12:00 AM
so...

so 1100 / 1 * 1/60 is 1100/60? this part is kind of confusing me
is it 1/60 nothing or is there like 1/60 sec or something like that in the denominator or numerator

so 1100 / 1 * 1/60 is 1100/60 yes
means the same thing

"A total of five players is selected at random from four sporting teams.
Each of the teams consists of twelve players numbered from 1 to 12 ... What is the probability that the five selected players contain at least four players from the same team?"

basically you need to find rev/minute now...

so 1100 ft / 1 sec is same as 1100 ft / 1 min?

12:05 AM
no

shouldn't it be diff

1100 ft / 1 sec is same as 1100 ft / 60 sec
the value changes!

oh we can reduce?
nvm I thought we couldn't do it

1100 ft / 60 sec = 55/3 ft per minute

[This current version of FAIL](http://chat.stackexchange.com/transcript/message/36474020#36474020) fails, thus I need to include one last rule:
$F a \# 0 \# \dots \# 0 \# 0 \# b \# \dots \# c = F a \# 0 \# \dots \# 0 \# a \# b - 1 \# \dots \# c$

12:07 AM
its not 55 feet per 3 seconds?
nvmmm I was thinking it differently

im not saying that!
should make more sense like this then..
(55/3) ft per minute

The answer I got was 11*10*9*44/(48C5), but the answer given is 4653/107019, which was simplified from 74448/(48C5), which could be further simplified to 1/23

so we multiply 55/3 ft/ min with 2 pi / 60 seconds or can it be 2 pi / 1 min ?
@IPAddress I just got confused cause of ft/sec thought same was with the fractions

no as this is the only hw my teacher really checks for he wants us to do it and get the answr

12:15 AM
you may be able to take out a factor of 2pi providing that is the correct speed
actually no
heres the solution...

I don't think i can multipy because I would get min square or sec square

First find the ft per minute: 1100 * 1/60 = (55/3) ft per minute

Secondly find the circumference: 2 * pi * r = 2 * pi * 1 = 2pi ft

12:22 AM
Notice: every time the blade moves 2pi ft it has moved 1 rev
Thirdly: (55/3)/(2pi) = 2.91784062 rev/minute
Notice (update): every time the blade moves 2pi ft it has moved 1 rev unit (also 2pi rev)

So @MATHASKER, I haven't read everything you guys have said, but you set it up correctly. So the angular speed is 1100 radians/sec. If you want to convert to revolutions per second, then you divide by $2\pi$.

ohh @IPAddress thanks a lot for helping me

@TedShifrin He was given rev per sec

Oh, we want it in revolutions/min.
So we divide by $2\pi$ and multiply by $60$.

oh why divide by 2 pi? so 2.9178 rev / min is not right?

12:31 AM
One revolution is $2\pi$ radians.
Speed of sound is a ridiculously fast speed for a lawn mower blade to be going :P
This isn't too realistic.

lol true

So, I get $1100/(2\pi)$ revolutions/sec, so $1100\cdot 30/\pi \approxeq 10504$ revolutions/min. Absurd.

Hi

Cars redline at about 6000 rpm :P
heya Zach

12:36 AM
I'm doing a bit better, thanks :) You?
Do you agree with me, @MATHASKER?

Still sick.

@MeowMix Get well soon

This is going around everywhere, it seems. But my doctor says it's almost always viral. Just have to take care of yourself. I recommend Mucinex DM for cough/throat.

I kind of don't understand how u got 1100/2 pi

Because I had 1100 radians ... and it takes $2\pi$ radians to make one revolution.
Remember that when you use $s=r\theta$, $\theta$ is unitless in radians.

12:39 AM
its 1100 feet

@Ted Have you seen that weird problem where one circle rotates around another fixed circle?
Like, the fixed circle is twice the radius.

so if ft cancels we just have radians left
oh

And you'd think it would take 2 rotations of the outer, but it takes 3 for some reason. something like that

Yes, Zach, that's a problem in my book :)
My story that I tell in class is that the SATs totally got it wrong.

