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00:00 - 19:0019:00 - 00:00

00:00
I think he means Hatcher's a bitch as in difficult. Which I don't think most people will disagree with...
00:29
Hi @Chris'ssis version: lemon green :)
@Charlie hehehe :-)
@Charlie I was watching this and had some fun :-) youtube.com/watch?v=9qaaklixcec
@Chris'ssis hahaha that's mean
@Chris'ssis I have fun with those stuff xD t
@Charlie it's the first time I look at these. Some are really nice. :-)
00:46
@Chris'ssis I.was watching some like that on tv, earlier
That's not nice. Now everyone has a cat avatar. I used it before it was cool.
woof @charlie
@Charlie I was also reading " reasons not to take life too seriously". :-) I think it's good to laugh a bit (at least) every day, to find reasons to do that.
@DonLarynx meow
@Chris'ssis I'm a compulsive laugher
@DonLarynx don,.is don your name? If so,.that's.the.diminutive of Donald, no?
01:10
Does anybody (with the ability to answer) think this question is suitable for MO? math.stackexchange.com/questions/568961/what-motivated-rademachers-contour-along‌​-the-ford-circles-when-he-used-the-circ
01:20
My name is resultant of a random string of thought: My thoughts are, 100% of the time, awesome.
The result? Don = the boss. Larynx? = voice box.
So...Awesome voice box. I profess my language of awesomeness to everyone. @Charlie
(this is assuming bosses are awesome)
question, if you have an x:y position of a point on a circle's circumference, and you know the circle's radius, how do you find the x:y position of the very centre point of the circle?
@Dave You can't. You need at least two points. The question you ask gives two pieces of information: a point and a distance. This happens to narrow down the possible solutions to a circle
(which is sort of funny because this is vaguely recursive)
so you need two angles
two points at least
hmm what if i use "position and position - 180 degrees"
01:33
well, actually, two points will do. Not "at least"
@Dave then it's fine - the center will be the midpoint of the line segment bounded by those two points
well i know how to do straight line calculation but not how to get the midway point =/
add the two x coordinates, divide by two. add the two y coordinates, divide by 2
yeh but lets say im at : x:5 y:7 (in a 2D grid) and the center is 150pixels away im still at an angle and so need to alter x:y accordingly as it wont simply be x-150 and y-150 unless im directly north of center point
I don't understand. I was under the impression that you were going to give me two points.
yeh the second point was a total distance of 300 pixels
/2 = 150 pixels from me to center of circle
let me do a doodle to explain better :P
01:38
ok
ok so basically im trying to find the x:y of the center from only knowing my own x:y, my angle and circle radius
Can anyone here say something interesting about tensor products of modules?
I need some food for thought
$M \otimes N$
:|
@Dave In your picture, you don't give me the coordinates of the second point. If you do, then this is simple. But if you know a point, the radius, and the angle from the center to that point, then this is also doable
i know the angle of x:5 & y:5 yes @mixedmath
In particular, an angle gives you a slope. The radius tells you how far to go in that direction.
01:51
what formula should i be using to get the correct co ordinates
@EnjoysMath @nimza was teaching me about it days ago :P
Teach me about i
*it
@DonLarynx then hows your name?
@Dave I'll tell you what. You give me a point, a radius, and the angle, and I'll do one out for you. But I'll do it in about 45 minutes, as I have to go and pick up some groceries
so ping me, and I'll get back to you in a bit
@EnjoysMath ask him
01:56
The way I understand it is you take the free group gen'd by $M\times N: F = \Bbb{Z}(M\times N)$
Declare elements like $(m + m', n) - (m, n) - (m', n)$ to be zero, or generate a group with them and take the quotient with F
@mixedmath okay!
So then you have a bilinear map $\otimes : M\times N \to F/H$
Prove that class. Kthxbye
I suck at explicating
@EnjoysMath I see...
back to studying
@EnjoysMath good luck
02:01
Hiya
thx
ganja is life
@EnjoysMath what?
trollolol
@EnjoysMath what?
Good Morning! :D
02:07
@Nick did you sleep?
