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7:02 PM
Oh, sorry, so you're right. The $N$ kept messing me up.
But what is the surface area of the boundary sphere? It's $na(n)r^{n-1}$. So everything is perfect.
 
Jam
?? sorry im missing something
i forgot to write on the left hand side the denominator
$$\frac{\int_{\partial B} u(y)dS(y)}{a(N)r^N}=\frac{N}{r} u(x_0)$$
 
You don't want that coefficient. You want to divide the integral by the surface area of the sphere, which is $na(n)r^{n-1}$.
 
Jam
this is what i got after differentiating
 
That's exactly what you get from your equation.
Average value on a sphere is different from average value on a ball.
 
@TedShifrin I guess $\varnothing$ is the best option. Thr best I got fiddling with \llap is $\small/\llap o$
 
Jam
7:08 PM
ohh
right
i need a different coeficient
really really thanks
 
@PM2, yeah, that's what I was thinking of. I just forget the Latex command.
 
Jam
so many ways to make a mistake
 
You're welcome, @Jam.
My original path was messier.
 
Jam
unbelievable that it actually worked
 
LOL, nah.
If you're going to work through Evans, you need good multivariable calculus skills.
 
Jam
7:10 PM
cant wait for some1 to ask me the same question and say "just differentiate " hjahaha
 
Well, the definition of average value is the key!!
 
Jam
true somehow i thought it was the same denomiator which is obviously not true
and wouldnt make sense
actualy its the derivatie of one another
derivative*
 
this evans guy sounds like trouble.
 
Jam
pdes is trouble
stay away
literally we should do pdes at mult calculus
feels like i dont know any mult calc hahaha
 
alright, i'm back. took a tea break.
@TedShifrin we said $[0,1]$, not $[-1,1]$, i think we can't say $|x|>1$ when $y=0$.
 
7:18 PM
You're right.
 
but i see why that two cases, alright.
and then, when $y=0$, we should say $x<0$ or $x>1$.
 
You're right. You win! ;)
 
ok, let me first finish this.
for $y=0$ and $x<0$, i simply choose $\epsilon=|x|/2$.
for $p=(x,y)\in\mathbb{R}^2-\mathcal{G}(g)$.
 
Yup.
 
now i'll show $B(p,|x|/2)\subset\mathbb{R}^2-\mathcal{G}(g)$.
 
7:29 PM
Better just to write it explicitly as $p=(x,0)$; easier to keep score.
 
oh, right.
 
🙏🏼
 
i'm an insurance policy with a high deductible. i don't kick in until google has done its job.
 
just stick with whole differential equations, don't mess with partial stuff, just leads to trouble.
 
evans, always trying to recruit more people to the dark side.
 
7:37 PM
@PM2Ring Googling led me to this: \usepackage{wasysym} \diameter
 
that is very characteristic of evans
 
@TedShifrin mathjax doesn't support wasysym evidently
 
I had never heard of it before.
 
the great pumpkin pie has investigated the issue, thanks for giving it your attention sir
 
@copper To do any interesting geometry (unless it's just curves), one must have PDE.
 
7:46 PM
let $b=(b_x,b_y)\in B(p, |p_x|/2)$. then i should show $b_x<0$, that suffices for $B(p,|p_x|/2)\subset\mathcal{G}(g)$.
 
@TedShifrin I did not realise the connection was so strong. Most of my PDE exposure was engineering, the usual suspects.
 
Yes, the stuff I taught in my year-long applied math course. But if you think about parametrizing $k$-dimensional manifolds in $\Bbb R^n$, then once you get past tangent planes to curvature, etc., you're at second (and sometimes higher) partial derivatives.
@sevdaicmis You're writing incorrect things. Unless that notation is the complement of the graph, which would be weird notation. Again, pictures should lead you to a valid proof.
 
i am always surprised how reluctant folks can be to using pictures as a way of guiding proofs & problem solving.
 
me too
 
oh, sorry. i meant "...for $B(p,|p_x|/2)\subset\mathbb{R}^2-\mathcal{G}(g)$".
 