12:41 AM
FAIL - SBA's Fast_Array_Iteration_Leaping

Epicycloid, Zach.

Hey everyone!

Ah yes
What happens when you have a rotating circle
and theres a point on that circle which is the center of a smaller circle
and that smaller circle is rotating, and on the smaller circle is the center of a SMALLER circle, and so on.
What does that draw?

OH, hush up, Zach :P

mmm, epicycles on epicycles

12:43 AM

@MeowMix The real question is if we had an infinite amount of circles, with the radius' in, say, geometric progression. That would be beautiful

Hi, DogAteMy.

Hey @Akiva!

not really still confused, so if the ft cancels we have radians left and u did 1100 / 1 sec * 2 pi
mhh still confued :(

Hi @Akiva

12:45 AM
So we have 1100 radians/sec. We multiply by 1 rev/(2 pi radians) to get rev/sec.

ohh nvm when u put it that way it makes a looot more sense

Zach, what you described is exactly how people used to attempt to model planetary motion on when they realized that standard motion wasn't working. So the planet would stack on epicycles until they realized that if you just make things circle the sun it all checked out

so whenever ft or lets say inches cancel out we have radians left?

ehhhh

Keeping track carefully of units is a good habit. :)

12:46 AM
That's an exaggeration. Epicycles were a thing, yes, but not epicycles upon epicycles.

Well, when there's an angle in the formula like $s=r\theta$, yeah.

I like how units work nicely with dimensional analysis.

That's tautology, Zach.

r in this case was 1 and theta was? @TedShifrin

12:47 AM
@Semi Huh, I thought they actually went a couple rounds

"I like how useful dimensional analysis is."

The manufacturer of Zbars estimates that 300 units per month can be sold if the unit price is ​$230 and that sales will increase by 10 units for each​$5 decrease in price. Write an expression for the price​ p(n) and the revenue​ R(n) if n units are sold in one​ month, n≥300.
help...

Oh, wait, that doesn't quite answer the question

Typically they'd just use one per planet. @Daminark

@MATHASKER: Well, here we really have $\omega$ (which you were writing w) ... angular speed.

12:48 AM
I think the idea is that, with infinitely many epicycles where you can choose the radii, speeds, and phases, you can theoretically draw anything.

@AkivaWeinberger Sure. It's basically just Fourier analysis.

The corresponding formula is $v=r\omega$ (linear speed = radius x angular speed).

I vaguely remember seeing an example which drew Homer Simpson's face.

@AkivaWeinberger That's the drawing!
It's a fourier series.

12:49 AM
So if I have to chose five players from four teams of twelve players each given at least four are from the same team there are $4 \times {12 \choose 4} \times 44$ possible choices right?

Right...?

I want to see that

But while astronomers prior to Kepler etc. did use epicycles, they didn't use epicycles upon epicycles.
I mean, it really overestimates the computations that they were capable of.

Lmao @AkivaWeinberger

12:51 AM
It took a lot of work just to do one epicycle. Doing more than that would quickly become too difficult.

@WillNjundong: You have a linear function. You are given a point on the graph and you have the information to know the slope. Write down the function.

@MeowMix so the answer to my previously asked question should be $55\frac1081$

Wouldn't all you have to do for that is create a function for the $x$ and $y$ values over time and find fourier series which match those?

@Semi I guess that's fair, I'm not familiar with how bad the computations get with epicycles, though I remember rotational mechanics being a pain

Back in a few.

12:52 AM
See you @Ted!

@TedShifrin Your back is in a few? Let me pull you out.

@MeowMix point is they didn't have the fourier series stuff back then

Hi all.

why did u do 1100 * 30/ pi @TedShifrin

Ok@Semiclassical

12:54 AM
Well, good night!

