Mathematics - 2014-11-19 (page 2 of 4)

robjohn

08:02

@RandomVariable would you rather I contact a CM or dev?

Random Variable

@robjohn Yes, I think I would prefer that. Thanks.

I mean I'd prefer if you could contact someone.

Random Variable

08:14

@robjohn It's not just the upvotes. I just received 2 badges for that answer.

robjohn

@RandomVariable okay, I talked to a CM and they don't think this is something to worry about. You gave an answer, some people found it useful. So be it. Forget it and move on to other answers. As I said, the person who answered the old question would not have gotten the points you've gotten had you not posted your answer.

GBeau

how might it be proven that $\gcd(\gcd(a_1,\cdots ,a_{n-1}),a_n)=\gcd(a_1,\cdots ,a_n)$

robjohn

@RandomVariable who cares about extra badges? don't sweat this. It is only some points. You probably didn't get some votes or badges that you did deserve on other answers.

It all averages out :-)

@RandomVariable Besides, you didn't know about the other answer when you answered this question. No real foul has been committed.

Random Variable

@robjohn I know this really isn't a big deal. But it's going to keep bothering me.

Twink

¬¬

robjohn

08:29

@RandomVariable The only thing I can do is change the answer to CW and the CMs don't want to do anything more. So either convince the OP to unaccept your answer, or just try your best to forget about it.

Twink

@RandomVariable or convience people to downvote you

Random Variable

@robjohn Then I guess change it to a CW.

Jasper Loy

@RandomVariable I hope your being bothered will pass soon. I am disturbed by all kinds of disturbing thoughts all day long.

Twink

@JasperLoy did you call your psychiatrist?

Random Variable

@JasperLoy Thanks.

robjohn

08:33

@RandomVariable There. Now move on.

Jasper Loy

@Twink No, I am fine on my own, they cannot help me anymore.

Twink

why not?

trust them

Jasper Loy

There is nothing they can teach me that I don't already know.

Twink

you're being arrogant

even psychiatrists may need to go to the psychiatrist

Random Variable

@robjohn Thank you.

Twink

08:37

it doesn't matter if they know everything @JasperLoy

@RandomVariable I love your profile pic, that cat is very cute.

Random Variable

@Twink It's Mooch from the comic strip Mutts.

Twink

08:52

I didn't know that comic strip

Random Variable

@robjohn I'm trying to find something interesting that I could answer to get my mind off of this. But I haven't found anything. I haven't been paying much attention to what's being posted on main for the past few weeks. I haven't been feeling well.

Twink

@RandomVariable do you have an OCD?

Random Variable

@Twink If I do, it's very selective.

Twink

@RandomVariable what do you mean?

Random Variable

09:09

@Twink I mean that if I do, it centers around very specific things.

Twink

@RandomVariable like upvotes on old answers?

Committing to a challenge

What drama occurred while I was on exam period? I am done btw

Gustavo Montano

If curvature is 0, what is the value of torsion?

Is torsion also equal to 0?

Random Variable

@Twink I used to not care at all about reputation. Then people started to say that they like the answers I post here and elsewhere. So I started to care about reputation. I want to go back to not caring.

Balarka Sen

@GBeau just use three variables first

GBeau

09:24

@BalarkaSen and how would you suggest that be proved?

Balarka Sen

assume that some $d$ is the gcd and use the definition of a gcd...

it's pretty straightforward

a cool-boy way to do it would be to compare gcds with intersections of multisets.

GBeau

@BalarkaSen I...am not a smart person

Balarka Sen

i'd let you think. (partially because i'm eating atm)

GBeau

09:40

@BalarkaSen the definition I had in mind is an intersection of multisets?

$\max (\mathrm{Div}(a_1) \cap \mathrm{Div}(a_2)\cap \cdots \cap \mathrm{Div}(a_n))$

as the gcd

Committing to a challenge

So apparently Chris's got temp banned? Is that the only relevant drama?

Twink

@Committingtoachallenge why was she banned?

Committing to a challenge

@Twink From what it seems, she swore at someone using GFY, and then claimed it meant something else(even though it was impossible under the context and even gramatically incorrect in her second 'apparent' form), oh and she has told me to f*** myself on this chat before in full, so I would believe it xD.

