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00:07
at last we get some closure.
i thought they had some medications for floppy discs?
0
Q: For each positive integers $n$, $C_n=\{(x,y)\in \mathbb R^2:x^2+(y-n)^2=n^2\}.$ Let $C=\cup_{n\in \mathbb N} C_n.$ Find $\overline{C}$.

Unknown xFor each positive integers $n$, $C_n=\{(x,y)\in \mathbb R^2:x^2+(y-n)^2=n^2\}.$ Let $$C=\cup_{n\in \mathbb N} C_n.$$ Find $\overline{C}$. My attempt:- $x^2+(y-n)^2=n^2$ is a circle centred at $(0,n)$ with radius $n.$ $x^2+(y-n)^2=n^2 \implies x^2+y^2-2yn+n^2=n^2\implies x^2+y^2+y^2-2yn=0 \implies...

00:24
Hi all, I have a silly question: Can I ask a question?
Don't ask to ask. Just ask.
i need to ask, why do people feel the need to ask?
00:41
i want someone to sign an NDA before i ask. where do i go for that
@copper.hat I imagine it takes the pressure off of asking. Your question can't be that annoying if the person you're asking has the ability to say no.
i think the problem on the receiving end is the person who may say 'yes' to 'can i ask a question' worries about whether they have signed on to a commitment of undetermined scope, when they only intended to provide the yes-no answer.
my preferred opener is "hey, can i ask a question? i will not take no for an answer."
i have lost all my legal nous from years of inactivity
i will sign anyone's nda now
@copper.hat Can I ask you a question?
@XanderHenderson okay
00:51
@Unknownx You aren't @copper.hat hat. My question is for him.
the standard response is "yes, you can even ask me another. LOL LULZ LMAO" and emojis if you got 'em.
@XanderHenderson sorry for the misunderstand.
@XanderHenderson I swear I did not kill that Mojave rattlesnake, it was dead when I found it.
I might add there was an unusually high density of corporate lawyers in the vicinity.
@copper.hat That wasn't my question. I just wanted to know if I could ask you another question.
01:11
@XanderHenderson I was just deflecting.
I did risk the CURED gods by answering a convex PSQ.
@copper.hat So... my question is: would it be okay if I asked you a question?
@XanderHenderson Excellent question.
Would that be two questions?
@copper.hat No, I just wanted to ask if it was okay to ask you a question.
Have you signed Leslie's NDA?
@copper.hat No. But I'm willing to.
01:23
I'm trying to prove this: Every integer $n > 1$ can be written in one and only one way in the form $n = p_1 p_2...p_r$ where $p_i$ are positive primes such that $p_1 \leq p_2 \leq ... \leq p_r$.


But it kind of seems like it's just restating the Fundamental Theorem of Arithmetic. Is there some subtle difference that I'm missing? Or is it valid to just say it follows from the theorem?
That seems to be the fundamental theorem of arithmetic. Do you have a proof of that theorem?
Looks like the fundamental to me.
That's exactly what I thought. And yes, I do have a proof for the theorem.
Then you're done.
My text just says that the above follows from the theorem and to prove it lol
Aight, then
01:25
Weird. I wonder what the author(s) were thinking.
Idk, maybe I'll ask my professor about it
I mean, the FTA does not say anything about the order of the prime factors... like, there is a two line statement about the commutative nature of multiplication.
That's true
The version of the FTA my text uses points out that if you were to find two different prime factorizations for a number, then they can always be reordered to show they are equivalent. Now looking at the above, it seems like a slightly more strict version of this statement by not mentioning an ability to reorder the integers.
 
