Mathematics - 2016-02-06 (page 1 of 3)

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12:17 AM

Can someone help me understand the last step the user used in this answer?

@Michael Do you still need help?

0

Q: Need help with a certain integral

Sorry, this is my first post ever and formatting is bad. I appreciate all assistance Given the equation y= $ 3.8x^2-4.4x+1444 $ Find the definite arc length integral between 0 and 1200. I am not sure on how to format $(\frac {dy}{dx})^2$ ,so I left it alone. $$\int_0^{1200} \sqrt{1+ (\frac...

definite-integrals quadratics

@PedroTamaroff ya

So what is troubling you?

How he turned 1/2sinh(2z) into this (2ax+b){1+sqrt(2ax+b)^2}

@PedroTamaroff I tried doing $sinh(2arcsinh(2ax+b))$

but I don't know how to simplify that

If anyone wants to help, I would be very grateful

12:32 AM

OK.

Let me read.4

Do you understand up to the moment when the user gets to $\int \sqrt{1+t^2}dt$?

I underestand everything till "back to x, this gives"

Well, you "just" need to unwind the substitutions.

Do it carefully.

How can you unwind 1/2sinh(2z)

I know that z is $$sinh^-1(2ax+b)$$

Use a "double angle" formula.

mathworld.wolfram.com/Double-AngleFormulas.html

facepalms

Thank you so much hahaha I don't know why I didn't think of that

12:37 AM

No problem.

How are you @PedroTamaroff?

@guest How am I?

@PedroTamaroff sorry to annoy you again, but even after using the double-angle identity, the expression remains unpleasant

@Michael Yes, expressions are unpleasant.

So are raisins in cookies.

Agreed

But the way he (Claude) made it was not unpleasant. :)

12:47 AM

You should be able to get to his expression.

In fact it seems he did use the double angle formula.

Oh wait I just realized. Ok I won't waste your time anymore. I found it

You're not bothering anyone, really.

Graduate students/mathematicians sometimes get frustrated if I don't get something

Well, you shouldn't generalize.

The state of "not getting something" is probably the most common of them all.

Yah I shouldn't :) . There are alot of helpful people on this site

1:22 AM

Yep, and the amazing part is that they help for free.

1:40 AM

frustration is part of the game

Does anyone know a good book on elementary probability?

@ForeverMozart good your here

I wanted to ask a topology question

ok

fire away

Let us say we initially have a cycle that doesn't enclose anything

then after that we fill the hole that the is around a cycle with something with some orientation

i'm feeling very french with my belgian beer and croissant

1:46 AM

the reason now this cycle is homotopically equivalent to a point is because we can just squueze the ball to a point which would in turn make the cycle be also a point

right ?

yes I believe so

1 sec I wanted to verify something

user image

what do they mean here by "sliding over A"?

hey @TedShifrin

They just mean that you can contract a disk to a point, Karim.

They = Hatcher.

yeah that is what a figured

I figured *

gotcha

@Forever: You have a Belgian croissant? :D

2:02 AM

Belgian beer and croissant @TedShifrin

I love French croissant

I don't think I like croissants with beer ...

me neither

i dont know but its the biggest croissant I've ever seen

LOL, @Forever :)

I don't like beer in general I like wine

beer have weird taste to me

2:03 AM

OK, Karim, I'll serve you wine when you graduate :)

:D okay we have a deal

my topology professor told me he want me to read some book called stefan whitney classes in my spare time

it seems interesting

no, Stiefel-Whitney classes ... you are nowhere near ready

It is called characteristic classes

my topology professor said he used to drink beer with his students if they came to his office

you need cohomology for that ...

