Mathematics - 2017-05-26 (page 1 of 6)

Akiva Weinberger

00:07

Who wants to hear some sax

playing a song that was not written for sax

Well, really a medley of songs

https://youtu.be/RkD_ynqh2VM

Akiva Weinberger

00:19

@TedShifrin I found your name "in the wild":

83

A: Books you would like to read (if somebody would just write them...)

I don't know for certain that this doesn't exist, so I'm in a no-lose situation: if this is a rubbish answer then it means that a book that I want to exist does exist. Many mathematicians of a pure bent have taken it upon themselves to get a good understanding of theoretical physics. And many hav...

"Physics for Mathematicians: Mechanics I" is apparently a reworked and expanded version of these notes: math.uga.edu/~shifrin/Spivak_physics.pdf. Now that I know about it, I'm really looking forward to reading it!!! +1 — Vectornaut Jan 24 '11 at 15:36

Ted Shifrin

00:35

DogAteMy: Yeah, Spivak asked me to post a chapter for him when he first published his book. I didn't put that link on my updated webpage, but it's still on the old webpage if people stumble over it.

Daminark

01:05

Huh, this book sounds interesting, to say the least

Akiva Weinberger

@TedShifrin Did I ever share with you the proof of the Pythagorean theorem based on an annulus?

Wait, never mind, I don't actually know if I can explain it with words without drawing stuff

Wait, Wikipedia has a thing (that's slightly different than what I would have drawn but whatever)

There's this old puzzle that asks you to show that, if length of the "inner tangent" of an annulus (the line segment tangent to the inner circle with endpoints on the outer circle) is $d$, then the annulus has the same area as the circle of diameter $d$.

This can be shown to be equivalent to the Pythagorean theorem, through a very short calculation

So if you prove that puzzle, you've proven the Pythagorean theorem.

Daminark

Rehi @Ted

Ted Shifrin

Demonark, one of my former students found errors in it, but it's got interesting stuff, yeah.

Eric Silva

hi folk

Akiva Weinberger

The puzzle is true for polygonal approximations of annuli (and here is where the illustrations would come in), so taking limits gives you the puzzle for actual annuli, whence comes the QED.

Ted Shifrin

01:17

The 957th proof of Pythagoras, DogAteMy?

Rehi Eric

Daminark

Yo Eric

Akiva Weinberger

...Only 957?

Ted Shifrin

LOL, hell if I know.

Eric Silva

I feel like that's actually a plausible number for how many distinct proofs there are

not to say how many are the same with different dressing

Daminark

Lol was about to ask

Dodsy

01:18

I thought there were like 400.

Akiva Weinberger

But, like, imagine if you take a circle and divide it into $n$ sectors (in my mental image, $n\approx6$ for whatever reason)

@Dodsy Yeah, there's a book that collects 400

Daminark

There is a book with 367

Ted Shifrin

My favorite proof, I think, is the similar triangles proof.

Daminark

Wai does this keep happening?

Ted Shifrin

Hi Nate.

Eric Silva

01:19

@Dodsy tbf same order of magnitude so like to me it's basically the same

Daminark

physicist intensifies

Akiva Weinberger

And then you translate each sector in the direction of one of the radii bounding it

Dodsy

Hello Ted. :)

Eric Silva

@Ted did Mike show you that problem with the forms

Ted Shifrin

DogAteMy: It's somehow related to the wonderful calculus exercise that when you drill out a cylinder of radius $r<R$ from a sphere of radius $R$ you get the volume equal to that of a sphere of radius $R-r$.

nope, Eric.

Akiva Weinberger

01:21

(cont'd) so it kinda looks like the Aperture Science logo but the outer "circle" should be a lot bumpier

Dodsy

I found an article suggesting an infinite number of variations of proofs for pythagoreans theorem.

Akiva Weinberger

In any case, that has the same area as the original circle, and in the limit, it becomes the annulus, so...

Eric Silva

It was to classify manifolds with $d_{\nabla}^{3} = 0$

or $F^{\nabla} \wedge d_{\nabla} = 0$ by Bianchi

Akiva Weinberger

The Pythagorean Theorem is actually a lot cleaner if you assume that $c=1$.

Which you can do, wlog.

