Mathematics - 2018-08-28 (page 2 of 4)

Abdullah UYU

12:00

Ok, btw after concluding that it's strictly increasing, you say that it has to have exactly one real root.

micsthepick

a challenge question (one that does not count towards marks) from my assignment is to prove that there does not exist a rational number $x$, where $x^3+x+1=0$

and I am supposed to do it by contradiction

Tobias Kildetoft

@micsthepick Start by showing that it would in fact then have to be an integer

Abdullah UYU

The very theorem might help. The same method, I mean.

micsthepick

I was thinking that it would probably be a proof by infinite descent (right terminology?), but I wondered if there was a way of figuring out it's root

the thing is that that theorem is not part of my course

@TobiasKildetoft do I need that theorem for this step?

Tobias Kildetoft

what theorem?

user1732

12:06

mathworld.wolfram.com/RationalZeroTheorem.html

micsthepick

the Rational Zero Theorem ^

user1732

it'll probably be apart of some future course

Jasper Loy

I am natural, integral, rational, real, and complex, but I ain't transcendental.

user1732

are we supposed to guess what you are?

micsthepick

@JasperLoy you are not irrational

Jasper Loy

12:11

@user1732 No, I am just listing my character traits.

micsthepick

there are an infinite number of numbers that share those traits

Jasper Loy

@micsthepick If I am rational, then I am not irrational.

micsthepick

that fact that you are rational means that you are not transcendental also

Jasper Loy

@micsthepick I am not a number but a person.

user1732

The Prisoner 1967 I'm not a number, I'm a free man

►

Jasper Loy

12:13

How appropriate!

@user1732 Hello, pal.

user1732

hello, pal

user91500

@JasperLoy Some people here don't upload their profile picture on stackexchange server. What should we do with them?

Jasper Loy

@user91500 We should just let them be. I only uploaded a blue square myself.

user91500

@JasperLoy So you are a regular person!

Jasper Loy

@user91500 Yes.

@user1732 Why don't you use your original username?

user1732

12:19

i'm protesting 0celo's one year suspension

user91500

@JasperLoy There is just one person here that doesn't like to upload on SE server, So I'm very sad :(

Jasper Loy

@AnotherJohnDoe There is no pace you need to set. Just make sure you understand every sentence. Take your time. Exercises are important but if you don't have time skip them first.

@user91500 Who is this person?

user91500

@JasperLoy J S

Jasper Loy

@user91500 Don't know who that is.

user1732

more importantly, why would it make you very sad?

Jasper Loy

12:24

@user1732 What happened, in one sentence?

Oskar Tegby

Anyone here familiar with the Weierstrass approximation theorem?

user1732

in Physics Meta, May 18 '17 at 23:01, by Shog9

@heather I think he could benefit from some time not in chat. And I think chat can benefit from some time without him. We'll see how that works out. Beyond that... I don't know there's much of value I can say.

user91500

@user1732 Since his profile picture doesn't load in my browser, So an irregularity has been appeared here!

Jasper Loy

@user91500 You mean has appeared and not has been appeared.

@user1732 Doesn't say anything, but OK, that will do.

Hello @AlexClark.

user91500

@JasperLoy I mean "has been appeared".

Jasper Loy

12:34

@LeakyNun To reconcile what different people have told you, although there are people working in set theory, of course these are much fewer than people working in algebra, or analysis, or geometry, or topology. That's all.

user1732

@JasperLoy here

Leaky Nun

@OskarTegby oh we haven't talked since a long time ago

I hope he's still doing well

Jasper Loy

@LeakyNun Did you reply to the wrong line?

Leaky Nun

@JasperLoy no

Jasper Loy

@LeakyNun Are you referring to ocelot?

Leaky Nun

12:48

no

i was making a joke

AnotherJohnDoe

13:08

@JasperLoy, thank you for your advice. I do plan to solve the exercises.

Oskar Tegby

@LeakyNun: Really? You and Weierstrass go way back?

Leaky Nun

no, just his theorem

i'm friends with his theorem

Secret

$💥(Weierstrass)$

Nope I have no idea what happens in a Weierstrass category

Jasper Loy

Hello @Secret I am still thinking what your secret is.