2:04 AM

but he stopped doing it

Karim, wait a year.

oh his grad student explained cohomology for me but I want to cover it by myself

smart move, @Forever

yeah

he was just being nice though

2:05 AM

cohomology goes in the other direction he mentioned

he came from a different country

poland I think

With all the scandals of physicists, biologists ... seducing their graduate students ...

for example we know homology goes from C_n --> C_{n - 1} --> ... --> C_1 --> C_0

Karim, yes, I know what cohomology is :P Trust me, Milnor's book is for when you've learned algebraic topology very thoroughly.

yeah but he wasn't creepy or anything like that

2:06 AM

@TedShifrin I was just testing myself

:D

I noticed when I explain stuff to people I understand it better

and have it more clear in my head

@Forever, these days I am totally in shock about the inappropriate things that are surfacing.

I got some grad student email I will bug him every day so I could drink topology from his brain !

@TedShifrin people aren't professional

They're mentally f*****-up, Karim. It's not just unprofessionalism.

yeah

I was quite shocked with what happened to that physics prof from mit

I used to look up to him

Seems it's not really us gay folks who are causing all these problems, despite what the religious folk think.

Well, there are Cal Tech and Princeton and other examples, younger guys, Karim.

2:09 AM

oh

I didn't know that

Lots of big name professors being fired ... Oh yeah, and Berkeley.

that is stupid

I guess being smart doesn't make you really not be a *** person

right :D

You know I am taking anthropology this semester as elective

well, maybe you'll understand human beings better after that.

2:11 AM

that subject goes in like very big circular logic which annoys me

LOL, you can't expect everything to be like math!

Morning.

like it mentions we should understand other cultural and not *** with them and then it mentions you can't really be relativistic with your reasoning

very weird

good night, @MikeM

morning @MikeMiller even though here is night

2:12 AM

Sure, that too.

Karim, you come from a very different culture ... anthropology should be about comparing/contrasting different cultures.

Home sick today. Bleh.

Yeah, but they mention that advancement in science doesn't mean that a society is advanced

well, @MikeM ... I had to cancel a dinner date because one of my zillions of temporary crowns came loose. So maybe I can eat or maybe I can't ...

They also mention that people shouldn't change other people culture.

2:14 AM

Karim: Even though you want to dismiss this course as worthless, pay attention to it. The world is a messed-up place.

yeah

One society eats the bodies of people who died is one example !!

In order to erase the memories away

which is very stupid

Ah, interesting ... When I was in high school, we read about the Spelunkean explorers ... who would have starved to death if they hadn't eaten the guy who died. Interesting ethical/pragmatic questions.

yeah, but they were given no choice

Who was given no choice?

the people who starved to death

2:17 AM

well, there are still ethical questions ... maybe they should have all starved and died instead of eating the body of one who had died.

yeah

I don't know that I could bring myself to do that ... but one becomes irrational after starving for a week.

hi @robjohn

@MikeM: Hope you recover soon.

you know one vision I have for humans is that intellectual creatures who have like intelligent conversations with each other questioning about philosophical issues and exploring the world through math and sciences.

well, the majority of humans are illiterate (basically).

1

Q: Homogeneous space minus a point

If $X$ is homogeneous and $p\in X$, then is $X\setminus \{p\}$ necessarily homogeneous? This seems to work with all the simple examples I've tried. I would be interested in any counterexamples. Or if there is a theorem here, you can assume we are dealing with metric spaces. EDIT: Greg's pointe...

general-topology homogeneous-spaces

advertising

2:20 AM

@Tobias Could we say that $A$ is a full subcategory of $A^*$ with the inclusion map $I:A\to A^*$?

Definitely false @Forever.

but what about for connected spaces?

Still connected. Well, maybe I spoke too quickly.

I guess the group can be very different.

oh thats a related question

But if you remove the origin from $\Bbb R^n$, you can no longer translate, and so you only have rotational symmetries. So obviously I have an underlying group in mind (preserving a metric).

2:22 AM

is a connected topological group minus a point homogeneous?

what is homogeneous?

