Mathematics - 2019-06-11 (page 1 of 3)

Ted Shifrin

00:00

You can claim that exhaustion made you slow.

Daminark

Ah perfect

anakhro

00:24

@TedShifrin Am I crazy in thinking this is the proper way to calculate $\iota_X\beta$ for $X = (xy - z)\frac{\partial}{\partial y} + (yz - x)\frac{\partial}{\partial z}$ and $\beta = x\,dy\wedge dz$:
$$\iota_X\beta = \iota_X(x(dy\otimes dz - dz\otimes dy)) = x(dy(X)\otimes dz - dz(X)\otimes dy) = x((xy-z)\,dz - (yz-x)\,dy) = \dotsc$$

Because I have a more extensive example calculation I am trying to work out but it's not getting anything remotely close to the right answer.

Ted Shifrin

You can't do interior product with a general tensor. Leave it alone with the differential form.

So you'll get $x(xy-z)dz - x(yz-x)dy$.

Yeah, that's what you had.

Why is @Eric here when he's supposed to be dining in splendor?

anakhro

Hmmm. So suppose I then took $d$ of that. I get
$$2xy\,dx\wedge dz + x^2\,dy\wedge dz - (yz-1)\,dx\wedge dy - xy\,dz\wedge dy ?$$

Ted Shifrin

What happened to the $z\,dx\wedge dz$?

And where's the $2x\,dx\wedge dy$?

I give up.

anakhro

Ah yes I messed up that.

Ted Shifrin

Where did the $+1 dx\wedge dy$ come from?

anakhro

00:35

$\frac{d(xyz-x^2)}{dx}dx\wedge dy$, right. It should have been a $(yz - 2x)$ not $(yz - 1)$.

Maybe I am just doing this part too quickly and need to take my time and make sure I am not making errors like that.

Ted Shifrin

OK, that addresses my second complaint, then.

anakhro

The general principle I think I have right, though. So that's good to know. Just sloppy.

I will do it after I make my calzone dough and see if it works out better this time. Thanks @TedShifrin

Ted Shifrin

I've spent 45 years using differential forms, so I'm semi-competent.

Send me a calzone.

anakhro

Uncooked, or cooked?

Ted Shifrin

Hmmm ... depends how fast the stagecoach will be.

anakhro

00:39

Heh. I am pretty sure it would spoil before it got there in either case.

Stan Shunpike

01:18

hola

Ted Shifrin

heya @Stan. I seem to recall I answered a ping of yours.

Stan Shunpike

Yes that was very helpful

Ted Shifrin

Cool.

Stan Shunpike

So I tried to read more and from what I understand, a tensor product is a quotient vector space

is that accurate?

Ted Shifrin

Yes, but rather than thinking about the construction, you should understand the universal mapping property it has and just use that most of the time.

So, before I disappear to cook dinner, have you any questions to raise tonight?

Stan Shunpike

01:30

Yeah that was my first, so I guess I should start by reading about that then?

I learned about like equivalence classes and relations and quotient sets

and feel more comfortable with those now

Ted Shifrin

You've never learned modular arithmetic?

Stan Shunpike

uh i did but like idk

thinking of it like in these terms

anakhro

Stan, you are more from a physics background, si?

Stan Shunpike

with basis vectors

and coorodinate free viewpoint

hasn't come easily to me

Ted Shifrin

Hmm ... thinking of $\Bbb Z$ mod $m$ as a ring is all about equivalence relations.

I should send you the appendix from my algebra book.

Stan Shunpike

01:31

yeah that made it clear

but then I wasn't sure like, how that relates to the tensor product being an equivalence relation

didn't quite get that far

Ted Shifrin

It's not per se.

OK, I'm gonna send you the appendix anyhow. Not sure if I have a separate file for that.

Stan Shunpike

@anakhro sort of. I read it on my own but have almost no formal training. I was an econ major

Ted Shifrin

Stan just chose to hang around here and get bullied.

Stan Shunpike

@TedShifrin hahahahahaha basically

willing to hang in there

@TedShifrin so should I read about this universal property first?

the quotient vector space approach left me still confused how wedge products work.

