Mathematics - 2019-12-18 (page 2 of 2)

user

20:17

How does one prove that sqrt(z) has only two branches?

(z is some complex number)

If f(z) is any other branch then it must have absolute value abs(z)^(1/2) and argument theta+2kpi?

so it must be one of the two branches -r^(1/2)*e^itheta or r^(1/2)*e^itheta

Leaky Nun

because $x^2 - z$ only has two solutions

Balarka Sen

h-principle doing it's job

(@anakhro, @Akiva)

user

For two branches of the same multifunction must their derivatives be the same?

Leaky Nun

no

@BalarkaSen puzzle battle?

Balarka Sen

No man

I have to write

Leaky Nun

20:30

write what

user

Do you have a counter example?

Balarka Sen

proof of something i want to prove

Leaky Nun

@user sqrt(z)

user

really? Both branches have derivative 1/(z^1/2)

Balarka Sen

@user $f(x) = x$ and $g(x) = 2x$ jointly define a multivalued function on $\Bbb R$

user

20:32

sorry, 1/2(z^(1/2))

Leaky Nun

the z^1/2 are different

Balarka Sen

One branch is $f$, the other branch is $g$ :P

Leaky Nun

for z=-1, you get 1/2i and -1/2i

user

ah right so although the formula is the same we have to specify which branch we are lookin at?

i.e. specify restrictions on the argument?

Leaky Nun

high school maths focus on formulas instead of meanings

I don't care about the formula

user

20:34

ok...

thanks for the help anyway

Leaky Nun

or rather, "1/(2 z^1/2)" is ambiguous, to be less cynical

precisely because z^1/2 is ambiguous

user

yes, I see that now. Thanks!

Semiclassical

to render it non-ambiguous, parametrize your path as $e^{i\theta}$ with $\theta$ ranging from $\theta=0$ to $4\pi4

then you've got $e^{i\theta/2} = \cos\frac{\theta}{2}+i\sin\frac{\theta}{2}$, which is $4\pi$-periodic

Leaky Nun

$2 \tau$-periodic :P

Semiclassical

So you do end up back where you started

Balarka Sen

20:37

Two $\tau$'s make $\pi$ so $\tau$ should be $\pi/2$

$\tau\!\tau$

Semiclassical

Sep 7 '18 at 21:05, by Semiclassical

(2 flips of the bird, at that)

Balarka Sen

I have a smooth fiber bundle $(E, M, p)$ and I have smooth sections $s_1$ and $s_2$ of $p$ over open sets $U$ and $V$ of $M$. It should be possible to glue $s_1$ and $s_2$ to a section over $U \cup V$, right? I don't see how to argue this.

Alessandro Codenotti

What do they do on the overlap?

Leaky Nun

use the short exact sequence of sheaf

Balarka Sen

There is no sheaf, @Leaky. Fiber bundles, not vector bundles.

@Alessandro Ah, ok, without having some regularity there it's likely not possible. In my case $s_1$ and $s_2$ are very close on $U \cap V$

So I can argue by saying locally the fiber bundle is an affine bundle $\Bbb R^n \times \Bbb R^k \to \Bbb R^n$

By choosing charts

Then a bump function in that box neighborhood explicitly. Very messy.

Leaky Nun

20:44

why is there no sheaf?

Balarka Sen

What is the sheaf bro

Alessandro Codenotti

It's not clear to me what do you expect $s_1\cup s_2$ to be on the overlap

Leaky Nun

smooth sections

Balarka Sen

@LeakyNun Well what the fuck are those? They are not R-algebras or anything

Leaky Nun

sets

I just mean that you can always glue two functions together as long as they agree on the overlap

and the new function will be smooth

Balarka Sen

20:45

$s_1$ and $s_2$ do not agree on the overlap bro

Leaky Nun

and a section

Balarka Sen

@AlessandroCodenotti You can always choose charts near eg $s_1(M) \subset E$ such that the bundle looks like $\Bbb R^n \times \Bbb R^k \to \Bbb R^n$ by a change of coordinates, and your sections are graphs of functions $\Bbb R^n \to \Bbb R^k$, so we can do a bump function argument there

Alessandro Codenotti

I see

Balarka Sen

I think that works at least. In my context $U$ and $V$ are actually chart neighborhoods not massive open sets, so local interpolation by a bump function should suffice

Very messy coordinate-wise argument though

Adam

So what do you do if the accepted answer for a question is wrong?