"homogeneous"

what do you mean by that

homogeneous is weaker than group

it means you can map any point to any other point via a homeomorphism of the space to itself

Well, we need a structure. There is a structure-preserving map that sends any point to any other point.

oh oke so it is like a automorphisms of your space

basically every point plays the same role

2:24 AM

I see

So, if I put the metric structure on, then it's false for $\Bbb R^n$. But just looking at homeomorphisms, $S^{n-1}\times\Bbb R$ is still homogeneous.

yes those spaces minus a point are homogeneous

the non-connected example was $2\times S^1$.

cause if you remove a point then you have $S^1+(0,1)$

some points are cut points and others are non-cut points

so it is non-homogeneous

@Ted: I hope I feel better soon too! Sorry to hear about your teeth...

Why is it important to check the linear independence of functions when solving systems of differential equations?

@ThomasAndrews thanks, that's an easy example

2:40 AM

@TedShifrin I bought your multi var calc book

@ForeverMozart: I don't believe there are metrizable examples.

that would be a nice theorem then

assume connected also

Right.

assume compact also it would probably make things easier

Eh, I'm being silly. You should note that a homgeneous space $X$ such that $X \setminus \{p\}$ is homogeneous for all $p$ is equivalent to being strongly 2-homogeneous, meaning for any two pairs of distinct points $(x_1, x_2)$ and $(y_1, y_2)$, there's a homeomorphism $f$ sending $f(x_i) = y_i$. This phrasing is probably more amenable to a literature search.

2:47 AM

@ForeverMozart As mentioned in comments, I'm basically lifting the answer from an old question of mine.

@ThomasAndrews: Looks like we had the same idea.

It's mostly unrelated but still cute that there's a classification of manifolds with an effective 2-transitive Lie group action; they're spheres or projective spaces $\Bbb{RP}^k, \Bbb{CP}^k, \Bbb{HP}^k, \Bbb{OP}^2$.

Is it really equivalent? What if removing p still lets you send any $x_1$ to any $x_2$ but without always sending the neighborhoods of $p$ to neighborhoods of $p$? @MikeMiller I think $2$-transitive is stronger, but I'm not sure of that.

@ThomasAndrews: You've got me.

I blame the cold.

I can't think of an example, but it's actually hard to think of, say, Euclidean spaces that are homogeneous but not $2$-transitive.

I still haven't found one.

(Connected, of course...)

@ThomasAndrews: Do you agree that my 2-homogeneous is the same as what you mean by 2-transitive?

2:54 AM

That's the definition of $2$-transitive, so yes. :)

It is true that any connected maniold is n-transitive for all n, as long as it's of dimension at least 2. The proof is to show that the action of the group of compactly supported homeo/diffeomorphisms is transitive; such a *eomorphism extends to the whole thing.

And to do that, you draw a path from $x_1$ to $y_1$, pick a small (precompact) neighborhood of the path, and show that there's a homeomorphism taking two points on the interior of the ball to one another whilst keeping the boundary fixed.

3:08 AM

i've defined some interesting stuff

now I have to see it it does what I want

What is a super informal description of Higman's Lemma.

Like given the alphabet a through z, we may concatenate these letters together to form words and thus we are given the dictionary. However, if all of these letters are related in a certain way, then all words part of the dictionary are related in the same fashion as the letters that make them.

^ that is my attempt but I think that others can tweak it or come up with better.

I am open for your attempts and suggestions. :)

Here is Higman's Lemma if no one knows what it is: en.m.wikipedia.org/wiki/Higman%27s_lemma

3:51 AM

hi chat

h

how're you feeling? i remember you saying you weren't feeling well earlier @MikeMiller

Still hopped up on cough drops

Hello lads

4:16 AM

I'm trying to make sense of the following proof that $\displaystyle \text{lcm}(m, n) = \prod_{i}p_i^{\max(e_i, f_i)}$

Proof: if $p_i^{e_i} \| m$ and $p_i^{f_i} \parallel n$ and $p_i^x \parallel \text{lcm} (m,n)$ then $x \ge \max \{ e_i,f_i \}$ since if not, $x < \max \{ e_i,f_i \}$ then either $m \nmid \text{lcm} (m,n)$ or $n \nmid \text{lcm} (m,n)$. Thus, $x \ge \max \{ e_i,f_i \}$. Since we need $\text{lcm} (m,n)$ to be as small as we can, so $x= \max \{ e_i,f_i \}$.