I just got sick of the psuedo vector concept and want like the full version. None of the cool physics books use psuedo vectors

they all use differential geometry

Ted Shifrin

They don't know what diffferential geometry is :P

Stan Shunpike

01:35

im gathering

should I watch your lecture on it? you started off with boundaries

so i wasn't sure if that would lead to the tensor product

Ted Shifrin

On what?

Stan Shunpike

differential forms

Ted Shifrin

No, no, I never did tensor product, purposely.

Stan Shunpike

ok

Ted Shifrin

Boundaries?

I defined wedge products just by determinants, no fancy stuff.

Stan Shunpike

01:36

ok, i'll watch it and come back with questions

Ted Shifrin

Do you want the appendix on equivalence relations, or are you cool?

Stan Shunpike

yes definitely

thanks ted!

i think i have too many books written by physicists

Ted Shifrin

Make sure you do the exercises :)

Stan Shunpike

claiming to be mathematicians

hahaha i will and then come here when i inevitably get stuck

anakhro

@StanShunpike for the coordinate free def. of the tensor product $V\otimes W$ as the "free-est" vector space where bilinear maps on $V\times W$ are viewed as linear maps on $V\otimes W$.

Ted Shifrin

01:40

Stan: sent.

Of course, this requires double-duality to untangle.

anakhro

Or in other words, the universal property just says that the tensor product factors bilinear maps into linear maps.

Stan Shunpike

@TedShifrin you're the best!

Ted Shifrin

Yup, anakhro tells the truth (for once).

Stan Shunpike

@anakhro so does "free" mean like 'coordinate free'? i saw this notion of a free module, but i didn't really get what "free" meant

Ted Shifrin

no, nothing to do with coordinates.

Just means the things have no relations.

Stan Shunpike

01:43

is the tensor product in your differential geometry book?

Ted Shifrin

Noooo.

I think of tensor product as definitely graduate stuff.

Stan Shunpike

oh really? I assumed that was undergrad

wow ok

idk why. its just everywhere

so i assumed you had to know it

Ted Shifrin

Well, all sorts of crazy things are undergrad at UC, but not elsewhere.

Daminark

Lmao

Ted Shifrin

I did learn tensor product as an undergrad, but it's far from typical.

Well, Demonark stirred to laugh, rather than issue another horrid pun.

Stan Shunpike

01:45

and u ended up being a pro differential geometer

anakhro

@StanShunpike take a look at Gauge Fields, Knots and Gravity by Munian & Baez.

It might be more up your alley if you are a physicist.

Stan Shunpike

@anakhro oooo sounds very intriguing

anakhro

pg. 169 for tensor products

Ted Shifrin

Stan is not a physicist.

Stan Shunpike

i will definitely look

Daminark

01:45

Yeah here the honors algebra class talks about tensor products toward the end but regular doesn't

Stan Shunpike

I tell jokes and make puns

anakhro

@TedShifrin maybe he wants to be one

Ted Shifrin

It does show up in Spivak's Calculus on Manifolds, so fancy honors calculus courses sometimes do it.

Stan Shunpike

and drive the Knight Bus

Ted Shifrin

Stan, you and Demonark should get together for a drink.

anakhro

01:46

Tensor product is done in the second linear algebra class at my university

The module theory one.

Daminark

Should be soon if we do, I won't be around Chicago much longer

Ted Shifrin

Artin's algebra book doesn't do tensor product, for example, and it's damn sophisticated.

anakhro

Does Artin do modules?

Ted Shifrin

Good point.

Of course.

anakhro

Weird.

Daminark

01:47

Does he do STFGMPID?

Ted Shifrin

But tensor product normally is the purview of commutative algebra, not first algebra.

anakhro

I like Berberian for linear algebra.

Ted Shifrin

Yes, of course, Demonark.

I've never heard of it, anakhro.

But lots of people haven't heard of my books, so it's fair.

Stan Shunpike

Are tensor products a requirement to understand wedge products?

Ted Shifrin

OK, time for me to disappear and go cook dinner. Bye.