2 hours later…
03:29
$\det\left(I_n+\begin{pmatrix}A_1 & B_1 & 0 & \cdots &0 \\
A_2 & B_2 & 0 & \cdots &0 \\
\vdots & \vdots & \vdots & \ddots &\vdots \\
A_n & B_n & 0 & \cdots &0 \\\end{pmatrix}\right)=\det\begin{pmatrix}1+A_1 & B_1 \\
A_2 & 1+B_2\end{pmatrix}.$
Any clever way to eyeball this
does a general version of this exist? what happens if we add more columns to it?
have you considered where "cofactor expansion across the first row" gets you, both in that case and in potential generalizations
03:56
Quick question, if we have $f'(t)=g(t)+h(t)$ for distinct functions $g$ and $h$, does that imply $g$ and $h$ have antiderivatives?
nah. g can be arbitrarily shitty and h can be -g.
sorry, meant to say "$g≠-h$" instead of "distinct"
g can be "nice - shitty" and h can be "shitty" without g being -h and the conclusion still not working out right
h could even be "different nice + same shitty"
i do what i can to elevate the discourse around here
ah dang. well thanks tho
if you have a general function, can you decompose it into "something nice plus something as shitty as possible" is roughly akin to the ideas that underlie things like the lebesgue decomposition of a measure, maybe there is something to look for there
but that's semi orthogonal to the query above which imposes specific g, h
04:24
love the technical terms.
Technical hogwash.
04:44
such language deeply offends me, of course.
Hi. Suppose that D is any function D: $M(2, R)\to R$ that satisfies D(AB)=D(A)D(B) for all A, B in M(2,R). Given that: $D(\begin{bmatrix}0&1\\1&0\end{bmatrix})\ne D(I)$, then it is to be shown that for all A in M(2,R)-1)D(0)=0, 2) D(A)=0 if $A^2=0$, 3) D(B)=-D(A), if B is obtained by interchanging the rows of A, 4) D(A)=0 if a row of A is 0, 5) D(A)=0 whenever A is singular.
I showed 1) to 3) but having difficulty in 4).
For 3), $B= \color{blue}{\begin{bmatrix}0&1\\1&0 \end{bmatrix}}A$, so $D(B)=\color{blue}{D(E)}D(A)$. Also, $A= EB$, so D(A)=D(E)D(B). Hence, $D(E)^2 D(A)=D(A)$ for all A. So $D(E)=1,-1$. If D(E)=1, then D(I)=0 (D(E) can't equal D(I)). It follows that 1= D(E)=D(EI)=D(E)D(I)=0, contradiction. So $D(E)=-1$. It follows that D(A)=-D(B).
nvm, I found the answer here.
2
Q: Hoffman Problem. Section 5, Problem 13.

ViKaNLet $\mathbb{R}$ be the field of real numbers, and let D be a function on 2x2 matrices over $\mathbb{R}$, with values in $\mathbb{R}$, such that $D(AB) = D(A)D(B)$ for all $A, B$. Suppose also that $$D\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \neq D\begin{pmatrix} 1

05:15
@leslietownes right. Actually I was wondering about an alternate method where we don't need to expand, but seems like that might be the only way to go. I will try that by expanding along the first row
05:47
I think I have problem with integration by parts
Is there a good resource for that?
 
2 hours later…
07:50
user image
2
08:36
0'RLY?
08:56
I think the last time I saw someone say "O'RLY" was when I was in elementary school
good nostalgia tho
 
4 hours later…
13:18
How is average of sinx over a long interval solved?
How is it zero
13:29
How do I find the average of a function in a given time interval
I meant in a given interval*
14:08
What is your definition of average?
If we all talk at once it's confusing
Hii, I want to know how can I write $\sin(2 \cos^{-1}(x))$ in simplest form.

