Mathematics - 2024-03-16

Joe

00:00

Well, it's not easy to draw $\mathbb R^3$, but if you can do that, then I would simply plot the three vectors in your basis.

As arrows emanating from the origin, I should add.

Then I guess you could think about how any point in $\mathbb R^3$ can be reached by scaling and adding those vectors in the appropriate way (so the set of vectors span $\mathbb R^3$), and how the way the point can reached is unique (so the set of vectors is linearly independent).

@Pizza: Does that answer your question?

Pizza

.

I was trying to see the graph on Python

Joe

Hmmm maybe it is easier on a 3D graphing calculating like Geogebra (although Geogebra does annoy me a lot)

Pizza

cdn.discordapp.com/attachments/702893241691013182/…;

Is it something like that by any chance?

(the link Is the figure)

Joe

Those just look like the ordinary unit vectors of $\mathbb R^3$ to me.

Namely, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

Two of the vectors in your basis should you have a $y-$coordinate of $1$, so it doesn't look correct to me.

But maybe I am just incapable of visualisation.

Pizza

00:15

No , you are right

I thought I had to do that

I'll try to edit

Joe

I need to go to bed, sorry.

Hopefully someone else will be able to help.

Good luck with your linear algebra!

Pizza

cdn.discordapp.com/attachments/702893241691013182/…;

@Joe ok thanks Anyway , good night !

Koro

06:07

How to prove this?

If X is any left invariant vector field of G, then I have shown that $D\phi(e)(X_e)= D\psi(e)(X_e)$.

leslie townes

4

Q: Induced Lie algebra homomorphisms are equal

Let $\varphi,\psi:G\to H$ be Lie group homomorphisms, with G connected, such that the induced Lie algebra homomorphisms $d\varphi,d\psi:g\to h$ are identical. I want to show that then $\varphi=\psi$. A proof can be found in Frank Warner's book, but it uses differential forms and I would like to...

Koro

Thanks a lot !! :-)

I searched but probably not with the right key words.

they are using exponential map there.

Perhaps there is a way without the exponential map.

leslie townes

well, warner apparently does it with differential forms :)

user 85795

06:35

knowing the right key words is half the battle

More Anonymous

Hey all!

user 85795

🔑⌨️📖🔓🗝️

Hi pal❗

Soumik Mukherjee

How to do well in interviews? any suggestions are welcome.

2

user 85795

Rule #1: Tell them what they want to hear™

More Anonymous

Exactly! Be the person they want u to be! Cooperate training :p

Soumik Mukherjee

06:49

I am doing decent in written qualifiers, but when it comes to interview, my brain just freezes :"(

More Anonymous

That's all of us :p just be a nice guy and a team player ... Worked for me

Soumik Mukherjee

team player?

More Anonymous

Someone they think can fit in the team

Soumik Mukherjee

Not sure if that applies here, I am talking about PhD entrance interviews, not job interviews.

user 85795

You still have to be willing to collaborate in a group with other researchers.

More Anonymous

06:57

Sometimes I'm convinced Go is way more beautiful than chess :p

Koro

@leslietownes I looked at it. Now, I want to do it using exponential maps :-)

But I don't understand what EW is doing in the third line of their answer.

Why is the derivative at 0 equal to d\phi(X)?

leslie townes

i don't know what definitions you're using, and in my experience all of this subject boils down to which non-equivalent definitions someone is using

math.stackexchange.com/questions/3814534/… is someone posting a near-dupe of that answer, is it better for your purposes?

Soumik Mukherjee

@MoreAnonymous Go seems way more hard than chess

More Anonymous

@SoumikMukherjee yea .. but I may have a bias

leslie townes

math.stackexchange.com/questions/2461873/… is related, but specific to 'matrix groups' as defined in some unknown way in some undergrad text

More Anonymous

07:08

@SoumikMukherjee also I completely get ur concerns

leslie townes

see also the first comment "There are many ways to define the exponential map and the differential of a function and how you answer this question is gonna depend on which of those definitions you are using. I don't have access to Tapp's book, can you say what your definitions are?"

