Mathematics - 2022-09-04

Curious Mind

00:37

I will try.

(Diagram 0)

Recall the question was "Suppose I have a coin, a bag, and a bunch of red and blue marbles. I flip the coin twice and for each head, I put a red marble in the bag and each tail, I put a blue marble in the bag. Question 1: I look in the bag and tell you that I see a red marble; what is the probability that both marbles are red? Question 2: I draw a red marble out of the bag. what is the probability that the remaining marble is red?".

Question 1 mentions that the bag has at least one red ball which means we rule out BB.

As we see in Diagram 0 RR, RB and BR have the same probability to occour.

Question 2 mentions that you take out the red ball.

Here your possible outcomes are $R_1R_2$, $R_2R_1$, $RB$, $BR$. We got 2 RR and one RB and one BR so half of them are RR.

@TedShifrin

In my diagram those 16 possible outcomes' order matter so if you see $B_2B_1$ then first ball is $B_2$ and second one $B_1$.

Ted Shifrin

01:08

So you label the Rs in the second case but not in the first. Can you justify that?

Curious Mind

@TedShifrin which case are we talking about?

Ted Shifrin

Both :)

I’m asking why you distinguish the two Rs in question 2 but not in question 1.

I think robjohn’s version (having an independent observer report in question 1) is essential.

Curious Mind

Ah I see I was showing the order.

Ted Shifrin

If you yourself are looking at the bag, how do you see both balls without knowing what you saw?

Curious Mind

RR came from coin flip and $R_1R_2$ to distinguish that both balls are not same.

Ted Shifrin

01:17

As you’re stating it, if you yourself look and see at random only one of the two balls, the two questions are equivalent.

Wording is crazy important in this stuff.

Bob

Hello Ted

Ted Shifrin

Hi Bob

Bob

anything new with you? lately the news has been bad

perhaps you have some good news you would like to share

leslie townes

that is an understatement.

Curious Mind

@TedShifrin But you are not looking at the bag. You are told there is at least one ball in the bag that is red.

Ted Shifrin

01:22

@CuriousMind As I said, wording is super important. This is not how you phrased it.

@Bob Personally, I’m healthy and very fortunate. Globally, the US and the world suck. So no.

Bob

@TedShifrin Glad to hear you are healthy and fortunate. I have been having a bad few weeks in the stock market. I do not like that.

Ted Shifrin

@leslie I have given in with 90s and turned on the AC (set at 80).

Curious Mind

"I look in the bag and tell you that I see a red marble; what is the probability that both marbles are red? " I think you are talking about this ted

Bob

I noticed that the rocket launch to the moon has been delayed again.

Ted Shifrin

Oh, the stock market should have bombed badly during Tromp. We’re paying now.

Curious Mind

01:25

Oh so by "I" it means RobJohn was talking

leslie townes

we just got back from the pool (weirdly nobody was there) and have it on 75.

Ted Shifrin

@CuriousMind yes

@leslietownes You will crash the electricity grid.

Bob

@TedShifrin I gather you are not a fan of Donald Trump. I have two good friends that are. I am not.

leslie townes

our house doesn't get any late-day sun, so the AC actually is off right now, despite being set at 75.

Ted Shifrin

That is the understatement of the century. He has led to what is likely the end of our democracy.

@leslietownes mine is off, too. Specious ….

Bob

01:28

@TedShifrin I do not think that our democracy will survive.

Curious Mind

ok so I think everything is clear now...

Ted Shifrin

So your friends are partly to blame.

Bob

correction

@TedShifrin I do think that our democracy will survive.

leslie townes

that's quite a correction.

as someone said, wording is crazy important in this stuff.

Ted Shifrin

No, it is likely not to. Pay attention to what the MAGA people have done to voting. And that Graham threatens violence if T is brought to account.

@leslietownes Verily.

Curious Mind

01:30

that is pretty rare way of thinking me see

Ted Shifrin

The GOP has turned to autocracy, anti-constitution.

@CuriousMind Huh?

Bob

not all of the GOP and the GOP is currently out of power

leslie townes

so were the nazis in 1930. we're in the middle of volume 1 of a history of the third reich.

Ted Shifrin

All but a set of measure zero. You are not paying attention, truly.

Leslie and I agree again, very sadly.

leslie townes

i hate agreeing with you, ted.

Ted Shifrin

01:32

I know.

Curious Mind

@TedShifrin Nevermind :)

Bob

German did not have a 200 year history of free elections. We do

Ted Shifrin

You are totally naive. The process has been undermined in plain sight. Republicans can change results in many states if they don’t like them. Have you been following the news?