I can think like this,
$\sin(2 \cos^{-1}(x)) =2 \sin(cos^{-1}(x)) \cdot \cos(\cos^{-1}(x)) = 2\sqrt{1 - x^2}\cdot x$

Is it right?
looks OK to me
@HelpMeToUnderstandContours Without taking the time to draw some pictures or working through it myself, I have no idea. Have you tried graphing both sides of that equation using, for example, GeoGebra? You should be able to check your work without requiring others to go through it for you. ;)
(But it looks superficially correct to me---it has the form that I would expect.)
i'll draw the picture: the relevant angle is /_ in /_|, where _ is x, and / is 1. and its sine is indeed sqrt(1-x^2)
i have a phd in art, a lot of people don't know this
14:18
@leslietownes Oddly enough, that is very clear. O_o
But, mostly, I don't want to do someone else's homework.
helpme did the homework! we're just drawin' pictures, man! on a monday morning!
@leslietownes I've spent the last 90 minutes grading other people's work.
I need to write a quiz now. :/
And an exam.
I probably should have done all of this over the weekend, but I am trying very hard not to work on the weekends.
I didn't entirely succeed this weekend (I did a bunch of committee work). :\
boo to committee work.
i hope you did it badly, in revenge for doing it on the weekend.
14:37
@XanderHenderson Thank you very much! :)
@leslietownes No. My long term goal is to go into administration, so I feel compelled to do committee work well.
@leslietownes on a Monday morning in Mammoth Lakes... Just back from a long walk with my dog.
robjohn is showing us all how to do it
What happens if I take away the first set, instead of the middle one in the construction of the cantor set? Do I get a set with the same properties?
14:46
@robjohn That is a lovely area.
And there is a pretty significant obsidian source out there.
It is a supervolcano. There are lots of igneous rocks at the side of the road here.
Also, I am being very, very lazy. I am copy-pasting the most of last year's exam, and changing some numbers around.
That could be called efficient ;-)
@robjohn Indeed. But Mammoth Mountain obsidian tends to have swirls of black and red. It is a really pretty and distinctive stone.
@robjohn What do you think?
14:48
@XanderHenderson I will keep my eyes open for that. I will be going on a hike tomorrow.
@robjohn Awesome!
My recollection is that there is some BLM land north of Mammoth Lake, which includes part of Obsidian Dome. You are permitted to collect obsidian from BLM land, if you desire.
You are not allowed to take anything from Forest Service land, which is what most of that area is.
(Most of that area is in the Sierra NF, right? Or does the Stanislaus extend that far south?)
Also, I left my dice at home, so randomizing the numbers on this exam is being annoying.
15:04
@leslietownes what’s your opinion on my comment above ?
@Shinrin-Yoku have you tried a few iterations?
Yes @robjohn a similar set appears
@Shinrin-Yoku It is not clear to me what this means. Let me see if I follow: you start with $C_0 = [0,1]$, then define $C_1 = [1/3,1]$.
Do you then take the "first third" from each of $[1/3,2/3]$ and $[2/3,1]$? Or do you remove the first third of $[1/3,1]$?
Yeah take the first third of each
Ah. Okay.
You are going to end up with another Cantor set. Just not the "usual" Cantor set.
15:12
Yeah does it have a name
Cause properties seem to be same as normal
Although there is a bit of a complication: is 0 a member of this set?
or 1/3?
In any event, I would probably model this as the attractor of an iterated function system with two maps: $x \mapsto (1/3)x + (1/3)$ and $x \mapsto (2/3)x + (2/3)$. The attractor is a Cantor set. I doubt it has any other "special" name.
Just remove the whole thing. But I don’t see a problem with letting them be.
Hmm thanks @XanderHenderson
@XanderHenderson The real numbers with only 1 and 2 in the trinary expansion.
@robjohn Yup.
Does there exist a decomposition of $M=(0,1)^3$ into $F_{\alpha} \cong S^1 \times (0,1)$ accumulating at only two points each? By decomposition I mean $\bigcup_{\alpha} F_{\alpha}=M.$ Say $F_{\alpha}$ is assumed $C^{\infty}.$ What might a proof sketch look like?
One problem with existence, is I think the surfaces $F_{\alpha}$ will tend to get arbitrarily close to being $C^0$, where if the cube-like boundary $[0,1]^3$ was included you would get a non-differentiable surface
15:39
Have a great hike, @robjohn!!
16:22
It has been a very long time since I have taken probability, but this answer strikes me as wrong:
1
A: A fair coin is flipped $100000$ times and you get $100000$ Heads in a row. What is the probability that you get Heads on $100001$th flip?

