Mathematics - 2014-05-19 (page 3 of 5)

Chris's sis

16:00

@N3buchadnezzar One way is to split the integral at $x=1$.

N3buchadnezzar

@Chris'ssis That does not work

G.T.R

@BalarkaSen @N3buchadnezzar I got $O(1/\sqrt{n})$ with Bernouilli inequality, how do you refine it ?

N3buchadnezzar

$$
0 \leq
\int_0^\infty \frac{dx}{(x^2+1)^n}
\leq \int_0^1 \frac{dx}{(1+0)^n} + \int_1^\infty \frac{dx}{(0+x^2)^n}
\leq 1 + \frac{1}{2n-1}
$$

Balarka Sen

:15603833 Please go ahead.

Kill yourself.

N3buchadnezzar

@Chris'ssis Boom headshot

Chris's sis

16:02

@N3buchadnezzar Just do what I said!

OK, let me show you the way!

G.T.R

@N3buchadnezzar oh nice, thanks

N3buchadnezzar

@G.T.R Note this only show that $0 \leq I \leq 1$

It was just a quick stab at Chris,'sis :p

Balarka Sen

@G.T.R Not yet very nice.

@N3buchadnezzar It was pretty quick though.

G.T.R

@N3buchadnezzar that was sarcasm (@Chris'ssis)

N3buchadnezzar

@BalarkaSen Gotten pretty nifty with latex p

Balarka Sen

16:05

If it's anything, @N3buchadnezzar and I are better with hard analysis than @Chris'ssis. We are number theorists man!

N3buchadnezzar

You used latex a lot ?

user63181

@G.T.R You can interchange limit and integral due to the monotone convergence theorem.

Balarka Sen

@N3buchadnezzar Well, not much.

G.T.R

@N3buchadnezzar I still can't get that $O(1/n)$. But $O(1/\sqrt{n})$ is fine

Balarka Sen

@G.T.R Yeah. Any thing like $O(1/n^\epsilon)$

N3buchadnezzar

16:06

When I cite things in my papers you would often go See equation ("4") where 4 is a hyperlink. Now I was able to get "equation (4)" hyerlinked, but still only 4 colored.

@G.T.R Yeah, who cares about precise bounds anyway ;) Stab at Balarka!

Balarka Sen

@N3buchadnezzar Eh, never hyperlinked stuff.

N3buchadnezzar

Hey :p

Balarka Sen

@N3buchadnezzar stab stab stab

N3buchadnezzar

If it is only the number that is colored, and the color is dull. It just barely slides

G.T.R

@user63181 yes, you're right, thanks for the headsup

Balarka Sen

16:08

@N3buchadnezzar I never even used colors.

Text and Tex are sufficient.

N3buchadnezzar

I try to use two colors, blue and red and shades of gray

mirgee

Does anyone know how to prove $\|\mathbb A\vec x\|=\| \mathbb A^H\vec x\|$?

if $\mathbb A$ is normal

Balarka Sen

@mirgee $\Bbb A$ is a normal subgroup of $H$ which acts over $\Bbb A$ by some topological thingy?

That means both are Lie groups?

N3buchadnezzar

thingy ma jiggy

user63181

$\langle Ax,Ax\rangle=\langle x,A^HAx\rangle=\langle x,AA^Hx\rangle=\langle A^Hx,A^Hx\rangle$ @mirgee

Chris's sis

16:12

@N3buchadnezzar $$0\le\displaystyle \int_0^{1} \frac{1}{\left(x^2+1\right)^n} \, dx + \int_1^{\infty} \frac{1}{\left(x^2+1\right)^n} \, dx$$
$$\le \int_0^{1} \frac{1}{\left(x^2+1\right)^n} \, dx+\underbrace{\int_1^{\infty} \frac{x}{\left(x^2+1\right)^n} \, dx}_{\displaystyle\frac{1}{ 2^n (n-1)}}$$
Take the limit and you're done.

Balarka Sen

@Chris'ssis Tremendous amount of work.

Chris's sis

@BalarkaSen Gezzz, all that can be placed in a single line. :-)

Balarka Sen

@Chris'ssis Compute the integral on the left.