More Anonymous

I think there should be a better way for this ... Just don't know what

Soumik Mukherjee

First time when I learned the rules, I thought it would be easier than chess, well... It wasn't

I lost my first game 1-25 or something like that

Then lost two more games without scoring a single point:"

I realized that I was wrong, Go is much more harder than chess.

Koro

@leslietownes nvm, I figured it out :-)

The definition that I'm using involves integral curves:

\exp: Lie(G)---> G is defined as exp (X)= \gamma_{X,e}(1).
Here, \gamma_{X,e} is the maximal integral curve of left invariant vector field X at e.
The definition makes sense because the maximal integral curve of a left invariant vector field is complete (that is, domain R)

More Anonymous

Yea ... I think it's more important to ask the question than know the answer. Like is Go better than chess ;)

Koro

07:25

I also use the fact: X, D\phi(X) are \phi- related to transition to my definition of exp.

Fix an X in Lie(G). If it is shown that $\phi(\exp(X))= \psi (\exp (X))$, then by the fact that $\exp$ maps an open nbd. of 0 to an open nbd. U of G, it would follow that $\phi(U)=\psi(U)$. Since G is connected, $\langle U\rangle = G$, whence $\phi= \psi$ as $\phi, \psi$ are homomorphisms.

So it is enough to show that $\phi(\exp(X))= \psi (\exp (X))$.
Claim: $\phi\circ \exp_X: \mathbb R\to H$ is the maximal integral curve to $D\phi(X)$ at $e\in H$.
Proof: $\phi\circ \exp_X(0)= e$ and $(\phi\circ \exp_X)^{.} (t)= D\phi(\exp(tX))(X\exp(tX))=D\phi(X)_{\phi\circ \exp_X(t)}$. So the proof is complete.
So by definition of exp, $\exp(D\phi(X))= \phi\circ \exp_X(1)=\phi(\exp(X))$
$\square$

user 85795

07:45

Coolio, proof by definition :P

Koro

:-)

robjohn

@EE18 It can be done a bit easier:

Rewriting Pascal's Rule, we get
$$
\binom{m+i-1}{i}=\binom{m+i}{i}-\binom{m+i-1}{i-1}\tag1
$$
Summing equation $(1)$ telescopes:
$$
\begin{align}
\sum_{i=0}^n\binom{m+i-1}{i}
&=\sum_{i=0}^n\left(\binom{m+i}{i}-\binom{m+i-1}{i-1}\right)\tag{2a}\\
&=\binom{m+n}{n}-\binom{m-1}{-1}\tag{2b}\\
&=\binom{m+n}{n}\tag{2c}\\
\end{align}
$$
Explanation:
$\text{(2a):}$ apply $(1)$
$\text{(2b):}$ the sum in $\text{(2a)}$ telescopes
$\text{(2c):}$ $\binom{m-1}{-1}=\binom{m}{0}\frac0m=0$

user 85795

08:22

(removed)

leslie townes

arr matey! it telescopes! 🦜🏴‍☠️🔭☠

shiver me timbers! sums off the port bow

user 85795

👏🏻😁👏🏻

7

Q: 10th Anniversary of Math Educators SE!

This month (March) Math Educators SE turns 10 years old. Congratulations on asking, answering, voting, and building this community for a whole decade! Whenever a Stack Exchange site turns 10 we like to celebrate by encouraging users to reflect on their experiences here. How did you first discover...

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user 85795

09:19

Professor @TedShifrin would you like to share any of your anecdotes from your 36 years of teaching?

Sine of the Time

@SoumikMukherjee "you can't be rejected if you kill everyone who rejects you" - Sun Tzu, The Art of War

user 85795

An eye for an eye...

-Bible

Soumik Mukherjee

@SineoftheTime ☠️☠️

I can't be rejected if I don't apply for it in the first place

Koro

10:03

If $\phi: G\to H$ is an injective Lie group homom., then $\phi$ is an immersion.

How to prove this?

As before, I have $\phi(\exp(X))= \exp(D\phi(X))$. So if $D\phi(X)=0$, then I want to show that X=0.

I have $\phi(\exp(X))=0$, whence $\exp(X)=0$.