Leaky Nun

20:50

you ask Joe to delete the answer

Balarka Sen

The key point is $s_1$ and $s_2$ are $\varepsilon$-close on $U \cap V$, I guess. It's of course false otherwise: take eg a disconnected cover of $X$ and two sections on two different components

You can even do that with connected fibers but whatever

If they are "far apart" I can't interpolate

Adam

who is Joe

Balarka Sen

Hi @Ted

Ted Shifrin

rehi, a @Balarka

What is @Alessandro bundling today?

Alessandro Codenotti

Hi @Ted

I'm looking at websites of universities to find places to apply to for a phd

Ted Shifrin

20:54

Oh, I saw mumbling about fiber bundles.

Alessandro Codenotti

I'm also reading some stuff about C*-algebras

That was by Balarka

Balarka Sen

@TedShifrin I was wondering if a non-messy argument exists to interpolate between two sections of a smooth fiber bundle $(E, M, p)$ over open sets $U, V$ which are very close in the overlap $U \cap V$

Ted Shifrin

Oh, it's Balarka's fault, as usual.

Balarka Sen

Lol

Ted Shifrin

What are the fibers and what does "very close" mean?

Leaky Nun

20:55

@TedShifrin would you have a picture of Mobius strip x [0,1]?

Ted Shifrin

LOL, are you still beating that dead drum?

It's not going to be embedded in $\Bbb R^3$, of course.

Leaky Nun

it also does not have an immersion to R^3

but maybe we can have something close to being an immersion

Mike Miller

Didn't Balarka already talk to you about folds and cusps and whatnot

Personally I don't see that those help me think of the Mobius strip x I

Ted Shifrin

You can take the usual parametrization of a Möbius strip and add a parameter for an interval perpendicular to the ruling.

Mike Miller

I just think of a bunch of parallel shifts of the Mobius strip, where I forget that they intersect

Ted Shifrin

20:58

For the longest time, Leaky wouldn't admit it was non-orientable.

Balarka Sen

I do the same

Ted Shifrin

Ted just doesn't think.

Mike Miller

It's hard to come up with a definition of orientability for which the Mobius band isn't obviously non-orientable

Balarka Sen

@TedShifrin Fibers are smooth manifolds. Assume $U$ and $V$ are trivializing open sets; by very close over $U \cap V$ I mean the maps $U \cap V \to F$ are $\varepsilon$-close.

Mike Miller

I guess the Hatcher definition would be hard

Unless you are also allowed PD with it

Leaky Nun

20:59

by picture I just mean jpg

Balarka Sen

I can draw a bunch of parallel Mobius strip as a clusterfuck in paint and give you a jpg if you want Leaky

Ted Shifrin

@Balarka: What if the diffeomorphism group of $F$ is very small?

Leaky Nun

that actually doesn't work because the Mobius strip is non-orientable... (which direction would you start?)

Ted Shifrin

Can you write a proof in the case of a vector bundle or principal bundle?

Mike Miller

What

Diffeomorphism groups are never small

Ted Shifrin

21:01

I guess I was thinking in the complex analytic category, where they are typically very small.

Balarka Sen

@TedShifrin yeah vector bundle is easy and you don't need the epsilon-close condition

you can trivialize and then patch by a bump function

Leaky Nun

for usual shapes you can put a copy "slightly above" the original and think of it as the shape x [0,1]

Mike Miller

@BalarkaSen Put a fiberwise Riemannian metric on the total space and use logarithm to interpolate

Leaky Nun

there's no "above" for Mobius strip

Balarka Sen

@MikeMiller Ah, that's cute.

Mike Miller

21:03

There is a map Nbhd(Diag) -> M on any Riemannian manifold M sending a pair of points to the midpoint of a geodesic between them

Seems smooth

Might need some fiddling since the metric is fiber dependent

Balarka Sen

Yes this should definitely work

@LeakyNun sure it works i just stack them one upon another mercilessly

there is above in the page in ms paint

get used to thinking of the y-axis as the extra 4th dimension

Leaky Nun

I found this online

and is probably what you mean

Balarka Sen

Here's another puzzle: What does suspension on the Mobius strip look like?