Secret

lol

I like learning secrets

Stone–Weierstrass theorem

In mathematical analysis, the Weierstrass approximation theorem states that every continuous function defined on a closed interval [a, b] can be uniformly approximated as closely as desired by a polynomial function. Because polynomials are among the simplest functions, and because computers can directly evaluate polynomials, this theorem has both practical and theoretical relevance, especially in polynomial interpolation. The original version of this result was established by Karl Weierstrass in 1885 using the Weierstrass transform. Marshall H. Stone considerably generalized the theorem (Stone...

$💥(\text{Stone-Weierstrass theorem})$

Leaky Nun

13:17

who's gonna stone weierstrass?

Secret

he's ded

stones are useless

projecteuclid.org/download/pdf_1/euclid.pjm/1103037116

mick

13:28

2

Q: Prime twins and $ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $

Let $p$ be a prime such that $p+2$ is Also a prime. Define $$ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $$ For $z \neq 1 $ and $re(z) > 1$ this $f(z)$ always converges. If there are infinitely many prime twins , and prime twins grow like $O (n * ln(n)^2 ) $ as i...

Any ideas ?

usukidoll

Hiiii again

user1732

hello

usukidoll

I'm taking a break. That problem is still driving me nuts

😥

Semiclassical

@usukidoll this is regarding the "set of points in exactly k subsets" problem?

What I find weird is that, if k>1, then there's seemingly no guarantee that the sets A_i will have any points in common

as such, it hardly seems guaranteed that the resulting space will be nonempty

(e.g. take the integers and partition them into evens/odds. Then the set of integers which are either even or odd but not both is just the integers themselves. So the space of integers which are both even and odd is empty and it seems like that can't be an event space)

Oskar Tegby

13:44

What does $\|\cdot\|_\infty$ mean?

Semiclassical

'Infinity norm'

Oskar Tegby

Why would it appear in the sense of a contraction map?

$$|f(x)|=\left|\int_0^1K(x,y)f(y)dy\right|\leq\int_0^1|K(x,y)||f(y)|dy\leq\frac{‌1}{2}\int_0^1|f(y)|dy\leq\frac{1}{2}\|f(y)\|_\infty$$

Semiclassical

it's a norm, and I think one often wants to show that a contraction decreases norm

also, the reason it's called the infinity norm is that you start with the p-norm $(\sum_i x_i^p)^{1/p}$ and take $p\to\infty$

which if you think about it will just give the value of the largest $x_i$

usukidoll

@Semiclassical yeah that one. I can't even. . ughhhh what is it asking?!!!!! 😠

Semiclassical

As such, the infinity norm is just the largest value on the set

Oskar Tegby

13:48

I don't really see why it would be the largest value on the set, but I see why it's used.

Semiclassical

Suppose you pick an element $x_k$. then $(\sum_i x_i^p)^{1/p} = x_k (\sum_i (x_i/x_k)^p)^{1/p}$

usukidoll

I did the rest of the problems but only a select few gets turned in and this is unfortunately one of them. Like there's 6 problems and I did 5 sooooo this one thing. I can't ... More likely who is the m= 2 and k=1??

Oskar Tegby

Yeah

Semiclassical

If $x_k$ is the largest such element (and for simplicity I'll take it to be unique)

then $x_i/x_k\leq 1$ with equality only when $i=k$

as such, $(x_i/x_k)^p\to 0$ as $p\to\infty$ if $i\neq k$. if $i=k$, then instead $(x_i/x_k)^p = 1^p \to 1$ as $p\to\infty$

usukidoll

I wish exercise 1.8-1.10 can be turned in. I understood those. Also got the Venn diagram and proved the P( A or B or C) . It's that one problem left that's making my brain tired 🙄😫😓

Semiclassical

13:51

so in the limit $p\to\infty$ one just gets $(\sum_i x_i^p)^{1/p}\to x_k$

so the infinity norm is just another name for the supremum norm

Oskar Tegby

Okay. Cool! That's a really good explanation.