How to show that X=0?

one potato two potato

11:22

what is the meaning of $E_xY$ here? What is that subscript $x$?

Jakobian

12:00

Is this about discrete processes?

$x$ is just $\delta_x$ I think, the starting point

Just how $\mu$ is the initial probability measure

Pizza

i need a new pfp , advice ?

Soumik Mukherjee

@Pizza Pizza with pineapple toppings.

one potato two potato

13:47

@Jakobian I think you're right.

PM 2Ring

14:11

@Pizza A diagram of a pizza, with radius = z and thickness = a

one potato two potato

14:31

Why I still can't understand $P_x(X_{n+m} = z) = E_x(P_x(X_{n+m} = z\mid\mathcal{F}_m))$

one potato two potato

15:24

there're three "modes" of M.inside_view() in snappy 'material', 'ideal' and 'hyperideal'. Do you know what are those? @BalarkaSen

and what's the "fillings" option for?

Jakobian

I'm so tired today

maybe its because of my coffee addiction

I've been pumping two coffees a day, yesterday I had just one, and not started my coffee today yet

John Zimmerman

15:41

I'm trying to curb my coffee intake as well

Jakobian

I'm not trying to curb my coffee intake, coffee is good for you, especially consumed regularly

at least black one, with milk you got milk in it so of course, additional calories and what not

just minimally, but its good

John Zimmerman

I'm trying to just have one cup in the morning

Jakobian

I usually drink just one

watch out when you drink it though, immediately after waking up is bad because your organism has to make you less tired in the morning naturally

0.5-1.5 hour after waking up is OK

John Zimmerman

16:00

true

Balarka Sen

@onepotatotwopotato Seems to me that what you're looking at is the universal cover of the manifold with its hyperbolic tessellation, and material is when your eye is in the bulk, ideal is when your eye is on the ideal boundary (eg cusp) looking inside and hyperideal is seeing the entire $\Bbb H^3$ as a whole

@onepotatotwopotato Dehn filling

You input the slope, it spits out new hyperbolic manifold (if it's hyperbolic)

There's a SnapPy documentation you can read

Jakobian

@onepotatotwopotato This seems to be the formula $E(E(X|\mathcal{H})) = E(X)$

Note that $P(X\in A|\mathcal{H})$ is $E[1_{X\in A}|\mathcal{H}]$ by definition

John Zimmerman

16:21

math.stackexchange.com/q/4882018/460999 I'm interested in black scholes model

I did watch the veratasium youtube video

pricing options and asset allocation schemes are neat

Jakobian

huh. You should have read a book on stochastic calculus instead

John Zimmerman

stochastic calculus

Jakobian

that's right. Black-Scholes equation is an SDE, a stochastic differential equation

John Zimmerman

yeah

the thing I need to understand better is boundary conditions and initial conditions on that SDE

but I am slowly grasping that

Jakobian

maybe it isn't an SDE, I guess I'm wrong

John Zimmerman

16:28

it is a SPDE

stochastic partial diff eq.

user10478

Are there any probability distributions with the property that as you increase at least one of it parameters, there is at least one point in its CDF which does not increase or decrease monotonically (has a possibly-local extremum)?

This appears to be false of the Gaussian and Poisson, according to an eyeball test.

John Zimmerman

um

how about a compact distribution whose CDF attains a global max at (1,1)?

Then $F(1)=1$ would remain invariant wrt. the parameter of the distiribution

also $F(0)=0$ would be a global min

user10478

16:44

hmm

I might have asked the wrong question.

I think I want an "interesting" maximum as the parameter varies (i.e., left and right neighbors along the dimension of varying the parameter are both strictly less), not just an invariant.

Alessandro Codenotti

17:07

@Jakobian are you around? I need a stupid sanity check

Jakobian

@AlessandroCodenotti physically at least

Alessandro Codenotti

Say $X_n$ is a sequence of uniformly bounded metric spaces (they all have diameter $1$ for simplicity). Is $d((x_n),(y_n))=\sup_n\{d(x_n,y_n)\}$ a compatible metric on $\prod X_n$?