7:50 PM
Leslie's evil influence on the world.
I even drew pictures when I taught abstract algebra (as you will eventually see if you don't give up on my book) :D
 
but, aiming to show $b_x<0$ is reasonable, right?
 
Absolutely.
 
i wonder if legal documents have pictures other the vile approximationsfound in patents?
 
i drew a picture and following the proof in there by the way @TedShifrin
 
For lawyers who are paid by the word, they do not subscribe to "a picture is worth a thousand words" at all.
2
 
7:51 PM
@user178758 are you @leslietownes's alter ego?
 
Good, @sevdaicmis :)
@copper Do you make a return trip to Santa Cruz this weekend?
 
a thousand words is a billable hour
 
or two
 
@TedShifrin Most likely I will bring him back on Monday.
He cooked a version of my mother's stuffing recipe last night to bring to some friends.
 
copper: no. and i've always been surprised how little patent figures matter. i've spent years on a single patent case without ever looking at the figures.
 
7:53 PM
I am glad his rudimentary culinary skills vastly outperform my own already.
 
in the beforetimes when people would hand me printouts of patents the first thing i would usually do is tear out the figures to save me the trouble of flipping past them every time.
 
I am visual. Words are a hindrance until you need precision.
 
I think my culinary skills exceeded my parents' at some point, but they were both pretty good.
 
We were not a culinary family growing up.
 
@leslietownes Of course. It's because you are a vigilante against all pictures/geometry.
 
7:55 PM
don't you think of symbols as small pictures?
 
You probably want the Texas law to apply to geometry as well.
 
pictures are poetry, they contain the truth but you need to unveil
besides, truth rarely lives up to the fanstasy
 
ted: that and there are legal doctrines which basically say that the figures don't matter if they aren't supported by text. i'm kind of surprised that there are so many technical drawing shops whose only business is generating figures for patents. i think they prey on the small inventor who thinks they need a lot of them.
 
I would think it would help in enforcing the patent down the road.
 
it is to placate the engineers who are writing the patents to get the $5k bonus.
 
8:00 PM
they are legally close to irrelevant to the scope of a patent right. sometimes it does help to have a visual aid, but if there's no supporting text for it, you wouldn't get to the point where you could use it.
so often you find yourself not looking at them until very late in a dispute.
 
Lawyers who are paid by the word will lack conciseness..
 
lawyers aren't paid by the word. i've never heard of that billing model. charles dickens was paid by the word, at least apocryphally.
the people who do technical drawings for patents do sometimes get paid on a per-drawing basis. half of my work spam email is for that kind of stuff.
 
@TedShifrin i have $(b_x-p_x)^2+b_y^2<p_x^2/4$, and i can't derive $b_x<0$ from here.
i feel dumb.
this should be straightforward.
 
Throw away the $b_y^2$.
Again, the picture should guide you. If $\|(x,y)-(a,b)\|<r$, then $|x-a|<r$ and $|y-b|<r$.
 
oh, i see.
 
8:04 PM
We're doing negatives here, which makes it slightly more confuzling.
 
i mean, i see it from the inequality.
 
(removed :)
 
it becomes $b_x^2-2b_x p_x+3p_x/4<0$.
this quadratic will probably give us $b_x<0$.
 
Your algebra is not thoughtful enough.
You should get $|b_x-p_x|<|p_x|/2$.
 
try slowing down
:)
 
8:08 PM
or down
or should it be slowing up?
Every language is so bizarre with prepositions and idiomatic expressions.
 
hold up now; we need to slow down
 
that is why we need lawyers
 
@TedShifrin most probably. but, i've managed to derive $b_x<0$. the quadratic would be zero at $b_x=p_x/2$ and $b_x=3p_x/2$. we are looking for $b_x$'s that make the quadratic negative, that would be $3p_x/2<b_x<p_x/2$, which shows $b_x<0$.
 
But you should learn the insightful way to approach things, @sevdaicmis, not always bludgeoning.
 
alright, i've also read ones you've written above, it should be thought like that.
@TedShifrin you're right. i shouldn't be solving a quadratic inequality for this.
 
