Mathematics

William Sun

00:01

(At the last line) the author writes it out like it’s a very obvious fact but I have to verify it separately.

Ted Shifrin

@DarkRunner You can turn this into algebraic equations. Write $\sec\theta$ in terms of $UI$ and $DE$ with the first equation, then substitute into the second equation (replacing $\tan^2\theta$ with $\sec^2\theta-1$).

DarkRunner

I just asked a question; If anyone could take a look, that would be great. math.stackexchange.com/questions/3387557/…

@TedShifrin Ok, let me try that

@TedShifrin Oh; That's the problem. I actually meant $(\frac { 1 }{ 2 } -\tan { \theta } )^{ 2 }=DE(DE+IE)$, not $(\frac { 1 }{ 2 } -\tan { ^{ 2 }\theta } )=DE(DE+IE)$

Sorry about that bad latex

Ted Shifrin

I played with your original one, and it comes out with no unique answer, for sure, and it's a total mess.

OK, this is going to be a horrendous mess. I am not going further with it.

Konformist Liberal

is the derivative of $f(x) = x / ||x||_2$ defined from n-dimensional real space without zero, just zero?

Ted Shifrin

Good luck.

Of course not, @KonformistLiberal.

The unit vector varies as you change direction.

DarkRunner

00:17

Oh no. It's that bad?

Konformist Liberal

Could you tell me what it's?

DarkRunner

OK let me see what I can do

@KonformistLiberal Me?

Ted Shifrin

@KonformistLiberal: What precisely are you trying to do?

Konformist Liberal

I am trying to show that that function is differentiable

Ted Shifrin

Without using theorems?

Konformist Liberal

00:20

With anything really, but actually I want to find the derivative itself too

Ted Shifrin

So work it out in $\Bbb R^2$. You can take partial derivatives of the component functions by formulas. Or you can use the quotient rule. The beginning is know the derivative (or gradient) of $\|x\|$, which is one of my favorite examples in multivariable calculus. You can reason it out geometrically or, of course, you can just do the easy calculation of partials.

Konformist Liberal

I did that and found that derivative of $\|x\|$ is $x / \|x\|$

sorry $x^{T} / \|x\|$

Ted Shifrin

OK, good! (Points radially outward because that's the direction in which $\|x\|$ increases the fastest, and the maximal rate of change is $1$, so a unit vector!)

OK, so now just do quotient rule.

Konformist Liberal

I did and got 0

Ted Shifrin

Oh no you didn't :P

It's a bit tricky because this is a vector-valued function.

schn

00:28

When finding the matrix of a linear transformation, one can find out where the transformation sends the basis vectors. Will any basis vectors do here or is the matrix specific to a given basis? (i.e. can one try the standard basis)

Ted Shifrin

The derivative is an $n\times n$ matrix.

@schn: Depends what you're trying to do. Unless otherwise specified, usually the standard basis is intended.

Konformist Liberal

oh is it $(\|x\| - 1) / {\|x\|}^{2}$?

Ted Shifrin

How is that an $n\times n$ matrix?

schn

@TedShifrin Okay.

Konformist Liberal

I honestly get 0 haha

Ted Shifrin

00:33

You're messing up the matrix form of the thing. One term is $(1/\|x\|)I$, where $I$ is the identity matrix. What's the other term?

Semiclassical

Hi

Ted Shifrin

Hi @Semiclassic

Konformist Liberal

The other term is $-(1/\|x\|)I$?

Ted Shifrin

Nope, nope.

If you think about taking the directional derivative in direction $v$, the other term will give you $x(-x^\top v/\|x\|^3)$.

It's very important not to confuse $xx^\top$ (a matrix) with $x^\top x$ (a scalar).

Konformist Liberal

Is the answer messy?

Ted Shifrin

00:40

I don't consider it messy.

But it has two terms :)

Whatever the answer is, it will not be $0$, but it should give $0$ if you apply it to a vector in the direction of $x$ (because as you move in that direction, the function is in fact constant).

Konformist Liberal

$(1/\|x\|)I - (xx^\top)/ \|x\|^{3}$?

Ted Shifrin

Yes, which you can write a bit more nicely as $$\frac1{\|x\|}(I-xx^\top/\|x\|^2).$$

Note you're subtracting from the identity a projection onto $x$.