Jakobian

compatible meaning?

agrees with product topology?

the answer is no iirc

Alessandro Codenotti

@Jakobian yes

Jakobian

it should be box topology

Alessandro Codenotti

17:13

Hm. Annoying but reasonable

Jakobian

yep

Alessandro Codenotti

But $\sup_n\{d(x_n,y_n)/n\}$ works, right?

Jakobian

@AlessandroCodenotti no sorry, its not box topology, its the so called uniform topology

which is different from box topology and from product topology (in general)

oh no you don't

@AlessandroCodenotti sure yeah, you just need to make $r$ small enough for all the balls to be contained and the ball is open so it works

I'm saying, for $B(x, r)\subseteq U_1\times U_2\times ...\times U_n\times X_{n+1}\times ...$

You just need $r$ to be small enough so that $B(x_i, ir)\subseteq U_i$ for $i = 1, ..., n$

and since $i$ has upper bound of $n$ this is possible

nickbros123

given $\sum a_n$, we suppose a new "series" $\sum b_n$ gets constructed by removing the 0s from $\sum a_n$, keeping the same order. I want to prove $\sum a_n$ converges iff $\sum b_n$ converges. my approach: supposing $b_n$ series converges, say that for each n, there exists $0 \leq k \leq n$ such that $\sum_{i=1}^n a_i =\sum_{j=1}^{k(n)} b_j$.

I want to assert, the LHS here is a subsequence of the main series $\sum_{j=1}^n b_j$, and since the main converges, the subsequence converges too. is this argument fine?

Jakobian

since we know our metrics are uniformly bounded we also know that $B(x, r) = B(x_1, r)\times B(x_2, 2r)\times ...$ will have $B(x_m, mr) = X_m$ for big enough $m$

nickbros123

17:24

to get rid of k actually being 0, perhaps we can assert 2 cases: finitely many 0s and infinitely many 0s?

im having a particular discomfort making this argument for, k(n) is actually less than (or equal to) n here, for every n.

Alessandro Codenotti

Right. Thanks!

I was hoping to be able to just take the supremum. This is going to cause me much suffering in the form of having to juggle $\varepsilon/i$ everywhere

There is something funny going on here. Suppose that the $X_n$ are also compact. Then there is a unique compatible uniformity on $\prod X_n$, and that includes sets that have "width" $\varepsilon$ in all coordinates

Jakobian

17:45

not sure what you mean by includes

I suppose you're going off of the entourage definition

Alessandro Codenotti

I just mean it is an element of the uniformity

Jakobian

Not sure what you mean

Alessandro Codenotti

The set of pairs $((x_n)_n,(y_n)_n)$ for which $\sup_n\{d(x_n,y_n)<\varepsilon\}$ belongs to the uniformity

Pizza

@SoumikMukherjee Okay!

Alessandro Codenotti

So the uniformity contains sets that have diameter $\varepsilon<1$ when projected to all coordinates

Soumik Mukherjee

17:53

@Pizza Nice pfp

Jakobian

18:10

@AlessandroCodenotti that doesn't sound right

Alessandro Codenotti

All sets containing the diagonal belong to the uniformity

EE18

Bad question I know, but I can't get a bound on the denominator

Any hints?

Jakobian

@AlessandroCodenotti surely not all sets

just the neighborhoods of the diagonal

EE18

That is, I can't relate $1 + |x|$ to $1+|a+b|$ ($x=a,b$) in a useful way

Alessandro Codenotti

Ah yes of course

Jakobian

18:13

and I don't think $A = \{(x, y) : \sup_n d(x_n, y_n) < \varepsilon\}$ is one

Alessandro Codenotti

No it isn't

This is all very annoying

Pizza

desmos.com/calculator/vjp9ga5fvn?lang=it

Lol

ZaWarudo

I think that is true that if $f:\mathbb{R^n} \to \mathbb{R}$ is continuous and $A \subseteq \mathbb{R}^n$ is bounded, then the infimum and the supremum of $f$ on $A$ are finite. I tried to prove it by this: since $A$ is bounded, there exists a ball $B_R(0)$ such that $A \subseteq B_R(0)$. Hence, $\overline{B_{R+1}(0)}$ is a compact set and $f$ is continuous on $\overline{B_{R+1}(0)}$.

