12:45 AM
@TedShifrin My solution still works

Was it an actual solution?

Well unless you specify the problem more yes
I doubt it's what you want
or I've misunderstood the statement

Semi had a picture from DogAteMy.

Oh no I see the problem with what I said
Despite all the angles and edges of my hexagons being equal, its not regular
Anyway, I have no idea how to tackle such a thing

I will think about some spherical geometry later.
Hmm, looks like we can’t have a spherical equilateral triangle with a 60$^\circ$ angle.
End of story. Law of cosines for the win. @Semiclassic

1:22 AM
oh yeah

1:51 AM
I'm having two issues with understanding the proof: 1) where did the idea to use $T(x) = f(x) - f(0)$ come from? what kind of construction was done to get the "idea". 2) it says that $\|T(x) - T(y)\|^2 = \|x-y\|^2$, but all we have shown is that $\|T(x)\|^2 = \|x\|^2$. I can't use linearity because it hasn't been shown yet for my $T$.

linear maps send $0$ to $0$, so you modify $f$ to a map $T$ that sends $0$ to $0$ by simply subtracting $f(0)$

1) To make a map that sends $0$ to $0$.
Deferring to Thor.
2) rigid motion says that equality

@TedShifrin an equilateral spherical triangle with a $\pi/3$ angle would have $0$ area.

I have this proof in my diff geo text.

for 2) it's because it was established $T$ is the composition of rigid motions hence rigid motion.
well I had to prove it in a previous exercise so it applies here

2:00 AM
@robjohn I didn’t see that without law of cosines, as it requires deducing equiangular. More thought.

@Thorgott Why do we want to do that?

we want a linear map

@TedShifrin Spherical excess of a spherical triangle is equal to the area of the triangle over the square of the radius of the sphere

But how do I know the other angles?

it is equilateral?

2:04 AM
@D.C.theIII Read the statement of the theorem?

so you are trying to show that equilateral means equiangular?

You are, yes.
I was avoiding that. It follows from the law of cosines, but I got there in a more elementary fashion.
But I admit my instinct originally was angles and area.

IT seems orthogonality "fell out" of how I defined the map to make sure it is linear. I was thinking that we would have to construct it with the purpose of orthogonality

It fell out of being an isometry.
I don’t know what your sentences meant.

In a paper I wrote a long time ago, I worked with dual triangles. For a given triangle, there is a dual triangle where each side is supplementary to the corresponding angle of the orginal and each and each angle is supplementary to the corresponding side of the original.

2:09 AM
When I went into trying to decipher the proof I was of the idea that whatever object I construct, my intention would be to get orthogonality to come from the creation. But what you and Thor said is that the main concern was making sure my map sends $0$ to $0$.

I believe that this can be used. I don't think the Law of Sines or Law of Cosines is needed for the dual triangle stuff.
let me look.

I think I see why we needed to do that to make it a linear map. In the definition of rigid motion it isn't mentioned that the function has to necessarily be linear

@robjohn I actually have this duality in my algebra book, although I’m confused about the supplements.

I am wrong. I use the Law of Cosines in my paper.

@D.C.theIII of course.
@robjohn All clear from the doc?

2:14 AM
I think so.

🤘🏿🙏🏿🤘🏿

Prove that, for any r.v.'s $X,Y$ we have $\ \mathbb{E}|XY| \leq \norm{X}_{\infty} \norm{Y}_{1}$. Any hints.

@TedShifrin I do have it here, but I wrote that paper before I went to college. My LaTeX needs reworking.

@Roby5 do you mean $\mathbb{E}|XY| \leq \|X\|_{\infty} \|Y_{1}\|$?

2:17 AM
I rewrote the paper in LaTeX while I was in grad school.

@D.C.theIII Yes.

try \|X_\infty\| for $\|X_\infty\|$

I thought anything beyond one character didn't get captured by subscripts or postscripts?

\infty is one "object"

is it because of the "\" forward slash?
@Roby5 I don't remember the exact derivation but that smells of Cauchy Schwartz needing to be used

2:21 AM
H_\junk gives $H_\junk$
so it considers the \junk one command
even though it does not know what to do with it

there you go. ...the more you know ✨✨✨

@Semiclassical I see a pentagon

@AkivaWeinberger the one in the center has a hexagon

Yes
Assuming this object exists physically, there should be exactly 12 pentagons.
There can be as many hexagons as desired.