8:15 PM
If you're going to mess with analysis and topology and deal with inequalities, you have to do so thoughtfully.
2
 
@TedShifrin i comprehend this. when reasoning with it, should i do any justification for it? i assume not.
@TedShifrin you are so right.
i gotta go now. i'll be back though.
thank you for the guidance today @TedShifrin, i appreciate it very much.
 
You're most welcome, @sevdaicmis.
@sevdaicmis It comes right out of the algebra. If $x^2+y^2<r^2$, then $x^2\le x^2+y^2<r^2$, so $|x|<|r|$.
 
@copper.hat no, i'm just an innocent bystander
 
No one here is innocent.
 
a not-guilty bystander?
 
8:21 PM
the end of the innocence
 
. o O ( Don Henley )
 
and bruce hornsby
 
indeed. I had forgotten that
 
😎
🙏🏼
 
There was a follow-up question to that Moon's orbit precession question I answered a couple of weeks ago. I initially misunderstood the new question. Oops. And then it took me a couple of hours to realise it could be solved with a simple cross-product. Even so, it was still a bit tricky. It's so easy to make a sign error when doing 3D trig. :) I expected the equations to become a hellish mess, but remarkably everything simplified right down. astronomy.stackexchange.com/a/47554/16685
The 3D diagrams were actually quite helpful in verifying that my equations were (probably) correct, and that I hadn't made some silly blunder.
 
8:33 PM
Lots of spherical geometry/trigonometry boils down to cross products, yup.
 
Traditional navigation doesn't use them though, since there's no sane way to do them using latitude & longitude spherical coords, unless you translate back & forth to Cartesian coords, and navigators never did that.
 
@user178758 bystanders are always guilty...
 
how did leslie pass trig without diagrams :P
 
the world would be so much simpler if things just stopped rotating.
 
In Scotland, they have a 3rd verdict: not proven.
 
8:39 PM
stop the world, I want to get off
 
they toss telephone poles & eat haggis
and terrorise irish giants
 
They're also very fond of fried food.
 
i had fried pizza for the first and only time in edinburgh
once was enough
 
"Scotch salad" is a slang term for chips (fries).
 
it's probably still working its way through my system
 
8:46 PM
Let $E = \{(x,y) \in \mathbb{R}^2: y = \sin(\frac{1}{x}),\ \text{for}\ 0 \lt x \leq 1\} $

Showing it's not compact. Means to show that it is either not closed, not bounded or both.

I can see visually that it is not. In particular it is not closed. I'm having an issue to do this formally. After a bunch of messing around I've settled on the idea of the ball around $(0,0)$ and showing that the complement of my set $E^c= \{y = \sin(\frac{1}{x})\} \cup \{y \neq \sin(\frac{1}{x}) \forall (x,y) \in \mathbb{R}^2$ is not open. Visually what is going to be happening is that any ball around $(0,0)$
 
maybe consider x of the form 1/(2 pi n), n a positive integer
 
I was thinking about trying to find a sequence that converges to $0$.....but couldn't figure one out
 
with similarly well chosen sequences you should be able to convince yourself that {(0,y): -1 <= y <= 1} is a subset of the closure of E
 
@copper.hat A month or so ago on the Space Exploration stack, there was a question about the feasibility of using tidal power on the Moon. The OP thought it'd work well because the Earth's gravity is so much stronger than the Moon's. We pointed out that the Earth barely moves in the Moon's sky, and although there would be tides from the Sun, they'd run on a monthly cycle.
 
thanks Leslie
 
8:50 PM
waiting for some kind of joke premised on monthly cycles. this is risky territory with someone like copper around.
 
ever since my daughter asked me to buy her supplies the repressed catholic fascinations evaporated.
 
Let's not go there...
 
of course, the book she gave me with the mathematics of cunnilingus went a long way too...
 
mathematics of cunnlinigus?...........this sounds interesting.....
 
@PM2Ring i am sure i am as guilty as the next person, but sometimes it seems that common sense is in shorter supply
 
8:53 PM
dc3rd in fact i think closure(E) = E union that segment mentioned above.
 