So, by the extreme value theorem, there exists max $M$ and min $m$ of $f$ on $\overline{B_{R+1}(0)}$. But since $A \subseteq \overline{B_{R+1}(0)}$, this means that inf and sup of $f$ on $A$ are bounded too b…

Jakobian

@ZaWarudo $\inf_{x\in A} f(x) = \inf_{x\in \overline{A}} f(x)$ and similarly for supremum, since $A$ is bounded, $\overline{A}$ is compact

so $\inf_{x\in A} f(x) = f(a)$ for some $a\in \overline{A}$

your solution is fine too, sure

its basically the same proof, mine could have been applied a little bit more generally, say when $\overline{A}\subseteq U$ and $U\subseteq \mathbb{R}^n$ is the domain

but its not much of a difference

psie

18:35

Consider the conditional probability $P(E|F)=P(E\cap F)/P(F)$. Why doesn't it make sense to speak of $E|F$ as an event? The notation suggests that $E|F$ is something we take the probability of, just like $F$ and $E\cap F$, so it seems like it should be a subset of the sample space $\Omega$, but I don't know which subset it would be.

At least, no text I've come across speaks of $E|F$ as an event...

ZaWarudo

@Jakobian Thanks! Can I also say that, being $\{B_r(x)\}_{r>0}$ decreasing, we have $\lim_{r \to 0^+} B_r(x_0)=\bigcap_{r>0} B_r(x_0)=\{x_0\}$ so by continuity of $f$ in $\mathbb{R}^n$ we can deduce that $\lim_{r \to 0^+} \inf_{x \in B_r(x_0)} \{f(x)\}=f(x_0)$ and $\lim_{r \to 0^+} \sup_{x \in B_r(x_0)} \{f(x)\}=f(x_0)$?

Jakobian

The equalities you mentioned hold of course, but I am not sure about your justification

ZaWarudo

With justification you mean when I take the limit of the infimum and the supremum invoking the continuity of $f$?

Jakobian

I mean whatever you wrote

the entire thing

ZaWarudo

Okay, I am actually not sure too. I was trying to use this result math.stackexchange.com/questions/748639/…

Jakobian

18:44

@psie I'd think about it as a function $E\mapsto P(E|F)$

@ZaWarudo you're not sure what you're trying to achieve in your own argument?

EE18

Shamefully gonna perseverate :( any chance you can comment or give a hint to me on the above Jakobian?

Jakobian

@EE18 on what

EE18

The picture above re: that identity for ordered fields

Jakobian

oh okay. I didn't notice that

psie

ok, makes sense Jakobian. I guess | is not a set theoretic operation, so hence it makes no sense to speak of $E|F$ as a set, let alone an event

EE18

18:50

No worries at all!

Certainly no obligation :)

Jakobian

@EE18 Two things, $x\mapsto \frac{x}{1+x}$ is increasing for $x\geq 0$, and $|a+b|\leq |a|+|b|$

@psie it would make sense if you are willing to change your event space $\Omega$ to something else. But no one does that - I don't think there's a need to

ZaWarudo

I am not sure if that is a correct application of that theorem in the question I linked before, because it's the first time I apply continuity to infimum and supremum seen as functions of the sets we are taking supremum and infimum on; in particular, I am not sure if the result here https://math.stackexchange.com/questions/748639/prove-functions-defined-by-sup-and-inf-are-continuous?noredirect=1&lq=1 holds on a ball in place of an interval (I believe it holds). I will try to formalize it: since $f$ is continuous on $B_r(x_0)$, by this result the functions $m(r):=\inf_{x \in B_r(x_0)} \{f(x)…

EE18

@Jakobian I'm not quite sure I can see how I can use that first part but I will think about it. Thank you!

Jakobian

@ZaWarudo it should, but why do you want it to hold if there's a simpler argument for why $\lim_{r\to 0^+} \inf_{x\in B(x_0, r)} f(x) = f(x_0)$?