@AkivaWeinberger How does one show $\mathbb{E}|XY| \leq \|X\|_{\infty} \|Y_{1}\|$?

2:26 AM
There can’t be hexagons. Robjohn and I have different proofs.

No. 12 pentagons and howevermany hexagons.

Huh?

Not regular hexagons, mind.
Deformed slightly.

Oh? Not regular?
By continuity, you can’t be anywhere close to regular.

@TedShifrin even with Miralax?

2:28 AM
smacks robjohn

The straight line segments pictured are what?

@D.C.theIII Is the result not just straight up Holders inequality?

Intersect half-spaces to get the faces to be flat
if that's what you're worried about
but the original was on a sphere anyway

I’ve been on a sphere. So these are geodesic polygons.

2:35 AM
@Roby5 Holder's and Measure THeory is beyond my pay grade at the moment homie.

3:19 AM
Hölder's inequality seems like an excellent place to start, but I'm not knowledgeable enough to know what the expectation of a random variable in $\mathcal L^{\infty}(\mathbf P)$ would be

3:39 AM
How did you get involved in this, Novice?

What, this $\mathbb E \left[ \vert XY \vert \right] \leq \Vert X \Vert_{\infty} \Vert Y \Vert_1$ business?

Wow, it now takes a verification of non-bot-ness to post an answer?!!
@Novice yes

I just happened to see the question posted in the chat here, and it's closely related to stuff I'm trying to learn, so I chimed in

Ah. @Roby never specified which spaces the rv actually live in. Care to clarify?

I can't actually clarify, but I just assumed that what was written was valid, so $X \in \mathcal{L}^{\infty}$ and $Y \in \mathcal{L}^1$

3:56 AM
Sorry, that was addressed to @Roby.

i suggested this up above, but maybe it got lost in scrollback. you may wish to explore and/or understand the relation between the expectation and what you are writing as the 1-norm.
oh, "you" or whoever asked the original question.

It was Roby.

the p = infty case of Holder's inequality is barely worthy of the name. it is just positivity of the integral, as a linear map on a function space.
well, maybe not "is" just that. it boils down to just that.

Right. And he did have the expectation of the absolute value.
Which Novice omitted.
So, as Leslie suggest(s)(ed), it is immediate.
Assuming $X$ and $Y$ live where it makes sense. That’s why I done asked.

I might have goofed. This is a little above my pay grade. Maybe I shouldn't take Roby's problem away from him

4:01 AM
no harm done. we spoiled it already.
4 hours ago, by leslie townes
not sure what EE is, but maybe establish the inequality |XY| <= |X|_infty |Y| of random variables, and take expectation.

Ah, yes, I berember you said that.

after a fashion. in "lesliejax"
the la in latex was also named leslie. i'm really just bringing latex back to its roots.

Grrr … You do your best to be incomprehensible, yes. Not to mention inscrutable.

4:44 AM
Unless I've missed something, assuming $X \in \mathcal L^{\infty}$ and $Y \in \mathcal L^1$, the desired inequality really is just Hölder's inequality. There's hardly anything to do except confirming that the 1-norm is the same as the expectation of the absolute value

5:23 AM
Right, but as leslie points out, it’s rudimentary. We don’t need to cite Hölder.

6:08 AM
I don't immediately see that, so I'll think about it a bit

1 hour later…
7:33 AM
hi, at this point in the video youtu.be/… ; can someone pls explain to me about how the two integrals merge into one double integral? Any theorems etc?