@dc3rd xkcd
 
let me go work on this sequence instead of being distracted....
 
that is something i excel at
distracting that is, to avoid any confusion
 
The Moon's motion is scarily complicated, if you want precision. Just calculating the time of the full & new moons to within 30 seconds involves series with ~15 trig terms. That's good enough for predicting eclipses. By the start of the 20th century, the full lunar theory involved around 1400 terms. And they worked out all those terms by hand, doing calculations using log & trig tables.
 
Does this crop up a lot in problems? Is it worth remembering? $u^2+v^2+w^2−vw−wu−uv=\frac{1}{2}((u−v)^2+(v−w)^2+(w−u)^2)$
 
8:59 PM
Do you think I have to argue that $(0,0)$ is not in the set @Leslie or could I drop in a "It is clearly evident that........"
 
did they use analog computers for any of that? i've seen something for trig series with small numbers of terms (maybe a dozen) in a museum.
dc3rd: if the thing above is the definition of E, it's evident. E contains points with x coordinate in (0,1] only
 
yup that is the definition of the set.
 
@LearningCHelpMeV2 I wouldn't bother memorising it. Just remember that it exists. You can derive it (and similar expressions) fairly quickly when you need them. OTOH, if you're onto doing those crazy competitions where you have to do high speed algebra, then sure, memorize it.
@leslietownes Analog computers were fairly crude back then. I doubt they could compete with 7 figure trig tables. But I guess they could be helpful in verifying that 2 trig series were (approximately) equivalent.
 
learning: i second pm2ring's answer.
 
@PM2Ring at least there are no stupid butterflies flailing their wings and affecting the precision of computation. i mean, for god's sake, if we just eliminated the entire butterfly population on the earth we could predict the weather so much better. i'm surprised no one has thought of this.
i love analog computation. the idea of using on-the-fly reconfigured hardware is something that needs a lot more exploration i think, not just gpu silliness.
 
9:13 PM
And that Moon stuff wasn't just an academic exercise. Because the Moon's motion relative to the stars is so fast (and it's large parallax) it was the best timekeeper that we had, until atomic clocks came along. Official / legal time was ultimately based on time from astronomical observations, and that meant lunar observations.
 
yeah, atomic ruined it all
i think i satisfy the legal definition of curmudgeon now.
 
yeah, it put a bunch of nerdy moon men with tables and slide rules out of work. good riddance.
 
@leslietownes If @dc3rd is working on my book still, he doesn't get to closure until a later exercise. :P
 
what did you do as a young man, grandpa? i sat down with some tables and worked out exactly what time it was... a few hours before i finished.
 
breaking out my faber castell for a quick three-in-a-row computation...
 
9:15 PM
ted: i heard something about closed and bounded so i felt free to go there.
 
No I did the closure stuff from Ch.2....I was working on the compact stuff from 5.1
 
compactly put, Leslie
compactness is an amazing concept. (serious for a moment.)
 
Ohhh, duh, my fault. I'm not thinking ....
 
Hmmmm.......the tables have turned....
 
I guess >6 years away from teaching the course has an effect.
 
9:16 PM
copper there is a writer calvin trillin who used to write a lot on food and other issues in the new yorker. he has a memoir of his young days at princeton where apparently he was friends with a faber of the faber pencil fortune. they called him "the pencil king."
 
Apologies to all.
 
now ted's making insincere apologies. it's contagious.
 
That still puts me miles ahead of Trompolini.
 
whoa, i missed something...
 
Even now, although UTC is derived from TAI (atomic time), it's still kept in synch with the Earth's rotation, but we now use a network of extragalactic sources (quasars), as well as the Moon. We regularly observe the Moon & a bunch of satellites to get the fine details of Earth's rotation. That's the job of the IERS. iers.org/IERS/EN/Home/home_node.html
 
9:17 PM
scrolling back frantically
 
i use a network of extragalactic sources to keep the government from stealing my thoughts.
 
@leslietownes we forget how things were.
i have my trusty hat.
its actually a tilley, not copper
in chernoff's class i spent too much time trying to use compactness in various guises (weak, weak*) to prove homework exercises. it drove the ta mad.
and then i discovered convexity.
 