Semiclassical

one does have to be a bit careful: what if multiple $x_i$ were the same?

e.g. $x_1=x_2$ both gave the same max value

it turns out not to be a problem. in that case you'd just end up with $(\sum_i x_i^p)^{1/p}\sim x_1 (1+1+0+\cdots +0)^{1/p}\to x_1=x_2$

so it does work regardless, but it's a bit more annoying

Now, this of course doesn't tell you why they want the infinity norm in your particular problem

But that's not something I'd know the answer to :P

Oskar Tegby

Okay. :)

Jasper Loy

@usukidoll For the m sets you just need to split them into 2 groups, one group of the k sets and the other group of m-k sets. Then you can take intersection over (not of) the k sets and intersection over (not of) the m-k sets. Then there are m choose k ways to choose k out of m sets. Then you can take union over (not of) this set of m choose k ways. This is how to write your answer in short.

usukidoll

I'm gonna cry. This is worse than the heads and tails problem attempt I did tonight 😭

$( m \choose k) (k \ choose m-k)$ ajdhsjshf

Jasper Loy

13:57

It's OK to cry. I hope you feel better soon. =D

usukidoll

Draw picture
Explain like I'm five. I've been at this thing for 2 days x.x

Semiclassical

huzzah for inclusion-exclusion

Jasper Loy

I think if you read some proofs they have this kind of thing also for general numbers m and k and what not. I guess you just haven't come across them.

usukidoll

I still don't get itttttttttttttttt 😲😭

Semiclassical

Well, here's a common way this kind of problem would show up

Jasper Loy

13:59

Just expand what I wrote above and you get the proof you can turn in.

Semiclassical

Suppose you've got a collection of university students, each taking a set of courses

usukidoll

Inclusion-exclusion is like $P( A \cup B)= P(A)+P(B)-P(A \cap B)$

Semiclassical

Then you could certainly talk about the set of students which are taking 2 out of m possible courses.

or 3 out of m possible courses, etc.

usukidoll

Oh I remember the students taking courses problem

It's like some take English ... Others take Math and some are in both English and Math

Semiclassical

Right.

So in principle you could partition the students into event spaces based on how many courses each of them is taking.

usukidoll

14:02

P(A) is the event that students take English and P(B) is the event that students take Math

Semiclassical

Right. And P(A Delta B) would be the people who are taking exactly one of the two

What bothers me, going to what I said above, is that there's no guarantee that these spaces are nonempty

usukidoll

Exactly one English or exactly one math

Semiclassical

Right

usukidoll

If there's no guarantee that the spaces will be nonempty that means that either A will be empty or B will be empty and that breaks one of the definitions of an event space

Semiclassical

I mean, suppose you have different versions of the same course (e.g. two different profs teaching first-semester intro physics)

then the event space of students taking both versions of the course is presumably empty

in which case it's...not much of an event space

Jasper Loy

14:04

Let x_1,...,x_k be the indices such that the points belong to Ax_1,...,Ax_k and not to the others of A1,...,Am. This is how you can start off @usukidoll, something like that.

user1732

perhaps, some quick review of high school is in order?

Semiclassical

So I feel like there's some assumptions being made in order for the problem to make any sense.

Jasper Loy

We don't need any combinatorics to solve that problem, just need to take complements and unions and intersections @Semiclassical.

Semiclassical

Regardless, I hope the example is clear enough: You'd have $m$ courses being offered, and each student would take some $k$-subset of these courses.

@JasperLoy true enough. I just like being able to ground problems like this in examples.

usukidoll

@user1732 highschool was 12 years ago :p

Jasper Loy

14:07

@Semiclassical Oh yeah, that's an exact parallel

usukidoll

So if m= 3 and k=2 then out of three courses students take 2 of them.