GBeau

@BalarkaSen oh...that's just obvious...I'm dumb

Twink

10:04

@Committingtoachallenge really? wow that girl gotta be crazy

GBeau

wait is th at even true in general?

if A_i are sets, then is

$\max (A_1\cap A_2\cap \cdots \cap A_{n+1})=\max(\max (A_1\cap A_2\cap \cdots \cap A_n),A_{n+1})$

can somebody explain how this could be true?

$\max (A_1\cap A_2\cap \cdots \cap A_{n+1})=\max(\max (A_1\cap A_2\cap \cdots \cap A_n)\cup A_{n+1})$

or is it max of both

$\max (A_1\cap A_2\cap \cdots \cap A_{n+1})\stackrel{?}{=}\max(\max (A_1\cap A_2\cap \cdots \cap A_n),\max(A_{n+1}))$

dies

brain broke

user2179021

10:24

hi

hi @BalarkaSen

@Committingtoachallenge what is GFY?

oh I got it :)

go find yourself

GBeau

Let's just go straight to the question: $$\max (\mathrm{Div}(a_1) \cap \mathrm{Div}(a_2)\cap \cdots \cap \mathrm{Div}(a_{n+1}))\stackrel{\text{why}}{=}\max (\mathrm{Div}(\max (\mathrm{Div}(a_1) \cap \mathrm{Div}(a_2)\cap \cdots \cap \mathrm{Div}(a_{n}))),\mathrm{Div}(A_{n+1}))$$

where Div is the set of divisors

and $a_i\in\mathbb Z$

Committing to a challenge

@user2179021 ;)

GBeau

does anyone have a hint I've been working on this for quite a while and this is the last piece I need for a proof

:(

Committing to a challenge

@GBeau What level of math is it? I don't recognise the notation, so I need to know if learning the notation will be sufficient to help you or not

GBeau

@Committingtoachallenge What notation don't you recognize?

it's abstract algebra

(introductory-level)

$\cap$ is the intersection of sets

I guess best written:

$$\max (\mathrm{Div}(a_1) \cap \mathrm{Div}(a_2)\cap \cdots \cap \mathrm{Div}(a_{n+1}))\stackrel{\text{why}}{=}\max (\mathrm{Div}(\max (\mathrm{Div}(a_1) \cap \mathrm{Div}(a_2)\cap \cdots \cap \mathrm{Div}(a_{n})))\cap \mathrm{Div}(A_{n+1}))$$

since the comma before is ambiguous

user2179021

10:40

what is Div and what are you generally talking about? :)

and is A_n+1 secretly a_n+1 ?

GBeau

yeah that was just a typo

as I mentioned above, Div is the set of divisors

user2179021

ok so what are a_i ?

GBeau

as I mentioned above, a_i are any integers

user2179021

basically, can you give an example?

so why is it abstract algebra?

seems to be elementary number theory

GBeau

what?

I didn't name the material

user2179021

10:43

"
it's abstract algebra

(introductory-level)"

Cortizol

@robjohn What is asymptotic formula for $\sum\limits_{n \leqslant x} \frac{1}{\ln n}$?

Committing to a challenge

What are the set of divisors? Prime factors?

user2179021

@GBeau what the divisors of 10? 1,2,5,10 ?

GBeau

@Committingtoachallenge $\mathrm{Div}(a_i) = \{d\in\mathbb N : d\mid a_i\}$

user2179021

so 1,2,5,10 ?

Committing to a challenge

10:45

So what is your question in english?

Why does the gcd of a bunch of numbers = the gcd of the gcds?

GBeau

is the greatest common divisor of a set of numbers equal to the greatest common divisor of a subset of those numbers without one element and the last element

it is, but the question is for a proof

@Committingtoachallenge proving $\gcd(a_1,\ldots,a_n,a_{n+1})=\gcd(\gcd(a_1,\ldots,a_n),a_{n+1})$

Committing to a challenge

Have you considered the unique prime factorisation?