@copper.hat When I was in my teens, I had access to a pretty groovy electronic analog computer, built in the late 1960s. It was in a local science museum. maas.museum/powerhouse-museum
 
@PM2Ring i used the (limited) control lab stuff, mostly 741 amps.
 
I typed an edit that took forever to make. And it got rejected... How can I get the code so I don't have to do it all again?
 
9:22 PM
around then i discovered discrete systems (as in $x_{k+1} = A x_k + B u_k$) and was able to escape the real world for a while on the digital computer.
 
The machine I played with predated integrated circuit op-amps. But it had a nice multi-trace blue oscilloscope screen.
 
my big goal in college was to get time on the storage scope
then a tektronics guy came by with the latest stuff that could be uploaded to your computer (around 1982). blew me away.
i still have a 4 ch. 250Mhz scope in my basment.
bit of overkill for debugging arduino designs.
i just wrote two, not just one, answers to a convex question. have instituted lock down and have my taser at the ready. (no, i don't really have one.)
 
Nvm
IF someone comes up with an answer, THEN please email it to TheGreatJRB@Berkeley.Edu
 
an answer for what?
 
9:39 PM
I was hoping to answer this question about inscribing an equilateral polygons in an ellipse. math.stackexchange.com/q/4301282/207316 Unfortunately, the OP declined to improve their question. My latest algorithm is quite fast, if the eccentricity isn't too extreme. It gets tricky with really skinny ellipses. You have to feed it a reasonable initial approximate solution, and then procede with caution. Otherwise, it converges on bogus solutions.
@TheGreatJRB Sorry, I don't think you can access rejected edits.
 
@PM2Ring Thank you
 
It's not in the edit history accessible to us?
 
It is but I haven't found a way to get the code out. Maybe it's because I only have 29rep
 
I am sure PM2 knows best, but what's the link?
 
@TedShifrin Talking to me?
 
9:43 PM
Yes.
 
@Link to what? The post?\
 
Duh.
 
:D
-1
Q: Density of a quotient. Let $A,B\overset{\tt iid}{\sim} \mathcal{N}(0,1)$, $X=ae^A$ and $Y=be^B$. What is the density of $Z=X/Y$?

TheGreatJRBLet $A,B\overset{\tt iid}{\sim} \mathcal{N}(0,1)$, $X=ae^A>{0}$ and $Y=be^B>{0}$. What is the density of $Z=X/Y$? I've tried $\texttt{change of variable}$, $\texttt{convolution}$, and $\texttt{CDF}$ I'm just lost. I made an edit to an answer so that the one who gave the hints would get the credit...

I'm in the process of retyping it
 
So you have an answer posted and then deleted. Where is the edit I'm supposedly looking for?
 
I edited Minus One-Twelfths' post.
 
9:47 PM
Presumably that person rejected your edit. I don't see any trace of an edit.
 
No some reviewers did
Rejected 2 hours ago:
Harish Chandra Rajpoot reviewed this 2 hours ago: Reject
This edit was intended to address the author of the post and makes no sense as an edit. It should have been written as a comment or an answer.
Leucippus reviewed this 2 hours ago: Approve
user2715281 reviewed this 2 hours ago: Reject
This edit deviates from the original intent of the post. Even edits that must make drastic changes should strive to preserve the goals of the post's owner.
 
Interesting. I thought I could see that stuff, but I can't.
The way I would suggest handling this is to ADD to your original question in accordance with the comments after Minus's answer, comment asking him if that edit is correct, and then accept his given answer.
 
@TedShifrin So don't make a new answer with their suggestions?
 
Right. Not if you want that person to get credit for the ideas.
 
@TheGreatJRB Is it this edit suggestion? math.stackexchange.com/review/suggested-edits/1705673 You can't easily extract your stuff, but at least it's not totally lost.
 
9:54 PM
Okay, I type it and then send you the link. And then you tell me how to put it in the original. IF you'd be so kind :)
@TedShifrin Yes it is. I can't get all the LaTeX code out but I can see what I typed
 
Ah, right. I've had that happen. Usually I just copy/paste and then rectify the LaTeX code.
 