2 hours later…
9:14 AM
(int a..b f(x) dx) (int c..d g(y) dy) = int c..d (int a..b f(x) dx) g(y) dy = int c..d (int a...b f(x) g(y) dx) dy is the identity k int p..q h(t) dt = int p..q k h(t) dt applied twice, first with t = y as the variable of integration and k = int a...b f(x) dx the 'constant' (ie something that does not depend on t) and h = g the one variable function, and then again, for arbitrary fixed y, with t = x as the variable of integration and k = g(y) as the 'constant' and h = f the one variable function
that this 'integrated integral' can be computed for that particular f and g and bounds as a 'double integral' over a 2D region in the plane is a special case of something usually called "fubini's theorem"
and its realization as a double integral is what he's using when he computes the value of that using polar coordinates on the previous slide
and fubini's being used again to express the polar double integral as an iterated integral
the multivariable change of variable theorem is where you use the fact that those things have second lives as double integrals

Let S be an ordered set. Let A ⊂ S be nonempty and bounded above. Suppose sup A exists and sup A ∈ A. Show that A contains a countably infinite subset.
I don't know where to even begin...
I just want a bijective function mapping from either N to subset of A or subset of A to N.
Or Injective one.
|N|<=|subset of A| and |subset of A|<=|N|.
The statement isn't intuitive to me.
I think integer match the statement but I don't know whether it has countable infinite subset.

9:35 AM
Good morning everyone, the author in Example 7 in the following picture appears to be trying to explain something specific. Why the standard basis is not orthogonal?
Am I missing something here?

10:22 AM
They are not orthogonal with respect to the inner product written down in the same example (but they are orthogonal with respect to the usual inner product)

@AlessandroCodenotti so does this mean we have several definitions for the dot product?

"dot product" usually refers to the standard inner product on $\Bbb R^n$. An inner product on a vector space $V$ however is just a function $V^2\to\Bbb R$ satisfying some properties, and there's many different ones

To begin by defining the dot product and then providing that example seems like an awful pedagogical strategy.

10:38 AM
@NotTfue is that statement even correct? Perhaps it is supposed to say $\sup A\notin A$ rather than $\sup A\in A$?
If I take $A=\{a\}$, then $\sup A=a\in A$ and $A$ is finite.

@MartinSleziak yes not it
not in*

Ok, in that case this seems like something reasonable.

I don't have any idea what that thing is how do you even intuit it?
I was thinking to begin from something simple but don't have idea

S is a totally ordered set or a partially ordered set?

my course hasn't introduced us to this partially ordered set and total ordered
it just says ordered
An ordered set is a set S together with a relation < such that
(i) (trichotomy) For all x, y ∈ S, exactly one of x < y, x = y, or y < x holds.
(ii) (transitivity) If x, y,z ∈ S are such that x < y and y < z, then x < z.
We write x ≤ y if x < y or x = y. We define > and ≥ in the obvious way.
here is the definition
I would appreciate any advise you give

10:42 AM
Ok, this makes things simpler. BTW many people call this total order or linear order: en.wikipedia.org/wiki/Total_order
You can simply try adding new elements to A.
If we start with some $a_0\in A$, then $a_0$ is not a supremum. So there exists $a_1\in A$ such that $a_1>a_0$.

In this way you can inductively get $a_0< a_1< \dots < a_n < \dots$
Exactly!

yes this was the step where i was lost

So is it clearer now?

yes but after i thought about it i didn't knew where to go
may be i will take time

10:45 AM
If you already have $a_0< a_1< \dots < a_{n-1}$.
You know that $a_{n-1}$ is not an upper bound for $A$. (Otherwise you would get $a_{n-1}=\sup A$, a contradiction.)
So there exists some element in $a$ which is strictly larger than $a_{n-1}$.

oh think it is clear now 0u0
i can go ahead from this thank you very much

This is not exactly the same - but it is similar: Prove that the intersection have infinite elements.
The linked post deals with real line - but I guess the idea is the same.

proof by contradiction is timesaver in this case :)

11:14 AM
prove that there is infinite element then prove that subset can't be finite

11:26 AM
oh sorry i was wrong
i read it wrong i just need to construct a countable infinite subset my bad
direct proof is enough

11:41 AM
@TedShifrin English is confusing. For example, why's it called evening when it has already happened? It should just be called even.

3 hours later…
2:22 PM
@AkivaWeinberger huh, you’re right
@TedShifrin law of cosines seems like overkill here: the sum of angles in any spherical triangle exceeds 180 degrees

But how do you show that all the angles are greater than $\pi/3$? We need to show that an equilateral triangle is equiangular.

1 hour later…
3:40 PM
@robjohn How about the spherical Law of Sines?