News according to FOX, maybe.

I’m going to cook dinner. BBIAB.

Bob

Nice chatting

I am going to bed

good night

robjohn

02:48

Interesting that Bob has two friends that are supporters of Mr T. I have two that I have to contend with on a regular basis at the dog park.

Ted Shifrin

03:04

I won't even inquire.

robjohn

Sometimes they spout such garbage.

Ted Shifrin

So they’re fine with destroying democracy and installing a lying thug dictator.

robjohn

The problem is that they don't see him as a lying thug dictator.

For some unknown reason, they ignore the evidence, or have drunk the koolaid and think that the evidence is fabricated.

Ted Shifrin

It’s the fault of Graham, McConnell, McCarthy et al for all rolling over.

i game on a mac

03:20

For the theorem above, does the converse hold true as well?

if T is a normal operator and eigenvectors $u,v$ are orthogonal, then will they correspond to distinct eigenvalues?

I believe it to be the case however I'm just looking for confirmation

Ted Shifrin

No. Of course not.

Look at the most well-known matrices.

Ted Shifrin

03:40

@robjohn Any thoughts on this? It's a bit unusual.

robjohn

03:54

If I am reading this properly, the inner product would be $0$ if the first derivative has constant magnitude.

but I am not sure I understand their question

user4539917

About

> Master student. Now I am taking a research in geometry and analysis.

> Bachelor in Top 2 from China.

Ted Shifrin

@robjohn The fact that $f$ is vector-valued makes this a little subtle.

robjohn

Oh, I guess I didn't notice that. It does complicate things.

Ted Shifrin

Yeah, I missed it when I made my first comment, to which he responded by making the derivative nonsingular.

robjohn

The derivative will be a matrix, I guess

Ted Shifrin

04:02

Yes, $n\times n$.

So the hessian is a vector-valued bilinear form.

robjohn

yeah, I haven't thought about those.

user4539917

(for a long time :)

Ted Shifrin

I have in some contexts (like second fundamental form of a non-hypersurface).

@user4539917 About?

user4539917

@TedShifrin From the user's profile, sir.

Ted Shifrin

I do not understand. What is your comment/question?

user4539917

04:07

Just providing some background context, sir.

robjohn

for whom?

user4539917

The OP

Ted Shifrin

No clue what you’re talking about.

robjohn

me neither

user4539917

threeautumn

157 9

About^

Ted Shifrin

04:09

Oh ….

robjohn

ah, the author of the question

user4539917

:-)

Sorry for being too terse.

Ted Shifrin

He’s asked some not-so-good questions before. But there’s nothing wrong with this question. I just don’t know the answer yet.

robjohn

I have no thoughts on it either.

Ted Shifrin

I suspect there’s a counterexample, but who knows.

robjohn

04:25

Got a downvote on an answer today that I think is fine (and so have the 138 upvoters). I have no idea if they just don't understand the answer or one of the steps. Without comment, I can't help.

once in a while, I get someone who downvoted when they thought they were upvoting

after 5 minutes, they can't change their vote unless the answer is edited.

user4539917

138 to 1 sounds like a random act of unkindness...but, yeah it could be just a misunderstanding

Ted Shifrin

04:40

@robjohn I misvoted once like that.

user4539917

04:53

Honest mistake.

Drew Brady

Suppose I have a positive semidefinite operator P on l^2 (self-adjoint and bounded). Let S_j denote projection from l^2 onto first j coordinates.

I am wondering if the supremum sup trace([S_j (P + L^{-1})S_j^*]^{-1}) converges to sup trace((P + L^{-1})^{-1}) as j \to infinity, where the suprema range over L, which are self-adjoint positive operators with tr(L) <= 1.

Thomas Peng

09:18

Hi everyone.

Could anybody take a look at my question plz?