It's 3 cats, beats 10 geometric diagrams
@Charlie: yes, It's 6Am now. A beautiful, wonderful new day
Any hints? Find a sequence of closed, connected subsets $C_1, C_2, \dots$ of $\mathbb{R}^2$ such that $C_1$ contains $C_2$ which contains $C_3$ (and so on) and $\bigcap_{n = 1}^{\infty} C_n$ is not connected.
I don't even know because isn't this intersection necessarily a point?!
@Nick :) you slepr only 5 hours
pm?
02:13
@charliie: well, what's wrong with that :D
@EnjoysMath@ meow meow meow!
@Nick nothing
@Charlie Jossie
@DonLarynx really? Sounds like girl name
@Ethan hi ethan
hi
@Don: Jossie and the pussycats?
02:18
@Nick that movie is nice :)
@Ethan hows it going?
horrible
Wow you guys have it right
;)
@Ethan :( what happened?
No one anymore knows what Josie and the pussycats are
@Charlie: that was a cartoon too, ya know :p
02:19
look me up on fb. I am a skydiver dude
@Nick why you are using a catvatar too?
@DonLarynx I know
@DonLarynx really?
@charlie: I was inspired by Gustavo's idea that there's a surjective relation between humans and cats.
@Nick oh no
@Charlie: Yes
@ethan: oh no!
02:23
yeah it's bad
@charlie: oh no, hoh no.
@Ethan :/
@DonLarynx hmm jossie...hmm.....
@Ethan ohh !!!!
@ethan: and you're not too fond of that?
no
@DonLarynx Donald is way nicer
02:26
I only have like 3 realistic ones so far
@ethan@ sometimes, it depends on the nature of the institution.
@ethan: probably.
@Ethan good luck :)
@ethan: Chatting with you was like talking to a ghost.
@don: He did it so no one will know what he said about you.
Stopping removing!!!
02:39
@DonLarynx there are too many.jossies on fb...
Oh, guys it's late night, I need to sleep. Good night you all @Don @nick
@charlie: Good night
@Nick hahaha :D have a good day, Nick
@charlie: ya know I will
@mixedmath i solved it in the end im now working out a way to get the radius from 2 points
@Nick excellent
02:57
Good night dorothy
@enjoys: wizard of oz reference?
yes
good night catness evardine!!
03:09
is it possible to work out a radius with just two points on a circle ? without knowing its circumfrence?
@Dave Hello!
hello :)
and good job!
im still stuck though lol
no - it's not possible to determine a circle with only two points
any two points are on infinitely many circles
03:10
@Dave obviously there are infinitely many circles through those points
ok then ive no idea how im guna implement a orbital path between two orbiting objects =/
is there a minimum size circle that can fit between the two points ?
yes
the circle of radius $d/2$, $d$ the distance between them
no idea what $ represents
centered at the average of the two points
see to the right
click on$LaTeX in chat
Read the instructions
gah
03:14
that's what she said
lols i do most of maths in programming so im used to using words xD
Use words with lots of dollar signs
then you'll be good
what's this circle thing have to do with orbital paths?
because to go from inner orbit to outer orbit you go in a circular path
rather than straight
i thought you go in an elliptical path
im not fully replicating space
just want a circular path so it looks about what you'd expect :P
like this
green line - the path. the blue points are the start and end
i might just make the path's circumfrence the same as the larger of the two orbits involved
03:21
@Dave Yes, that's a hohmann transfer orbit
what are you doing that makes orbits?
what do you mean by what am i doing
@Dave Well, you're trying to figure out Hohmann transfer orbits. Why?
im making a solar system animation with a ship going from different planets
but im not incororating complex physics im trying to keep it simple just making it reasonable
it seems to me - the easiest method will be to offset the larger orbit like this:
bit cheating but itll do :P
that will look weird since the centers are different
no, the easiest method is to look at the line connecting the initial point (the blue point on the inner circle) to the center of its circle (the center of the planet, or whatever)
the planet will orbit its normal orbit but the ship will follow that offset one until they intercept
wait what do you mean ?