Dodsy

n = 1

?

Akiva Weinberger

01:23

Because then, dropping the altitude onto the hypotenuse divides it into two pieces of length $a^2$ and $b^2$.

Daminark

"but for the purposes of this book on mechanics the material in *A comprehensive Introduction to Differential Geometry*, Volumes 1 and 2, will generally be regarded as prerequisite"
On second thought...

Akiva Weinberger

Without that rescaling at the start, they're $a^2/c$ and $b^2/c$, which is distinctly less pretty

Dodsy

I'm starting back on my homework tomorrow, I've been slightly depressed these past couple of days, but it's time to get going again.

Ted Shifrin

LOL, you're not quite ready yet, Demonark, but you could read a fair amount of it, modulo notation.

What homework, Nate?

Dodsy

I have to study for exams

and finish 2 units of chemistry and 2 units of English.

Ted Shifrin

01:25

Eric, manifolds or bundles on manifolds?

Dodsy

I've only finished the bare minimum to send midterms, thus far.

Ted Shifrin

Oh, yeah, I guess you'd better, Nate.

Dodsy

I have until august first, but the procteur only does an exam every monday.

Akiva Weinberger

I wonder if you could prove the Pythagorean theorem by first showing the equivalent in spherical geometry and then taking the limit as the radius of the sphere goes to infinity

Dodsy

Which means it will take up an entire month.

Ted Shifrin

01:25

That's not as far as it seems ...

Eric Silva

I assumed on bundles

Dodsy

You are correct, Mr. Shifrin.

Ted Shifrin

DogAteMy: it seems ridiculous to use trigonometry (needed for spherical geometry), which relies on Pythagoras rather seriously, and then "deduce" Pythagoras.

Right, @EricSilva, so I'm not sure what we're actually classifying.

Dodsy

I am hoping to hear back from the University of Western Ontario either next week or the week afterwards. I'd rather have only exams to complete. If I get accepted to Western I'll need to maintain an average of 83%.

Daminark

So, spherical reasoning?

Ted Shifrin

01:27

smacks Demonark

Akiva Weinberger

You'll realize I completely forgot how the...

groans at Daminark

Ted Shifrin

we should just ignore him

Mike Miller

It's me, it's bundles, of course. I was thinking U(1) and SU(2).

Akiva Weinberger

...how the proof of the spherical Pythagorean theorem goes. Or even what the statement is.

Dodsy

Stop abusing Daminark, guys.

He's soft.

Akiva Weinberger

01:28

I dunno. Amin, are you soft?

Dodsy

He's a sensitive soul.

Daminark

denies being soft while a tear comes down

Eric Silva

I havent thought about it much though

Ted Shifrin

DogAteMy: It's $\cos a = \cos b\cos c$.

Dodsy

Alright, cya lads.

Ted Shifrin

01:29

So what are we classifying, @MikeM?

Mike Miller

Fix a bundle E over a manifold M. There is a corresponding moduli space $F^{(2)}(E)$ of connections that satisfy $d_A^3 = 0$ modulo gauge. Tell me about that moduli space.

Ted Shifrin

(DogAteMy: I guess I should have been consistent with the usual and written $\cos c = \cos a\cos b$.)

It depends where the right angle is, DogAteMy. Yeah.

So from Taylor polynomials you get the usual the way you suggested. But I think it's crap.

Mike Miller

i've come to suspect that the tangent space defined by the linearization should be infinite-dimensional at an honest flat connection, but most of those deformations should be non-integrable. at a 2-flat but not flat guy though maybe it's actually finite dimensional

Ted Shifrin

Oh, so the bundle is fixed and we're just studying connections on a fixed bundle. OK, that makes sense.

Akiva Weinberger

By the way, I realized something — You know how any angle on a diameter is a right angle?

Ted Shifrin

01:31

Um, yes, DogAteMy. You obviously didn't do one of my early homework problems with vectors.

Akiva Weinberger

That wasn't the finished thought

Ted Shifrin

LOL

Akiva Weinberger

If we take a semicircle built on any line, that'll have area $Kc^2$, where $K$ is probably $\pi/8$ or something

Ted Shifrin

Huh?

