go to grad school in pure math had better get an A in his course.

D.C. the III

During the time of the book being a red book?

Ted Shifrin

It was a xeroxed version pre-publication.

D.C. the III

It's not a far off statement though is it

@TedShifrin Leslie gonna comment on something about dinosaurs in a minute....

Ted Shifrin

It was a bit self-indulgent.

@D.C.theIII Or cavemen.

copper.hat

@TedShifrin no engine braking on the 96 if i recall correctly. beautiful car. saabs were underappreciated, not quite sure why. fashion, i suppose.

Ted Shifrin

02:50

Free-wheeling was the usual (to save gas), but one could disable it — and I certainly did that day — for hilly/mountainous driving.

The engine did get noisy under stress.

But that was my first car. Dead fuel pumps caused me nervous breakdowns. In the end, I sold it in 83 in GA because leaking transmission and no AC.

copper.hat

It had character.

copper.hat

03:29

mildly amusing comment stream here math.stackexchange.com/q/4639399/27978

Silly Goose

03:49

does anyone happen to have an example of taking the limit of a sequence of matrices :P i can't seem to find one online for some reason. I think I am thrown off because let's say we are considering the space of all 2x2 matrices over the reals. Let us then consider some sequence $A_m$ of 2x2 matrices. Let us then consider the sequence of the top left entries of each matrix. Couldn't we end up with a sequence like ${1, 10, 2, 11, 114, 0, 28}$ of which how would one take the limit?

copper.hat

A matrix has a limit iff each element has a limit. No magic.

Silly Goose

so the sequence i provided above has no limit?

but a sequence defined by $1/n n \in \mathbb{N}$ does, for example

copper.hat

well, you would need to provide the rest of the sequence before I could say that :-)

Silly Goose

oh i just meant for this to be the sequence ${1, 10, 2, 11, 114, 0, 28}$ is this unusual notation

leslie townes

unless you are playing with some weird topology, which it very much sounds like you aren't, lack of convergence in the calc 1 sense in any entry would prevent convergence of any sequence of matrices having that sequence in that entry.

copper.hat

03:52

if the matrix sequence $A_n$ has a limit then so does $u^T A_n v$. Then pick $u,v$ to be suitable unit vectors.

Silly Goose

i see okay

i am looking at hall's book on matrix lie groups

(is the context)

copper.hat

its all a lie

Silly Goose

xD

leslie townes

yeah, nothing weird going on in lie groups. finite dimensional ones anyway.

Silly Goose

i am trying to understand taking the limit of a sequence of heisenberg group elements. i think intuitively the 1s and 0s do not change under taking a limit. and sequences of real entries converge (if they do) to real entries, so the structure of the element is preserved in that way.

copper.hat

04:00

the set of upper triangular matrices is closed, so any limit will also be upper triangular.

Silly Goose

i guess i am trying to prove that it is closed though

since i am trying to prove (well just this quality since the others are more straightforward) that the heisenberg matrices form a matrix lie group

which is defined in this book as a closed subgroup of the general linear group (of n dimensions over the complex field)

copper.hat

Each entry must lie in a closed set $\{0\}, \{1\}$ or $[0,\infty)$, hence the matrix must lie in the 'product' which is also closed.

Ted Shifrin

04:40

@copper.hat smack

copper.hat

"I am a mathematician."

Ted Shifrin

not a very sophus-ticated one.

copper.hat

A few times in my life I have mentioned that I am an electrical engineer only to have to eat crow shortly afterwards.

Ted Shifrin

On the hot wire?

copper.hat

:-).

user10478

05:12

Is there a test that can take temporally ordered samples and tell you whether they're I.I.D. w.r.t. whatever alleged sampling distribution is formed when the samples are graphed?

copper.hat

i suspect that is a better question for the stats perverts.

user10478

kk

copper.hat

but i am often wrong.

one potato two potato

0

Q: The Ricci tensor is a contraction of the curvature tensor

In Petersen's Riemannian geometry, there is a statement that the Ricci tensor is a contraction of the curvature tensor. We use the isomorphism $TM\simeq T^*M$ given by $E_i\mapsto g_{ij}\sigma^j$ and $\sigma^j\mapsto g^{ij}E_i$ where $(M,g)$ is a Riemannian manifold and $E_i$'s and $\sigma^j$'s a...

Is there anyone who can explain this nonsense?

Ted Shifrin

The Ricci tensor assigns to a unit vector (a scalar multiple) of the average sectional curvature of all planes containing tgat unit vector. You can then get the $2$-tensor by polarizing. I don’t read Petersen.

one potato two potato

05:47

@TedShifrin Then would you recommend any Riemannian geometry textbook you like?

Ted Shifrin

Unless you’re heavy into the analysis side, start with Boothby, Spivak, DoCarmo. I don’t know it personally, but for the analysis emphasis, try Jost.

one potato two potato

06:02

@TedShifrin Thanks!

Not Tfue

07:28

A coin is tossed twice. Alice claims that the event of two heads is at least as likely if we know
that the first toss is a head than if we know that at least one of the tosses is a head. Is she right?
Does it make a difference if the coin is fair or unfair? How can we generalize Alice’s reasoning?

This is clearly wrong for fair dice.

P(HH|First toss H)>=P(HH|at least one of the toss is head) for fair dice.

But the problem is saying <

Also if it is unfair then it depends on how you assign probably init?