Jasper Loy

I think the problem with writing that proof for @usukidoll is that she hasn't seen proofs involving unions and intersections taken over sets we define, only the usual 2 or 3 set unions and intersections.

usukidoll

There we go

Semiclassical

Right. Of course, you could have a school where some students take 1 course and other students take 3

Jasper Loy

That's why although the idea is simple, it is hard for her to write the concrete proof.

usukidoll

14:08

I usually see a bunch of A and Bs and ABCs things

Semiclassical

Yeah

usukidoll

But for this it seems like a string ... A string from the index set

Jasper Loy

Yeah, right, index set.

usukidoll

Of unions... Like in definition 1.4

That's what's driving me nutZo

Semiclassical

It still bothers me that, taking the definition as they've written it, there's no guarantee that "the set of all elements in exactly k subsets" is nonempty

Jasper Loy

14:10

Whether that set is empty or not doesn't matter right?

Semiclassical

It does matter for the definition of an event space

usukidoll

If A is in F then the complement must also be in F. The empty set and Omega must be in there which is what 1.2 says. Like it needs something. Not blank space

Semiclassical

That's the first condition: that the collection of subsets of the sample space be nonempty

usukidoll

Yup

Meaning our sample space must have elements

Semiclassical

So if you take the definitions as written, then the problem is wrong on trivial grounds.

usukidoll

14:12

1.3 is if A is in F then the complement must also be in there so not A is in F

Jasper Loy

@Semiclassical That is saying there is at least one set in the sample space, not that there is at least one element in the set of all elements in exactly k subsets, right?

usukidoll

Wut da

Semiclassical

that's requiring that the collection of subsets be nonempty

but if there's no such subset, then that's empty

usukidoll

I'm using Grimetts's book. probability an introduction 2nd edition. Got both an ebook and a paperback

Jasper Loy

Now there is a difference between the collection of subsets being empty and an empty set that is a subset itself.

Semiclassical

14:13

oh, good point

usukidoll

🙃

Semiclassical

So F would just contain the empty set and therefore be non-empty itself

Jasper Loy

The empty set is always in F, so F is nonempty.

usukidoll

The proof to 1.4 was super condensed. I was trying to replicate the proof in the book but only understood 1.2 and 1.3 so when I saw that string in exercise 1.11 it was like related to 1.4 and I was like ... 🙃

Semiclassical

hmm

That sounds like a reasonable resolution, but it's not obvious to me that "the set of points in the sample space which belong to exactly k of the subsets" would contain the empty set

Jasper Loy

14:16

@Semiclassical No, the set of points being discussed here is just considered as a subset of omega.

Semiclassical

what?

Jasper Loy

And you just wanna show that this set of points is in F.

Semiclassical

ugh, you're right

you're not trying to show that "the set of points which are in exactly k subsets" are an event space

you're trying to show that "the set of points which are in exactly k subsets" are a subset of the event space

usukidoll

Isn't that the ... I have the book . One of the examples is $\Omega$ is any nonempty set and $F= \{ \emptyset, A, \Omega \backslash A, \Omega \}$ where A is a given subset of $\Omega$

Jasper Loy

So we just need to take the complements and unions and intersections and we are done.

usukidoll

14:18

Oh crap

Jasper Loy

So only the writing is hard because it involves index sets.

usukidoll

Flash floods warning on my phone

I'm indoors

Semiclassical

yikes

usukidoll

But yeah we need the union, intersection, and complement

Just that when I did it to 1.8-1.10 I was able to see it... But I'm dealing with a string so I'm like ughhhhhhh

Jasper Loy

So you just write things like x1,...,xk and things like take the union over this index set and bla bla bla.

usukidoll

14:20

Which is painful

X1 U X2 U,..., U xK and then take complement?

Jasper Loy

If you understand the idea completely, then it's only tedious, not hard, now that we know how to write using index sets.

Semiclassical

"tedious but straightforward" as one textbook would put it

Jasper Loy

Therefore left as an exercise for the reader.

Semiclassical

lol

usukidoll

😥

Jasper Loy

14:22

If I didn't have to type on a keyboard I could just show you the proof on the blackboard now.

But typing in this chat is hard enough, lol.

Semiclassical

This is the kind of problem I have a hard time sustaining interest in nowadays

usukidoll

Cool a blackboard. Pictures help me visualize things

Jasper Loy

But there is a caution.

Sometimes we think that something is obvious and we don't bother writing it out.

usukidoll

I know what it is.