I'll type a proof up, one sec

GBeau

@Committingtoachallenge soooo, the set of divisors without a_i and 1?

(with the prime ones)

?

Committing to a challenge

I am not sure what you mean with that comment, but I will prove $gcd(a_1,\dots,a_n,a_{n+1}) = gcd(gcd(a_1,\dots,a_n),a_{n+1})$ in a sec

GBeau

that's all I want

the set is just one of the effective definitions of gcd, so it must be true as well

but really I'm trying to prove the former

I thought there might be a reason the set intersection defintion was true with a logic argument and I was being dumb

user2179021

10:58

the collective math brain of math.SE is not doing so well with my question math.stackexchange.com/questions/1017169/…

:(

Balarka Sen

er hi, @user2179021

user2179021

hi

Balarka Sen

@MikeMiller I proved it. It's just rotation. $x \stackrel{H(\bullet, t)}{\longrightarrow} e^{2\pi t}x$ establishes a homotopy.

Committing to a challenge

@GBeau meta.math.stackexchange.com/questions/4666/…

user2179021

@BalarkaSen please excuse the spam but I was wondering if math.stackexchange.com/questions/1017169/… interested you?

Balarka Sen

11:03

it doesn't in the least, @user2179021

user2179021

@BalarkaSen thanks for the clear answer :)

@BalarkaSen can I ask why? It doesn't seem to interest anyone sadly

but it is a math question, right?

Balarka Sen

for me, i am not interested in linear algebra. i don't see how you can say it doesn't interest anyone seeing that your question has 10 upvotes already.

user2179021

@BalarkaSen oh there is no linear algebra there!. I should remove that tag

Balarka Sen

well in particular i am not interested in matrices.

Vincenzo Oliva

Speaking of which, why is it that my latest questions related to my paper (or at the very least, to superabundant numbers) are quite ignored?
http://math.stackexchange.com/questions/1028868/on-asymptotically-equal-functions

GBeau

11:06

@Committingtoachallenge I see the way: $gcd(a_1,\dots,a_n,a_{n+1}) = gcd(gcd(a_1,\dots,a_n),a_{n+1})$, ὅπερ ἔδει δεῖξαι

Committing to a challenge

So you are good?

GBeau

@Committingtoachallenge Yes :)

that is a much easier way to prove than what I was trying

Committing to a challenge

Glad to have helped(assuming it was me xD)

Balarka Sen

LOL @AlexanderGruber the new gravatar is hilarious.

Vincenzo Oliva

@BalarkaSen You seemed interested back then, you might want to give it a look? I'd be very grateful, but no worries anyway

Committing to a challenge

11:09

I shall go to sleep now, girlfriend still has exams to do(9:09PM). Goodnight

Balarka Sen

@VincenzoOliva Well, superabundant numbers do not appeal to me much. I was interested in your proof(correct?) of RH.

user2179021

@BalarkaSen ok.. are you interested in coin weighing puzzleS?

:)

or polynomials or fourier analysis?

Balarka Sen

polynomials, sure.

i am interested in mainstream analytic number theory and algebra, alongside.

Twink

I wanna cry

:'(

Vincenzo Oliva

@BalarkaSen If the lemma in the body of the question is correct, the proof itself most likely is

The limit equality at the bottom is the onpy thing remaining of what caught the attention of who read the paper.

Balarka Sen

11:19

It's undoubtedly true.

Vincenzo Oliva

@BalarkaSen ?

user2179021

@BalarkaSen ok so my question can be entirely rephrased in terms of polynomial multiplication if it helps

Balarka Sen

@user2179021 man, i already said i am not interested in your question whatever it is. why don't you link it here and wait for someone to respond than spam?

@VincenzoOliva (1) is true.

user2179021

@BalarkaSen Thanks. Sorry I wasn't trying to be a pain. I suppose I took the view that if people are hanging out here they can't be too busy :)

3

Balarka Sen

i am not busy, just not interested

user2179021

11:22

sure

I suspect a problem with my question is that it's not clear which part of math is relevant

so specialists aren't looking at it

Balarka Sen

well it has a lot of upvotes

someone will look eventually

user2179021

thanks.. I hope so. I am seeing a lot of tumble weed recently. The related question math.stackexchange.com/questions/1017402/… has no answers either

it would be interesting to see a plot of time to accepted answer

i.e. do you either get an answer in a few days or never?