That goes beyond a single line.

Chris's sis

@BalarkaSen No, that is elementary.

N3buchadnezzar

@Chris'ssis What about bounds on the integral from 0 to 1 ?

Balarka Sen

16:15

@Chris'ssis OK, show me how you compute it in a single line.

Chris's sis

@N3buchadnezzar I took the limit under the integral sign.

N3buchadnezzar

@Chris'ssis Then why not do that from the start?

Chris's sis

@N3buchadnezzar I still think of doing that integral elementarily.

Balarka Sen

@Chris'ssis No, he asks why not take the limit under the integral sign from the beginning.

Chris's sis

@BalarkaSen Because I wanted to do all elementarily.

Balarka Sen

16:17

What's the need for all this works?

Chris's sis

brb

Balarka Sen

@Chris'ssis It's tremendously tedious that way.

mirgee

@user63181 But how would you prove the first equality? My definition of $\mathbb A^H =\overline{ \mathbb A^T}$

G.T.R

@BalarkaSen show me your way. Is it shorter than Bernouilli ?

Balarka Sen

@G.T.R Take limit under integral sign.

Chris's sis

16:19

@BalarkaSen I say it again: I wanted to do it elementarily (with high school knowledge only).

I'll do it that way.

Balarka Sen

@Chris'ssis Taking limit under integral sign is elementary and highschool.

I think.

Chris's sis

@BalarkaSen Sorry? I didn't learn that in high school ...

Balarka Sen

Oof, another power cut.

user63181

Yes $A^H$ is the adjoint of $A$ in the hermitian space so we have the first equality.@mirgee

Balarka Sen

@Chris'ssis Right, that's true. Agreed.

I never took calculus on highschool.

I was forgetting that. =P

mirgee

16:22

@user63181 How?

user112495

If f is a function such that f(sin(x)) is continuous, must f be continuous?

Balarka Sen

But Bernoulli is elementary, @Chris'ssis

That can be applied too.

mirgee

@user63181 What you say works for linear operators. But $\mathbb A$ is simply a matrix

G.T.R

@user112495 if f(sin(x)) is continuous, so is f(sin(arcsin(x)))

Chris's sis

@BalarkaSen Yeah, I know, but I think of something different now ...

Balarka Sen

16:24

@Chris'ssis Something tedious =D

Chris's sis

@BalarkaSen :-))))

Balarka Sen

But, ok, I am interested in the computation of the integral there elementarily and non-tediously.

user112495

@G.T.R Oh yeah :p. Thanks!

user63181

If it works for linear operators as you said then it works also for a simple matrix @mirgee

G.T.R

@BalarkaSen @Chris'ssis the next step is to determine an asymptotic equivalent of this integral. Now it's getting tedious

mirgee

16:26

@user63181 OK, so I have to make a map between operators and matrices, then prove it's bijective...?

Balarka Sen

@G.T.R Which integral?

mirgee

@user63181 Isn't there another way?

G.T.R

@BalarkaSen $\int_0^{\infty } \frac{1}{\left(x^2+1\right)^n} \, dx$

Balarka Sen

Ah, that one.

user63181

No there's a simple proof to it@mirgee

mirgee

16:28

@user63181 Would you please show me?

@user63181 To be clear, I know definition of operator adjoint and proof of it's uniqueness using Riesz theorem

@user63181 No matrices there

user63181

Simply $\langle Ax,y\rangle=\overline{(Ax)}^Ty=\overline{x}^T\overline{A}^Ty=\langle x,\overline{A}^Ty\rangle$@mirgee

mirgee

@user63181 And relationship between matrix of operator and its adjoint in ON basis

Studentmath

I never would've imagined probability requires such deep knowledge of calculus and set theory

mirgee

@Studentmath I've recently found to my surprise that probability makes heavy use of calculus, but don't know about set theory...

@Studentmath Heard set theory is hard!