Akiva Weinberger

$c$ being the length of that line.

Ted Shifrin

01:32

Line = line segment of length $c$?

Lines are infinite.

in a pedantic mood

Mike Miller

The tangent space is the set of End(E)-valued 1-forms $\eta$ such that $d_A \eta \wedge d_A \sigma + F_A \wedge \eta \cdot \sigma = 0$ (the object here is an E-valued 2-form) for every $\sigma$.

Akiva Weinberger

Dammit. So if we take any point on that semicircle, we get two new line segments with a right angle between them

Ted Shifrin

What is $\sigma$, Mike? What does that dot mean? I'm lost.

Sure, DogAteMy.

Mike Miller

section of E

Akiva Weinberger

And then we can build semicircles on those

Ted Shifrin

01:33

So dot is tensor?

Akiva Weinberger

(Preferably in such a way that they don't overlap each other)

Ted Shifrin

The hypotenuse hasn't changed, DogAteMy. I'm confuzled.

Mike Miller

I write dot to say that I'm applying $F_A \wedge \eta$, an End(E)-valued 2-form, to the E-valued $\sigma$, to get an E-valued 2-form.

Akiva Weinberger

Call the original line $AB$. Call the new point $C$.

Mike Miller

I had to denote the module structure of E over End(E) somehow.

Ted Shifrin

01:34

Oh, I see, @MikeM.

Akiva Weinberger

Then $ABC$ has a right angle at $C$, since $C$ is on the semicircle built on $AB$.

Mike Miller

In the first term it's hidden in the wedge product.

Akiva Weinberger

Build semicircles on $AC$ and $BC$ (and say those have lengths $b$ and $a$ respectively).

Ted Shifrin

Oh, I see, DogAteMy. You were totally not clear the first time.

Akiva Weinberger

Then the area of those two new semicircles will be $Ka^2+Kb^2$, which equals $Kc^2$.

Mike Miller

01:35

Modding out by gauge amounts at the linear level to adding a Coulomb gauge fixing condition $d_A^*\eta = 0$

Ted Shifrin

Sure.

OK, Mike.

Mike Miller

So combine those two equations and tell me if the solution space is finite dimensional. :)

Akiva Weinberger

And then we can repeat this ad infinitum I guess, "popping" each semicircle into two
(not necessarily equal-size) smaller semicircles, keeping the total area the same

and I guess I don't really have anywhere else to go with this, other than to say that I think it kinda looks cool

It's fractal-y.

Ted Shifrin

But total turns into longer sums, DogAteMy.

Yeah, I see ...

Akiva Weinberger

@TedShifrin What do you mean?

Ted Shifrin

01:38

From one stage to the next, the sum of two is one. But to keep the sum equal to $Kc^2$ you need a bunch of terms.

Like $2^k$ or something.

@MikeM: I truly haven't thought about this stuff in 30 years or so. So it's all going to depend on compact/Fredholm stuff, I assume.

Mike Miller

When working with $U(1)$-bundles and a connection $A$ that is flat to start with, this reduces to "$\eta \in \Omega^1(i\Bbb R)$ such that $d^*\eta = 0$ and $d\eta \wedge d_A \sigma = 0$ for all $\sigma$"

Akiva Weinberger

Yeah, I guess. It's just the total area of the semicircles.

Ted Shifrin

Agreed, DogAteMy, by transitivity.

Mike Miller

reducing even further to the trivial bundle and trivial connection, the last equation is just $d\eta \wedge df = 0$ for every complex function $f$.

Ted Shifrin

BTW, DogAteMy, what is the locus of points $C$ so that $\angle ACB$ is fixed? ($A,B$ fixed.)

If you're doing $U(1)$, isn't $f$ pure imaginary or something, Mike?

Do we know anything about the base manifold?

Akiva Weinberger

01:42

@TedShifrin It's a larger circle, yeah? Well, more like two circles stuck together since it's symmetric about $AB$

Ted Shifrin

A bit sloppy, DogAteMy.

Not sure exactly what you mean.

Akiva Weinberger

Angles on a chord are constant

Ted Shifrin

When the angle is $\pi/2$, then we get a circle, I agree.