Vincenzo Oliva

@BalarkaSen Hooray, I just was looking for someone else to agree with my thought on it

user2179021

I also like math.stackexchange.com/questions/1023351/… which has no answers

Vincenzo Oliva

@BalarkaSen Hooray, I just was looking for someone else to agree with my thought on it

user2179021

11:24

I can't believe there aren't any probabilists around !

Vincenzo Oliva

@BalarkaSen Originally I wasn't even planning to use that lemma, I thought it was obvious, but I guess it's better to use it

Balarka Sen

@VincenzoOliva It can be proved as follows : $g(m) \sim f(m)$ implies $g(m) = f(m) + o(f(m))$ and thus $$\frac{\prod_{n \leq g(m)}f(n)}{\prod_{n \leq h(m)} f(n)} = \frac{\prod_{n \leq f(m) + o(f(m))}f(n)}{\prod_{n \leq h(m)} f(n)} = \prod_{n \leq o(f(m))}f(n)$$ and the rightmost guy vanishes as $m \to \infty$.

@VincenzoOliva I am still not convinced about a proof of RH though :)

UserX

Any chemistry lover?

Balarka Sen

erm, replace some $f$ by $h$s above. typo.

Vincenzo Oliva

11:42

@BalarkaSen Dò you think yours is a more rigorosi and/or straightornare prof than mine? Both I cannot properly read LaTex from my mobile and I haven't deals with o-notation yet, lal

Sure you aren't! :D And you haven't Even read the pa per, would you like to?

Balarka Sen

i believe mine is more rigorous, yes.

Vincenzo Oliva

*straightforward

I see

Can I use yours then, and acknowledge you?

Balarka Sen

no need to acknowledge. mine is essentially what you did, with a bit clearer notations.

Vincenzo Oliva

Fair enough. Willing to read the whole thing?

Balarka Sen

only if your are willing to post it publicly.

Vincenzo Oliva

11:47

I have submitted it to viXra before arXiv precisely for this reason

Balarka Sen

ok

link?

Vincenzo Oliva

Just a moment

(My mobile is going crazy, I had posted it but it seems it's not here)

http://vixra.org/abs/1411.0206

Version 5 will have Lemma 2

Now I have to go, I'll be back in an hour or so

Balarka Sen

apparently that uses a lot of results on superabundant numbers i am unfamiliar with

user2179021

12:15

I have an order of precedence question... are all the brackets needed in 4^n*(2*n)!/(n!)^2 ?

Integrator

@N3buchadnezzar Deleted my answer :p

robjohn

@Cortizol It goes as $\frac{x}{\log(x)}$, but unless you use $\mathrm{li}(x)$, all of your early terms go to approximating $\mathrm{li}(x)$ instead of getting to smaller terms. $\mathrm{li}(x)\sim\frac{x}{\log(x)}\left(1+\frac1{\log(x)}+\frac2{\log(x)^2} +\frac6{\log(x)^3}+\frac{24}{\log(x)^4}+\dots\right)$

Integrator

@robjohn Finally removed my answer!

robjohn

@Integrator the OP finally unaccepted your answer?

Integrator

@robjohn I guess so!

user2179021

12:29

it's annoying when an incorrect answer sits on a question math.stackexchange.com/questions/1017169/…

robjohn

@Integrator Yep... they unaccepted it about 48 minutes ago

Integrator

@robjohn she!

@robjohn Do you think something's wrong with Calculating $i^i$

robjohn

@Integrator Do you see something wrong? It looks okay to me.

I would have approached it slightly differently, but it is essentially the same.

Integrator

Have you seen @Did 's comment under Question? I'm unable to understand that!

@robjohn What? Complex logarithms?

Cortizol

12:44

@robjohn Thank you.

robjohn

@Integrator For complex numbers $z^w$ is defined as $e^{w\log(z)}$ where $\log(z)$ requires a branch cut to be well-defined; that is, it can't be well-defined and analytically on all of $\mathbb{C}$.