The probability of choosing such a team should be $\frac{55}{1081}$ then?

bubye@Simply

@MeowMix I think you'd interpret the curve as a path in the complex plane, and use the $\sum_{n=-\infty}^\infty c_ne^{2i\pi n}$ form
(if I have the formula right)

@MATHASKER: I wanted to divide by 2pi and multiply by 60, so that's ...

I am verifying the correctness of my recursive definition for a set. But I have no idea how can I verify it, please take a look at math.stackexchange.com/questions/2216115/…

12:57 AM
why multiply by 60 doe?

Because we want revolutions/min, and there are 60 sec/min.

ohhhhh @TedShifrin got it..now we good man thanks a lot for helping I should've really paid attention in class lol

@AkivaWeinberger Looks right.

LOL, @MATHASKER. Yeah, you might actually learn something in class. My students did from time to time :P

@TedShifrin hahah will try lol

1:05 AM
@TedShifrin use equation of the line?

Why do I keep seeing "Sacred Geometry" in my recommended, seriously, is there a way to remove that crap?

i ended up with 290=n225
which im certain is wrong

@Daminark The author I learned it from was Owen Gingerich. He gives a very good long presentation here: youtube.com/watch?v=qHmjGUxWRAI

@WillNjundong: You are told that $p(230)=300$ and that if $p$ goes down $5$, then $n$ goes up $10$. This says that the slope of the linear function $p(n)$ is $-1/2$. So use point-slope.

2:03 AM

@Daminark what's up

Also hey @Mike!

hi @Daminark

How's it going?

good just about to mark some papers

2:12 AM
@Daminark I'm modeling a network protocol design as a differential equation, and trying to find a good stable regime to let it optimize on the conditions I want it to work well on

Neat! To both lmao

2:24 AM
@Adeek When in grad school, what proportion of classes would you say are more rando elective?

2:51 AM
Hey @arctic! How's it going?

hello
okay

Cool!

how are you?

I'm doing pretty well, thanks!

@Daminark I like knowing a lot of math
so I would say noone of the classes is electives

2:58 AM
I'm thinking more, is it just the general breadth classes and then straight into more specialized stuff?
Or do you have a few classes that are more miscellaneous, like logic?

straight into more specailized stuff
there is no like Miscellaneous stuff like logic

Hmm, so then I'm prob gonna try to take logic while an undergrad

@arctictern I was proving bott theorem today was very fun

periodicity of pi_k(O(inf))?

of $SO(inf)$
I don't know what is wrong with imgr.com
I want to show you the problem

3:04 AM
well, to get pi_0=pi_8 you need to use O(inf) right?

@arctictern share.pho.to/Af7ge
yeah
@Daminark I always wanted to take logic class

What I'll be able to do as an undergrad in large part depends on whether I do grad analysis now or hold it off until grad school
That'll take up a year and squeeze things a bit

@Adeek looks cool. did you do it?

yeah @arctictern
took me the whole afternoon but yeah

So I might not be able to do everything I'd hope to in that case

3:08 AM

Otherwise, I'll prob have enough time to do most of what I hope if I play cards right

I ran today for 15 km was awesome

Nice

running helps with thinking it is quite nice

Huh, maybe I'll start trying that

3:36 AM
Hello math people

yeah @Daminark running is awesome

Hey @mez! And also @Akiva!

@Daminark What is up?

Not too much, how about you?

Working

3:46 AM
Nice, on what?

On a research question. and you?

Right now I'm doing analysis, later I'll get back to my difftop work

homework?

Yeah, the problem is to show that there's no finitely additive measure on $\ell_{\infty}$ for which the measure of every ball is positive and finite
Haven't thought about it much yet, just got started

What is \ell_{\infty} ?

3:48 AM
Space of bounded sequences

real sequences?