Studentmath

I actually have easier time with set theory, I don't really have the required knowledge in double integrals and so on in calculus... I have a habit of taking courses together with their required pre-courses..

mirgee

16:35

@Studentmath Why? Are you that impatient to learn? :)

Studentmath

I am a bit short on time and want to cover as much knowledge as possible :P Plus in chemistry I managed to handle it easily, in math it's a bit harder

user63181

To find an elementary proof to $\lim_{n\to\infty}\int_0^1\frac{dx}{(x^2+1)^n}=0$ take $\epsilon>0$ and write $$\int_0^1=\int_0^\epsilon+\int_\epsilon^[email protected]

user112495

@G.T.R Can you tell me what's wrong with my reasoning here:

I have to decide whether sin(f(x)) being continuous implies f(x) is.

I considered f(x) = 0 for rational x and pi for irrational x. Then sin(f(x))=0 and so is continuous but f(x) isn't. But then couldn't you do arcsin(sin(f(x))) is continuous by composition of continuous functions which means f(x) is continuous?

sdf

is there a situation in which it is possible to deduce that a map is continuous because its kernel is open?

mirgee

@user63181 But that's assuming we are in a euclidean or unitary space

@user63181 With usually defined inner product... Which I'm allowed to assume :D Thx

user63181

16:41

So you're allowed to assume! Fine;-)@mirgee

G.T.R

@user112495 arcsin(sin()) is a wild function

@user112495 sin(arcsin()) is the identity, no problem. arcsin(sin()) is some kind of piecewise function

user112495

@G.T.R I still don't get why sin(arcsin()) is that different to arcsin(sin())

Chris's sis

@user63181 This is what I was preparing to type. It's the best way. @G.T.R @BalarkaSen

user112495

@G.T.R have you got an example where it doesn't work properly just so i can visualise it

robjohn

@N3buchadnezzar Wasn't that computed either in chat or an answer recently?

G.T.R

16:43

@user112495 see here google.fr/… , and think for a moment

Chris's sis

@G.T.R $$\int_0^{1} \frac{1}{\left(x^2+1\right)^n} \, dx=\int_0^{\epsilon} \frac{1}{\left(x^2+1\right)^n} \, dx +\int_{\epsilon}^1 \frac{1}{\left(x^2+1\right)^n} \, dx$$ This way is simply at the high school level.

user112495

@G.T.R Thanks!

user63181

The problem is that $\arcsin$ function is only defined on the interval $[-1,1]$ so $\arcsin(\sin x)\ne x$ for $x$ not being in this interval.@user112495

Chris's sis

@G.T.R This reminds me of a very beautiful question from a high school contest!

@G.T.R It's this one $$\lim_{n\to \infty} n 2^n \int_1^n \frac{1}{(1+x^2)^n} \ dx$$

robjohn

$$
\int_0^\infty\frac1{(1+x^2)^n}\,\mathrm{d}x=\frac12\int_0^\infty x^{-1/2}(1+x)^{-n}\,\mathrm{d}x=\frac12\mathrm{B}\left(\frac12,n-\frac12\right)=\frac{n\pi}{(2n-1)4^n}\binom{2n}{n}
$$

Chris's sis

16:58

@robjohn Right, this can also be expressed in terms of beta function. I only wanted to use some elementary tools. Of course, it's nice to have many tools at hand. :-)

G.T.R

@robjohn $$
\int_0^\infty\frac1{(1+x^2)^n}\,\mathrm{d}x=\frac{\sqrt{\pi } \Gamma \left(n-\frac{1}{2}\right)}{2 \Gamma (n)}$$ ???

N3buchadnezzar

@robjohn How to determine the asymptotic growth of that one ?

G.T.R

Wow the last equality is super cool

N3buchadnezzar

@robjohn Also just the wallis integral, use $\tan y \mapsto x$ to see it =)

robjohn

@N3buchadnezzar aymptotically, it is $\sqrt{\frac\pi{4n}}$

N3buchadnezzar

17:04

You mean 1 over as it tends to zero but ok.

robjohn

@Chris'ssis what is that?