5

Q: Numbers of ‘Equilateral Division’ of 2-$\sqrt{3}$-$\sqrt{7}$ Right Triangle

Problem: A right triangle has two right angled sides with lengths $2$ and $\sqrt{3}$. Cut this right triangle twice (into three pieces) and put the three pieces together to form a equilateral triangle. Solution: Now my questions are: Is there another way? How do you prove that? More generally, i...

geometry triangles

one potato two potato

11:18

I'm trying to prove the following: Let $\Gamma = \{|z-z_0|=r\}$. Then for any $z\in\Bbb C_{\infty}$,
$$z^* = {r^2\over \bar{z}-\bar{z}_0}.$$
I already know that if $\Gamma$ is the unit circle, then $z^* = {1\over\bar{z}}$. For general case, I used the fact that cross ratio is invariant under Mobius transformations. First note that $|z_i-z_0| =r$ for $i =1,2,3$.
\begin{align*}
\overline{[z:z_1:z_2:z_3]} & = \overline{[z-z_0:z_1-z_0:z_2-z_0:z_3-z_0]}\\
& = \overline{\left[{1\over r}(z-z_0):{1\over r}(z_1-z_0):{1\over r}(z_2-z_0):{1\over r}(z_3-z_0)\right]}\\

Something is wrong in the middle of the argument but I can't find

Here, $z^*$ is a symmetric point relative to $\Gamma$ i.e., $\overline{[z:z_1:z_2:z_3]} = [z^*:z_1:z_2:z_3]$ for $z_1,z_2,z_3\in\Gamma\subset\Bbb C_{\infty}$: circle.

Jam

11:45

a prime element of a Ring R does it remain prime element on the ring R[a]={m+na|m,n in R}

2 is a prime in Z. But 2 is not a prime in $\mathbb{Z}[i\sqrt{17}]$ since it divides -16 but neither $1-+i\sqrt{17}$

Thorgott

@Jam depends on what a is supposed to be

robjohn

an article, usually representing a single item ;-)

leslie townes

what a is supposed to b? color me confused

Jam

12:33

Let $I=<x+3y,x-y>$ and $\mathbb{C}[x,y]$ is I prime ? is I principal?

can i say x=y so x+3y=4x=0 so $\mathbb{C}/I \cong \mathbb{C}$

Thorgott

you have to formulate those equalities with residue classes, but yes the conclusion is correct

Jam

i want to write it in amore rigorous way will you help me?

i gues ill need a couple of ismoprhism theorems

$\mathbb{C}[x,y]/<x-y> \cong \mathbb{C}[x]$ trivial isomorphisms

now $\mathbb{C}[x] \ <4x> \cong \mathbb{C}$

whats the argument now with I and c[x,y] to combine them

Thorgott

$\mathbb{C}[x,y]/I=(\mathbb{C}[x,y]/\langle x-y\rangle)/(I/\langle x-y\rangle)$ and under your isomorphism $\mathbb{C}[x,y]/\langle x-y\rangle\cong\mathbb{C}[x]$ the ideal $I/\langle x-y\rangle$ corresponds to $\langle 4x\rangle$, so this quotient is $\mathbb{C}[x]/\langle 4x\rangle\cong\mathbb{C}$

Jam

12:54

ohh thanks !!

Thorgott

note that the projection $\mathbb{C}[x,y]\rightarrow\mathbb{C}$ inducing this isomorphism is given by $x\mapsto0$ and $y\mapsto0$, so its kernel is $I=\langle x,y\rangle$ a posteriori

you could also argue $I=\langle x,y\rangle$ for an alternative argument, but it's more or less equivalent

Jam

yes that was my initial argument

<x,y>=<x-y,x+y> and thats where i stuck

im trying to use the following property

<m,n>=<m+rn,n> for any r in the ring

Thorgott

yeah, you can use that as well

Jam

didnt find a clever way to get it 3y

if you do id be glad to tell me. I tried reversing the argument and solve for some r but no luck

Thorgott

$\langle x+3y,x-y\rangle=\langle (x+3y)-(x-y),x-y\rangle=\langle 4y,x-y\rangle=...$

Jam

13:08

omg

that was redicussly easy

i forgot i can multiply by different units its generator

tho i shouldnt rely on memory on something like that

<u1x,u2y>=<x,y> for unitis in R

robjohn

@leslietownes I c...

Jam

is principallity of ideals invariant under isomorphism?

i know it preserves the p.I.D structure

but for 2 specific ideals ?

Thorgott

what do you mean

Jam

if I and J are isomorphic

and I is principal

is J princiapl?

well isomorphic as rings

Thorgott

13:23

so I and J are rings?

Jam

is there any other notion for ideals to be isomorphic

i was thinking <x+3y,x-y>=<x,y> so the first one is not principal. but this is not an isomorphism they are actually equals as sets.

but suppose they dont belong as subsets to the same set. I and J and I is generated by one element and is isomorphic to J

is J generated by the image of that element too?

i think yes just like vector spaces

bases go to bases

Thorgott

what does it mean for ideals to be isomorphic

note than ideal is not an independent notion, it lives inside some ring

so keep that in mind when you try formulating what you want

Jam

yes sometimes i think of ideals as rings on themselfs

but it is no right

cause rx in I for every r in R

Shinrin-Yoku

It is well known that SSA is not a valid congruency rule; however if we are given two sides and an angle opposite to the longer side then it becomes valid, how to prove this? Is there a geometric proof? Also is it always the case that it fails if the angle is opposite the shorter side?