Real is most common
You can put a norm on it as being the supremum of the points in the sequence, and then it becomes a Banach space

remind me what finitely additive means for a measure

If you have $A_1,\ldots,A_n$ which are pairwise disjoint, then $\mu(\bigcup_{i=1}^n A_i) = \sum_{i=1}^n \mu(A_i)$

Are the measurable sets specified or can be arbitrary (but including all balls)?

3:58 AM
This is one that's defined on all subsets of $\ell^{\infty}$

Ok, good, let me think

What we did in class was Banach-Tarski, which proved that if you have a finitely additive measure which was invariant under isometries, you couldn't define it on all subsets of $\mathbb{R}^3$, now we're no longer assuming that

Hmm interesting, so maybe axiom of choice is needed for this proof

Banach-Tarski does need the axiom of choice for sure

Yes I was trying to construct some explicit contradiction, so maybe this approach won't work
it seems like a nice problem. but i am so out of touch for measure theory, afraid cant be much help here

4:03 AM
It's fine, I do want to keep working on it

yea, maybe relook at the banach-tarski proof and see what would apply if your space is L_infty with this norm

"The story starts in enumerative geometry, a well-established, but not very exciting branch of algebraic geometry that counts objects. "

nice@Secret

haha, "not very exciting".... who is the author to say that

4:05 AM
I imagine it'll have something to do with the unit ball not being compact
OH WAIT A SECOND
Hold on hold on
This might work

The counts produced this way are often virtual counts and needs to be adjusted to be the real number of curves on the variety.... and although Gromov-Witten theory is motivated by physicists, algebraic geometers can certainly produce these numbers without knowing any physics. On the other hand, it is indeed impressive that physicists, long before the intersection theory was properly developed, could predict counts that are verified to be true. All these are still quite olds results though.

I see, well... I am a bit behind on the latest trends in maths currently, and I am not aware of any good sources to keep up to date in the maths field
I tried reading AMS journals but it does not seemed board enough

It's a good article I think

4:47 AM
if you like the AMS page in fb,you may be in touch @Secret
or MAA

How do I model a multi-objective optimization problem as a single objective problem?

5:19 AM
@Secret u could browse relevant categories of arxiv everyday

sounds like a good idea. I sometimes came across interesting abstract algebraic structures in arxiv but I have not checked other branches of maths there yet

5:39 AM
Thundercat sleeps :)

@TedShifrin Thank you, i got it now :)

5:56 AM
Semi philosophical question: Why is the thought process and the actual writing of $\epsilon-\delta$ proofs ran in opposite directions?
e.g. thought process is we try to find the form of the epsilon given some delta, but when writing out the proof, we say what we need to pick in order for a given epsilon to give the required delta

Well, in a proof, you need to demonstrate that there exists the correct delta
But to see how small of a delta you need, you'd need to ask yourself how much leeway you get with a given delta

Anonymous
6:48 AM
Hello people. Anyone good at 3D Geometry here? I wanted to know how to find the incentre, circumcentre and orthocentre of an irregular tetrahedron. Suppose the position vectors of the vertices are $\vec{a},\vec{b},\vec{c},\vec{d}$.

7:34 AM
@Daminark Definitely, it is consistent with not AC that all subsets of the reals are Lebesgue measurabke
That was shown with the now called "Solovay model", which has a few more weird properties

2 hours later…
9:43 AM
Hi @Alessandro

Hi @Balarka
How are you?

Still trying to understand pictures :)

10:05 AM
Well they do say math is art

@Balarka Do you try to understand things via trying to understand a picture or do you use being able to understand a picture as a test to confirm that you have understood something?