Chris's sis

@robjohn let me check one more time ... (there was a small mistake)

robjohn

@N3buchadnezzar yeah, I multiplied by $\sqrt{\pi n}$ instead of dividing by it

@N3buchadnezzar $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$

Chris's sis

@G.T.R by the way, don't miss my question ... it's very nice :-)

N3buchadnezzar

@robjohn From stirrling

robjohn

17:10

@N3buchadnezzar yeah

N3buchadnezzar

Thats neat

G.T.R

@Chris'ssis I'm still thinking about that $\epsilon$ expression. How do you choose the $\epsilon$ ? I'm tempted to say it's related with continuity of the function at $0$

Chris's sis

@G.T.R You choose $\epsilon$ very small (sorry for my non-mathematical language), $\epsilon>0$. The rest is a piece of cake.

robjohn

@N3buchadnezzar I dropped a factor of 1/2 (now fixed)

G.T.R

@Chris'ssis how do you get a bound for this $$\int_0^{\epsilon} \frac{1}{\left(x^2+1\right)^n} \, dx$$ ?

N3buchadnezzar

17:13

@robjohn From where?

robjohn

17 mins ago, by robjohn

$$
\int_0^\infty\frac1{(1+x^2)^n}\,\mathrm{d}x=\frac12\int_0^\infty x^{-1/2}(1+x)^{-n}\,\mathrm{d}x=\frac12\mathrm{B}\left(\frac12,n-\frac12\right)=\frac{n\pi}{(2n-1)4^n}\binom{2n}{n}
$$

N3buchadnezzar

@robjohn Yeah I saw just now you fixed the integral

Chris's sis

@G.T.R Isn't this $$0\le \int_0^{\epsilon} \frac{1}{\left(x^2+1\right)^n} \, dx \le \epsilon \cdot M$$? Since $\epsilon$ is arbitarily small, you're done.

robjohn

@N3buchadnezzar The integral was fine, I forgot the $\frac12$ in front of the Beta function

N3buchadnezzar

Mmm

G.T.R

17:17

@Chris'ssis but then you get a problem taking also $\epsilon \to \infty$ in $$\int_{\epsilon}^1 \frac{1}{\left(x^2+1\right)^n} \, dx$$

@user63181 How should I choose $\epsilon$ in the expression you suggested ?

N3buchadnezzar

you forgot the dollar signs in front

Chris's sis

@G.T.R I'm only referring to this integral $$\int_0^{1} \frac{1}{\left(x^2+1\right)^n} \, dx$$. The other part from $1$ to $\infty$ is clear.

N3buchadnezzar

@robjohn This is a nice question btw

@Chris'ssis How is $\epsilon$ to $1$ bounded ?

Studentmath

Refreshing my memory here a bit and being completely unsure of myself: $\int e^{f(x)}dx=\frac{1}{f'(x)}*e^{f(x)}+C$ iff f'(x) isn't 0 for every x, and in case of definite integral for every x the definite integral is over, correct?

G.T.R

@Chris'ssis yes, I got that. But you can't just let $\epsilon \to 0$ in both $$\int_{\epsilon}^1 \frac{1}{\left(x^2+1\right)^n} \, dx$$ and $$0\le \int_0^{\epsilon} \frac{1}{\left(x^2+1\right)^n} \, dx $$

user63181

17:20

You have not to choose $\epsilon$, it's arbitrary and you should find $N$ such that $\forall n\ge N$ the integral is less than $C\epsilon$ where $C$ is a [email protected]

N3buchadnezzar

@Studentmath Only if $f'(x)$ is constant.

Studentmath

Darn.

N3buchadnezzar

say $e^{x^2+1}$

Ted Shifrin

Heya @Studentmath, @N3, @robjohn

Studentmath

How do I go about it otherwise? Hey @Ted

N3buchadnezzar

17:22

You simply dont :p Except if you have some junk in front of the exponential, or have some very special limits.

Studentmath

Jee..

user63181

Hello @TedShifrin

Studentmath

Not even for a definite integral? @N3buchadnezzar

N3buchadnezzar

$\int f'(x) e^{f(x)}\,\mathrm{d}x = e^{f(x)} + D$

Mike Miller

hello

Ted Shifrin

17:23

Hello, @user63181: Do I know you?