For a particular branch cut, the value of $\log(z)$ needs to be defined at one point since for any choice of $\log(z)$, $\xi=\log(z)$ and $\xi=\log(z)+2\pi i$ satisfy the equation $e^\xi=z$.

What Did is saying is that $z^w$ cannot be defined unambiguously on all of $\mathbb{C}$ unless you introduce multiple-valued functions, which are problematic in themselves.

Integrator

@robjohn that means I cannot say $$i^i=e^{-(4k+1)\pi/2}$$

robjohn

@Integrator strictly, no. What needs to be said to be completely rigorous is that depending on the branch cut (part of $\mathbb{C}$ excluded) and branch of log (what value $\log(z_0)$ has for some $z_0$ not in the branch cut), there is a $k\in\mathbb{Z}$ so that $i^i=e^{-(4k+1)\pi/2}$. Using the most standard branch cut, $i^i=e^{-\pi/2}$

But most people would look at what you've written and, not being overly pedantic, would say that you were okay.

UserX

13:00

Hey @Integrator @robjohn Chris'ssis left me with an integral I couldn't solve elementarily, and now she's gone. Will you help me? $\displaystyle\int_0^{\infty} \frac{x}{\sqrt{e^x-1}}\mathrm{d}x$

robjohn

Most people could pull out the essence of what you've written and understand what is true.

Integrator

@robjohn So, should I edit that answer and add depending on the branch cut (part of $\mathbb{C}$) and branch of log (what value $\log(1)$ has), there is a $k\in\mathbb{Z}$ so that $i^i=e^{-(4k+1)\pi/2}$. Using the most standard branch cut, $i^i=e^{-\pi/2}$

@UserX What's in a name? That which we call a rose. By any other name would smell as sweet

robjohn

Well, Did might object to $\log(1)$ since $1$ could be in the branch cut...

UserX

@Integrator ?

robjohn

@Integrator I've modified my statement so that it should be okay, even for Did.

Integrator

13:07

@robjohn Thanks I'll add it as a note.

Wrzlprmft

Hello, can anybody tell me what the name of the following theorem is: If $a$ and $b$ are coprime, then $ra \bmod b \neq sa \bmod b$ for all $r \neq s$; $r,s < b$.

Integrator

@Wrzlprmft first tell me how to pronounce your name!

Wrzlprmft

@Integrator The first r, the l and the m are vocalic.

(apart from it, read as written)

Vincenzo Oliva

@BalarkaSen The references are there for a reason. Also, I might well link you what you need, but if you're not interested, no problem.

Wrzlprmft

@Integrator In IPA: [vʀ̩ʦl̩pʀm̩ft]

Venus

13:21

@robjohn @DanielFischer I wanna ask, according to Ramanujan, if
$$f(x)=\sqrt{ab+(a+c)^2+b\sqrt{ab+(a+c)^2+(b+c)\sqrt{ab+(a+c)^2+(b+2c)\sqrt{ab+\cdots}}}}$$
then
$$f(x)=a+b+c$$
How does one proof this one?

user2179021

nice :)

Venus

I meant 'prove'. Sorry, bad grammar.

user91374

PLEASE WORK OUT THIS QUESTION... PLEASE
http://math.stackexchange.com/questions/984216/condition-for-trigonometric-inequality

user2179021

what does !! mean as in x!! ?

is it the same as (x!)! ?

Wrzlprmft

@user91374 You know that you can set a bounty on it?

user91374

13:32

yes.

Venus

@user2179021 No, they are different. See this: mathworld.wolfram.com/DoubleFactorial.html

Wrzlprmft

@Integrator Do you (or anybody else) have an answer to my question or do I need to post it as an actual question?