It's some clustermess which has $S^2$ as "boundary" (note that it's not a manifold by the way)

If you fill that $S^2$ with a $3$-ball you get $S(\Bbb{RP}^2)$

Alex

hello, can someone help me understand where my mistake is in implementing a formula? apparently I don't understand how to find the pressure gradient at a point in my fluid simulation

the spatial derivative is giving me weird numbers

or is this not the place for this kind of question?

Leaky Nun

21:23

@SimplyBeautifulArt long time no see

Mike Miller

@BalarkaSen What does an answer look like here

Simply Beautiful Art

@LeakyNun I would hope you see more often then

Leaky Nun

have you brought new collapse functions

Simply Beautiful Art

I have, of course

Leaky Nun

lol

it's 7 days too early

Ultradark

21:24

Simply Beautiful you answered or at least gave me a hint on my question this afternoon

Simply Beautiful Art

indeed it is

Balarka Sen

@MikeMiller I have no clue, I just said it to stop Leaky from pestering

Simply Beautiful Art

I might've but I don't remember you so...

Balarka Sen

I was googling and found some massive paper which computes the homotopy groups of $\Sigma^n \Bbb{RP}^2$

hopf.math.purdue.edu/WuJ/mod2Moore2-2.pdf

Mike Miller

lol

nerds

Simply Beautiful Art

21:29

If anyone's bored, try making the biggest number you can in only 69 symbols using:
    0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
    +, -, *, /, ^,
    ?, : (ternary),
    &, | (logical operators),
    <=> (comparisons/assignments),
    () (parentheses),
    and variable names (1 symbol per variable name)

Mike Miller

@SimplyBeautifulArt nice

Simply Beautiful Art

and yes definitely nerds

Balarka Sen

id put this in the same category as homotopy groups of suspensions of RP^2

Simply Beautiful Art

a(x, y=x, z=x) = x*y*z<0 ? z+9 : a(x-1, z, a(x, y-1, a(x, y, z-1)))
a(a(a(a(a(9)))))

Balarka Sen

why 69 symbols

Simply Beautiful Art

21:31

>.> that being the biggest I came up with

Balarka Sen

what was going on in your mind when you thought of this

Mike Miller

Because it's the sex number bro

What's the largest number you can make in only 42069 symbols

Simply Beautiful Art

What was going on was it needed to be around 50 so it wasn't too long but more than 50 so that you could actually do stuff with it

Balarka Sen

LOL

Ultradark

cmon algebro

Balarka Sen

21:32

@Simply that was not what was going on clearly

Simply Beautiful Art

@MikeMiller <spam ordinal collapsing functions here>

@BalarkaSen sure it was

$\newcommand{cof}{\operatorname{cof}{}}$

Cofinalities assuming normal forms:

$\cof(n)=\min\{n,1\}$

$\cof(\Omega_\alpha)=\cof(\alpha),~\cof(\alpha)>1$

$\cof(\Omega_\alpha)=\Omega_\alpha,~\cof(\alpha)\le1$

$\cof(\psi_\pi(\alpha))=\omega,~\cof(\alpha)\notin\pi\setminus\{0,1\}$

$\cof(\psi_\pi(\alpha))=\cof(\alpha),~\cof(\alpha)\in\pi\setminus\{0,1\}$

Fundamental sequences assuming normal forms:

$\Omega_\alpha[\gamma]=\Omega_{\alpha[\gamma]},~\cof(\alpha)>1$

$\Omega_\alpha[\gamma]=\gamma,~\cof(\alpha)\le1$

Also speaking of ordinal collapsing functions, there's a quick thing involving ordinal collapsing functions

also that it uses some not-so-standard notation but it's surprisingly simple

@Simple somehow I remember this other simple person

Also is there any benefit to exponentiation by squaring using $a^{2n}=(a^n)^2$ instead of $(a^2)^n$? The latter is better for tail recursion or loops

Astyx

Hi chat

Simply Beautiful Art

Hi, but I'm simple not chat

Also I recently learned about complimentary linear programming and how to convert equality constraints into minimization subproblems which was cool I guess.