Daminark

01:48

Same. I used and liked Hoffman-Kunze though it's fairly old school

Stan Shunpike

@TedShifrin ciao

Ted Shifrin

@Stan: Of course not. My course/book/videos don't use tensor product to do wedge products.

Daminark

I hear good things about Linear Algebra Done Wrong

Stan Shunpike

@TedShifrin yeah i'm gonna try to learn it your way then

Ted Shifrin

LOL.

OK, have fun :)

anakhro

01:52

@TedShifrin this is how Berberian does it.

Which roughly follows what my university did, except my university introduces modules in the second course.

When you start doing ring stuff

Stan Shunpike

@anakhro that book looks so great

anakhro

Berberian's linear algebra book. @StanShunpike it is quite cheap through Dover if I recall correctly.

Daminark

It feels similar to H-K but more geometric

anakhro

However, the biggest part of math is to not worry too much about books, but more about doing exercises and getting appropriate help when needed.

Daminark

Maybe covers a bit more actually, H-K doesn't really do much tensor products or PIDs

~~It's the instructor's job to worry about books~~

anakhro

01:58

I have one professor who always teaches brand new courses and does the lecture notes ad hoc, compiling a usually 100pg textbook at the end.

If I could be as half a good teacher as him, I would be happy, I think.

Math geek

0

Q: Continuity of the improper integral $\int_{0}^{x}y^{-1/2}dy.$

The improper integral $\int_{0}^{x}y^{-1/2}dy$ is $1.$ Continuous on $[0,\infty).$ $2.$ Continuous on $(0,\infty).$ $3.$ Continuous only on $[1/2,\infty)$ It is cleat that option $3$rd is wrong one. I am confused about $1$st and $2$nd option. I tried it as $$\int_{0}^{x}y^{-1/2}dy=2\sqrt{x}$...

How it is continuous at $0$?

in the negative real axis. the function doesn't exist in real.

Stan Shunpike

@anakhro so what's ur specialty?

/ area of interest

anakhro

@StanShunpike I like contact and symplectic geometry.

But I am pretty newbie.

I am a masters student, though, if that indicates any level of expertise to you.

Stan Shunpike

i still don't understand what symplectic geometry is

mainly because

i still haven't grasped the wedge product

anakhro

Well it's differential geometry, but with a tool for measuring area.

Stan Shunpike

02:06

cuz i thought i needed the tensor product for that

OHH

really?

wait how is rimenanian geometry different then?

anakhro

Oversimplified, but basically.

Riemannian geometry measures angles.

Like an inner product space.

Stan Shunpike

hmmm

anakhro

angles/distances are to Riemannian, as areas/volumes are to symplectic.

Stan Shunpike

so why does this topic come up in relation to hamiltonians?

in physics

anakhro

The area/volume idea I highlight is not so impressive with respect to its origins in Hamiltonian mechanics AFAIK.

Do you know anything about Hamiltonian mechanics?

Like the main thing to keep in mind is you want to determine the dynamics from the Hamiltonian (a function on your phase space).

Stan Shunpike

02:15

From what i've read

and i don't understand a ton, symplectic geometry is one of the benefits of using hamiltonians vs lagrangians

i have done some simple problesm with hamiltonians in classical mechanics and a few standard ones for quantum

but not a lot of experience beyond that

anakhro

I am not very good at physics, but as far as I know, the benefits of using the Hamiltonian vs. the Lagrangian is based on the symmetries you know about your problem.

Physics effectively is about solving physical problems using symmetries (analogously, math is about studying symmetries, despite no apparent application to physical problems).

So some symmetries are easier to take advantage of when dealing with $H = K + U$ and some are easier to take advantage of when dealing with $L = K - U$.

@StanShunpike the main point is you want to take a Hamiltonian $H$ to a vector field $v$ whose dynamics (the flow lines of $v$) describe the dynamics/physics of your Hamiltonian.

Does that sort of make sense?

Simone

08:40

I'm having a tough time with improper integrals, I own a copy of Spivak's calculus but he only dedicates a couple of exercises to them.
Is there a textbook that deals with them more comprehensively?