Jasper Loy

And then when we actually write it out we realise we actually don't know the proof.

That means we thought we solved the problem when we actually did not!

Semiclassical

14:24

Point.

usukidoll

Or I can throw this problem in a furnace... Nah. It's starting to sink in a little

😓😥

Jasper Loy

So although many answers on Math SE are hints, I wonder if the answerer actually knows the proof or not.

usukidoll

Someone suggested induction. And I tried to do it in my head

Semiclassical

induction is not always the most intuitive approach, but it does keep you honest.

Jasper Loy

Induction would be the more formal way to do it.

usukidoll

14:26

Base case

K case

Jasper Loy

But I think in this case, using index sets and ... will suffice.

usukidoll

K+1 case

D:

The burnout is real

Jasper Loy

Honestly I can't stand using induction here.

It is highly non-intuitive and non-natural to think this way.

user1732

for you :-)

usukidoll

Sleepy

user1732

14:28

sleep

Semiclassical

i'm reminded of the old anecdote

"The mathematician is lecturing about a topic. He writes a preliminary lemma on the board and then says “This result is trivial to prove.” At that point, he stops, pausing pensively. After a few minutes, he says “Excuse me” and leaves the room. Ten minutes later, he returns and says “Yes, it is trivial, so let’s move on,” continuing the lecture from where he had left off."

6

abenthy

If $M$ is a surface of genus $g$ with $k$ punctures (i.e. $k$ points removed), and $N \to M$ is a $d$-sheeted covering, can we say something about the genus and punctures of $N$?

usukidoll

K uses user1732 as a pillow

user1732

:D

Semiclassical

(full disclosure: I blatantly c/p'd that statement of the anecdote from the internet)

user1732

14:30

we know

usukidoll

XD

Semiclassical

lol

Jasper Loy

Let the k sets be Ax_1,...,Ax_k and the remaining m-k sets be Ay_1,...,Ay_(m-k). Now consider Ax_1 intersect...intersect Ax_k intersect (Ay_1 complement) intersect ... intersect (Ay_(m-k) complement) @usukidoll for example.

Call this set B_i. Now there are altogether (m choose k) ways to choose k sets out of m sets, and we now let C be the union of the B_i's where i goes from 1 to (m choose k). @usukidoll

Since F is closed under intersections and complements, each B_i is in F. Since F is closed under unions, C is in F. QED. @usukidoll

Just tidy up what I wrote above a little bit or two and you will be fine.

Semiclassical

in other news, the complementary error function has a surprisingly-nice continued fraction representation:

$$\sqrt{\pi}e^{x^2}\text{erfc }x=\frac{2x}{2x^2+1-}\frac{1\cdot 2}{2x^2+5-}\frac{3\cdot 4}{2x^2+9-}\cdots$$

Jasper Loy

@usukidoll Are you still here?

Maybe fell asleep already.

Oh she left the chat room. Hope there are no flash floods there.

I am leaving this room too, bye bye.

Semiclassical

14:44

later

Oskar Tegby

Maybe it wasn't that trivial if it took him ten minutes to figure out why.

https://chat.stackexchange.com/transcript/message/46461396#46461396

Okay. That doesn't work.

Jake Rose

When doing Gaussian e

oops

when doing Gaussian elimination, let’s say on a real 3x3 matrix, what stops you from doing the following row operations and concluding the rows are linearly dependent. R1 +R2 and R2+R1

Then followed by R1 x R2 or vice verse to get a 0 row

@TedShifrin this seems up your alley

@Semiclassical also just because you’re my fave physicist

Is it because ultimately what I am doing is the original r1 minus itself,

just hiding it behind a veil?

Semiclassical

R1+R2 isn't a row operation. Do you mean R2->R1+R2?

the point will be that, once you add row 1 to row 2, you've changed what row 2 is

i.e. you go from rows (R1,R2,R3) to (R1,R1+R2,R3)

and if you now add row 2 to row 1, you'll get (2R1+R2,R1+R2,R3)

Jake Rose

15:01

yes sorry you’re right

Semiclassical

so the new first row is 2R1+R2, not R1+R2

Jake Rose

what about adding the origins, row 2 to row 1

original*

Semiclassical

You do successive row operations on the rows you have, not on the rows you started with

Jake Rose

why can’t you do it?