4:01 PM
@PM2Ring is that any less involved than the Law of Cosines?
We were hoping for some way without spherical trig

@robjohn Oh, ok.
The spherical law of sines is just like the flat one. In a sense, it's simpler.
Another related structure is the buckyball. Here's a stereo view I raytraced several years ago.

4:46 PM
@PM2Ring I wrote a paper on Spherical Trig about 45 years ago, and latexified about 38 years ago (so the LaTeX could be improved a lot), and I posted a link to Ted yesterday. The first and easiest thing to show was the Law of Cosines.

@robjohn Looks good. I'll have a closer look later. I have a slight headache, so reading PDFs is a bit painful. (I mostly browse in dark mode, but my phone can't do dark mode on PDFs).
I learned spherical trig a few decades ago, and proved all the basic stuff, but I don't remember the proofs offhand. I remember that proving the angle excess = area theorem was a little tricky, but very satisfying. :) I often use the spherical cos law for astronomical stuff, although these days I also "cheat" and convert to cartesian.

in what seems like a usual proof of $\mathbb{Q}$ being the same cardinality of $\mathbb{N}$, you partition the rationals into sets like $A_1 = {0}, A_2 = {1/1, -1/1}$, ... . When the set elements are ordered as such, it seems like there is a pattern of what rational is attached to what natural (for odds, and for evens). And then you can pair each natural number to an element in each set. Is it possible to generate the rational corresponding to each natural?

how would I get the mean of a grouped frequency distribution given by class boundaries & frequency. I tried using the midpoint of each class * frequency / # of classes
nvm I was meant to divide by total frequency

5:05 PM
@SillyGoose IIRC, Cantor used a simple diagonal pattern, but that was only for pairing naturals with positive rationals.
There's a nice algorithm to generate ordered subsequences of the rationals as Farey sequences. But I guess that's not so helpful when you want all rationals. ;)
But if you are intrigued by Farey sequences, check out Hatcher's The Topology of Numbers. He explores them extensively in the early chapters.
Sep 15, 2022 at 4:05, by PM 2Ring
Hatcher has taken an interesting approach with The Topology of Numbers. He makes the PDF of it available while he's still working on it. So he gets free proof-reading & feedback. https://pi.math.cornell.edu/~hatcher/TN/TNpage.html

5:33 PM
@SillyGoose Sorry, I just realised that Cantor's enumeration is useless for you because he just jumps over rationals that aren't in their lowest terms, so you can't easily map them to their index number. Farey sequences only produce rationals in their lowest terms, but (AFAIK) there isn't a simple way to calculate them from the index number, you just have to chug through the algorithm.

5:53 PM
That was a silly "fact." Is it likewise true that any series of positive terms must diverge?

Wow how did you read that? I removed my message.

I am all-powerful.

He's considering $\lim_{n\to\infty}(1+1/n)^n$ as an infinite product like taking product $1+1/n$, $n$ times and letting $n$ to infinity.

Yeah, which isn't even an infinite product. An infinite product typically is $\prod a_n$.
In his case, the terms keep changing.

6:30 PM
i'm forever frustrated that \det exists but not \tr
obviously you can add it as a command, but it means that $Tr(A)$ keeps showing up forever and ever on the main site
it taunts me

6:50 PM
$\det(A)$
$\tr(A)$
hmmm never knew that

let out Semi.........pour out the emotions

it's not hard to fix, but it's annoying to have to fix it at all

is it difficult to create the new command into latex?
as in implement it?

nope. very easy. but if you're just somebody typing up their problem on the main site, you won't know about that
so it shows up over and over
whereas \det doesn't, because it's a built-in command

6:54 PM
can't petition to the MSE Gods to do something about it?

doubtful.
also it fits in that awkward spot of
it annoys me enough to grumble about it, but not enough to actually think it's worth fixing

I'm guessing, but this is my naivete on the subject matter, but I'm guessing trace is not used as much as det in the "world of math" so it was a process decision

blame Knuth for it

@Semiclassical take a sip of peppermint tea and come back down...

it is a bit funny tho. historically, determinants mattered a lot more than traces

6:57 PM
what makes R(+,\exp) so innacessible?

but in QM, traces are just basically more important than determinants

ah...and here is the personal bias coming out... 😄

(e.g., trace is how you deal with density operators rather than pure states. det only shows up because you sometimes need to diagonalize stuff)
@geocalc33 how do you mean "inaccessible"?
as in, hard to prove whether various elements are irrational/transcendental?