Jam

13:40

im trying to prove that an epimorphism from k[x] where k is a field to an integral domain must be 1-1. But i dont want to use the fact k[x]/kerf integral domain iff kerf is prime.

i want to prove it more elementary

i know f(1)=1 since i go to an integral domain

f(g(x))=0 =>$ a_nf(x)^n+...+a_0=0$

since $f(x)^n cant be zero in an integral domain all ai=0 and im done?

ohh right i have to exclude i cant divide with f(x) to say all ai=0

ok solved it

so either it is a field or the map is 1-1

Thorgott

that initial statement is very incorrect

the alst statement is correct

though I do not follow your argument

Jam

since i have an epimorphism to an integral domain f(1)=1

now take any g(x) f(g(x)) to be zero is equivalent to $a_nf(x)^n...+a_0=0$

since f(1)=1 f(a)=a

hm.. nope thats wrong

i dont know is ai exist in the image as elements

if*

Shinrin-Yoku

@robjohn could you take a look at my question?(if you have time)

Jam

it could be K[x]--> Z

Thorgott

well, $f(1)=1$ holds for any ring homomorphism, nothing to do with epimorphism or the codomain being an integral domain

Jam

13:54

isnt the zero function also an homomorphism?

Thorgott

not a ring homomorphism

be precise about what you mean by "epimorphism"

Jam

in my notes we count the zero function as a homomorphism

say Z to Z f(m)=0

point is my argument is not valid for many reasons

Thorgott

ok, then define what "epimorphism" means here

you have a map from k[x] to some integral domain, what properties are you assuming this map to have

Jam

f(a+b)=f(a)+f(b)

f(ab)=f(a)f(b)

kai is onto

Thorgott

kai?

Jam

13:58

and*

damn kai=and but in greek

i think i got dislexia hahaha

i wanna prove the result in an elementary way

Thorgott

oh, I see

Jam

not using R/I integral domain iff I is prime

Thorgott

so a surjective ring homomorphism except you are not requiring that $f(1)=1$

Jam

ye but to be onto it has f(g(x))=1 and u can prove that g(x) is 1

Thorgott

however, as you say, in this case because the codomain is an integral domain, either $f(1)=0$ in which case $f$ is the zero homomorphism or $f(1)=1$

Jam

14:00

ye

Thorgott

for the record, when you are talking about homomorphism of rings, epimorphism =/= surjective

anyway, primality is a very elementary notion. I don't think you'll have success formulating this argument without (implicit) reference to primality.

Jam

ye ok

ty

Tryst with Freedom

14:46

halo my beautifuls

I have a deep mathematical question : Who is the most attractive mathematician of all time?

Sourav Ghosh

15:07

"Attractive" in which sense?

Koro

15:55

Climate has huge effects on electronic devices. I think in very humid climates, laptops start to malfunction.

kuspia

The probability that the median of these N numbers chosen from the [1, M] range is an integer. can somebody suggest something

graffe

16:28

I feel this must be all known... If I repeatedly add a uniform random number from (0,1) what is the expected number of numbers before you get to 1 or more?

leslie townes

16:46

63

Q: Choose a random number between $0$ and $1$ and record its value. Keep doing it until the sum of the numbers exceeds $1$. How many tries do we need?

Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?

probability probability-theory expectation

Shinrin-Yoku

Any help with this: chat.stackexchange.com/transcript/message/61929344#61929344

@leslietownes

?

If you have time

leslie townes

mm, i dunno. interesting question though. it probably follows from some geometric characterization of the ambiguity (i.e. the set of possibilities) that having the longer side then resolves.

ted: our coyote (lounging next to a tree outside) says hi.

Ted Shifrin

Thanks. Is he overheated?

leslie townes

he looks hot, yes.

Ted Shifrin

It’s always good to ask for an expected value when you

know not what it be.

leslie townes

16:58

my wife and daughter are now outside watering some plants, in full view of the coyote. the concern is that if we don't do this, the plants will die

Dr. Danial Sadeghi

17:29

Hi

Jakobian

My book didn't prove the James characterization of weak compactness in full generality, but restricted the proof to the separable case (using Godefroy theorem) because it's apparently hard in the non-separable case. I wonder if I should look up the proof in full generality, or maybe I should just give up?