Ultradark

Would it be desirable to have a space of points...transcendental points...of the form $(\Bbb T, \Bbb T)$ s.t. these particular points...can be multiplied to remain in such space?

Simply Beautiful Art

Lol it was in a linear optimization class, and for the final project, my group had to be that one group to pick a problem from like the last chapter of the book. The problem looked like a linear problem but became quadratic x.x

@Ultradark isn't that just part of the definition of a vector space?

Ultradark

21:46

@SimplyBeautifulArt I was thinking of adding an identity element of $(1,1)$ and trying to make a group or something

and of course (1,1) would be the only non transcendental number

Simply Beautiful Art

uh sure I guess

@Simple wait didn't you ask me about tetration and stuff?

Also recently got into Baba is You, would highly recommend for anyone into puzzles.

Ultradark

okay I got it

So we have points of the form $(e^{\Bbb {-A}}, e^{\Bbb {-A}})$ and also points of the form $(e^{\Bbb A}, e^{\Bbb A})$ and the identity is $(1,1)$

binary operation is multiplication

associativity holds

anything wrong so far?

J. Doe

22:16

@Ultradark How can it be a group or whatever if it doesn't contain the identity.

Ultradark

I guess the only point not being transcendental would be the identity?

Adam

why do people use exp(x) for $e^x$?

it's like the most annoying thing in the world

Simply Beautiful Art

emphasis on it being a function opposed to an operation

Also sometimes makes it clearer to write (if you have a lot in the exponent)

J. Doe

@Ultradark also how do you define multiplication? if component-wise wouldn't you get something like $(a,b)$ where a is transcendental but b=1 etc.

Adam

if you have a lot in the exponent that's a good time to just declare another function and write it out in the following line

Ultradark

22:25

@J.Doe hmm good question... here's my set: My set consists of points $X=\{(e^{\Bbb {-A}},e^{\Bbb {-A}}), (e^{\Bbb {A}}, e^{\Bbb A}),(1,1)\}.$

and let me see about multiplication...

you think I could do the Cartesian product?

and combine the points that way?

Simply Beautiful Art

@Adam no it doesn't? Sure there's a point where you should move it all out, but the point is that $e^{-x^{2n+1}\arctan(x)}$ is hard to read while $\exp(-x^{2n+1}\arctan(x))$ is not, due to sizing.

Also if you have several superscripts/subscripts, it can be really unclear.

J. Doe

@Ultradark Are you familiar with direct products? I think you might be interested in something like that. But anyway transcedental numbers don't have closure, so you wouldn't have closure (or identity).

Simply Beautiful Art

But there's still no need to rewrite it into another function

Simple

@SimplyBeautifulArt I don't remember that

Ultradark

@J.Doe but all of the transcendental numbers are of the same class so they would have closure I think

J. Doe

22:31

I like $\exp(X)$. It has certain class and sophistication to it.

Simply Beautiful Art

@Simple hmmmmmm but I remember you >.>

J. Doe

@Ultradark I don't know what that means. But if $a, b$ are transcendental is $ab$ transcendental?

Ultradark

so $\Bbb A$

is any algebraic number

by exponentiating with the same base $e$

it means we can keep $e$ as the base and add the algebraic numbers

meaning the result will still be $e^{\Bbb A}$

which is transcendental

by Lindemann-Weirstrauss theorem

Adam

@SimplyBeautifulArt you know you have a zoom function on most applications right? people using that will make exp usage go extinct and increase adam happiness

J. Doe

@Ultradark Damn I'm out of my depth now lol

Simply Beautiful Art

22:41

@Adam you know that doesn't change the fact that the superscripts and subscripts will have different sizes than the base(?) and zooming in will just make it harder to read everything else instead?

J. Doe

@Ultradark Teach me the proof of this theorem

Ultradark

I mean the statement of the theorem is just that the exponential raised to a non zero algebraic number is transcendental

I'm pretty much just using the result I don't understand the whole theorem in all its depth

transcendental number theory is very difficult

JBis

If you have a lottery card that has a 50% chance of winning you $200, is it worth $100?

Ted Shifrin

22:57

@JBis: Well, your expected pay-off is $100, so would you pay $100 for that?