Venus

@robjohn @DanielFischer Also this one, is it possible the following problem has a unique solution
$$x=\sqrt{i+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+\sqrt{i^6+\cdots}}}}}}$$
where $i=\sqrt{-1}$.

robjohn

@UserX I get $2\pi\log(2)$

UserX

@robjohn me too numerically. Did you get it analytically? How?

robjohn

13:38

@UserX First analytically, then verified by Mathematica, which can do it symbolically

Let me write up my approach

UserX

I got bio class, can you tag me so I can see it later or post it on mathb.in?

user2179021

@Venus thanks

Vincenzo Oliva

@BalarkaSen Also, in your proof you always wrote $\le$, whereas you were supposed to prove (1) in which the numerator has $\le$ but the denominator $<$. As I'm unfamiliar with o-notation, I can't tell if your proof is still correct. (Plus, I assume $o(h(m)) \to 0$ as $m \to \infty$, so that the rightmost product becomes 1, but is that valid for all $h$? )

Venus

13:57

@UserX Setting $y^2=e^x-1$ then $x=\ln(1+y^2)$, we get
$$\int_0^{\infty} \frac{x}{\sqrt{e^x-1}}\mathrm{d}x=2\int_0^{\infty} \frac{\ln(1+y^2)}{1+y^2}\mathrm{d}y$$
From this should be easy.

@UserX The latter expression is solved here: math.stackexchange.com/q/358386/146687

robjohn

14:10

Start by substituting $x\mapsto\log(1+x)$:
$$
\begin{align}
\int_0^\infty\frac{x}{\sqrt{e^x-1}}\mathrm{d}x
&=\int_0^\infty\frac{\log(1+x)}{\sqrt{x}}\frac{\mathrm{d}x}{1+x}\\
&=\left[-\frac{\mathrm{d}}{\mathrm{d}\alpha}\int_0^\infty x^{-1/2}(1+x)^{-\alpha}\,\mathrm{d}x\right]_{\alpha=1}\\
&=\left[-\frac{\mathrm{d}}{\mathrm{d}\alpha}\frac{\Gamma(1/2)\Gamma(\alpha-1/2)}{\Gamma(\alpha)}\right]_{\alpha=1}\\
&=-\frac{\Gamma(1/2)\Gamma'(1/2)\Gamma(1)-\Gamma(1/2)\Gamma(1/2)\Gamma'(1)}{\Gamma(1)^2}\\
&=-\frac{\sqrt\pi\sqrt\pi(-2\log(2)-\gamma)-\sqrt\pi\sqrt\pi(-\gamma)}{1^2}\\

Random Variable

14:28

Hello @Venus

Venus

Hi @RandomVariable

user2179021

can anyone with lots of rep remove the linear-algebra tag from math.stackexchange.com/questions/1017169/…

the question has nothing to do with linear-algebra

Wrzlprmft

@user2179021 Can’t you suggest that as an edit?

user2179021

@Wrzlprmft I did but it was cancelled

I think they ignore low rep people :)

Nick

I was doing a diff eq when I realized something

Doesn't the fact that you've divided by $x$ mean that $x \neq 0$ ?

user2179021

14:32

it does!

Nick

Yeah, but people rarely mention it.

user2179021

well.. in my experience every time you divide by x you always look at what happens if x is zero

Nick

Infact for $\frac{\mathrm dy}{\mathrm dx}$ to exist doesn't technically mean x can't be zero. It just means x can't be some constant. Right?

user2179021

oh sorry

I thought you meant something else

can you write down an explicit equation?

Nick

$$(2x+y)\frac{dy}{dx} + x = 0$$

Integrator

14:35

@user2179021 oops, I've approved edit, instead of improving it! It still needs one approve vote!

Nick

Easy homogeneous order 1, deg 1 diff. eq.

Random Variable

@Venus I saw that you recently posted that integral that sos440 posted on the other forum. I wonder how sos440 approached it because I think he was saying that he had an evaluation.

user2179021

@Integrator thanks :)

@Integrator of course if you wanted to answer it too that would be even better :)

Nick

Putting $y = vx$, $$(2x+vx)(v'x+v) + x = 0 \\ \stackrel{\div x}{\implies}(2+v)(v'x+v)+1 = 0$$

Now, it isn't until you divide by that $x$ are you saying that $x \neq 0$. You're constricting the domain of $x$ to get the solution of the HDE

Eh, maybe my question is some naive stupidity (ah yes, another one)

Venus

@RandomVariable Yes, I posted his problem but without his permission. I got lots of points & two badges because of posting it. Hehe

I am sure he has an answer but maybe he doesn't want to post it because he doesn't want to get any credits of his own problem. Did you ask him about this?