Heya @Mike

N3buchadnezzar

@Studentmath Well you can compute an integral on the form e^f(x), but the rule above does not hold.

user63181

I'm Sami I changed my look;-)@TedShifrin

robjohn

@N3buchadnezzar and the answer seems about the only closed form.

N3buchadnezzar

I was more interested in the proof using CA

Chris's sis

@user63181 Well, the idea is that one can choose any $\epsilon$ arbitrarily small such that the integral is smaller than $\epsilon M$, $\epsilon>0$.

Studentmath

17:24

Hmm.. I am getting this from probability over normal distribution

Ted Shifrin

I prefer real names :(

Studentmath

It's really interesting

Mike Miller

@Ted Do you/does one think much about the "shape" of tensor bundles, in the sense that one does for the tangent bundle (I.e. parallelizability?)

N3buchadnezzar

@TedShifrin I prefer names living in $\hat{\mathbb{C}}$

Ted Shifrin

slaps @N3

N3buchadnezzar

17:26

:p

Chris's sis

@N3buchadnezzar How? Just think about it! It's easy! :-)

N3buchadnezzar

That would be some pretty complex names

G.T.R

@user63181 It's clear now, thanks

r9m

@Cortizol $\sum\limits_{cyc} \dfrac{a}{b+2c} = \sum\limits_{cyc} \dfrac{a^2}{ab+2ac}$, Using the Cauchy Schwarz Inequality, $(\sum\limits_{cyc} ab + 2ac)\left( \sum\limits_{cyc} \dfrac{a^2}{ab+2ac}\right) \ge (\sum\limits_{cyc} a)^2 \ge 3(ab+bc+ac)$, thus proving $\sum\limits_{cyc} \dfrac{a}{b+2c} \ge 1$ :)

N3buchadnezzar

@Chris'ssis Imagine

G.T.R

17:27

@TedShifrin Salut, c'est GabrielR

Mike Miller

I've got no real geometric intuition for them and I don't know if that's -bad-

Chris's sis

@G.T.R did you get my point?

r9m

@Chris'ssis did ya ping me ? (IDK .. but I can't find where)

Ted Shifrin

I don't know that triviality of a bundle is "shape," @Mike, unless we're talking connections/curvature.

Chris's sis

@r9m Yeah, I have a surprise to you. Are you ready?:-)

Ted Shifrin

17:28

Salut @Gabriel: Pourquoi tout le monde se change de nom?

r9m

@Chris'ssis If you are gonna breathe dragon fire on me ... I am always ready :P :D

G.T.R

@Chris'ssis hmmm it was quite unclear. Underlining that $\epsilon$ is fixed, and that we're seeking some $N$ is good

N3buchadnezzar

@robjohn The proof might be done using the mean value theorem

Mike Miller

@Ted I dunno, I'd say that's part of a bundle's geometry... a simple part, but still not an easy one

user63181

Chaque saison a son propre nom;-)@TedShifrin

N3buchadnezzar

17:29

$$
u(z_0) = \frac{1}{2\pi} \int_0^{2\pi} u( z_0 + r e^{i\theta} ) \,\mathrm{d}\theta
$$

G.T.R

@TedShifrin I wanted to roll it back to GabrielR, but I have to wait 30 days :(

r9m

@Chris'ssis WoW !!!! good morning :D

Chris's sis

@G.T.R I said " for any $\epsilon$ arbitrarily small".

@r9m Yeah, indeed! I'm in love with it :-)

Ted Shifrin

Particularly interesting in the holo category, @@Mike. A beautiful paper by Griffiths years ago on positivity of bundles. And alg geo is full of such issues (e.g., Kodaira vanishing thm).

Je ne me rappellerai jamais de vous tous ...

Mike Miller

@Ted Oh, I'm fascinated by the complex/algebraic lives of these things. But I'm trying to grasp smooth, first.

@Ted Anyway, forms feel like cheating.

Ted Shifrin

17:32

Cheating? Hell no. They rule.

robjohn

:15605487 was there something wrong with that integral? (the one you just deleted)

Mike Miller

It makes me uncomfortable that they trivialize $H^i=0$ for $i>n$, @Ted. We didn't have to work hard for that topologically but we had to work for it.