Konformist Liberal

Thank you!

Ted Shifrin

You're most welcome.

Rithaniel

00:58

I hate that uncertainty after you've just finished a test where you weren't sure how you did

Ted Shifrin

Are you one of those people who moans "Oh I flunked it so bad" and then gets a 97?

Rithaniel

Maybe? Usually I like to think that my predictions are accurate.

I just hate when I could see it going either way

Ted Shifrin

Hi @Eric.

Rithaniel

Like, I'm pretty sure I did the math correctly, but did I use the correct rules in the correct places?

Ted Shifrin

Yeah, with probability you always have to wonder about setting up the sample space correctly.

Rithaniel

01:06

I would actually like if the class were more theoretical and less computational. (On that note, I need to brush on my calculus. Got stuck on one integral for 15 minutes)

Ted Shifrin

I taught a pretty computational class, but we discussed theory and proved a lot of stuff (and I even made them write a few proofs on exams).

Rithaniel

Some people hate proofs. I've grown to where I actually enjoy them

schn

@TedShifrin Regarding my question of finding the matrix of a linear transformation earlier...say one is given the equation of a plane in 3-space, but the basis isn't explicitly given except that it's orthogonal. To find the matrix of the transformation of projection onto the plane, one can first find the transformation onto the line with the normal of the plane as a direction vector, but which vectors are at hand to plug into the transformation to see where it sends them

if the basis vectors aren't given?

Rithaniel

Though, my proofs are very "stream of consciousness" style. I start writing what I am thinking and then just follow where my mind goes

Ted Shifrin

What basis is orthogonal?

You need practice writing good proofs, @Rithaniel.

schn

01:10

*orthonormal

Ted Shifrin

@schn: I do not know what your exercise is. I have stuff in my course where we use a convenient basis (namely vectors in the plane together with the normal) and write down that matrix, and then use the change of basis formula to give the standard matrix. You can do it all directly, if you think.

Rithaniel

Well, obviously I plan the proof somewhat, or at least go back and clean it up after getting it all down

schn

@TedShifrin Okay.

JavaScriptCoder

okay, so i'm doing a little boolean algebra for computer science, and i'm stuck on this one question. basically, it's asking me to prove A or ((not A) and B) is equal to A or B using only properties. does anyone know where i should start? i've been trying to use Double Complement and Distributive Property, but i'm not sure that's the right way to go.

Sorry for interrupting, i just have absolutely no idea where to go.

Karl Kronenfeld

@WilliamSun Ha, I remember having trouble with that line the first time I read it some years ago!

The idea shows up frequently, so it's not a bad thing to spend some time internalizing it anyway.

Ryan Unger

01:47

@Semiclassical arxiv.org/abs/1907.03873

JavaScriptCoder

02:03

any answers for my question?

Karl Kronenfeld

@JavaScriptCoder Yeah, the distributive property is a good idea here. Different people list different axioms to work with. Do you have some sort of way to derive A = A or (A and X) for any X?

JavaScriptCoder

yeah i think i did, isnt it
A or (A and X) = A and (1 or X) = A and 1 = A?

Karl Kronenfeld

yep

JavaScriptCoder

I'm not sure how that helps though...

Karl Kronenfeld

Let X=B

JavaScriptCoder

02:18

yeah, but that's not A... oh wait I think I might get it, give me like 15 seconds

wait no, I don't think I get how that works with not A

Karl Kronenfeld

The distributive property helps; think in reverse.

You might have to use the commutative property.

Joe Shmo

i dont know what properties youre working with, but did you manage to convince yourself that in fact the equality holds?

JavaScriptCoder

I used truth tables, so I've convinced myself in a most thorough way

it's just that I'm not allowed to use truth tables in real life

I mean for this assignment at least haha

Joe Shmo

truth tables are ok, but here's better -- suppose A is true, does equality hold?

well also good luck using truth tables when you have even 4 variables

JavaScriptCoder

yeah that's true. i've already imagined that, but i don't see how that will help me get to a final answer

Joe Shmo

02:26

so wait. indulge me for a second. suppose A is true, does equality hold? why?