Is it correct? Why can't we have $x > \max(e_i, f_i)$? The role of $x$ here confuses me a bit.

the way i'd read that is that you can perhaps find $x$ which exceed that maximum, but they won't be suitable for the lcm

though i'm presuming that the double bars means 'divides', which I perhaps shouldn't?

4:37 AM

@Semiclassical That makes sense. Yeah, the double bars are meant to be divides.

okay. wasn't sure since you used a single bar with a slash for 'not divides'

Does the general solution to an nth order linear differential equation capture all solutions?

I just assumed it was meant to be the plural of divides. xD I've never seen it used before.

@user276387 though I'm a bit perplexed by that proof as well upon considering an example---say, with $(m,n)=(12,8)$ and $lcm(m,n)=24$

we have $2^2| 12$, $2^3|8$, so $2^3|24$. nothing wrong there

but what's wrong with $2^3|24$?

only thing that i could see is that $a\|b$ means "a is divisible by b", which is the opposite of the notation as I'd remembered it

Nothing, since it's true.

4:45 AM

wrong in the sense of being less than $\max\{3,4\}$, though

Ah, I see.

Does it work when you read it the opposite?

you mean, as $a$ is divisible by $b$?

in that case, i think so

Yeah. Why would anyone use that notation! Madness! xD

@Semiclassical thanks for clarifying this.

if one reads it in that way, one has (as an example) $8$ having $2^3$ as a factor, $12$ having $2^2$, and then the lcm indeed must have $2^3$ as a factor since otherwise there's no way for the factors of two to work.

Yeah, that makes sense.

4:52 AM

so i'd suggest looking at your source and seeing how they define that symbol

Perhaps I need to invest in a good book on elementary number theory, instead of learning it from online sources.

5:06 AM

What does the arrow symbol mean?

5:28 AM

Does there exists metric spaces such that given any sequence there will always exist a subsequence of the sequence which converges to a point in the space. I know that a compact metric space has this property ... Are there spaces which are not compact but having this property?

I figured out that to have this property and infinite set in that metric space has to have a limit point. So is a compact metric space only such space having this property

holas

3 hours later…

8:22 AM

Hey math whizzes, would somebody mind checking if my question makes sense?
http://math.stackexchange.com/questions/1642767/determine-equation-of-a-continuous-function-by-value-at-axis-of-symmetry-and-are

8:46 AM

@Lovsovs?

3 hours later…

11:46 AM

Hello, I've been wondering about a problem for a while and can't find a good solution. It's too complex to google, unfortunately.
I want to find the amount of integers in a range that are (not) divisible by any prime number in a list.
A common example is to find the number of integers divisible by either 3 or 5 in the range [0,10000). The solution is the number of integers divisible by 3 + number of integers divisible by 5 - number of integers divisible by 15 $= 3334 + 2000 - 667 = 4667$. (for non-divisible, $10000-4667=5333$)

11:58 AM

@robjohn Have you an idea: math.stackexchange.com/questions/1642354/…

Okay, I've just realized this is the number of integers in a range (not) coprime to the product of the primes.

Just found this: math.stackexchange.com/a/218911/218455 - seems to have the same issue, however mentions the euler function. I'm not familiar with it unfortunately

Okay, amazing. I found the solution

Unfortunately, I didn't, the euler function is the number of integers coprime to $n$, less than or equal to $n$, not any other number.

2 hours later…

Mithlesh Upadhayay

2:03 PM

Hello everyone, would you like to check this question

http://math.stackexchange.com/questions/1643149/new-coordinates-after-clockwise-rotation-of-triangle

1 hour later…

3:29 PM

@robjohn hey. Is the calculation of the following series obvious to you $$\sum_{n=1}^{\infty} \frac{\displaystyle \binom{2n}{n}}{4^n n^5}$$?