Nick

14:40

@user2179021 See, we were talking about the same thing alright :D

user2179021

@Integrator now it has got confused

@Integrator oh.. maybe it is just waiting for another vote.. my mistsake

Random Variable

@Venus No, I haven't asked him. I was aware of that theorem, but I didn't know it was applicable.

Venus

15:02

@RandomVariable I've never known that theorem until Oloa posted his brilliant answer and achille hui gave mathematical foundation of it. Anyway, I love your posts here and I&S. Very good & nice works!

Nick

@JasperLoy: Greetings conquistador Jasper! =)

Jasper Loy

@Nick Hi, I am a bit troubled these few days.

Nick

@JasperLoy What's the trouble troubling your troubles, are you in some troublesome trouble?

Random Variable

@Venus Thanks. I'm trying to dissociate myself from one of my answers on here because it turned out to be a duplicate and I don't want the upvotes or the 2 badges.

Jasper Loy

@Nick Well, I won't mention it here. I hope I will overcome them all one day.

user2179021

15:08

@RandomVariable given your name, you seem not to care much about probability :)

Nick

@JasperLoy We shall overcome, one day!

Jasper Loy

@Nick Sorry, don't speak in foreign tongues.

Teddy

hey

could anybody help me in algebraic topology?

Random Variable

@user2179021 I studied stats in college. But I've been fixated on other things since then.

Venus

@RandomVariable You're so humble :-)

Mike Miller

15:13

What's your question, @Teddy?

Teddy

i have to do the same homework math.stackexchange.com/questions/1027392/…

i think the author of this question is in the same course, but i dont know who it is

Mike Miller

The hint given in the answer is very nice.

Nick

@JasperLoy That song has been translated into so many languages. A dyslexic muslim man from my country who barely knew any English went to the US for employment. One day, he broke down crying and, unable to find a mosque, he went into a church where he sang this song along with the church gospel. he didn't know English but he still knew this song =)

Teddy

yes i agree, and i understand it more or less

Mike Miller

@Teddy A counterexample if you don't consider rational coefficients: $S^2 \to \Bbb{RP}^2$.

Teddy

15:16

thank you. i have to wah me dishes. oh no

my*

@mike miller ok thank you i will think about it. where in this proof do you need that the coefficients must be rational, do you know it?

Mike Miller

@Teddy To get rid of the $n$. You can't divide with integer coefficients.

Teddy

ahh, ok thank you very very much!

Nick

@MikeMiller: I'm currently facing a problem. I have a table with two columns, let's say $x$ and $y$, the $x$ part is what I put into a machine (I control these values) and the $y$ is what the machine spits out. I've prepared a table of $100$ such values and what I can observe is that as $x$ increases, $y$ also increases. How do I prove $y \propto x$ besides saying it's obvious from the table. I want mathsy way of saying it.

Hippalectryon

@Nick Hullo !!

Jasper Loy

@Nick That means $y=kx$.

Teddy

15:24

this helps me a lot

Mike Miller

@Nick You don't, unless you know a formula. Who's to say the machine won't start spitting out bizarre numbers later?

Hippalectryon

Guys

axisthegame.com

Try it, love it, star it !!

Mike Miller

meh

Teddy

i wish to be more intelligent:/

is is frustrating

Nick

@MikeMiller Okay, ... I've made a recursive formula: $$x_n = x_{n-1} + 1 \iff y_n = y_{n-1} + 6$$

Mike Miller

15:33

I don't think you mean iff there... perhaps you mean and?

Teddy

hahaha

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today is the world toilet day :D

Mike Miller

It would be more reasonable to just write that as $f(n)=f(n-1)+6$. Anyway, from this formula and the value of $f(0)$, you can calculate $f(n)$. Ideas how?