MattYou are correct, the answer is 1/2. Ask your friend this question: If $2^n$ people were flipping a fair coin $n$ times each, then the probability $p(n)$ of at least one of them getting $n$ heads in a row is $$p(n) = 1 - \left(1-\frac{1}{2^n}\right)^{2^n}$$ Observe that $$\lim_{n\rightarrow\infty}...

It seems to assert that if you run the experiment an infinite number of times, then the probability of getting 100,000 heads in a row is only around 0.63.
It should be 1, no?
If an "infinite number" of people each toss a fair coin 100,000 times, then, with probability 1, an infinite number of them should get 100,000 heads...
16:36
I have no idea what "simple Bayesian analysis" is.
Given Cayley Hamilton theorem for diagonal matrices, how can I extend it to any square matrices?
Either we know the coin is "fair" or we don't.
@Koro Work over $\Bbb C$. Diagonalizable matrices are dense.
yes, professor Ted. I'm having difficulty understanding proving this statement. How would you suggest proving it?
that diagonalizable matrices are dense in M(n, C).
Every matrix is arbitrarily close to a matrix with distinct eigenvalues.
Does that work if I replace C (algebraically closed field) by R?
16:40
No, because there are real matrices with non-real eigenvalues. But if the polynomial identity holds working over $\Bbb C$ for a real matrix, it obviously holds working over $\Bbb R$.
17:06
@robjohn The Iva Belle hot springs are in the vicinity. Nothing spectacular, but if you are near.
@Koro Maybe look at the Jordan normal form and add a matrix of the form ${1\over n} \operatorname{diag}\{1,2,...\}$?
I was trying to avoid JCF ... which also only works over $\Bbb C$, anyhow. :)
@copper.hat: I'm trying to avoid JCF :(.
I think my sketch is the neatest way to do it.
34
A: Diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices.

Martin ArgeramiI cannot really follow the reasoning you are hinting in your question, but here's my take: To talk about density you need a topology. Since $M_n(\mathbb{C})$, the space of complex $n\times n$ matrices is finite-dimensional, a very natural notion of convergence is entry-wise; so we can consider ...