Or, $$\sum_{n=1}^{\infty} \frac{\displaystyle \binom{2n}{n}}{4^n n^6}$$

@robjohn they are not hard, but I discovered new ways of calculating them I think.

I'm out for some hours

4:24 PM

Hi @L33ter.

4:52 PM

@BalarkaSen: you around?

@Huy I am.

@BalarkaSen: I have a silly question about something I thought was obvious but now that I'm thinking about it again I'm having doubts

OK?

@BalarkaSen: given an oriented closed curve $\alpha$ on some surface, let $\alpha_0 \in \alpha$ be some point. Let $x_0 \in S$ be a different point and assume path-connectedness, so obviously $\pi_1(S, x_0) \cong \pi_1(S, \alpha_0)$ by choosing some path $\gamma_0$ going from $x_0$ to $\alpha_0$ and applying conjugation.

Yep.

5:04 PM

this works for any path $\gamma_0$ we choose, and of course if we choose a different $\alpha_0' \in \alpha$, the same holds

Sure.

now if I want to "jump over" to the definition of a free homotopy class, why is this exactly the same as a conjugacy class in $\pi_1(S)$? the latter sounds like a set $\{hgh^{-1}: h \in \pi_1(S,x_0)\}$ for a fixed $g$. but these $h$ are all closed curves with the basepoint $x_0$. instead, I would have only allowed conjugation with paths starting somewhere on $\alpha$, but I think this would yield something different.

Right, so the surface thing is irrelevant here. It is true that $[S^1, X]$ correspond bijectively to conjugacy classes of $\pi_1(X, x_0)$ for whatever path connected $X$ you want. This can be done as follows.

You have the natural map $\pi_1(X, x_0) \to [S^1, X]$. Just forget the marking at the basepoint.

what precisely is $[S^1, X]$?

Free homotopy classes of maps $S^1 \to X$.

5:11 PM

ah, ok

OK, (1) The above thing is well defined (why?).

(2) It's surjective. For any $S^1 \to X$, choose basepoint to be image of $0 \in S^1$. Then join with $x_0$ by some fixed path. The thing you get is of course freely homotopic to the original $S^1 \to X$, and is also based at $x_0$.

@Huy On a CV, do grad school students usually have coursework on there?

@Clarinetist: define coursework

@Huy What classes they've taken in grad school

@Clarinetist: never seen anyone put that on their CV

5:14 PM

K, thanks

@Clarinetist: ask Mike and Ted maybe, they'll know better

Yeah, I'll see if they're on sometime and ask

@BalarkaSen: I agree

(3) It's injective. Assume $\gamma : S^1 \to X$, $\gamma' : S^1 \to X$ based at $x_0$ are freely homotopic by $\gamma_t : S^1 \to X$'s. We want to show they are homotopic rel basepoint $x_0$.

@BalarkaSen: how can it be injective if you're claiming that $[S^1, X]$ is bijective to the set of conjugacy classes of $\pi_1(X, x_0)$?

5:18 PM

I am being sloppy, sorry. What I meant was that they are homotopic upto conjugate. Let me correct myself. $\gamma, \gamma' : S^1 \to X$ are loops based at $x_0, x_1$, freely homotopic through $\gamma_t$'s.

I want to show there exists $p$ path from $x_0$ to $x_1$ such that $\gamma, p \cdot \gamma' \cdot p^{-1}$ are homotopic. Agree?