Nick

@JasperLoy Oh, so you're saying I should find $k = \frac{y}{x}$ and determine it is constant. Brilliant. I know $\frac{\Delta y}{\Delta x} = \text{constant}$ .... Mhh, maybe I can plot a graph and find $\frac{dy}{dx}$ which is a constant and integrate that equation to get $y = kx$. How do I prove every $\frac{dy}{dx}$ on my graph is constant without a general equation for it?

Jasper Loy

@Nick I just mean to say that that is what proportionality means. If one increases as the other increases, they might not be proportional.

Mike Miller

Your data is a set of discrete points. It doesn't make sense to take its derivative; it could come from many different continuous functions.

Nick

15:35

@MikeMiller I meant to imply correspondence meaning $f(x_n) = y_n$ in a way. Okay, and if it floats your boat =)

Mike Miller

@Nick yes, but you may as well define $x_n = n$.

At the very least, $x_n = n+x_0$.

@Nick The reason I object to iff is that, well, what if the first thing isn't true? what if $x_n = 17$ for all $n$? Then the latter thing isn't true and we know nothing! It seems to me you want to assume both.

Nick

@JasperLoy Oh yeah, whoops, that means I need to prove $y \propto \frac{1}{x}$ That's cool. My question about the tangent is still is open.

@MikeMiller Oh! yes, I see. Also, "use functions to write this sort of thing next time". I get the message ;D

user130018

@JasperLoy @Finn @Bal

Random Variable

@Venus I'd like to delete it. But I can't because it's an accepted answer. If I were to delete it I think I would keep the rep but at least lose the badges.

Jasper Loy

@user130018 Hello Bart.

Mike Miller

15:42

@Nick Now you lost me.

Balarka Sen

@MikeMiller I need your help for a bit.

Nick

@MikeMiller Ah it's okay. You've helped me out enough :D Thank you so so very much :D I'll use what I know about monotonicity to extract more information.

@MikeMiller I can graph it and smooth it out. Then apply calculus! ;D

@Hippalectryon Hoola Hoop Hello to you too :D

Balarka Sen

I am trying to prove that $\pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y)$. My approach is to take two loops $\sigma_1$ and $\sigma_2$ in $X$ and $Y$ resp and map them to $(\sigma_1, \sigma_2)$ in $X \times Y$. I am having troubles showing this is preserves path homotopy. Let $\sigma_1 \sim_p \sigma_1'$ in $X$ and $\sigma_2 \sim_p \sigma_2'$ in $Y$.

Nick

@MikeMiller yes, I wanted to assume both =)

@BalarkaSen: Heya Balsensai :D

Balarka Sen

Then you have homotopies $H_1 : [0, 1]^2 \to X$ and $H_2 : [0, 1]^2 \to Y$ which maps the two side edges of the square to base points in X and Y resp and the upper and lower edges of the square to $\sigma_i$ and $\sigma_i'$ resp

Nick

15:48

@user130018 Hola El-Barto!

Balarka Sen

My initial idea was to take $H = (H_1, H_2)$ but that has domain $[0, 1]^4$. Grmph.

@Mike^

Mike do you even help without being unhelpful?

Venus

@RandomVariable Which problem? I think it shouldn't be deleted. Just let it be.

Nick

@BalarkaSen If one were to be entirely helpful while helping, how helpful would that be?

Matt Rockwell

anyone know anything about this? math.stackexchange.com/q/1028206/193819

robjohn

@BalarkaSen That sounds pretty unfriendly... What did Mike do?

Nick

15:54

@robjohn Ermagash, you sound like a concerned parent XD

Balarka Sen

Oct 23 at 22:31, by Mike Miller

I think I can get Balarka to solve most problems just by giving him a hard time and being completely unhelpful.

:P

robjohn

@BalarkaSen Aha. That was a while ago. No wonder I missed the context.

Nick

@BalarkaSen Although that point of view is twisted and sadistic, it's quite actually effective and is a principle enforced by many a teacher.

Balarka Sen

@robjohn right. of course, i should have added "to me" at the end of the statement :P

it was nothing serious.

Nick

nothing Sirius Black here, nope. No sir.

@BalarkaSen: Are 2,3,5,7,11 emirps?

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$Mathematics$ Mathematics