Akiva Weinberger

Draw the circle such that the angle of the line on the origin is twice the desired angle

Ted Shifrin

I have no idea what you're telling me.

Akiva Weinberger

01:44

Say $AB$ is the line.

Ted Shifrin

segment.

Yes.

Akiva Weinberger

Choose an $O$ on the perpendicular bisector such that $\angle AOB$ is $2\theta$.

Draw the circle $OA$. (Is that the notation? Center $O$, radius $OA$.)

That should be the desired locus, except that it only works for the part of the circle that's on the same side of $AB$ as $O$

but you can mirror it across $AB$ to get the other half of the locus

Ted Shifrin

Oh, OK. I get what you meant now, but your communication skills were weaker than usual..

Akiva Weinberger

Sorry

Ted Shifrin

Yup, so it's the union of two congruent arcs of circles.

When $\theta=\pi/2$ those are semicircles gluing together to give a nice circle.

Cool. I have to go cook dinner. G'night all (for now).

Akiva Weinberger

01:47

And when $\theta=\pi$, it's the line segment itself.

Mike Miller

02:00

@TedShifrin No, eta is. f is a section of the associated complex vector bundle.

Base manifold isn't a surface.

Secret

02:55

In short, cleverly decomposing a non hermitian matrix into a symmetric formulation speeds up calculation without sacrificing much accuracy

Nick

05:02

Can someone help me with math

I have a test today in like 2 hours.

arctic tern

maybe

Nick

Could you explain why $P(a \leq X \leq b) = P( a < X < B)$?

Do you get my question. Sorry for the typo

arctic tern

presumably the probability of $X=a$ or $X=b$ is $0$

Nick

@arctictern Yeah, my question is why is that?

The chapter I'm looking at is probability density

arctic tern

if $X$ is a continuous random variable, the probability of $X$ being in an interval is the integral of a density function on that interval. but set with one element has measure zero, so the integral over it will be $0$

Nick

05:13

Why does a set with one element have a measure zero?

arctic tern

what is the measure of $[0,1]$? what is the measure of $[0,1/2]$? what is the measure of $[0,1/1000000]$?

you make an interval smaller and smaller, its measure (length) gets smaller and smaller

a single point has no length

Nick

Well, the interval $[x- \delta x, x+\delta x] \text{ as } \lim \delta x \to 0$ has zero length. You can't really say that's no length. It's some small $dx$

arctic tern

It's zero. Don't be silly.

Nick

What exactly do you mean by measure? Do you mean length?

arctic tern

for real numbers, measure = length

Nick

05:17

Ohk, the probability given by area will hence be null. I get it now.

arctic tern

more generally, "measure" is a mathematical way of assigning a number to subsets in a way that matches the properties of length / area / volume / probability / cardinality etc.

If you think $[0,0]=\{0\}$ has nonzero length $\ell>0$, then what do you think the length of $[0,\ell/2]$ is?

How can $[0,\ell/2]$ have less length than its proper subset $[0,0]$?

Nick

Ah, this is a nice discussion. It can't be. $$\frac{0}{n} = 0\times n = 0$$

I made a little website for my test today : goo.gl/YoIrZO

I hope editing it will help me ace the test.

Daminark

hears measure

You call?

arctic tern

no

Nick

@Daminark Yeah, I need help to pass my test today. (I will also be linking the transcript to this chat on my site)

Links to good videos on each of the topics would help me a lot

Daminark

05:28

Probability, I don't know any of that unfortunately

Nick

@Daminark how about laplace transforms?

Laplace transform

In mathematics the Laplace transform is an integral transform named after its discoverer Pierre-Simon Laplace (/ləˈplɑːs/). It takes a function of a real variable t (often time) to a function of a complex variable s (frequency). The Laplace transform is very similar to the Fourier transform. While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable. Laplace transforms are usually restricted to functions of t with t > 0. A consequence of this restriction is that the Laplace transform...

Fargle

I know some stuff about it, what do you need @Nick?

Nick

A gist of it. Some tricks on applying it.

This is the stuff I need to go through : sites.google.com/site/ktuma202/transforms

Fargle

What do you know so far, and what are you struggling with?

Nick

I just know the formulas. I haven't practiced yet.