JBis

Probably not.

Ted Shifrin

"On the average" if you do this a bunch of times, you will break even.

JBis

so it is worth the same (not including future value stuff)

I was looking at this relevant video on expected profit.

Ted Shifrin

I'm not going to watch a video.

JBis

$E(x) = P(win)(200-100) + P(loose)(-100) = 1/2(100) + 1/2(-100) = 0$

Simply Beautiful Art

23:02

Well technically speaking if you know the expected pay-off is how much you pay, and you can repeatedly play, then wouldn't it make sense to play it until you win more than lose and then call it quits?

JBis

@TedShifrin I was just providing link so people know where I got the math I just put

J. Doe

@Ultradark I kinda understand the statement of the theorem. Not sure if it helps you though. But it's also a waste of time since you don't even have an identity lol.

JBis

@SimplyBeautifulArt I agree with that.

Ted Shifrin

OK, @JBis.

Simply Beautiful Art

guhhh the sound of getting mentioned

Ted Shifrin

23:03

Of course, you could be in the hole thousands of dollars before your start to see pay-offs.

People often misunderstand the law of large numbers and say "I'm due" ...

Simply Beautiful Art

Yeah of course

but in the long run you expect to exceed 50/50 sometimes in something like this

just like how you can expect to go below 50/50

Balarka Sen

Also the longer you play the shorted the expected deviations from 50/50 become

by strong law

Ted Shifrin

So, if you're infinitely wealthy, would you pay for the privilege of breaking even? :D

Balarka Sen

it might not be worth being in the hole of thousands of dollars to finally get it equalize and never get more than thousand and one dollar

Shine On You Crazy Diamond

0

Q: Natural polynomial maps

Let $f,g : \Bbb{N} \to \Bbb{N}$ be given by polynomial maps such as $f(n) = n(n+2)$. How does one prove that if for sufficiently large $n$ if $f(n) = 1$ then $f(n) = 1$ is the constant function?

Simply Beautiful Art

23:06

lol, but what I mean is it's not breaking even if you quit while you're ahead

Shine On You Crazy Diamond

If $f(n)$ is a natural polynomial map and $f(n) = 1$ for some large $n$ then the function is constant, right?

J. Doe

77

Q: Why did my friend lose all his money?

Not sure if this is a question for math.se or stats.se, but here we go: Our MUD (Multi-User-Dungeon, a sort of textbased world of warcraft) has a casino where players can play a simple roulette. My friend has devised this algorithm, which he himself calls genius: Bet 1 gold If you win, bet 1 ...

probability stochastic-processes martingales

"...which he himself calls genius" lol.

Ted Shifrin

@Shine: It's not really a great question. A polynomial of degree $n$ can only take on a value $n$ times. If it happens more than that, the polynomial doesn't have degree $n$ at all — it's constant.

JBis

@TedShifrin If you loose 1000 times, you have no better chance of winning right?

Simply Beautiful Art

If everyone played the coin flipping game and quit once they had more heads than tails, would the global actual value be greater than 0.5, where heads = 1 and tails = 0?

Ted Shifrin

23:07

Nope, @JBis.

Shine On You Crazy Diamond

@TedShifrin do you have a link or proof for that?

Ted Shifrin

It's in every high school or college algebra book.

Shine On You Crazy Diamond

What proposition?

Is it named?

Ted Shifrin

Every root corresponds to a factor.

Root-factor theorem.

Shine On You Crazy Diamond

thx

Balarka Sen

23:09

@Ted did you just make that up

Ted Shifrin

I make everything up.

Balarka Sen

LOL

J. Doe

lol

Simply Beautiful Art

Factor theorem

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.The factor theorem states that a polynomial f ( x ) {\displaystyle f(x)} has a factor ( x − k ) {\displaystyle (x-k)} if and only if f ( k ) = 0 {\displaystyle f(k)=0} (i.e. k {\displaystyle k...

Ted Shifrin

My books are all full of untruths.

Simply Beautiful Art

23:10

It's called the factor theorem, not the root factor theorem :P

Ted Shifrin

The linear algebra answer I just posted I made up, too.