Ted Shifrin

Hardly work to have a cell/simplicial complex of finite dim?

Sab ಠ_ಠ

Hello world!

Hello Prof. @Ted

Ted Shifrin

Salut @Sab

Sab ಠ_ಠ

17:36

Je vois que le francais est devenu familier dans ce salon :D

Studentmath

I have managed to get to this expression:
$$\int_{279.792}^{320.208} e^{-\frac{(x-300)^2}{2\sigma ^2}}dx$$

Err that looks ugly.

Ted Shifrin

Le français et l'espagnol, oui ...

Alizter

ew

Studentmath

Better.

Ted Shifrin

Ugh @Studentmath

Sab ಠ_ಠ

17:37

Je veut apprendre l'allemand

Studentmath

I know it's equal to 0.6! now I just need to find that sigma.. easy peasy.

Mike Miller

@Ted Not every manifold is a simplicial complex... and the proof I know for manifolds is essentially inductive

Sawarnik

@Sabಠ_ಠ Ye kaun sa language hai?

Studentmath

@Ted you are going to teach this soon.

Ted Shifrin

Il y en a plusieurs entre nous qui parlent allemand, aussi ...

Sab ಠ_ಠ

17:38

@Sawarnik urdu or hindi?

Mike Miller

Prove for open subsets of $\Bbb R^n$, prove that if it's true for two sets and their intersection it's true for their union, patch all these things together.

Sawarnik

@Sabಠ_ಠ Hindi of course. You didn't understood?

Sab ಠ_ಠ

Someone told me the language used in bollywood movies is urdu :S

Sawarnik

@Sabಠ_ಠ Not really.

Sab ಠ_ಠ

"What language is this" I think

Sawarnik

17:39

@Sabಠ_ಠ Right.

Sab ಠ_ಠ

I learned the language from bollywood movies

Ted Shifrin

Yes, I love Mayer-Vietoris for deRham, @Mike ... Totally direct from partitions of unity :)

Studentmath

Is it even solveable...

Sab ಠ_ಠ

Exams start in 2 weeks :O:O

I need to plan my revision and ace this now :O

Sawarnik

@Sabಠ_ಠ I meant in which language were you talking with Ted?

Mike Miller

17:40

@Ted That's why it all feels like cheating! These topologically a bit difficult facts are now trivial.

Alizter

@Studentmath No. But in terms of error function it may be put.

Sab ಠ_ಠ

Oh @Sawarnik that was French.

@Ted What would be the best way to ace a math exam in 3 weeks? :3

Studentmath

Yeah. I dislike the error function.

Mike Miller

@Sab Know the math

Sab ಠ_ಠ

I need to know all the details inside out

Ted Shifrin

17:41

Understand the material? @Sab

Sab ಠ_ಠ

Thing is I'm quite bad with graphs

Sawarnik

@Sabಠ_ಠ Cheating? On your own risk though.

Mike Miller

@Ted Beat ya.

Alizter

I had my final German exam today!

Ted Shifrin

I guess I still have Sawarnik on ignore ...

Sab ಠ_ಠ

17:42

@Sawarnik Cheating is not an option. Most people fail the course, so the chance of failing is higher when cheating. This is pretty ironic.

Ted Shifrin

I'm super slow on iPad, @Mike

Sawarnik

@Sabಠ_ಠ Haha, pick some better students.

Sab ಠ_ಠ

@Sawarnik I can beat the best students if I put my mind to it.

Mike Miller

@Ted I posted a question you might like.

Sawarnik

@Sabಠ_ಠ That's a better approach.

Ted Shifrin

17:43

@Mike: It's not by accident that my students learn forms in my multi course ... :D

Sab ಠ_ಠ

I do pretty well learning in 1 day. Now 3 weeks will put me in a good position to get 90+

Sawarnik

Although cheating was the only thing which let me pass the Sanskrit tests last year :D

Sab ಠ_ಠ

However I need a good approach to grasping the material, thus my question to @Ted

Ted Shifrin

What does your course cover, @Sab?