Semiclassical

Because that won't be an elementary row operation anymore.

Jake Rose

15:02

(I see that it leads to the problem I have but is there a more concrete reason)

Leaky Nun

because it isn't reversible, if you're looking for a moral reason

Semiclassical

Do you know how elementary row operations are implemented in terms of matrices?

Jake Rose

Yes I think so

Leaky Nun

other than that, I don't know what kind of reason you're looking for

Semiclassical

Then what you should note is that, if you have a set of elementary matrices A1,A2,A3 acting on a matrix M

you would implement row ops in the order 1,2,3 as A3.A2.A1.M

and the point is that, when you do the third row op, you're really doing A3.(A2.A1.M)

which is to say, A3 is formulated as an elementary row operation on the rows of A2.A1.M

i.e. on the rows you have access to after having done the first two row ops

Jake Rose

15:06

@Semiclassical gonna change my previous answer to no

Semiclassical

lol

Jake Rose

Sorry :’)

Semiclassical

Suppose you want to implement the row operation R2-> R1+R2 on a matrix $M=\begin{pmatrix} R1 \\ R2 \\ R3\end{pmatrix}$

Leaky Nun

5 mins ago, by Semiclassical

You do successive row operations on the rows you have, not on the rows you started with

this is the reason, and Semi is just rephrasing this, which I don't see the need to

Semiclassical

the way you'd do that is to multiply on the left by the matrix $A_1 = \begin{pmatrix} 1 & 0 & 0\\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$

Jake Rose

15:08

@LeakyNun trying to figure out why you must do it this way though.

Leaky Nun

because it's how it's defined

Semiclassical

since then $$A_1 M = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}R1 \\ R2 \\ R3\end{pmatrix} = \begin{pmatrix} R1\\ R1+R2\\ R3\end{pmatrix}$$

siiigh

Leaky Nun

@JakeRose you do the row operations one by one

Jake Rose

@Semiclassical ah I see what you mean now

Mhmm I get it

Semiclassical

whew

yeah

once you write it like that, it's obvious that you lose access to the old rows once you do an elementary row op

Jake Rose

15:11

Yeah I get it

Semiclassical

right

Jake Rose

And if you do bring the old rows back in you’re essentially doing a row - itself which proves nothing

@Semiclassical thanks for the help @LeakyNun

Semiclassical

right

it's not that the old rows are no longer valid---they still represent sensible equations--- it's that they'll be redundant

and once you have redundant equations, you definitely risk doing the equivalent of equation - equation = 0

abenthy

16:19

Is a totally geodesic embedding of $S^1$ in a Riemannian manifold $M$ the same as what is called a "closed geodesic" in $M$ or rather what is called a "geodesic loop" in $M$ in the literature?

Or rather "simple closed geodesic"? I'm confused about these concepts.

Akiva Weinberger

16:39

A geodesic loop need not have the same tangent at the start and end @abenthy

Consider a cone (with a sufficiently small angle)

Jasmine

@IceInkberry
21:57
←
I found magic. Magic to use these: ∫ x² dx = αβγδ⁹⁸ ± ∮x⁴ dx please elaborate

Abcd

← 1 message moved from JEE/High School Chemistry Problems

Sorry not meant to be in this room. moved here by mistake.

Please ignore it.

Perturbative

17:18

Hey @TedShifrin :)

Ted Shifrin

hi Perturb

Perturbative

I have a quick question on the definition of a Riemannian metric. So Lee defines a Riemannian metric as the following: "Let $M$ be a smooth manifold with or without boundary. A Riemannian metric on $M$ is a smooth symmetric covariant $2$-tensor field, $g$ on $M$ that is positive definite at each point. " So I just wanted to spell out the details of that and make sure they're correct.