Shine On You Crazy Diamond

@TedShifrin that doesn't apply here I don't think

Ted Shifrin

Nah, it usually is called root-factor theorem in the US. I dunno about globally. Wiki is not a God.

Balarka Sen

I didn't know it had a name

damn

JBis

@TedShifrin I don't get it.

Simply Beautiful Art

23:10

lol

Ted Shifrin

Well, ask @Balarka, @Shine. He says I make everything up.

Shine On You Crazy Diamond

$f(n) = $ some polynomial that takes only natural values

Balarka Sen

@TedShifrin Hey, I was just surprised you came up with a name for that theorem so quickly!

Ted Shifrin

That actually is not relevant. No polynomial can have infinitely many roots.

JBis

I thought that each instance is individual and doesn't effect each other

Ted Shifrin

23:12

@Balarka: Truth be told — it's in my algebra book (because future teachers need to actually understand it ... not to mention math majors).

@JBis: That's right. So what happens on the 1000th time is not influenced by the fact that you lost the first 999 times.

Balarka Sen

True, people suck at basic algebra, and roots corresponding to linear factors is foundational

Ted Shifrin

You still have a 50/50 chance of winning.

Balarka Sen

I use it everyday while doing field theory exercises

JBis

5 mins ago, by JBis

@TedShifrin If you loose 1000 times, you have no better chance of winning right?

Ted Shifrin

@Balarka: The usual misapplication is to say that every reducible polynomial must have a root :P

JBis

23:13

you said nope

Balarka Sen

Hahah

Ted Shifrin

@JBis: This is an English issue.

Balarka Sen

@JBis You have no better chance of winning but no worse chance either :P

Ted Shifrin

When you ask a negative question, how one answers it is ambiguous.

JBis

what

Balarka Sen

23:14

The chance of winning is not affected by how many times you play the game, as pointed out by Ted

Ted Shifrin

You didn't eat lunch today, did you? No, I didn't. Yes, I didn't? No one says that.

2

Balarka Sen

It's always a constant 50/50

Hahah

Ted Shifrin

Language.

JBis

@TedShifrin Yes because the question itself is "did you eat lunch today"

> , did you [eat lunch today]?

Ted Shifrin

No, the question had a not ... That's entirely my point.

Rephrase your winning question, then, to make it clear.

And it's LOSE, not loose. :P

<--- in an assish mood now.

Balarka Sen

23:16

savage

Ted Shifrin

LOL, you know you miss me when you're gone, @Balarka :D

Balarka Sen

I do, this chat feels like home, tbh, however weird that sounds

JBis

so if expected profit is positive it is worth it, if it is negative it is not?

Ted Shifrin

That's what I would say, yes, @JBis.

And your question was right on the dividing line.

JBis

who chose this sound for pings

wth

@J.Doe good question

cool

thanks!

Ted Shifrin

23:18

Sure :)

JBis

Off to vegas with my newfound knowledge

cya o/

Balarka Sen

LOL

Gone

Alessandro Codenotti

Your newfound knowledge should tell you not to go to Vegas tho

Ted Shifrin

wow, it's getting late in Alessandro land.

Balarka Sen

After two probability courses a friend of mine says if you toss a coin 99 times and get tails it's very likely you'll get heads in the 100th toss

Ted Shifrin

23:21

Tell your friend he would have flunked my course :P

Balarka Sen

Hahah

To be honest I get where he's coming from; he's subtly thinking about a larger probability space in his head whereas he should really be thinking about the probability space of the next toss

The event that he got 99 consecutive tails is itself the unlikely one

That has nothing to do with the next toss ofc

Ted Shifrin

The other thing that some of my students never got, even though a student asked about it and I discussed it carefully (and then put it on the final) is: What's the probability that I draw the king of spades given that someone has already removed a card from the deck?

Thorgott

That's probably the most common mistake in all of probability theory.

Alessandro Codenotti

@TedShifrin I'm becoming a real mathematician, I read papers instead of sleeping now

Ted Shifrin

Wow, @Alessandro. Are you a chain-smoker like Balarka, too? :D

Alessandro Codenotti

23:24

Nah that's awful for your health, I just do methamphetamines to suppress the need for sleep instead

Ted Shifrin

Oh, that reassures me.

Mike Miller

That's what a real mathematician does?