The proposition states --
$A \ lor (\neg A \wedge B) = A \ lor B$

JavaScriptCoder

If A is true, the equality holds because (TRUE OR X) is always true, and if A is false, the equality holds because (FALSE OR X) is X.

lemme turn my MathJax on

but I still don't get how to get to a final thing

Joe Shmo

one sec, ill get it right

$A \vee (\neg A \wedge B) = A \vee B$

no, @JavaScriptCoder that wasn't quite convincing

JavaScriptCoder

hmm, how should I make it more convincing?

Joe Shmo

well, if $A$ is true, then both sides of the equation are true. If $A$ is false, then $\neg A$ is true, so that $\neg A \wedge B = B$, and so that both sides of the equation reduce to the truth value of just $B$.

anyway, the replacement rules you're looking for are here -- en.wikipedia.org/wiki/Distributive_property#Rule_of_replacement

the second rule is exactly what you need. show me your proof when youre done

JavaScriptCoder

02:49

okay!

so, here's what I got:
A OR ((NOT A) AND B) = (A OR NOT A) AND (A OR B) = 1 AND (A OR B) = A OR B

Interesting that my teacher didn't give me this part of the distributive property - to be fair it looks really familiar, almost like a De Morgan's Law sort of thing, but I don't know exactly where I last saw it

@JoeShmo

Joe Shmo

correct!

hm, if he didn't give you this distribute property, then you might need to prove it from your existing distributive properties (mind listing them all out here?) indeed, this property is at the heart of this exercise

Noob

hello how you all doing

just a simpe help

is the sentence

" the sun is shining " - a proposition ?

JavaScriptCoder

03:04

I was relatively easily able to prove it as a lemma using distributive property (A OR NOT A) AND (A OR B) —> (A AND A) OR (NOT A AND A) OR (A OR B) OR (NOT A OR B) @JoeShmo

And hello @Noob

I am pretty sure it is a proposition -- it is either true or false, no?

Noob

this can be true or false

Joe Shmo

why are you asking

Noob

03:22

i can not understand like i dont know whether the sun is shining or not then how can i assign a truth value to it . thus its not a proposition ?

Leaky Nun

that's the problem with defining a proposition as "something that can be true or false"

it's pedagogically (and semantically) ambiguous

Noob

so does this make it not a propostion or a propostion

for eg. x + y = 10 , we dont know any informatio about x and y so this is not a propostion . so can we use this same analogy for the above proposition ?

Noob

04:06

Ok i have figured it out

one more thing which of the given two is propostion

" will it rain today ?"

"today it will rain"

Semiclassical

hi @ted

Akiva Weinberger

04:57

I can't remember where I got this link from. Was it one of you guys?

The Future of Mathematics?

►

Neat lecture about formalizing mathematics, and the lack of people actually doing it

with a concrete plan and call to action

and also a lot of railing against the old guard and their egos

Leaky Nun

@AkivaWeinberger it was viral on fb etc

Akiva Weinberger

Several people on this chat belong to the intersection of "people who understand algebraic topology" and "people who understand type theory"

(I'm not one of them)

so his claim that there's basically no one in that space seems not very true

Leaky Nun

I guess he means professors

Akiva Weinberger

Fair

user76284

Does there exist $f : \mathbb{R}^{n \times m} \times \mathbb{R}^{n \times m} \rightarrow \mathbb{R}^{n \times m}$ such that $\operatorname{argmax}_{x \in \Delta_n} \min_{y \in \Delta_m} f(A,B) \cdot y \cdot x = \operatorname{argmax}_{x \in \Delta_n} (\min_{y \in \Delta_m} A \cdot y \cdot x + \min_{y \in \Delta_m} B \cdot y \cdot x)$?

Akiva Weinberger

05:00

You watched it, I'm guessing?

What's $\Delta_n$

user76284

where $\Delta_m$ is the standard simplex.

Because of this the $\min_{y \in \Delta_m}$ can be changed to $\min_{i \in [m]}$ and the $\cdot y$ to $\cdot \mathrm{e}_i$.

i.e. minimizing $y$ over a simplex means you just have to choose the best coordinate.

That's just an addendum in case it might be helpful to figuring this out.

Leaky Nun

@AkivaWeinberger not really but he told me those things many times lol

Akiva Weinberger

Oh you go to where he teaches?

Leaky Nun

yes

Akiva Weinberger

Where is that, again?