03:27
somewhere on that line (extended) is a point that will be equally far from both blue points
use that point as your center, and the resulting circle is the correct circle
oh, hmm - I'm cheating a bit
so i use all 3 points to get the circle ?
by assuming I know where the second blue point is
give me just a moment
nope, still okay - there is a point that will be equally far from the first blue point as to the outer circle (which has some radius)
so find the distance of green line / 2 ?
ive got the distance of the planets calculated but the arc is not decided
 
1 hour later…
04:45
oh, it got quiet in here
It sure does get frustrating when someone thinks they have proved something, you point out why it has an obvious flaw, and yet they continue to argue that the argument is sound by constantly invoking the exact same logical flaw.
@Zibadawa was it the riemann hypothesis
05:20
@Zibadawa LINK
LINK LINK
Zelda!
 
4 hours later…
09:28
@Pedro: Mario, Mario, Mario!
 
2 hours later…
11:48
Are there mathematical objects similar to matrices that deal with non-affine transformations?
12:11
@Alyosha: good question. Ask it on main.
 
1 hour later…
13:16
Are sets $\{a,b\}$ and $\{a,a,a,b,b\}$ equal by definition?
@Lazar: Yes, because the second set does not exist in that state.
The second set would be, inorder to be a set, {a,b}
I see, so elements have to be distinct by definition. Thanks!
@lazar: Exactly, the elements have to be distinct. Do you know what a group of elemets that aren't distinct are called?
@Lazar: yes or no.
I take your silence to mean a no.
Well, I don't know either, that's why I asked you.
13:56
Hi @nick
@charlie:hi charlie. I went back to being human.
@Nick :) that cartoon...
@Nick I'm so used to Charlie, that sometimes I really think that's my name xD
@Charlie: Yeah, I love it.
@Nick hehehehehe
@charlie: oh, yeah , I forgot you're name is evangeline or something like that... I never asked for your name.
14:02
@Nick evangeline is a beautiful name :P but my name is Marília
@charlie: sounds like someone put the name Maria in a shaker and served it with ice cream.
@charlie: but it's a beautiful name.
@Nick it's a poetical name. You can make a research on it };)
@Charlie: dang, you're a brazilian city with 200,000 inhabitants!!
@Nick yup :) keep reading
@charlie: in the meantime, have you ever seen breaking bad and malcolm in the middle
@charlie: Cool, you're name can be interpreted to mean " Divine, God, Goodness, Truth, Unconditional love, Gift, Free will, Ideal, Whole, Endless" ... that's like everything me and superman stand for.
14:33
@Nick :D
@Nick there is another author
@Nick no, I've never watched
Marília Laboissière?
um, idk who the other author is. :p
@Nick bocage
oh, Epistola a Marilia
cool.
@Nick :)
@charlie: random question, was what I said to lazar earlier, correct?
14:45
@Nick let me check
@Nick well, they are both "sets"
@charlie: see, the second one isn't technically a set. Cuz it doesn't have distict elements.
@Nick the definition of set is something yet to be defined.
O-o
it used to be defined. What happened? Did someone divide by zero?
@Nick in ZFC (if.I'm not mistaken) a set is a bag full of guys that obey a property
? sets don't need to obey properties each and every singly time though, right?
{Alice, Jim, Bob} is a set
No property there.
14:51
@Nick you must define what a set is
In mathematics, a set is a collection of distinct objects, considered as an object in its own right. For example, the numbers 2, 4, and 6 are distinct objects when considered separately, but when they are considered collectively they form a single set of size three, written {2,4,6}. Sets are one of the most fundamental concepts in mathematics. Developed at the end of the 19th century, set theory is now a ubiquitous part of mathematics, and can be used as a foundation from which nearly all of mathematics can be derived. In mathematics education, elementary topics such as Venn diagrams ar...
"a collection of distinct objects"
@charlie: I'd say that's a pretty good definition.
@Nick I don't think so }:)
@charlie: oh, I get it.