Simone

08:57

For instance, I see people using tests that I'd use for infinite series, like Abel's or Dirichlet's tests for these improper integrals. I understand it makes sense due to the definition of the integral, but I've never done this myself and I need practice which Spivak doesn't provide.

northerner

09:27

Does anyone have a reference for how to do the Lucas Probable Prime test? I find the wikipedia page hard to follow en.wikipedia.org/wiki/…

Micrified

09:38

This might be a noob-ish question, but I'm trying to understand the following statement: In general, if $q=p^{m}$ where $p$ is prime and $m \geq 2$, the element $p$ would not have a multiplicative inverse, since there exists no $b$ such that $b \times p = 1 mod p^{m}$.

I guess that since multiplying a prime by any number other than 1 means it is no longer prime, that I find it hard to understand why you couldn't get $1$ when modulo'ing the result by $p^{m}$ ...

Akiva Weinberger

09:59

$bp\equiv1\pmod p^m$ means that there exists some integer $k$ such that $bp=1+kp^m$

Now reduce both sides of that mod $p$

@Micrified

(As a concrete example: what are the multiples of $3\pmod9$?)

Buddhini Angelika

Let $G$ be a solvable, non-nilpotent group of order $p^2qr$, where $p,q,r$ are distinct primes, and let $F$ be a Fitting subgroup of $G$. Then $F$ and $G/F$ are both non-trivial and $G/F$ acts faithfully on $\bar{F}:=F/\phi(F)$ so that no non-trivial normal subgroup of $G/F$ stabilizes a series through $\bar{F}$.

Can someone please help me to understand how to write the group $G$ using notations , when $|F|=pr$? Is it correct if I say $G \cong (C_p \times C_r) \rtimes (C_p \times C_q)$ or $G \cong C_{pr} \rtimes (C_p \times C_q)$ ?

Micrified

10:23

@AkivaWeinberger If I reduce both sides mod p, I get 0 = 1?

Abdullah UYU

10:36

hi all

s.harp

10:48

I got a silly question, if I have two continuous functions defined on open sets, so that they agree on the intersection I can glue them together to get another continuous function

I can do the same thing if htey are defined on closed sets and agree on the intersection

The first property is something I would call "continuous is a local property", or just the "sheaf property"

what would be a name for the second property?

(as an example, differentiability satisfies the sheaf property, but not the second one)

Abdullah UYU

Sometimes I hear that some integrals don't have an analytic solution (also don't know a precise definition of this). But really, how they prove it, what's the procedure?

Mathphile

11:14

6

Q: Is $\lfloor \zeta(-n) \rfloor$ only prime for $n=23$?

I searched for primes of the form of $\lfloor \zeta(-n) \rfloor$, where $n \in \Bbb{N}$, for a range of $n \le 10^4$ on PARI/GP and found $\lfloor \zeta(-n) \rfloor$ is only prime for $n=23$. My PARI code: for(n=1, 10^4, if(ispseudoprime(floor(-bernfrac(2*n)/(2*n)))==1, print([2*n-1, floor(-be...

number-theory prime-numbers riemann-zeta conjectures

any thoughts?

Akiva Weinberger

@Micrified Exactly. So what does that mean?

Micrified

@AkivaWeinberger It means there is no such integer k?

I guess that proves it.

Thanks :)

Ryan Unger

11:48

@ÉricoMeloSilva the bathroom in the dorm we're staying in is absolutely gross

there's a massive fruit fly infestation

skullpatrol

eeew

Leaky Nun

fruits sound tasty

skullpatrol

+ flies?

Mathphile

12:15

how do we check if the following sum converges

$$\sum_{n=9}^{\infty} \frac{1}{\ln \left( \frac{-B_{2n}}{2n} \right)}$$

$B$ is the Bernoulli number

Thorgott

12:27

$B_{2n}$ alternates in sign, so every second term is undefined

Mathphile

what about $$\sum_{n=9}^{\infty} \frac {1}{\ln \left( \vert \frac{-B_{2n}}{2n} \vert \right)}$$

@Thorgott

Thorgott

12:43

Diverges according to WA.