This also uses JCF :(
I told you the proof without JCF.
17:14
@XanderHenderson The author seems to assume that so many people take part that the expected number of people actually throwing $10^5$ heads in a row is about $1$. Not sure however what this has to do with the question. And yes, in an infinite number of throws of a fair coin, the probability is $1$ that eventually $10^5$ heads will occur in a row.
Your other claim is also correct : If infinite many people participate (not just the number of people the author assumes) then infinite many people will throw $10^5$ heads in a row since the chance is greater than $0$.
hmm, By Schur's lemma every matrix over C is similar to an upper triangular matrix.
Changing diagonal entries of the upper triangular matrix slightly (such that all entries remain distinct) will make it diagonalizable.
Then, using continuity of polynomials and determinants, I'm done.
:-)
17:40
I never really got into linear algebra properly
@XanderHenderson Matt is saying that if $2^n$ people each toss $n$ coins then the probability of at least 1 person tossing a perfect run of heads approaches $1-1/e$ as $n\to\infty$. But if we let the number of tosses per person and the number of people both approach infinity independently, then his limit is no longer valid. But I Am Not A Probability Expert. :)
17:57
@PM2Ring Ah, right. But this has nothing to do with the problem. Okay. There's my confusion.
@TedShifrin I noticed you mentioned Bridge in the transcript. You may find this amusing: statnews.com/2016/06/24/trumps-bridge-games-biomedical-studies A few things have changed in the 6 years since that article was written...
People are generally not good with questions involving extreme probabilities. There's a strong tendency to go "That can't be right, so I'll answer this more reasonable related question instead".
We've had some great discussions about this topic on Physics.SE, eg Can 1 kilogram of radioactive material with half life of 5 years just decay in the next minute?
18:16
@PM2Ring Indeed.
But I strongly suspect that something like this came up on an intro prob class: "Okay, so the whole point of independence is that the previous results have no influence on the next result. For example, if I toss a fair coin eleventy-bajillion times and get heads every time, I still only have a 50/50 shot at getting heads on the next toss, because the coin is fair, and the tosses are independent."
We can all acknowledge that getting eleventy-bajillion heads in a row is very unlikely, and we would probably stop assuming that the coin is fair somewhere in that sequences of tosses, but that isn't what the problem is stating, nor asking about.
:/
Agreed.
@leslietownes I just noticed that your avatar appears to be someone who has drowned in a kiddy pool. Should I be reporting your actions to the police?
What did you do?
18:35
Jul 11, 2021 at 6:00, by leslie townes
it's a still from an irish public service announcement about water safety. the link is one of the featured comments.
@XanderHenderson I think those are ducks in the avatar.
@Koro They are clearly loafers.
Or, possibly, duck billed platypi. (Yes, "platypi". I don't care if the mainstream media insists that the "correct" plural is "platypuses".)
they could be platypuses.
@XanderHenderson I see the resemblance 😬
@Koro PLATYPI!
@TedShifrin hmm, is there an easy way to see this without triangularizing, actually
18:51
Here's the context for Leslie's avatar:
Jul 10, 2021 at 3:50, by copper.hat
ireland's version, the one we joke about is at 1:04: https://www.youtube.com/watch?v=NjWHrjCG1LQ
"Platypi" is common. "Platypodes" is technically more correct, but pretty rare.
@XanderHenderson he's just resting
@leslietownes His feet are nailed to the edge of the pool!
@PM2Ring Man... that is some dark stuff.
He's pining for the fjords peat bogs.
I thought Leslie's display picture was one of those new fangled AI generated images.
I wrote on a blackboard app today. Wish I could explain the question is about train 🚂 speeding in less time and moar distance. I actually forgot the question but this should still be right. Tomorrow, I'll be looking into how to solve this system using matrix method.
19:11
@XanderHenderson It is pretty dark. But some people deal with dark stuff by joking about it.
Personally, I think Leslie's avatar is in poor taste, but hey, it's his choice, and I don't think it violates any SE CoC rules.
One of my sisters lived on a rental waterfront property when she had two toddlers. The view was beautiful, but it was just too nerve-wracking, so they moved.
19:32
leslie is just an avid advocate for water safety
I was actually in the chat room when Dr Leslie made that avatar; the whole context of the discussion is what gives the image its flavor and it was definitely not made in bad taste.