So here's my attempt at that: Let $M$ be a smooth manifold and let $g$ be a Riemannian metric on $M$. The bundle of covariant $2$-tensors on $M$ is given by $$T^2T^*M = \bigsqcup_{p \in M}T^{(0, 2)}\left(T_p^*(M)\right) = \bigsqcup_{p \in M} \left(T_p^*(M) \otimes T_p^*(M) \right).$$

A smooth covariant $2$-tensor field on $M$ is a section of the smooth vector bundle $$(T^2T^*M, \pi, M)$$ so $g$ is a smooth map $$g : M \to \bigsqcup_{p \in M} \left(T_p^*(M) \otimes T_p^*(M)\right)$$ such that $\pi \circ g = \operatorname{Id}_M$.

Now note that $g$ is also symmetric, by which we mean that for each $p \in M$, $g(p) \in \bigsqcup_{p \in M} \left(T_p^*(M) \otimes T_p^*(M)\right)$ is a symmetric tensor and by positive definite we mean that $g(p)(X_p, X_p) > 0$ for all $X_p \in T_p(M)$ (again we make the usual identifications of tensors products with multilinear maps and elements identified with their counterparts in their biduals).

Ted Shifrin

Using $T^2$ for tensors is terrible. Is that Lee's notation? $T$ is for tangent bundle.

Some people use a script $T$.

Perturbative

Yep that's Lee's notation

Ted Shifrin

Ugh.

Perturbative

17:28

Ted Shifrin

I think that's horrible. He should have used a different font.

$T$ is the tangent functor.

But, yeah, it looks like you regurgitated fine.

Abdullah UYU

@micsthepick I'll be trying to solve the problem you asked. When it ends, I'll ping you.

Perturbative

Thanks @Ted. Also sorry about this question, our intro diff geom course (if you can call it that) at my university is taught by a physicist and done without any sort of formality and we're doing stuff on metric tensors and Riemannian metrics so I wanted to get a rigorous understanding of them

Daminark

Hey everyone!

Perturbative

Hey! @Daminark

Ted Shifrin

17:32

@Perturb: Lee's book should be fine. You can also look at Spivak's first volume (maybe eventually second and third).

yo Demonark

Jasper Loy

@Perturbative Since you mention Lee and Riemannian metric, I would like to inform you that Lee's Introduction to Riemannian Manifolds, second edition, will be published on 29 Sep 2018 by Springer.

Perturbative

@Ted Lee's book will be good for now, I'll take a look at Spivak's one too when I have some time

@JasperLoy Yeah I remember seeing you say something about that on the chat a day or two back

Jasper Loy

@Perturbative Yes, the date has changed many times on springer.com.

Ted Shifrin

I've gotten annoyed a bit with some of the grad students posting geometry questions on MSE. They have things wrong and then argue with me when I give them a correct answer :P Who do they think they are?!

Abdullah UYU

:)

Mike Miller

17:41

@TedShifrin some questions are very bad lately

but perhaps I have forgotten what it was like to be young

Ted Shifrin

LOL

Jasper Loy

If you have forgotten what it was like to be young, then what about Ted?

Ted Shifrin

Right.

Mike Miller

Ted has forgotten what it's like to be middle-aged.

Ted Shifrin

There was a good one about Chern's proof of Gauss Bonnet. Too easy, though ... :P

Remember that I'm teaching 3 10th graders and 1 11th. :P

Jasper Loy

17:43

Spivak's fifth volume proves the Chern-Gauss-Bonnet theorem.

Ted Shifrin

So do I in my courses.

Hi Oskar.

Oskar Tegby

Hi, Ted!

I'm nervous. My exam in real analysis is tomorrow morning.

Ted Shifrin

No point worrying.

Jasper Loy

A little stress is actually good for performance.

Oskar Tegby

No. It's just that there's a thing with the Weierstrass approximation theorem that I still have to figure out.

Ted Shifrin

17:45

You'd better not mess up uniform convergence or I'll smack you :)

Mike Miller

@TedShifrin I'm teaching 100 something-graders across 3 classes.

Just not in person.

Ted Shifrin

Mike, my calc was 1, so 4 is great. But I'd rather have 15.