Mathphile

WA wouldn't accept my query

Thorgott

Yeah, I don't know whether WA handles Bernoulli numbers, but you can use asymptotics.

Mathphile

can you send a link @Thorgott

or the query you used on WA

Thorgott

wolframalpha.com/input/…

Mathphile

thank you @Thorgott

@Thorgott could you take a look at this question?

2 hours ago, by Mathphile

6

Q: Is $\lfloor \zeta(-n) \rfloor$ only prime for $n=23$?

I searched for primes of the form of $\lfloor \zeta(-n) \rfloor$, where $n \in \Bbb{N}$, for a range of $n \le 10^4$ on PARI/GP and found $\lfloor \zeta(-n) \rfloor$ is only prime for $n=23$. My PARI code: for(n=1, 10^4, if(ispseudoprime(floor(-bernfrac(2*n)/(2*n)))==1, print([2*n-1, floor(-be...

number-theory prime-numbers riemann-zeta conjectures

Thorgott

13:03

I have no clue nor do I see a reason to suspect there is a feasible answer.

Mathphile

okay

thanks for having a look

CowperKettle

13:40

7

Abdullah UYU

13:53

chat.stackexchange.com/transcript/message/50635939#50635939 I meant elementary but not analytic, I guess.

I found this: math.stackexchange.com/a/1625620 It says differential Galois theory

Akiva Weinberger

14:29

@AbdullahUYU Yeah, "analytic" is not the right word.

https://en.wikipedia.org/wiki/Analytic_function

A function is elementary if it can be written as a combination of +, -, x, /, exponentiation, logarithms, and trig

Leaky Nun

@AkivaWeinberger we also say "this equation cannot be solved analytically"

Akiva Weinberger

15:21

Yeah but that would include weird things like Gamma and Bessel functions, no?

Secret

4 hours ago, by Mathphile

6

Q: Is $\lfloor \zeta(-n) \rfloor$ only prime for $n=23$?

I searched for primes of the form of $\lfloor \zeta(-n) \rfloor$, where $n \in \Bbb{N}$, for a range of $n \le 10^4$ on PARI/GP and found $\lfloor \zeta(-n) \rfloor$ is only prime for $n=23$. My PARI code: for(n=1, 10^4, if(ispseudoprime(floor(-bernfrac(2*n)/(2*n)))==1, print([2*n-1, floor(-be...

number-theory prime-numbers riemann-zeta conjectures

I have no idea how questions of this category can be answered without brute forcing all integers

Semiclassical

16:01

From the top of math.ucla.edu/~tao/resource/general/121.1.00s/…:

>Don't take these axioms too seriously! Math is not about axioms, despite what some people say. Axioms are one way to think precisely, but they are not the only way, and they are certainly not always the best way. Also, there are a number of ways to phrase these axioms, and different books will do this differently, but they are all equivalent (unless the book author was really sloppy).

There’s two parts of that claim, one of which is rather uncontroversial: not every author defined vector spaces in the same ways, so being too wedded to a particular set of vector space axioms isn’t productive.

Akiva Weinberger

I was gonna say something but then I realized "taking these axioms too seriously" probably meant something different than what I was thinking

Semiclassical

Well, that’s what I wonder

Akiva Weinberger

They probably mean, don't write proofs super-formally

Semiclassical

The narrow meaning I gave—different axioms can define the same object—is easy enough to accept

But he seems to be asserting something stronger that that

Thought it was interesting to see that attitude anyways

Especially since, judging from the html, that’s a Terry Tao page

Akiva Weinberger

Does Tao teach?

Secret

16:09

What is the alternate to an axiomatic system in defining a mathematical system?