}:)
In mathematics, Zermelo–Fraenkel set theory with the axiom of choice, named after mathematicians Ernst Zermelo and Abraham Fraenkel and commonly abbreviated ZFC, is one of several axiomatic systems that were proposed in the early twentieth century to formulate a theory of sets without the paradoxes of naive set theory such as Russell's paradox. Specifically, ZFC does not allow unrestricted comprehension. Today ZFC is the standard form of axiomatic set theory and as such is the most common foundation of mathematics. ZFC is intended to formalize a single primitive notion, that of a heredita...
@Nick I'll eat,.I'll be back
@charlie: so, zfc is the most accepted set theory.
.. but wait, what does the c stand for?
..choice?
XD
15:01
yo guys wassup
i have a question,,, is there a way to find how we transformed the rectangle from the left into the polygon on the right? puu.sh/5l3PS.png
like in terms of equations?
@Ivan: um, from the image it looks like you moved the top left vertice a bit more closer to the bottom left vertice ad then took the bottom right vertice and stretched it a bit mor further away.
Without the co-ordinates of these points. I'm not sure there's much that can be said about the tranformation in terms of equations.
i see @Nick
so with coordinate,, it is possible right
well, yes. The following may interest you - math.stackexchange.com/questions/69099/equation-of-a-rectangle
15:17
@Nick yes
@charlie: what's the difference b/w an element and a urelement
@Nick whats the latter?
In set theory, a branch of mathematics, an urelement or ur-element (from the German prefix ur-, 'primordial') is an object (concrete or abstract) which is not a set, but that may be an element of a set. Urelements are sometimes called "atoms" or "individuals." Theory There are several different but essentially equivalent ways to treat urelements in a first-order theory. One way is to work in a first-order theory with two sorts, sets and urelements, with a ∈ b only defined when b is a set. In this case, if U is an urelement, it makes no sense to say :X \in U, although :U \in X, is perfe...
ok, maybe sometimes elements ca be sets as well.
Urelements could be those elements which are absolutely not sets.
right?
@Nick hmhmhmhm
@Nick they are not sets
mhmh
@charlie: ok, so you're thorough with the ZFC definition of sets there, right?
15:29
@Nick yeah
@charlie: so, a set of fractions is cool, right?
{1/2 , 1/3 , 2/4, 7/6}
that's a set right?
@charlie: quickly, how many elements does that set have!
@Nick I think it is
@Nick 4
What is the cardinality of that set?
same, right.
4...
@charlie: you agree?
absolutely, positively agree
@Nick yes
@charlie: But Marilia, 2/4 is 1/2 is it not?
15:37
Is there any easy way to obtain the number of whole solutions to
$x_1 + x_2 + x_3 + x_4 + x_5 = 50$
@Nick if you had {2/4, 6/12, 50/100, 72/154,....} it is a set.
Given that $x_4 \leq 50$, $x_5 \leq 50$ and all the numbers are positive? I solved it by bruteforce, but there ought to be a better way presumably involving number theory.
sure but isn't it a singleton?
@Nick A={a} is a singleton
exactly
all those guys evaporate into 1/2
15:40
@Nick the set I have you is the sets that follow tje property a/b=1/2, therefore a set.
@Charlie Look at this:
@Nick I was "away", sorry. No idea, really.
>>Por que faculdades particulares são tão discriminadas?
>Porque em algumas faculdades as provas são fáceis. Nas federais as turmas de cálculo são lotadas de tantas reprovações.
>Então "lotadas de reprovações" significa "melhor"? Vou fazer uma faculdade em que os quesitos das provas serão assim:
-Antes da prova, foi sorteado um número aleatório entre 1 e 99999999999999999999999999. Escreva abaixo que número é esse.

Tenho certeza que será "lotada de reprovações"
@GustavoBandeira at what?
@GustavoBandeira é algo a se discutir
@Charlie Altas provas com números aleatórios.
15:45
@GustavoBandeira não é bem por aí, mas deixe pra lá
@Charlie É exatamente por aí. A heurística do cara empurra pra uma dimensão apenas. Tem uma caralhada de aspecto sendo esquecido.