Jasper Loy

I used to have 30 naughty kids in my classes.

It was more of managing their behaviour than proving the quadratic formula.

Oskar Tegby

May I bother you with one last question on real analysis?

Ted Shifrin

The only naughty students I had at university were the future elementary school teachers. They were horrible.

8

Oskar Tegby

17:48

"Show that if a continuous function $f:[0,1]\to\mathbb{R}$ satisfies
\begin{align}\label{1}
\int_0^1f(x)x^{1/(2n+1)}=0
\end{align}
for $n=1,2,\dots$, then $f(x)=0$ for all $x\in[0,1]$. Does the statement hold true if the interval $[0,1]$ is replaced by $[-1,1]$?"

Ted Shifrin

Bother everyone, @Oskar.

Jasper Loy

@TedShifrin OMG, they must be as naughty as the elementary school students.

Ted Shifrin

Worse, Jasper. Spoiled anti-intellectual brats. Trained by the education school faculty to believe that a teacher is a good teacher iff all the students "get" A's.

10

What's your proof, @Oskar?

Secret

Incredibly Satisfying Sphericons

►

lol, if not because of youtube recommendations, never heard of this shape before

Oskar Tegby

I don't know what the corresponding $g(x)$ will be in this problem. I asked a question here on StackExchange and got grasp of the most of the general method.

https://math.stackexchange.com/questions/2896001/showing-that-a-bounded-continuous-function-is-zero-if-an-integral-with-it-is-zer/2896239?noredirect=1#comment5984063_2896239

Semiclassical

17:50

@TedShifrin blinded leading the blind

Ted Shifrin

I'm not looking. We're talking. @Oskar

Oskar Tegby

Not looking on the link?

Okay.

I'll copy paste my proof from the other problem that I solved.

Semiclassical

I have a lot of cynicism with math education tho

Ted Shifrin

So, by Weierstrass, you approximate $f$ by a polynomial. What do you learn?

Semiclassical

at least at the K12 level

Ted Shifrin

17:51

Semiclassic: You may be more cynical than I.

Oh, a lot of college level sucks rocks.

Oskar Tegby

This is what I did for the problem when $\int_1^\infty f(x)x^{-n}dx=0$.

"If $\int_1^\infty f(x)x^{-n}dx=0$ for all $x\in[1,\infty)$, then we define $g(x)$ such that $g(x)=x^{-8}f(x)$. Thus, $g(x)$ is bounded and continuous on $[0,\infty)$, and consequently integrable on $[0,\infty)$ as well. Now
\begin{align*}
\int_1^\infty g(x)x^{-n}dx=\int_1^\infty x^{-8}f(x)x^{-n}dx=\int_1^\infty x^{-n-8}f(x)dx=0
\end{align*}
for all $n\geq0$. The Weierstrass approximation theorem now gives us that as this holds for all $n\geq0$ we have that

Ted Shifrin

@Oskar: I don't want to talk about that one.

Oskar Tegby

I'm trying to describe what I don't get, sir.

As you wish.

Ted Shifrin

You're thinking about the wrong thing.

Oskar Tegby

Okay.

Ted Shifrin

17:53

So what happens if you take any polynomial for $f$ on $[0,1]$?

Semiclassical

@TedShifrin yeah, but K12 is where the bad habits/mindset gets laid

Ted Shifrin

Oh, I see the issue.

Oskar Tegby

I'm not sure what you mean, professor.

Semiclassical

So I tend to take more issue with K12 than college

Ted Shifrin

Do you know a more general statement of Stone-Weierstrass?

Semiclassical

17:55

(though that probably also reflects my general disdain for high school vs. my general enthusiasm for college)

Oskar Tegby

Yes. It's more or less that you can approximate any function with polynomials under certain circumstances.

Ted Shifrin

Well, that's the usual Weierstrass theorem. Did you discuss something about approximating by other families of functions?

Oskar Tegby

I don't recall.

Ted Shifrin

Stone generalized to a collection of continuous functions that separates points.

If your class didn't do this, I don't think you can do this problem. However, you can answer the second question ($[-1,1]$) by inspection.

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