@EE18 apply it to the inequality in the second part

ZaWarudo

@Jakobian thanks for your help, I just wanted to be sure to not do logical mistakes. Of course a simpler argument would be better, but the former was the one that I came out with :)

Jakobian

18:59

You should just use continuity directly

Exists $\delta > 0$ such that $|f(x)-f(x_0)| \leq \varepsilon$ for all $|x-x_0|\leq \delta$

Soumik Mukherjee

@EE18 If d is a metric then d/(1+d) is also a metric

Jakobian

This makes it clear that $|\inf_{x\in B(x_0, r)} f(x)-f(x_0)| \leq \varepsilon$ as well, for $r\leq \delta$

@SoumikMukherjee I don't think EE18 knows what a metric is

Soumik Mukherjee

Ohh

Jakobian

besides the proof of your statement follows what I was saying above

EE18

I do but its not yet in this book

At any rate, I will use said tips

thanks very much to both of you

Soumik Mukherjee

19:05

@Jakobian Yes, I just gave EE18 a hint on why $x\mapsto \frac{x}{1+x}$ that you wrote is important.

EE18

Was this one that you two just knew from experience, or how does one see this?

I certainly have no intuition for it

Jakobian

I've been given this as an exercise in the past

Soumik Mukherjee

Same

Not exactly this, but the d/1+d being a metric

Jakobian

How do you see this? Well... you look at the function $\frac{x}{1+x}$ and its properties

and then you just figure it out

EE18

fair enough, i guess i just wouldn't have thought to use that lemma of that function increasing in this context

but i see it now

Jakobian

19:15

there's lots of things in math that you just need to notice

and a human being won't possibly be able to notice all of those things

but you have to try

sometimes you need to take a wild guess or make your hands dirty in some ugly argument

thats just how it is

Sahaj

20:05

Hi jakobian

good evening

leslie townes

EE18 there is a whole subfield of contest math that is increasingly elaborate variations on inequalities like the above. stuff like that is ideal for certain contest purposes because it's deep enough to allow for problems of arbitrarily high difficulty without drawing on anything more "advanced" than ordered field axioms, i.e., algebraic manipulations involving order that you might be able to expect a contest oriented preteen to know

robjohn

20:47

@EE18 The triangle inequality is a bit messy, but not too bad.

robjohn

21:04

Oh, I see that that is exercise 10, I just hadn’t looked back far enough.

EE18

21:17

No worries at all, thanks as always for your help!

My book has me prove the relatively weak claim that the homomorphism between the ring of polynomials and the set of polynomial functions is not in general injective if the field $K$ in question is finite, but it has me do this by example. I assume one can show that this is true for any finite field?

@leslietownes Touche to that, and I certainly got that sense from my brief tour with baby Rudin...it's necessary to be a bit "clever" sometimes

Thorgott

22:05

how many functions are there on a finite field? how many polynomials are there?

EE18

$|K|^{|K|}$ for the former, no clue for the latter

Thorgott

if $K$ is finite, what can you say about the former?

EE18

$|K|^{|K|}$, no?

Thorgott

of course, but what type of cardinality is this?

EE18

im not sure i follow the question, what do you mean by type of cardinality?

Thorgott

22:17

I'm looking for an adjective

EE18

finite? natural number?

Thorgott

yes, the former

EE18

Ohhhh I see now

whereas $K[X]$ infinite always

Thorgott

now tell me why there are infinitely polynomials over any field (hint: you can just write enough down)

EE18

$1X^n \in K[X]$ for all $n \in \Bbb N$

ergo, $K[X]$ not finite since it has an infinite subset

Thorgott

22:20

yup :)

EE18

thank you thorgott. Like the old Staples ads, "that was easy" (but it eluded me :/)

Jakobian

23:46

@EE18 yes, any finite field. And only finite fields

The homomorphism from polynomials to polynomial functions is always surjective but not always an isomorphism. In general, if your field, $K$, is infinite, then its an isomorphism, but if $K$ is finite and $|K| = q$, then the kernel of this homomorphism is $(x^q-x)$

@EE18 no really write $1X^n$, you can just write $X^n$

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