@Thorgott The alternative is to prove (say by Rouché, although Artin has a more naive argument) that the roots of a polynomial (as a multiset) vary continuously with the polynomial.
The avatar stems from an Irish water safety ad from the 70s.
yeah, I was thinking of that, but I didn't actually see how to finish the argument from there
19:42
Ah, right, cuz wiggling the roots doesn’t wiggle the matrix without triangularizing. I’ll ponder.
Mad
Mad
20:02
If i place my hands one metre far from each other. And then make the left one go closer to the right one, in finite time, i will reach half the distance. and if i keep going, i will keep reaching half and half. Eventually and obviously, i need finite time to clap my hands, however, i effictively kept halving the distance, which i can do infinitly.
Does this make sense to any of you
It did to Zeno.
Nah, it’s either that or prove the characteristic polynomial is an open map from matrices to polynomials.
Mad
Mad
So i guess, you can do an infinite process in finite time, which is funky
On a related note, I posted this comment yesterday. I thought it was reasonably clear, but (at least) one reader found it confusing:
"an infinite sum of numbers larger than 0 would always go to infinity" Not necessarily. Eg, $$\sum_{i=1}^{\infty}\frac1{2^i}$$ isn't infinite. — PM 2Ring 15 hours ago
20:36
wow, i've been using this avatar for apparently over a year now. i missed my anniversary party. (i agree with PM that it is in poor taste, concealed slightly by the fact that on many screens it is not recognizable until you click in.)
@leslietownes Right. Reported. A moderator will be along shortly to reprimand you.
sigh. i wish i could give those lousy mods a piece of my mind.
stupid diamond crew
Right... so I hear that @leslietownes has an inappropriate avatar?
Huh... I don't see the problem.
It's just a pair of platypi in a kiddy pool...
20:57
@copper.hat I will have to find them on the map and see. We are just settling in today. I took Rosie for an hour walk in the neighborhood this morning to get better acclimated to the altitude (I was hoping that going up and down the stairs 20 times unpacking our car last night would do it).
21:20
@TedShifrin something like this, or more general?
@XanderHenderson wearing robes, perhaps.
21:55
inf
xander: congrats on always existing
@robjohn For the denseness of diagonalizable matrices, we need to know that if we take distinct eigenvalues near our given eigenvalues, there’s a nearby matrix with those eigenvalues.
@leslietownes It is?
22:11
it's a matter of opinion. the coyote says hi.
@leslietownes bad boy. Now think on what you've done.
@leslietownes No coyotes here, but there have been many bear sightings recently, I've been told.
I think there was one in The Village near us. We saw an overturned trashcan and Rosie was very intrigued by a spot nearby.
@leslietownes It's nice that you support the local wildlife.
i like this one he mostly just sits there vibing
@TedShifrin That is a bit more involved.
22:21
@leslietownes Of course he does.
Hello
I have a question about the Laplacian in polar coordinates.
Namely?
Strauss, in his PDE book, writes
$$
\frac{\partial^2}{\partial x^2}=& {\left[\cos \theta \frac{\partial}{\partial r}-\frac{\sin \theta}{r} \frac{\partial}{\partial \theta}\right]^2 } \\
=& \cos ^2 \theta \frac{\partial^2}{\partial r^2}-2\left(\frac{\sin \theta \cos \theta}{r}\right) \frac{\partial^2}{\partial r \partial \theta} \\
&+\frac{\sin ^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2}+\left(\frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}+\frac{\sin ^2 \theta}{r} \frac{\partial}{\partial r}\right)
@XanderHenderson I was going to ask you a question but I forgot what it was.
22:31
Where do those last two terms come from? How can you even get 5 terms when squaring a binomial?
@copper.hat Can I ask you a question?
Always.
Right, so this is my question: how many questions am I allowed to ask?
You have 3 wishes.
That's straight from the book, but I don't understand it.
22:32
As kids we could never figure out why the last wish was not a request for an infinitive number of wishes.
Wishing for wishes is explicitly forbidden.
@DarkRunner Which term can you not account for?
What is $\frac d{dx}(\sin x\frac d{dx})$, for example?
In particular, how does this act on a function $f$?
@XanderHenderson Whenever my students asked this, I always said, "You just did."
@TedShifrin :D
23:16
0
Q: Find the order of convergence of the sequence $\{\frac{5}{6^{(\frac{7}{3})^n}} \}. $

Unknown xFind the order of convergence of the sequence $\{\frac{5}{6^{(\frac{7}{3})^n}} \}. $ My attempt:- Definition of Order of convergence We know that $\lim_{n\to \infty}\{\frac{5}{6^{(\frac{7}{3})^n}} \}=0$ Consider $$\frac{|p_{n+1}-p|}{|p_n-p|^{\alpha}}=\frac{\frac{5}{6^{(\frac{7}{3})^{n+1}}}}{(\fr...

order of convergence is 7/3. right?

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