Even categories and HoTT follow the axioms of categories and functors

Semiclassical

@AkivaWeinberger think do, yeah

Akiva Weinberger

By the way

You know how "micron" means "small" and "mega" means "big"

and how there are two letters in the Greek alphabet that make the /o/ sound

Οο, Ωω

Semiclassical

@Secret I think it’s a matter of mathematical practice. Ie, do you prove stuff by manipulating at the axioms until you get what you want

MatheinBoulomenos

that's why the plural of omega is omegala

megas has irregular declination

Akiva Weinberger

It's what now

I mean I guess I don't even know how to pluralize Greek words when they're regular

MatheinBoulomenos

16:13

I should rather say the pedantic plural of omega is omegala

Semiclassical

Or do you allow yourself more mental freedom, eg intuition etc

There are certainly areas of math where thinking about axioms alone is the right approach

Akiva Weinberger

@MatheinBoulomenos That's kinda like saying more than one aleph is alephim

Semiclassical

But I don’t think it’s always the right approach. If I’m trying to think of a good algorithm to solve a problem, for instance, I don’t think axioms will help you much

MatheinBoulomenos

@AkivaWeinberger d'accord

there exist some Hebrew plurals like seraphim, so why not with aleph?

Semiclassical

We are the alphema and the omegala

Akiva Weinberger

16:18

Japanese doesn't have plurals, so one katana, two katana?

(red katana blue katana)

Semiclassical

Red—damn

Akiva Weinberger

Or tsunami

Secret

I see, I often use intuition in geometric problems and some algebraic problems, and axioms when I need to prove something from first principles

Simone

You english speakers have superplurals!

I wonder if there are other languages with superplurals

like "peoples"

MatheinBoulomenos

@AkivaWeinberger one sushi, two sushi, so why not?

Akiva Weinberger

16:20

Hm

Sheep is borrowed from Japanese confirmed

ÍgjøgnumMeg

Bir adam, iki adamlar

Turkish plural!

Akiva Weinberger

I had an awkward, Google translate-mediated conversation with some Russian speakers earlier

trying to figure out how to get to the train station

MatheinBoulomenos

English sometimes borrows plurals from Latin or Greek and very rarely Hebrew. German even borrows some complete declinations for all cases from Latin (e.g. for Jesus, which itself has irregular declination, being the latinized version of the hellenized version of a Hebrew name)

Akiva Weinberger

(they were)

I've definitely heard both "siddurim" and "siddurs"

but that's within a community of people who know (at least some) Hebrew

Simone

I asked this earlier: do you guys have a suggestion for a textbook which goes into detail about the convergence (or divergence) of improper integrals?

MatheinBoulomenos

16:27

I'd say seraphim and cherubim are established even among the Hebrew-ignorant

Simone

or maybe a link

Akiva Weinberger

I don't, but I'm pretty sure a lot of the rules from series carry over

like, the comparison rule

Simone

I saw people using the tests from the series

and I'm familiar with those

but how do you use them when the integral is improper in the sense that the integrand is unbounded?

I see people use Abel's and Dirichlet's tests

Akiva Weinberger

With something like $\int_0^1\frac1{x^p}dx$, you can find an antiderivative

Simone

but I need some guided practice

Akiva Weinberger

16:30

(I think it converges for $p<1$?)

and then you can compare other things to those

Semiclassical

I don’t think merely being unbounded is enough to ensure divergence

Simone

yes it does converge for p<1

@Semiclassical It doesn't

the problem is: I'm having a tough time with the comparison test

Akiva Weinberger

You can prove a version of Dirichlet's test using integration by parts

Simone

I'll show you the integral that's giving me nightmares:

1 moment

$$\int_1^2 \frac{\sqrt{x-1}}{\abs{\sin(1-x)} \tan(2-x)} dx$$

I turned it into an integral whose limits are 0 and 1

and tried to find a smaller [bigger] function that is divergent [convergent]

but I failed to find one that is also easy to integrate

Semiclassical

If you want to talk about the version with endpoints at 0,1 then you should probably write it out so we can reference it

Simone

16:37

So, maybe I need to learn about how to apply the tests for the series

Semiclassical

Also, where are the absolute value bars supposed to be?