@Alessandro According to Poincare, math is the art of extracting good logic out of bad pictures
@s.harp The former, certainly

That's a great quote, I'd never heard it before

I think I didn't quote him exactly right, but it was something like that nonetheless

10:43 AM
For a given field $\Bbb F$ and an extension $X$, what exactly is the difference between $\Bbb F[X]$ and $\Bbb F(X)$? For all I'm aware, they're interchangeable (never delved too far into group theory)
Then again, the fact that I'm even asking this question probably means that I have no idea what I'm doing
I guess I'll just disappear

$\Bbb F[\alpha]$ is the smallest ring containing $\Bbb F$ and $\alpha$, $\Bbb F(\alpha)$ is the smallest field

1 hour later…
12:14 PM
$\Bbb Q[\pi]$ does not contain $\frac1\pi$. $\Bbb Q(\pi)$ does.
@LegionMammal978

Example why pictures are unreliable: Nobody can faithfully draw the rational line on a piece of paper without distortion of its meaning
Having said that, whether pictures are reliable almost surely, is still an open area of research
(and don't ask me what that measure zero set look like, because I have no idea yet)

12:36 PM
At one point at MathCamp this summer, a bunch of us were seated around a table and with a few teachers, learning about the Vitali set. We were asked to show why it was non-measurable. Our construction quickly became confusing, so I offered to draw the situation.
Which quickly elicited replies of (I don't remember exactly but something like) "How are you going to draw a Vitali set??" and a sarcastic "Yes, Akiva, draw a Vitali set"
(The drawing ended up fine though I think)

How did the drawing work?

I have faith in Akiva's pictures.

Lots of cloud shapes

So more a doodle than a diagram :P

It was like, a bunch of horizontal lines (representing $\Bbb R)$, each containing a translated version of the Vitali set (drawn like a long thin cloud of unit width)
The translations were like an enumeration of $\Bbb Q$ (the first was translated by $0$, the second by $1$, the third by $-1$, etc.)
I think the idea was that the union of all of these was supposed to be $\Bbb R$?
Wait, not exactly like that

12:44 PM
Hi chat

I think you also needed to take the fractional part of each element of each set, or something, so that the union (which is still a disjoint union) becomes $[0,1]$.
Whatever, I don't remember, but it made sense at the time

@AkivaW Suppose I have a smooth map $g : \Bbb R \to \Bbb R$. I want a $C^1$ map $f : \Bbb R \to \Bbb R$ which agrees with $g$ outside $\bigcup I_n$ where $I_n$ are intervals around the $n$-th rational (give $\Bbb Q$ whatever enumeration you want) such that $\sum_n I_n < \infty$. Can see why this can be done?
For $C^0$ it is more or less obvious by a uniform limit argument (modify it at each step with$\bigcup_{k < N} I_k$ and put some height conditions) I guess.

Hi

I don't know how to make this $C^1$. Initially what comes to mind is fitting in small bump functions at each $I_k$, but what's counterintuitive is this doesn't work for $C^2$ (by a theorem). So I have no idea.

I take it we can't take $f=g$? @BalarkaSen

12:57 PM
Sorry, I actually forgot a lot of hypothesis. $g$ is a diffeomorphism, and I want $f$ to be C^1-diffeomorphism too.
@AkivaWeinberger Right, I want $f$ to act nontrivially on $I_n$'s
Say, $f(I_n) = I_{n+1}$.

@BalarkaSen Is this equivalent to stating that $g$ be strictly increasing (and smooth)?
Well, monotonic

Yeah.

I have to go, but I'll think about it @BalarkaSen
Bye

Ok, thanks. Byes

How can we calculate the integral of a holomorphic function around a non-isolated singularity?

1:12 PM
Depends on the context.
For example, are the singularities just singletons accumulating to a single limit point?
Or, is it a path of singularities?

Are there any maths that allows you to directly operate on singularities as if they are algebraic objects?

Hi Zach

1:30 PM
Hi there
Is there ways to model the evolution of a scalar field over time ?
of any scalar field, rules on how any field evolves over time

Hi everyone

@Secret This might be relevant. It's partial fractions, but generalized; you essentially write meromorphic functions in terms of things of the form $\frac1{(x-p_i)^k}$, which can be thought of as describing the function's poles/singularities

> Show that if $x$ is an odd integer, then $x^2 ≡ 1(mod 8)$
Shouldn't there be restrictions like $x> 2$ ?