Not Tfue

07:44

may be language problem

I interpret a is at least as likely as b as a>=b

In this case I am okay but in case of unfair dice I am skeptical

P(AUB)>=P(A) is what Alice is sayin

Where A is first toss head

and B is second toss head

Koro

08:02

Let $g$ be a unit in $Z_n$, let g be of order m. $\sigma_g: x\to x.i \pmod n$ is a permutation. How can I find the cyclic decomposition of $\sigma_g$?

For example: if we take Z_5, the $\sigma_2=(1243)$

In $Z_6, \sigma_5=(15)(24)$

But I'm not sure how to do it for the general case.

@shintuku @DLeftAdjointtoU : you may find this interesting.

Curious Mind

08:26

why don't y'all ask chat gpt?

Chat gpt is smarter than all of you combined.

In few decade no mathematician will be needed. AI will solve all the math problem.

ペガサス Seiya

So true

Curious Mind

Thank you very much for accepting your fate.

Just typing this so that terminator won't terminate me.

Chat gpt literally helped me pass my real analysis.

Not not joking.

D Left Adjoint to U

@Koro

what is $x.i$?

Normally what you do, if $\sigma$ is your permutation

Is you compose $\sigma^i(x)$ until you reach loop back

That is one cycle

then for any $y \notin \{\sigma^i(x)\}$

you do the same thing with $\sigma^i(y)$

Do this until all elements fall into one of those cycles

These cycles commute with one another so it doesn't matter what order you write them int

@Koro heh

Koro

Hi

D Left Adjoint to U

High

:D

Koro

08:37

@DLeftAdjointtoU it's x.g

Sorry for the typo.

D Left Adjoint to U

$xg$?

With mult. we just usually juxtapose

Koro

yeah, just multiplication mod n

The problem is that g^k may not be 1 mod n.

D Left Adjoint to U

So you want a formula that gives the cycle decomposition?

Koro

Yes

D Left Adjoint to U

So you don't want an algorithm for finding it per se, but you want a freakin formula?

:D

Koro

08:40

So I want to rule out the possibility that $ (g, g^2,...,) $ is not an infinite cycle.

D Left Adjoint to U

How can it be infinite

$\Bbb{Z}_n$ is finite

and closed under its mult

so it has to loop back eventually

It doesn't need to be $1$ that it loops back to

Koro

(1,g,g^2,...)

D Left Adjoint to U

Pick any $x \in \Bbb{Z}^n$

Compute $\sigma(x)$

Koro

This cycle being finite means g^k =1 mod n for some k

D Left Adjoint to U

then compute $\sigma^2(x)$

Martin Sleziak

08:42

@Koro If $m$ is the order of $g$, then this cycle will have length $m$, namely $(g,g^2, \dots, g^{m-1},e)$.

Koro

@MartinSleziak m is the order of G should imply mg=0

Martin Sleziak

in Group Theory, 20 mins ago, by Martin Sleziak

Each cycle will be of the form $(a,ag,\dots,ag^{m-1}$, right?

Not $g^m=1$?

Koro

Yes

Martin Sleziak

I thought that you meant the order in the group of units (i.e., the invertible elements).

In any case, we can rephrase this: Let $m$ be the smallest positive integer such that $g^m \equiv 1 \pmod n$.

Koro

No, I meant the additive order. g is a unit to discard cases like 2 in Z_6.

Martin Sleziak

08:45

Ok, so do you want to use a different letter for the multiplicative order?

Koro

Yes

Martin Sleziak

Which one?

D Left Adjoint to U

$|5|_{\cdot}$

Koro

Let's say that mg=0, g^t =1

m, T are the smallest non negative integers which do that.

D Left Adjoint to U

The cycles are $\langle g \rangle a$. So what you want is a collection of residues $a$ representing uniquely each cycle.

Koro

08:48

Sorry, I am on mobile right now. So I may not have seen all the messages.

Martin Sleziak

Let $T$ be the smallest positive integer such that $g^T \equiv 1 \pmod n$.

One of the cycles is $(e,g,g^2, \dots, g^{T-1})$.

The cycle containing $a$ is $(a,ag,ag^2, \dots, ag^{T-1})$.

There are $(n-1)/T$ cycles of lenght $T$.

Koro

Yes

Ahh

D Left Adjoint to U

That's Lagrange's thereom

Koro

Is that all? I thought that m should also be brought to the mix somehow.

Martin Sleziak

@DLeftAdjointtoU Not exactly, since $(\mathbb Z_n^*, \cdot)$ isn't always a group.

@Koro That probably depends on what you're actually trying to achieve.

D Left Adjoint to U

08:52

Don't you mean $|\Bbb{Z}_n^{\times}|/T$?

Koro

@MartinSleziak I’m just trying to generalise the case for Z_5, Z_6.

Martin Sleziak

@DLeftAdjointtoU That would be a group and a subgroup. I though that Koro wants to apply $\sigma_g$ to all elements of $\mathbb Z_n$ (not only the units).

Koro

But it seems that m is in a way redundant information.

that g is a unit is just fine.

Martin Sleziak

In any case, I'll have to leave. I wish a nice day to all of you!

Koro

If g is not a unit, then consider 2 in Z_6. I think we never have 2^k ==1 here .

Thanks @MartinSleziak @DLeftAdjointtoU. Have a nice day!

I’ll think about existence of T that you mentioned. I am not sure why that will exist considering that Z* may not be a group.

If it does exist then I suppose it should have a relation with m.

@DLeftAdjointtoU yes, here you take <g> to be ‘multiplicative’.

I am thinking about ‘cyclicity’ of this <g>

Curious Mind