Simone

@Semiclassical ok

Akiva Weinberger

My first instinct is to graph it

My second instinct is to replace $\sin x$ with $\frac{\sin x}x\cdot x$

Simone

yes, I don't remember command for the absolute value

Semiclassical

There isn’t one

Akiva Weinberger

16:39

||

Semiclassical

Just use |

Simone

ah, just | |?

ok

Akiva Weinberger

because $\frac{\sin x}x$ is a bounded function that avoids zero for small $x$, and $x$ is a pretty simple function

Semiclassical

My instinct is to look at where the denominator vanishes

Akiva Weinberger

Similarly for tan

Semiclassical

16:40

Which I think is a bit less trivial than it may seem

Simone

$$\int_0^1 \frac {\sqrt(x)}{\sin x \tan(1-x)} dx$$

Semiclassical

You’re looking at tan(x) on the range 0 to 1

Akiva Weinberger

There's a neat notation you can use: $\operatorname{sinc}(x):=\sin(x)/x$

(No good LaTeX code for it)

Semiclassical

@AkivaWeinberger people occasionally mean sin(pi x)/(pi x) when they write that, lol

Akiva Weinberger

Hm true

Semiclassical

16:42

@Simone still need the absolute value bars. Do you see why? (And where?)

Simone

@Semiclassical I don't think they're needed

Akiva Weinberger

In any case: $\displaystyle\int_0^1\frac{\sqrt x}{x(1-x)}\frac1{\sin x/x}\frac1{\tan(1-x)/(1-x)}dx$

Note that, for $0\le x\le1$, $0<\sin x/x\le1$

(In fact, $\sin 1\le\sin x/x\le1$)

The point is, there exist positive numbers $C_1$ and $C_2$ (which I can compute later if I want) such that

Semiclassical

@Simone Yeah, ignore that

Silly mistake on my part

Simone

@Semiclassical ;P

np

Akiva Weinberger

$0<C_1\le(\frac{\sin x}x)(\frac{\tan(1-x)}{1-x})\le C_2$

Semiclassical

16:45

Tho in that case it’s mysterious why they’d include them, other than scaring students

Akiva Weinberger

for $0\le x\le1$

Simone

@AkivaWeinberger great, so we're trying with the comparison test

Akiva Weinberger

Yeah, we're comparing it with $\frac1{C_2}\int_0^1\frac{\sqrt x}{x(1-x)}dx$

and we can ignore the $C_2$ because it doesn't affect convergence

Simone

clearly

Akiva Weinberger

Hm

Graphing that

Semiclassical

16:49

It behaves like 1/(1-x) near x=1

So rip

Akiva Weinberger

Oh, yeah, $\int\text{your thing}>\frac1{C_1}\int_0^1\frac{\sqrt x}{(1-x)}dx>\frac1{C_1}D_1\int_0^1\frac1{1-x}dx$

where $D_1$ is the minimum value of $\frac{\sqrt x}x$ over $[0,1]$

…which is $1$, actually

Whatever, ignoring the constants

the point is

Semiclassical

Main point is that the numerator only affects the divergence on the left endpoint, not on the right endpoint

Akiva Weinberger

$\dfrac{\sqrt{x-1}}{\sin(1-x)\tan(2-x)}$ behaves like $\dfrac{C}{C(2-x)}$ near $x=2$

and that kills your thing

(This is the original integral, not the shifted one)

Simone

so since $\int_0^1 \frac {1}{1-x} dx$ diverges then so does $\int \text{my thing} dx$

Akiva Weinberger

In fact, if we saw this at the start we could have done it much quicker:

$\dfrac{\sqrt{x-1}}{\sin(1-x)\tan(2-x)}=\dfrac{\sqrt{x-1}}{\sin(1-x)\frac{\tan(2-x)}{2-x}}\dfrac1{2-x}$

${}=[\text{positive thing}](x)\dfrac1{2-x}$

It's not bounded near 1, hence the edits

Simone

16:57

by bounded thing you mean any bounded positive function

Akiva Weinberger

I think what I meant to say was

Semiclassical

It is bounded below tho

Akiva Weinberger

it's bounded below by some positive number

It's bigger than $C$ for some $C>0$

Semiclassical

Also, this is where you need to use absolute values or just write |sin(1-x)|=sin(x-1)

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