Ted Shifrin

LOL

Alessandro Codenotti

Let's say a real student

Ted Shifrin

But you're an imaginary student.

Balarka Sen

23:25

You'll be in a real hospital very soon

Akiva Weinberger

Can you think of any words with an "ibe" sound other than "fiber"

Balarka Sen

Meth is bad for you, just do plain amphetamines

Akiva Weinberger

Also "ige" other than "tiger"

Alessandro Codenotti

@BalarkaSen That's genius

Balarka Sen

Alessandro Codenotti

23:26

Oh lol, I read it as "math" rather than "meth"

Balarka Sen

Face of amphetamines

Akiva Weinberger

Oh, "gigantic"

Ted Shifrin

DogAteMy: libel and liable

Akiva Weinberger

Thanks

"(S)iberia"

"Cyber"

J. Doe

@BalarkaSen Aka Adderall?

Akiva Weinberger

23:29

"Migraine"

Balarka Sen

iger, eager

:3

Mike Miller

Jibe

Akiva Weinberger

Is that a word?

Balarka Sen

yeah

Alessandro Codenotti

Gy!be

Balarka Sen

23:30

close

loch

@BalarkaSen alternatively he could think that the coin is biased ----

Mike Miller

actually unrelated to jive afaik

Max

Balarka Sen

@loch Yeah that's what, the 99 consecutive tails is unlikely in the probability space of 99 coin tosses

Somehow the brain extrapolates biasedness from that I guess

Max

Would anyone be so kind as to give me a hint on part B of this problem? Part A is trivial if you recognize the combination on the bottom as chi-squared. But for part B, the same trick doesn’t work because the numerator and denominator are dependent.

Alessandro Codenotti

23:32

@BalarkaSen Now I need to listen to F#A#$\infty$ though

Max

I can’t use the mgf technique because I can’t manage to push $Z_1$ and $Z_2$ into separate fractions. The best idea I had was to use the cdf technique; so far I’ve gotten to $F(v) = P(Z_1 \leq \sqrt{(Z_1^2 + Z_2^2)/2 v} = 2P([1-v^2 /2 ]Z_1^2 - Z_2^2 =leq 0)$. On the left, we have a linear combination of squared standard normals. Chi-squared is a sum of squared standard normals, so is there a generalization of chi-squared that gives this … ? I think I must just be missing some insight.

Balarka Sen

Humans are statistical creatures

should write a paper on that

Max

Last eq should read: $2P([1-v^2 /2 ]Z_1^2 - Z_2^2) \leq 0)$

Balarka Sen

"Why students are bad at probability: A socio-statistical explanation"

Get a lot of data and manipulate the p-value a little above 0.05 just to be extra cheeky

Mike Miller

The socio statistical explanation probably comes down to the fact that they can't add fractions

Fix that and you're off to a good start

Balarka Sen

23:37

suddenly probability > 1

Thorgott

@Max it might help to note that $(Z_1^2+Z_2^2)/2$ is Exp(1)-distributed

Max

@Thorgott That's how you do part A, but I don't see how that helps in part B, given that the numerator isn't independent of the denominator.

Thorgott

hmm, maybe separate into positive and negative part and then rewrite $Z_1$ as $\pm\sqrt{Z_1^2}$ and crank the algebra machine?

Balarka Sen

Write down the join distribution and use the multiplicative convolution formula

There's maybe an easier way. Write Z_1 = R cos Theta, Z_2 = R sin Theta where R has the Rayleigh distribution and Theta has the Unif(0, 2pi) distribution, and R, Theta are independently distributed.

Then Z_1/sqrt((Z_1^2 + Z_2^2)/2) = sqrt(2) * cos(Theta)

Leaky Nun

23:54

31 mins ago, by Ted Shifrin

The other thing that some of my students never got, even though a student asked about it and I discussed it carefully (and then put it on the final) is: What's the probability that I draw the king of spades given that someone has already removed a card from the deck?

hmm...

Thorgott

(I haven't written down all the details, but the approach I suggested appears to work without too much effort)

Balarka Sen

It's direct from what I said, all you need to figure is pdf of cos(U) where U ~ Unif(0, 2pi)

You need to differentiate cos^-1, which is why you get square root etc in the answer

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