Also, is log-normal distribution the same as exponential distribution?

yesterday, by Akiva Weinberger

"Say minus one-twelfth one more time, I dare you, I double-dare you m*therfucker"

@AkivaWeinberger Challenge Accepted : math.stackexchange.com/a/642246/60900

arctic tern

05:55

that doesn't count as "one more time" since it precedes akiva's joke by years

it strikes me as strange that you haven't practiced with laplace transforms if you're being tested on it. wasn't there homework involving it or something?

Nick

06:07

Yeah, I constantly avoided doing the hw. Now, I must pay the price.

@arctictern I guess I must have time travelled. That's not cheating ;)

Daminark

06:33

Oh just saw this @Nick, and I don't really know about Laplace either :/

Nick

That's okay. I don't either and my exam is just an hour away from starting. My fate is in God's hands.

Daminark

Good luck to you

user314159

^

Zee

@Daminark I been thinking a lot lately on why people do mathematics

user314159

and?

Zee

06:37

Idk

But I don't buy the usual answer

I do it couse I enjoy it

user314159

why do you think people climb mountains?

Nick

@Zee Most do it because they have to

Zee

That's a nice analogy

I agree with it

Nick

@user314159 So they can write a travel blog and monetize on ad revenue. Also, so that journalists will write articles about them. Mountain climbers have a good life and they probably make more than you.

I'm 20. I'm not sure how math is going to get me money.

user314159

If you want to learn about money study economics.

Nick

06:51

No, I already learnt about Monte Carlo engines and market stock from NN Taleb in Fooled by Randomness. I'm not really interested in that.

My question is Are there any olympiads or papers I can write to get money from doing math?

user314159

perhaps you could tutor

Abdullah UYU

$A=1\cdot3^0+2\cdot3^1+3\cdot3^2+\dots+21\cdot3^{20}$ Find the units digit of the $4A-1$

Daminark

@Zee I've got a bit of a view along the lines of, I want to explore that which interests me. May sound somewhat selfish, but I dunno, while there are parts of math which don't have an immediately obvious pragmatic benefit for society, I also don't believe that the purpose of my life is to just go and provide that

Abdullah UYU

anybody could help me on this problem?

arctic tern

07:18

modular arithmetic

$3^k$ is periodic mod $10$

Captain Bohemian

07:34

I seem not to ever have an interest which can make money since childhood.

I had never thought about what to do as vocation when I was in school.

I just majored in the department I like the most and took any course interesting without considering their use to vocation opportunities.

I have never used Laplace transform in physics though I learnt it in the applied math course.

Abdullah UYU

$3^1\equiv3, 3^1\equiv9, 3^1\equiv1$ so it repeats for every $3k$. Do i have to perform a calculation for each term @arctictern

Daminark

08:16

@Captain I've got a similar-ish sentiment modulo the fact that if your field doesn't really have any jobs, you won't be able to engage in the activity of it either

That's somewhat my concern, like, if I were to focus all on traditional math, say going real deep into algebraic geometry, only to find little opportunity for a math career, I might have just lost out on something I would like almost as much, theoretical compsci, and will find myself having to work in something just to make a living which may pay well but not be as intellectually stimulating

Captain Bohemian

@Daminark I guess all things I have great interest can only be done in school.

Daminark

I don't particularly like the fact that this is the case, like I want that vocational concerns just fall in line when they come, but that isn't particularly realistic so unfortunately some mindfulness is necessary

As in, academia?

Captain Bohemian

I only feel university is an lively environment.

Astyx

08:32

Hi chat

user314159

hi

SteamyRoot

ohi

Captain Bohemian

some people, particularly philistines, think schools are for children.

but I found it's like nothing learnt outside schools are more profound than those learnt in schools

Abdullah UYU

08:50

For my question, first i calculated $A\equiv x \mod 10$ then i calculated for $y$ using $4x-1 \equiv y \mod 10$. The answer should be $y$ according to my calculations. Is these are valid for solving this problem? @arctictern

Astyx

09:00

@Abdullah Yes, these steps are valid

Abdullah UYU

$A\equiv 7 \mod 10$ and $y=7$ i got. but book says that answer is 3. I get calculated $$\sum_{k=0}^{20}(a+1)(3^a)=...1$$ directly on wolfram. So $4A-1 \equiv 3 \mod 10$ also according to wolfram. I am confused :=

Astyx

Perhaps you miscalculated some terms

Abdullah UYU

maybe

When i calculte $A \equiv x \mod 10$ i used the fact $3^1 \equiv 3 \mod 10 \dots$ like this: $$1\cdot 3^0+2\cdot 3^1+3\cdot 3^2+ \dots 21\cdot 3^{20} \equiv 1\cdot 1+2\cdot 3+3\cdot 9+ \dots 21\cdot 9 \mod 10$$. Isn't this valid?