Why?
It's true for $x=1$
Hint: if $x$ is odd, what are the possible values of $x\mod8$?

0=n(mod k) ? For any k?

1:43 PM
No
$0\equiv k\pmod k$, though

I did like if $x=2n+1$ then $4n( n+1)=0(\mod 8)$

How do you prove that?
Or is that just stating what you want to prove

Yes,I did till there
And don't see any way

Hint: if $x$ is odd, what are the possible values of $x\text{ mod }8$?

(Fawad can do it his way too, instead of listing the numbers mod 8.)

1:51 PM
True
but you'd need to already know some properties of $n(n+1)$

@AkivaWeinberger 1,3,5,…

A few more
Note that $9\equiv1$, so you don't need them both

@AkivaWeinberger 7,9?

@Fawad Every odd number is congruent to either $1$, $3$, $5$, or $7$ (mod $8$).
For example, $9$ is congruent to $1$. $~11$ is congruent to $3$. Etc.
Does that make sense?

No,still thinking

2:00 PM
That is, $x\equiv\text{1 or 3 or 5 or 7}\pmod8$

@AkivaWeinberger mod 8 ?

OK,how to prove it?
By taking $x=2n+1$ right?
@dogatemy

That's one way to start, but I don't think it's hard to see where to go from there
The way I'm thinking is: Every number is congruent to one of $0$, $1$, $2$, $3$, $4$, $5$, $6$, or $7$ (modulo $8$).
Since $8$ is even, two numbers that are congruent modulo $8$ have the same parity (they're either both even or both odd)
That means that every odd number is congruent to one of $1$, $3$, $5$, or $7$ (the only odd choices from the list above).
Once we know that, we only have four choices for $x^2$.
If $x\equiv1\pmod8$, for example, then $x^2\equiv1^2=1\pmod8$.
If $x\equiv3\pmod8$, then $x^2\equiv3^2=9\equiv1\pmod8$.
If $x\equiv7\pmod8$… I think you can finish from here

@AkivaWeinberger I didn't read residue system (I think this come in that)

2:19 PM
Serious question. How do I integrate $\int \text{sgn}(\sin (x)) \, dx$?

piecewise

@arctictern Can that be done also with: $\int \text{sgn}(\zeta (x)) \, dx$

sure

What's zeta, the Riemann zeta?
Isn't that like always positive

not for all s<1

2:21 PM
Oh, right
It alternates for every negative even number, right?

Holy shit
@MatsGranvik That kinda looks horrifying, good luck

Hmm
I was thinking: Given a real $x$, $0$, complex subtraction and exponentiation, and the principal complex logarithm, would it be possible to determine $|x|$?
Wait, no, I'm silly, it would just be $|x|=\exp(\exp(\ln(\exp(\ln(\ln(x))-(0-\ln(\exp(0)-(0-\exp(0))))))-\ln(\exp(0)-(‌​0-\exp(0)))))$
Lemme check it

2:40 PM
@AkivaWeinberger Ted just gave me one of his favorite problems
Also, I have an analysis question
I guess he's gone.

Er, wait, no, it would be $|x|=\exp(\exp(\ln(\ln(\exp(\exp(\ln(\ln(x))-(0-\ln(\exp(0)-(0-\exp(0))))))))-\l‌​n(\exp(0)-(0-\exp(0)))))$

but why tho

I guess just a puzzle?

Yeah, certain programming language which only has those three operations
Trying to see what I can do without loops
Most elementary functions should work actually (just not most polynomial roots)

if a language only has three operations, will it really handle complex numbers?
and seriously, subtraction, exp and ln as the only operations? what.

2:49 PM
Is that a code puzzle?

Planning on forking it to add complex arithmetic