Leaky Nun

05:06

imperial

Akiva Weinberger

Is that right next to Metric

Leaky Nun

yes

we fight every sunday

user76284

Posted the question here in case anyone's interested: math.stackexchange.com/questions/3387792/….

genescuba

05:29

If $D=[-1,1] \times[-1,2]$, show that $1 \leq \iint_{D} \frac{d x d y}{x^{2}+y^{2}+1} \leq 6$

so I understand that the max area can be 6, which is just the rectangle

but why $1 \leq$

Leaky Nun

$x^2+y^2+1 \le 1+4+1 = 6$

genescuba

sorry where are you getting that from?

Leaky Nun

from $|x| \le 1$ and $|y| \le 2$

genescuba

ah I see

and x is 0 y is 0 to minimize

Leaky Nun

right

yuvraj singh

06:10

Hi I was reading Tom apostol calculus book I came across these [[enter image description here][1]][1]

Where is how how to get a new step function by adding step up function, I solved the question by actually find the value of h(x) =f(x) +g(x) by putting x, but I didn't, t get how to get the function h(x) without plotting this can anyone help me

Laotse

Hello. I wonder, if the figure about vector fields on manifolds on Wikipedia is incorrect: en.wikipedia.org/wiki/Vector_field#Vector_fields_on_manifolds I thought that vector fields "live" on tangent bundles.

So if that's not a vector field on S2, then what is it?

Akiva Weinberger

It's a map that associates each point $p\in S^2$ with a vector $v\in T_p(S^2)$

so the image is right

They've drawn the image of each point on the point.

They've placed the origin of $T_p$ at $p$

maths student

06:25

$$
\text { (a) IF } \lim _{x \rightarrow \infty} f(x)=\infty \text { THEN } \lim _{x \rightarrow \infty} \sin (f(x)) \text { does not exist. }
$$

Laotse

Thanks for the reply! But I don't understand, how can the vectors on the equator be in T_p, if they point outwards? I don't see now something...

maths student

I think this statement is false but is there any way to prove this formally using limit defination

Leaky Nun

just take $f(x) = \lfloor x \rfloor \pi$

maths student

Can you please elaborate using limit definition?

Akiva Weinberger

@Laotse Oh yeah on looking at the image more closely the vectors do look almost perpendicular at places

They shouldn't be doing that

Could be a perspective thing

(or maybe a genuine mistake)

Laotse

06:33

That's what I wonder. So each vector v should be indeed at T_p? If the vectors are not on T_p, does that image then illustrate then a more general vector bundle than a vector field?

Akiva Weinberger

Maybe it's a map from $S^2(\subseteq\Bbb R^3)\to\Bbb R^3$ or something

Better picture of a vector field on a surface

Laotse

I'm very new at differential geometry, so is $S^2(\subseteq\Bbb R^3)\to\Bbb R^3$ a vector bundle?

Akiva Weinberger

I don't actually know what a vector bundle is, sorry

MatheinBoulomenos

@ÍgjøgnumMeg lol I didn't consider that this is not standard everywhere, I'm so used to it

Akiva Weinberger

A continuous (tangent) vector field on a sphere cannot be nonzero everywhere

Laotse

06:37

Now those vector fields look like to be on TM!

MatheinBoulomenos

@Laotse no

Laotse

Where do you refer to?

MatheinBoulomenos

you can click on the arrow to see which message I'm replying to

Sayan Chattopadhyay

@Laotse I don't get by what you mean by a more "general vector bundle". What you might want to say that is the Wikipedia image is a pictorial representation of a section. But as Mathein says S^2 is not a vector bundle of R^3

MatheinBoulomenos

I was refering to the question if $S^2 \to \Bbb R^3$ is a vector bundle

Laotse

06:48

Thanks, I'm a beginner at these topics. I still wonder, what is going on in the Wikipedia image, if it cannot be a vector field (some of the vectors are almost perpendicular to the manifold)?

Leaky Nun

I guess we're conflating vector bundle with vector fields, two very different (albeit related) concepts

@Laotse it's supposed to be a vector field

Sayan Chattopadhyay

It's a vector field and the picture on Wikipedia is wrong, as Wiki mostly always is wrong.