@GustavoBandeira exatamente
@Charlie Você leu o texto sobre meritocracia que eu postei?
@GustavoBandeira não
@nick
15:57
@lazar: There's no need to apologize. :D we're cool.
I find answering questions on math.stackexchange quite addictive. It is a bit like gambling. Especially because you don't know the outcome of your answers. You give a trivial answer like math.stackexchange.com/questions/567684/… and you get reputation.
hey Nick I posted a question :) math.stackexchange.com/questions/570526/…
you have any idea about that?
And math.stackexchange.com/questions/550925/… gets points out of all proportion to the effort put in.
But put a day's effort into math.stackexchange.com/questions/552767/…, and you get nothing!
@StephenMontgomery-Smith you got a 1 :)
Thanks. You just out my reputation over the 1000 mark!
@Nick Wow!!!!
@Nick :)
Anyway, I really do find math.stackexchange addictive. I know I should be writing papers and grants. But the reward with math.stackexchange is so immediate.
well, I gots to splt guys
/i mean, bye bye
@charlie: ttl, bye
16:06
@Nick bye nick
Yesterday some edited one of my responses. But they didn't upvote it! I felt rather cheated. math.stackexchange.com/questions/566681/…
16:59
Why are some properties called universal?
Like the for any $R$-balanced map $\phi: M \times N \to L$, where $L$ is an abelian group, $M$ a right R-mod and $N$ a left R-mod, there is a unique group hom $M\otimes_R N \to L$ such that $\phi = \Phi \circ \iota$ where $iota: M\times N \to M\otimes_R N$ is injection into the quotient.
Is it because of the universal quantifier at the beginnin?
@Charlie let's talk about $R$-module homomorphisms
:D
Loosely speaking, because they are the unique map which everything else can be factored through.
17:17
@EnjoysMath lets
What is an $R$-module hom?
hehe
@EnjoysMath I don't kniw about it, but you can explain to me
Hahaha
Exactly what it should be. Look it up.
Yeah, look dat shit up
@Charlie
we don't discuss math here
It takes a few seconds on google to find.
17:28
@EnjoysMath no, only drugs, sex and rock 'n' roll
If you can find.on google, no need to ask
Hello
Hello
@Charlie, an $R$-module homomorphisim is analogous to a vector space linear transformation
In this case though the vector space is over a ring not necc a field
So they're called $R$-modules.
An $R$-module hom then is $f : M \to N$ such that $f(r x + y) = rf(x) + f(y) \ \forall r,s \in R, x,y \in M$.
17:50
@EnjoysMath aaah nice!
18:06
i'm 12 years old
@what'sup me too
Does COPPA not apply because Math.SE doesn't sell anything?
;|
Does anyone know of a result on the locus of points on the sphere from which two given points subtend a given angle?
I've gotten a result, and I would like to see any previous work on the question
@robjohn Greetings!
18:21
@Chris'ssis Hey there... I've been investigating this problem
@robjohn Hi. Which problem?
@Chris'ssis the locus of points on the sphere from which two given points subtend a given angle
The answer is pretty neat
on the plane, it is a circle of course
@robjohn I think I saw your answer these days.
@Chris'ssis Oh, that answer is what spurred me to look into this, but that answer does not tell what the locus is
If you do a google search, you find some geometry exercises from before 1900 which contain similar words (like locus of points for which two different spheres subtend the same angle), but not the same problem I think
18:29
@MarkS. that is a different problem. This is applicable to locating your position if you know the bearing to a couple of known locations.
Sorry for not being clear, I didn't intend to imply it was the same.
I meant that the big list of many problems from that old book didn't seem to contain the same problem anywhere
@MarkS. No, you were clear, I was just agreeing with you
It was asked on sci.math in 2007, but no decent answer was given.
Ooh, interesting
@Charlie really ?
@what'sup no, I'm 21
18:33
:-) good for you
@what'sup not much
18:53
@robjohn you there
@Ethan yes
@Ethan why would that be insane? I applied to a lot of schools, just to make sure
They always hate my essays lol
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