R^2 viewed as a ring is nice. although its not a field or integral doamain

simply put, because there's nothing terribly nice about adding exponents of stuff

7:00 PM
yes
I just wonder if there's anything more to say than the simple answer

What does that notation even mean?

I wonder if there's a way to understand exactly why it's not nice and where the breakdown occurs

I truly have no idea what you’re talking about. Not unusual.

like sums of exponents of stuff. they're not nice at all
i can see that from working through examples

Never mind. I'll just ignore you.

7:36 PM
if a basis for $W$ is $\{w_1, \cdots, w_n\}$ and a basis for $W^*$ is $\{w_1, \cdots w_n, w^*\}$, then the dimension of $W^*$ is greater than the dimension of $W$. of course, the definition of dimension is the number of vectors in the basis
but uh, how come we can count the number of elements in a basis
it feels like i'm missing some axiom or something, we haven't taken a bijection with a natural number or nothing of the sort
do we usually just say, hey, listen, you can count the number of elements in a basis
shh

is your vector space fintie dimensional or infinite dimensional? ....that's how you know you can count the elements of a basis

right, yes, in that sense we know there is a bijection with some natural number, at least one
if my vector space is finite dimensional
but where does the order relation appear
why does it feel i'm doing black magic when i say the dimension of a vector space is greater than that of another
we haven't defined arithmetic on bases anywhere

you can count the number of elements in a finite set

what I'm getting at, is, if we say the cardinality of a basis is greater than the cardinality of another basis, now I understand
is this what we're saying?
i.e., a vector space has greater dimension than another if the cardinality of its basis is greater than that of the basis of that other space

If I had a quarter for every time I screwed up between even and odd functions
I'd be a millionaire

7:48 PM
the dimension of a vector space is the cardinality of any basis, so yes

cheers thanks

there is no such thing as "the basis" as vector spaces tend to have many different bases, but it turns out they all have the same cardinality (this is a theorem)

hm, worthwhile noting
by chance do you know an exposition of that theorem?

every textbook on linear algebra

it is usually covered by the second chapter
whenver dimensionality is introduced

7:50 PM
great

in the finite-dimensional case, anyhow
the infinite-dimensional case is usually postponed

Now, now.....why bring in the complexity of infinite dimensions...

@ペガサスSeiya Look at even and odd powers of $x$.

cause it requires a different proof

@TedShifrin I do, but sometimes my brain just turns off and I screw up the signs

7:51 PM
I actually once wondered whether the usual proof in the finite-dimensional case can be extended to the infinite-dimensional case in a somewhat straightforward manner and turned it into an MSE question. It remains unanswered to this day.

agreed (one I haven't learned yet) but if shintuku is just asking about basic dimensionality then he probably ins't aware of that
but for rigour you are absolutely correct. and I should apply this strictness
@TedShifrin so the $LDL^t$ decomposition is an application of orthogonal operators huh?....you sly devil you

Axler says length of a list

@shintuku simpler language than the fancy word "cardinality"

i don't know, the way he's using lengths of lists, i'm not sure it behaves like cardinality
it probably does

I would suggest Friedberg, Insel and Spence over Axler's book, but that's just personal preference

8:01 PM

@D.C.theIII How is that?

@TedShifrin Ah...I should be better with the language I use... Orthogonal Equivalence.

I still do not follow you.
The decomposition applies to bilinear (or quadratic) forms, not to operators.

I was just doing some exercises on quadratic forms and they revolved around putting the symmetric matrix into one that is diagonal and the change of coordinates
So exactly what we did in your chapter on second derivative tests.

Yes, exactly. But nothing to do with operators, as I said. Bilinear forms are a different breed.
These diagonal entries are not eigenvalues, in particular.

8:09 PM
So for future reference. The correct language I should have used is "showing the symmetric matrix is orthogonally equivalent to a diagonal matrix".
@TedShifrin The ones we got from your approach are not necessary eigenvalues?

No, most definitely not. Why are you saying orthogonally equivalent? $L$ is totally not orthogonal!

Ok....I didn't actually work one out from your text. I just jumped to the conclusion due to how similar the ideas were.