Astyx

Well yeah, except $3^{20} \equiv 1 \pmod{10}$

Abdullah UYU

Then i added every thing on the right side and found $70+3\cdot77+9\cdot84$

Astyx

09:16

But $3^3 \equiv 7$

Soham Chowdhury

@Balarka, is it true that $M/M' \otimes N/N' \cong (M \otimes N) / (M' \otimes N')$?

this is sort of a note to myself though (I have a feeling it's false, maybe the last otimes should be an oplus?)

Abdullah UYU

oh, i made mistake at the beginning :( i realized that after you write $3^3 \equiv 7$

user314159

:(

Abdullah UYU

i'll try again

Abdullah UYU

09:30

finally success. i got also $3$ :D

Alex K Chen

Oops something bad happened

^Is this error showing to everyone else too ?

Abdullah UYU

at where?

Alex K Chen

When you open the main site.

Abdullah UYU

yeah

Alex K Chen

It's happening for you too ?

Abdullah UYU

09:32

yes, but questions section is healthy

Astyx

Everything seems fine for me

Abdullah UYU

@Astyx Do you think my question have also a simpler solution? this is a little bit long tor a test(i took it from a test)

Astyx

Perhaps you could see that it's the derivative of a geometric series

So ${d\over dx} {x^{21}-1\over x-1}$ computed at 3

That is $$21\times 3^{20}\times2 - 3^{21}+1\over 4$$

BAYMAX

Any1 can help !if I give an integral please give me the closed form of the integral

Astyx

Times this by 4 and remove one to get, mod 10 : $$3^{20} (2-3)$$

BAYMAX

09:43

$x = \int(\frac{y}{(\sqrt{\frac{-c_{1}^2}{4.c_{2}}} + \sqrt{c_{2}}.y)^2 + (c_{3} - \frac{c_{1}^2}{4.c_{2}})}) dy + c$

Astyx

Wait there's something I did wrong

It's 22, not 21

So $$22\times 3^{21}\times2 - 3^{22}+1\over 4$$

BAYMAX

does wolfram doenot support Latex?

Astyx

Multiply by 4 and remove one to get $3^{21} (4-3) \equiv 3\pmod 10$

@Baymax Not directly

BAYMAX

then how can do it?

@Astyx

Astyx

Do what ?

Well you still can input it in Wolfram

BAYMAX

09:49

like how we can render in latex there?

in wolfram

Astyx

You can't

Well, not too easily at least

I'm not an expert on Wolfram really

BAYMAX

me too

ok

Astyx

@Abdullah See above

user8469759

Hi guys

If you had a set {a1,...,an} (consider a relative order on that set, ai < aj iff i < j)

how can I describe

the greatest subset such that

the elements in this set

have the monotonic property?

i.e. i < j implies ai < aj

Astyx

I don't understand what you're trying to do

user8469759

09:55

take this set

{1,7,4,2,5,6}

a monotonic subset (I don't know how else should I call it)

could be

{1,4,5,6}

or

{5,6}

or

or {1,6} or { 2,5,6} etc

how would you describe the greatest set you can build in the generic case

{a1,a2,...,an}

Astyx

So you're not talking about sets, but about sequences

user8469759

it's finite, it's not a sequence

Astyx

A finite sequence, still not a set

user8469759

I don't understand why it's not a set

Astyx

well with sets you have for instance $\{1,2\} = \{2,1\}$

user8469759

09:59

given that I'm defining my problem in terms of a set

ah ok

well anyway

for a given finite sequence

Astyx

What you're talking about is tuples

Transcript for

$Mathematics$ Mathematics