Leaky Nun

or rather the zero vectors are depicted as pointing outwards for ease of visualization or something

maths student

0

Q: Iterated Integral question and fubini theorem

$$ \begin{array}{l}{\text { (3) Let } f \text { be an integrable function. Express the integral }} \\ {\qquad \int_{0}^{1} \int_{y}^{2 y} \int_{0}^{x+y} f(x, y, z) d z d x d y} \\ {\text { as a sum of iterated integrals in the order } d x d y d z \text { . }}\end{array} $$ For this to be in orde...

Laotse

That's what I was thinking about. But if we forget now that it's a misleading vector field and say that we indeed assign R^# vectors on each point of the manifold (which do not belong to T_p), are we then handling with vector bundles?

Leaky Nun

06:53

no, it's not a vector bundle

it's closer to a section of a vector bundle, if that's what you mean

it's actually a section of a vector bundle

it's the tangent bundle on R^3 restricted to S^2

Laotse

Ok, thanks a lot!

I would still have another question regarding to differential geometry, but I think I have spammed stupid questions enough...

yuvraj singh

Actually the question was already posted on main site, but no one answer how to get the new function math.stackexchange.com/questions/2857132/…

ÍgjøgnumMeg

07:17

@Mathein :D It was funny hehe

yuvraj singh

@SayanChattopadhyay please have a look once

Laotse

08:12

Since it's quiet here, I might ask my second question.

I'm confused about symplectic forms and Hamiltonian dynamics. If our configuration space is manifold M, then we our phase space is TM. A point q in TM is a tuple (x,p). Now, TM is 2n-dimensional. The question is, where does the symplectic form belong to? Does it map tangent vectors from T(TM) to real numbers or tangent vectors from TM?

Ok, I cant type T^M as T star M.

A new try. I have seen notations for the canonical two form like d_xi /\ d_pi, so the two form is constucted from the cotangent bundle's coordinates. I don't see, how the two form can map vectors from TM (which are tuples (x,v)), if the two form is NOT constucted from T_x^M, but rather from T^M.

In an easier case, one has covector p from T_x^M and it can be paired with vector v from T_x M. But how do things work in the bundle level?

I know, my description looks messy. I'm just confused, how can we plug d_xi /\ d_pi (which looks to belong to T_(x,p)^T^M) with TM?

feynhat

08:56

Hi

s.harp

@Laotse In that case the symplectic form is a form on $TM$, that is a map $T^*(TM)\to\Bbb R$

check the side-bar for instructions on how to compile latex in this chat

ÍgjøgnumMeg

09:17

rip

Laotse

I think it's maybe best to ask this multi-part question in the Math.SE itself...

feynhat

Can this result be used to prove the Mayer-Vietoris sequence?

If we show the square consisting of $C_n(A\cap B)$, $C_n(A)$, $C_n(B)$ and $C_n(X)$ (with the obvious arrows) is both pushout and pullback, then we will get an SES in complexes whose LES in homology will give us the Mayer-Vietoris.

feynhat

09:45

*for singular homology (I am aware that not all homology theories are given by the kernel-mod-image of a chain complex, so clearly this proof of Mayer-Vietoris will no make sense there).

Noob

10:44

hello anyone here know something about logic

?

i want to translate a given sentence in to propositional formula

Secret

11:02

♦ $Mathematics$ Basic Mathematics

This room is meant for all basic mathematical discussion, incl...

user21820 the logician resides there

ÍgjøgnumMeg

11:28

@Alessandro you were soooo right about german bureaucratic efficiency being a lie

Alessandro Codenotti

11:51

I know

What happened in particular? @ÍgjøgnumMeg

ÍgjøgnumMeg

12:11

@Alessandro just having lots

of trouble opening a bank account

Lolol

MatheinBoulomenos

@ÍgjøgnumMeg why though?

it shouldn't be that hard

maybe I can help you

Alessandro Codenotti

Ah I see, luckily I didn't need to do that

Ryan Unger

math.princeton.edu/events/…

my talk still isn't short enough

how to get faster

MatheinBoulomenos

talk really fast and write really fast

Ryan Unger

can I ignore questions too

user193319

12:37

If $Y$ and $Z$ are subspaces in a Hilbert space, what does it mean to say that $Y$ is orthogonal to $Z$? That $Y$ is contained in the orthogonal complement of $Z$?