Mike Miller

here

Sab ಠ_ಠ

17:44

@Ted second

Mike Miller

@Ted how do you define forms for them?

Ted Shifrin

I thought it was always true @Mike.

Second what? @Sab

Enjoys Math

Can you turn $R[[X]]$ formal series over a Euclidean domain $R$ into a Euclidean domain?

Sab ಠ_ಠ

I'll post the list of the content

Sawarnik

@Sabಠ_ಠ Why? Isn't the book good enough? Plus you have this site to clear your doubts.

Ted Shifrin

17:46

In terms of determinants, @Mike.

Mike Miller

@Ted It should be, IMO, but I don't see a proof yet! There's an answer for smooth manifolds via Morsetheory.

Sab ಠ_ಠ

@Sawarnik I think the book is not good enough in terms of questions. I want to do all type of questoins I can get my hands on to master the topic. I actually feel I bombed my physics class test today, because I didn't practice enough question.

Mike Miller

I wonder if the answer is somewhere in Kirby-Siedelmann.

Ted Shifrin

Look in Spanier @Mike.

Mike Miller

@Ted Everything is done on open subsets of $\Bbb R^n$?

Ted Shifrin

17:47

Siebenmann?

No, @Mike, submanifolds. Eventually ...

Mike Miller

Thanks for correcting me, @Ted

Sab ಠ_ಠ

@Ted here are the topics based on Stewart:
1. Functions and models
2. Limits and derivatives(tangent, limit laws, continuity, etc)
3. Differentiation rules
4. Application of differentiation rules (including lopital :( and optimizatoin :(:( , antiderivatives :'( )

5. Integration(evaluating integrals, fundamental theorem of calculus :O :O , sustitution, by parts, trig etc)

6. Proofs (induction and contradiction)
7. Several proofs in differntiation and I guess some in integration will pop up as well

Ted Shifrin

I think it's more basic. Doesn't it follow from the Hopf trace lemma for any simplicial complex? @Mike

Mike Miller

In Spanier should I be looking for a proof that all manifolds are homotopy equivalent to a finite CW-complex, or the Euler char itself?

Ted Shifrin

i think it's just algebra. No topology.

Mike Miller

17:53

@Ted Not all closed manifolds are simplicial complexes, though.

Sawarnik

@Sabಠ_ಠ Want some questions?

Sab ಠ_ಠ

@Sawarnik if you got some with their answers :) I would appreciate it

Ted Shifrin

@Sab: If you email me, I can send you exams I've given in my Spivak course, if you want to choose some problems to practice.

Sab ಠ_ಠ

But don't give the answers yet, I'll do the questions first.

I can email you @Ted :)

Sawarnik

@Sabಠ_ಠ Ok. $f(x)$ and $g(x)$ be twice differentiable, non-decreasing functions. $f''(x) = g(x)$ and $g''(x) = f(x)$. $f(x) \cdot g(x)$ is a linear function.

Then show that $f(x) = g(x) = 0$.

Sab ಠ_ಠ

17:55

@Ted what's your email?

Sawarnik

@Sabಠ_ಠ [email protected]

Ted Shifrin

rsee my profile, @Sab.

you need some finiteness condition, @Mike. Does the Hawaiian earring have $\chi$?

Sab ಠ_ಠ

@Ted I'll send you and email to your uga address, that's the only one I found.

Ted Shifrin

Right @Sab.

Sab ಠ_ಠ

Sent @Ted :)

Sawarnik

17:58

@Sabಠ_ಠ One more. Does there exist a continuously differentiable function $f: [1,5] \rightarrow \mathbb{R}$, such that $f(1) \lt 0, f(5) \gt 3$ and $f'(x) \leq e^{-f(x)}$?

Mike Miller

@Ted The Hawaiian earring is not a manifold. I want to know if the statement is true for closed manifolds, and if not, somewhere it fails.

Sab ಠ_ಠ

I seriously need to learn some LateX

user116900

In Soviet Russia, LaTeX learns you.

Sab ಠ_ಠ

lol

user116900

I am so lame.

Transcript for

$Mathematics$ Mathematics