I think you’re doing $PDP^\top$ with $P$ orthogonal/unitary. This is doing transformations, not bilinear forms.
They’re not at all!

Yeah that's what I was doing.

You’re doing spectral theorem. As I’ve told you before, see the last exercise in my book. This is Sylvester’s law of inertia.
Relating $D$ and eigenvalues.

8:15 PM
Well the question asked: find new coordinates $x'$, $y'$ so the quadratic form may be written as $\lambda_1(x')^2 + \lambda_2(y')^2$, such as for $x^2 + 4xy + y^2$ as an example of one of them I did
Oh ok, so what you introduced using $LDL^t$ is not the same thing.

It depends what kind of change of coords you allow.

the next section after these exercises "officially" introduces the "spectral theorem" concept

I have the same stuff in chapter 9.
You need to understand that completing the square and rotating axes are totally different.

ok so there are some nuances between the two things. Well when I get to your chapter 9 I'll be getting a second explanation of things.....with more pictures!! 🎉🎊🎉
Thank you for the clarification by the way

8:37 PM
@geocalc33 i will also note that this sorta doesn't make any sense as written. addition is a binary operation: it takes in two numbers and spits out two others. but exponentiation x->e^x is unary
there's (x,y) -> x^y but that's not an associative operation

I learned of a new way to complete synthetic geometry proofs, using a regular n-gon. This is fun

since $(x^y)^z = x^{yz}$ isn't the same as $x^{y^z}$
so it doesn't form a ring and therefore there's no real reason to be shocked that it doesn't behave nicely
garbage in, garbage out

This is annoying. First it's closed; then he edits; then, apparently, he deletes when I told him that it was (mostly) correct. Hmmm.

2 hours later…
10:23 PM
@TedShifrin after the edit, it would probably have been opened again.

@TedShifrin he's a troll
Like me

Well, I think I was right and it was a homework/exam question, and now he wants no part of it.
@robjohn Yeah, as I commented, I upvoted and voted to reopen (and I did not vote negatively, despite my original comment). Actually, I bet he found his "attempt" by — finally — searching related questions on here. That problem has appeared in various forms before. His notation is, however, quite non-standard.

When I get back to my computer, I will undelete it and vote to reopen.

10:46 PM
There

11:00 PM
It's probably labour intensive, but is there a way to find out what institution somebody is asking a homework/exam question from and then report it to the respective prof?

Hell no.

🤷🏿‍♂️

11:25 PM
@TedShifrin the moment when you're trying to answer a good question and OP deleted it

consider the differential equation $y' = \frac{-x}{ye^{x^2}}$ with starting conditions $(0,1)$, which after using the separable equations method produces $y = \pm \sqrt{e^{-x^2}} + C$
how would know we need to take the positive root in order to obtain $C=0$?
seems like we would also succeed if we let $C=2$ and take the negative root

11:40 PM
@ペガサスSeiya Not what happened.
@shintuku This is incorrect. Be careful!

hm
I did it again and arrive at the same indefinite function

Where does the $+C$ go?

I don't understand, its interpretation on the graph?
or in the equation above?

In the equation.
You don’t get to put $+C$ at the end of everything.

from $y \frac{dy}{dx} = -xe^{-x^2}$ we get $\int y \ dy = \int -xe^{-x^2} dx + C_1$, no?

11:48 PM
OK.

then we might obtain a different $C$ as we compute those two indefinite integrals

Huh?

guess not, I don't have the formal theory for this step

That is fine. Now go to the next step.

we get $\frac{y^2}{2} = \frac{e^{-x^2}}{2} + C_2$

11:53 PM
That’s still $C_1$ :)

hm, noted, thank you
so there is only a single correct $C$
or at least, the problem proceeds with a given $C$

Yes. You can combine two additive constants into one.
So what’s $y$?

then we obtain $y = \pm \sqrt{e^{-x^2}} + C_1$

YELLING

hehe, well, i'll go review this in some book
thanks for the attempt

11:55 PM
aw come on
You’re making a sixth grade mistake

oh well, for starters, we get $C_2$

Sure.

oh, and inside the root

Right!
Big difference, as $\sqrt{}$ is not linear?

ah
now it makes sense

11:59 PM
Moral of the story ?

or uh