MatheinBoulomenos

it means that $\langle y,z \rangle =0$ for all $y \in Y, z \in Z$

but the thing with the orthogonal complement works as well

ÍgjøgnumMeg

@Mathein it’s okay it’s been sorted out now lol, the building I live in is really new so I guess the delivery got screwed up, damn Deutsche Post!

MatheinBoulomenos

4

ÍgjøgnumMeg

12:53

Hahaha

Balarka Sen

A+

YOUSEFY

13:18

Hi, I have something not clear about issue of Godel's incompleteness theorem and their argument.
I have the following sentence from book titled, "Quantum Computing since Democritus" by Scott Aaronson, chapter 11, p.151. He writes,
[But let's start by summarizing, in a few sentences, the Godel argument itself for why human thought can't be algorithmic. How about this: The first incompleteness theorem tells us that no computer, working within a fixed formal system F such as Zermelo-Fraenkel set theory, can prove the sentence: G(F) = "This sentence cannot be proved in F" But we humans can just…

Could someone explains to me what he means by [But we humans can just "see" the truth of G(F) while computer not]? If you have question/sharing points, anyone is welcome

Sayan Chattopadhyay

13:50

Wow @RyanUnger, you're talking about the positive mass theorem, that's pretty cool.

Ryan Unger

14:09

@SayanChattopadhyay it is

Sayan Chattopadhyay

I don't know anything about it but I really hope to learn something on it someday.

Leaky Nun

15:11

@MatheinBoulomenos $\Bbb Q_2(\sqrt{6})$ reaccs only

T_01

Why is $\mathbb{Q}[\sqrt{5}] \cong \mathbb{Q}[X]/(X^2-5)$ ? And is $(X^2-5)$ the ring-theoretical ideal in this case?

ÍgjøgnumMeg

15:30

@T_01 $(X^2 - 5)$ is the kernel of the evaluation homomorphism $\Bbb Q[X] \to \Bbb Q[\sqrt{5}]$.

and yes $(X^2 - 5)$ is the ideal of $\Bbb Q[X]$ generated by $X^2 - 5$

T_01

@ÍgjøgnumMeg so if $L/K$ is a field extension and $\phi: K \rightarrow L$ the evaluation homomorphism, then $L \cong K[X]/Kern(\phi)$ ?

MatheinBoulomenos

no

ÍgjøgnumMeg

uhh well no

T_01

:(

ÍgjøgnumMeg

sniepd

T_01

15:34

I'm sorry, I'm pretty new to algebra

MatheinBoulomenos

if $L=K[\alpha]$ for some $\alpha \in L$ and you consider the evaluation homomorphism $K[X] \to L$ that evaluates $X$ to $\alpha$, then we do have $L \cong K[X]/\mathrm{Ker}(\phi)$

ÍgjøgnumMeg

fs

@Mathein I took the Straßenbahn going in the wrong direction from Mannheim and travelled from Mannheim, through Heidelberg, Dossenheim, Viernheim, back to my flat, instead of just Mannheim $\to$ my flat

T_01

@MatheinBoulomenos Oh well I see my problem. Thank you!

MatheinBoulomenos

@ÍgjøgnumMeg lol

ÍgjøgnumMeg

#Moron

T_01

15:40

And then if I have the field extension $L/K$ the Kernel of the evaluation homomorphism $K[X] \rightarrow K[a]$ is an ideal generated by some polynomial $\mu_a(x)$ and THAT is the minimal polynomial of $a$, right?

MatheinBoulomenos

yes

T_01

So this minimal polynomial thing does only make sense if we are looking at polynomial rings i guess

MatheinBoulomenos

I don't really get what you mean, it's a polynomial so of course it only makes sense in the polynomial ring

T_01

Yes. I'm just looking at my lecture notes and everything seems like I'm reading it the first time because the professor is so fast, I have no time to think about everything in the lecture

Leaky Nun

15:58

@MatheinBoulomenos hast du mein Kommentare gesehen?

MatheinBoulomenos

@LeakyNun ja

Leaky Nun

@MatheinBoulomenos bin ich richtig?

♦ $Mathematics$ Basic Mathematics

Transcript for

$Mathematics$ Mathematics

♦ Basic Mathematics

Transcript for

♦ $Mathematics$ Basic Mathematics

$Mathematics$ Mathematics