Let S and T be nonempty sets of real numbers and define S+T={s+t|s e S, teT}. Show that sup(S+T)=supS+supT if S and T are bounded above and inf(S+t)=infS+infT . hmmm

Ignore my previous message please (i.e., the one which said homotopic and didn't say conjugate).

which one precisely? :D

just delete whatever I'm supposed to ignore

or edit, I have time (unless you don't)

I can't. It's the one marked by (3).

ah, ok

5:22 PM

So what I am really proving there is that the original map above, $F : \pi_1(X, x_0) \to [S^1, X]$ by forgetting basept satisfies $F([\gamma]) = F([\gamma']) \implies \gamma \simeq p \cdot \gamma' \cdot p^{-1}$ for some path $p$ from $x_0$ to itself.

yes, I agree

Can you guess $p$? $\gamma, \gamma' : S^1 \to X$ are loops based at $x_0$, with free homotopy $\gamma_t$ between them (i.e., which does not fix $x_0$). What is the most obvious choice for $p$?

Draw a picture.

If needed, that is.

Assume for the sake of simplicity that $\gamma(0) = x_0$ and $\gamma'(0) = x_0$. I.e., it's the image of $0$. Keep in mind that $\gamma_t$ need not send $0$ to $x_0$ for all $t$. So...

Keyword: "slide".

@BalarkaSen: sorry, you confused me a bit. above, $\gamma, \gamma'$ had different basepoints and now they suddenly have the same. I think the former is the correct version, right?

The latter is a special case of the former. The former is correct, the latter is correct.

Work with the former if you want to.

but the latter is obvious by taking the identity

because by assumption the paths are homotopic

5:28 PM

Um?

ah I need a homotopy with fixed basepoint?

Yeah.

You have a free homotopy. You want a homotopy between conjugates.

Write out what you want to prove if you're feeling confused.

I'm looking at my sketch now, trying to figure it out, just a second

No probs, take your time.

@BalarkaSen: let's start with the case that $\gamma, \gamma'$ have the same basepoint. without loss of generality, assume $\gamma_0(0)$ is precisely that basepoint. I'm guessing that $p = \gamma_t(0)$ will be the answer, right?

5:37 PM

:)

Indeed.

ok, let me try to understand the picture a bit better

I'm not quite convinced yet but now I know I'm not trying to convince myself of something that's wrong

Right.

The picture is better if you look at the case when gamma, gamma' have different basept, btw.

all the more reason to consider the picture with identical basepts. :D

Why? $x_0 = x_1$ is special case of arbitrary $x_0, x_1$.

If you prove it there, you prove it here.

@BalarkaSen: $\gamma_t(s) - \gamma_t(0)$ is the homotopy with the fixed point, right?

urm

wait

5:43 PM

I am not sure what you're referring to, but you're free to think. I am here.

@Huy Just start with the different basept case. $\gamma_0, \gamma_1 : S^1 \to X$ two loops, $\gamma_0(0) = x_0$ and $\gamma_1(0) = x_1$ basepoint. $\gamma_t$ is a free homotopy between them. You conjectured that $\gamma_0$ and $p \cdot \gamma_1 \cdot p^{-1}$ are homotopic fixing $x_0$, where $p:[0, 1] \to X$ is the path $p(t) = \gamma_t(0)$. Can you visualize how the homotopy should go?

@BalarkaSen: yea, don't worry, I'll ask if I can't get further, I'm just a slow thinker

I think you are pretty fast.

It'd have taken me more time to come up with $p$.

But think along :) I'll lurk here if you need me.

@BalarkaSen: ok, so basepoints $x_0, x_1$. say $\gamma_0(0) = x_0$, $\gamma_1(0) = x_1$, so $p = \gamma_t(0)$ is a path from $x_0$ to $x_1$, so obviously $p \gamma ' p^{-1}$ is a path based at $x_0$. I'm losing my aim, I want to show that this is now homotopic to $\gamma$, right?

Yes, you want to show $p \gamma_1 p^{-1}$ is homotopic to $\gamma_0$ fixing $x_0$.

Both are loops at $x_0$. We want a homotopy fixing $x_0$.

You already know there is a free homotopy between $\gamma_0$, $\gamma_1$. Use that. And remember "slide".

@BalarkaSen how r you

you have been not here lately

5:59 PM

Yeah, I had exams.

hi @TedShifrin

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