I had put all kinds of code in it that I run through a c++ program that colors the text. I was going to put it in Overleaf as a solution to an exam I reviewing for classmates
 
Oh good grief.
 
ikr
 
@TheGreatJRB You should be able to copy at least some of it. Where there are 2 columns, they get combined, but the stuff with only 1 column should be ok. Eg,
 
9:58 PM
:D I retype and save somewhere before posting :)
 
${V}\sim\mathcal{N} ({\mu},{\sigma}^2 )\Longrightarrow {f}_{V} ( {v} )=\displaystyle \frac{1}{\sqrt{{2}{\pi}}{\sigma}}{e}^{-\frac{1}{2}\left(\frac{{v}-{\mu}}{\sigma}\right)^2}$ Let ${Z}={k}{e}^{V}$ for ${k}>{0}$ \begin{align*} {F}_{Z} ({z} ) & = \mathbb{P} ({Z}<{z}) \\& = \mathbb{P} ({k}{e}^{V}<{z} ) \\& = \mathbb{P} \bigg({V}<\ln{\frac{z}{k}} \bigg) \\& = \int_{-\infty}^{{v}=\ln{\frac{z}{k}}} \left[ {f}_{V} ( {v} )=\displaystyle \frac{1}{\sqrt{{2}{\pi}}{\sigma}}{e}^{-\frac{1}{2}\left(\frac{{v}-{\mu}}{\sigma}\right)^2}\right] d{v} \\& = \frac{1}{\sqrt{{2}{\pi}}{\sigma}}\int_{-\infty}^{{v}=\
 
HOW YOU GET THAT!!
Come to daddy
The text ... not you lol
 
:)
 
I think PM2 gets a gold star good samaritan award today.
 
So do I just go through all of their stuff and uparrow it ? lol
 
10:07 PM
@TheGreatJRB No. Please don't do that. But it wouldn't give me any points anyway because the system automatically reverses serial voting on a daily basis.
 
Huh g2k
 
I'm totally lost.
 
So one per day :o for a couple of years?
@TedShifrin huh? Lost how? With me paying back PM 2Ring by giving them rep :) and THEN PM 2Ring informing me that that is a no go?
 
573
Q: What is serial voting and how does it affect me?

Cody GrayI just noticed that I lost a bunch of points from my reputation score, and I used the "reputation" tab on my user profile page to try and track down the cause. During my investigation, I noticed there was an unusual event of type "reversal". In the normal place of a question title, it says "voti...

It's a necessary policy, but it can also be annoying. Eg, if you're researching some topic, and 1 or 2 experts have written a bunch of excellent answers to several questions you can only upvote a few of them per day, or they'll get reversed. Conversely, if some clueless person goes on a spree of posting a bunch of really bad answers in a short space of time, you can only safely downvote a few of them.
 
10:23 PM
So...the more bad you spew, the higher the chance of some getting through :)
 
11:05 PM
@leslie This one is for you.
 
Not what I suggested, but OK.
 
Sweet
I hope I took those derivatives correctly or I'm gonna be roasted :/
Laters fellow nerds
 
11:24 PM
Ah right, obviously Leslie. I was trying to think of the simplest counter-example for this commutation problem and it didn't occur to me to pick an operator that doesn't commute with its own adjoint.
 
i was thinking the infinite dimensional unilateral shift when i realized that a finite dimensional shift would do :)
i added a comment about fuglede's theorem to close the loop.
for extra credit, let's all try to pronounce 'fuglede' correctly
 
11:40 PM
Yeah, the raising/lowering of the harmonic oscillator was the first thing that came to mind as soon as I saw you wrote U = H
But that is complicating things unnecessarily
You know what I just realized this very moment..... When you all were talking about infinite dimensional Hilbert spaces and the operator that sends $e_n \to e_{n+1}$... that's exactly the raising operator for the harmonic oscillator.
Never realized before that that "weird operator" that always gets brought up as a counter-example is also one of the most used operators in physics . . .
 
i'm convinced it's the only operator. everything that isn't the shift is an illusion
 
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