(give up for now at least)

leslie townes

give up unless you need the non separable case for something

Jakobian

Yes. That's probably the most sane thing to do. I did even search in the references but it looked crazy hard with my current knowledge.

Koro

What is this symbol $\ni$?

Is it 'such that'?

Jakobian

Isn't it $\in$ but backwards

Xander Henderson

17:42

@Koro Depends on context.

Koro

I have seen this usually being used as in $\mathbb Z\ni x$.

Xander Henderson

I am most familiar with it in the setting $A \ni x$, where $A$ is a set, and $x$ is an element of the set.

You could read it as "$A$ contains $x$ as an element."

Koro

But then I saw $\ni$ being used in class in the place of 'such that'.

Xander Henderson

Or, really, just $x$ is an element of $A$.

Koro

@XanderHenderson yes, that's what I'm also familiar with.

Xander Henderson

17:44

@Koro It can be used that way, yes.

See, for example, math.stackexchange.com/a/2777911 .

leslie townes

i've ranted about this before, but the 'such that' symbol can simply be omitted from the 'formal/symbolic logic' that people pretend they are doing when they write it.

Koro

@XanderHenderson thanks a lot :).

leslie townes

it's a really bad habit and adds nothing. imvho.

Koro

then, what's its subsitute?

leslie townes

writing nothing at all.

you can simply omit it, or if it's somehow ambiguous, it would have been ambiguous anyway.

Koro

17:47

Ohh.

leslie townes

prototypical example would be something like forall x exists y such that __. just erase "such that." it's nothing

i guess it allows people to think of symbolic formulae as more closely corresponding with spoken language, which isn't nothing.

Koro

(Trying to write without 'such that' and seeing how it looks.) Let $(X, M, \mu)$ be a measure space. Then I define $\bar M:=\{ E\cup F: E\in M, F\subset N, N\in M, \mu(N)=0\}$. Suppose $S\in M$, then there exists $E, N\in M, F\subset N, \mu(N)=0, S=E\cup F$.

hmm, looks okay. :)

@leslietownes sometimes I feel like not writing 'such that' or '$A\subset X$ with A having property so and so' at all.

leslie townes

18:11

if you were feeding that to a computer i might replace some of the commas with "ands" but that is definitely the spirit.

Thorgott

19:05

you can greatly simplify this by using outer measure

Ted Shifrin

The people I’ve seen insist on the s.t. Symbol put dots on either side. I personally have never used it …. Not once.

leslie townes

19:43

it seems to have cooled a little here. only about 90 now.

Ted Shifrin

It was 87 here at 8 AM. Good thing this is all just a libtard hoax.

Koro

20:05

Suppose that $B_n:=\cup_{i=1}^n A_i \in M$ for all $n\in \mathbb N$, can I then say that $\cup_{i=1}^\infty A_i\in M$?

Xander Henderson

Thoughts on how this question should be tagged?

Is it elementary number theory?

Koro

The problem is that: I can’t even use contradiction here. If $\color{red}{\cup_{i=1}^\infty\not \in M}$, I am having difficulty interpreting red part.

leslie townes

discrete something something.

koro: what's M? some results in measure theory do take the flavor of "there is no abstract or general reason why this union ought to be in __, but if it is, [conclusion holds]."

Koro

I’m trying to prove that the set of all v* measurable sets is a sigma algebra , which I’m denoting by M.

leslie townes

look for example at footnote 3 of page 9 of uio.no/studier/emner/matnat/math/MAT2400/v13/mathanalch6.pdf

Koro

20:23

Thanks Leslie. I understood.

For example: If I take the set of natural nos. N, and then I define M={ E (subset of N): E is finite or E^c is finite}

Then, it is an algebra but not a sigma algebra.

leslie townes

koro i am not presently recommending this as an option, but finitely additive measure theory is a good way of re-approaching measure theory after learning the countably additive case

so maybe a semester or two from now, check it out

it's very much not what you need to develop a well behaved integral but it sheds a lot of of light on why countably additive proofs are the way they are

Koro

20:41

@leslietownes yeah, the book also uses finite additive of outer measure for disjoint sets.

Koro

21:11

@leslietownes which book is this?

ObjectsMorphisms

23:57

Anyone know whether the "set of all propositions" is a small set categorically-speaking?

I need to know for some code

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