12:29 AM
@ペガサスSeiya I thought most people take the subway in Japan?

1:12 AM
@ペガサスSeiya I had a 1.5 hour walk today
with mah dog, for exercise

1 hour later…
2:32 AM
@ペガサスSeiya isn't it possible to go to a stationery shop and get the book printed with hardcover?

2:42 AM
resurrecting my sunnyvale-albany commute. its only 50mi but takes close to 2hrs.
transit in the bay area is a joke unless your work & home are on a single transit line.

@copper.hat Yikes. I thought BART was better than that...
If you drive it, it's... what... 90 minutes on the 880?
Or do you go through SF?

yup, for most commute hours, 90-120 mins typical

Yuck. It is sad that, even in Bay traffic, it is faster to drive. Public transit in the US sucks.

i went back for the first time in a long time today. i drove down (90mins). after work i drove to the amtrak station (20 mins) the train to berkeley took 80mins and then i cycled home from there about 20 mins (i left my bike in a locker last night).

Fun.

2:51 AM
if i was to take transit from sunnyvale to the amtrak station that would effectively be an additional 20 mins to the above estimate and this assumes connections are made. however, the connection arrives at x:11 and amtrak departs at x:10, so that is risky
since the train oiriginates fro a nearby station.

Because AmTrak always runs on time, right? :P

there are many transit options, but none connect in any usable way
amtrak is often very late, unfortunately. i do, however, like traveling on amtrak.

I love AmTrak. I really do. There is a station in Winslow that I have made use of many times to get to SoCal. The problem is that there is only one train per day, and it is usually at least two hours late.
:/

i understand the problem of transit in the usa, it does not have the population densities. but the bay area does, but is completely balkanised, transit wise

The problem is that the freight companies actually own the track, and freight trains are given priority over passenger trains. So, for example, you might just sit for an hour near Seligman, waiting for a stupid freight train to go by in the other direction.
I've also taken the California Zephyr into the Bay Area. I love that route so very much (from Iowa to SF).
My biggest complaint is that there is no train from SoCal to the Bay.

2:55 AM
yep, but understandable, freight makes money, passengers don't

You can pick up a train in Bakersfield, but you have to get to Bakersfield, first. :/

yeah, that's totally stupid
trains will not catch in in the usa in my lifetime

What is most frustrating is that, until WWII, trains were the way to travel long distance. And then the auto companies started lobbying.

true, but there is still the density issue.

Oh, sure.
But I don't think that is insurmountable. I mean, China and Russia have similar density issues, and seem to have workable cross-country trains.

2:59 AM
but it inescapable to me that it takes 2hrs from 50 miles between two population centers
yes, there should be cross country.
amtrak is really the only one on which you can really work properly.
bart means being 'aware' all the time, caltrain has little space.
if i could take my daughter's google bus (luxury bus that drives from 2 miles away to work in the morning, on the nights she stays at home)

@copper.hat Yeah, I've heard done tell of these private busses for Silicon Valley folk.
They seem like a special kind of evil.

good idea though, wish there were ones open to the public
since transit can't manage
from VTA, a south bay transit system, "Rapid buses may depart up to five minutes earlier than the time shown, if traffic allows."
kinds misses the point of a schedule.
amtrak has a similar thing, albeit more like 30s.

3:14 AM
@copper.hat So you left your car at work ?

i left my son's car at the great america amtrak station. ill go down again tomorrow, pick up the car and drive down to collect him after work.

Oh, clever. Collect him on Thurs? No classes Fri?

yep, i find driving tiring now, so splitting the trip is appealing

Wait til you’re my age!

:-).

3:28 AM
$3(y,x, z^2 - 1) = 2 \lambda (x, y, z - 1)$ with constraint $g(x,y,z) = x^2 + y^2 + z^2 -2z = 0$

its a bit sad that i can leave a car in the parking lot for many days, but the longest i can leave a bike is 3 days.

I taught you how to do that, DC.

solving I have the relationship $y^2 = x^2 = z+1 = 2 \lambda /3$, I put $z+1$ into the constraint, but I will be left with something not nice

Wrong?

$x=y$?

3:30 AM
well it simplifies to $z^2 + 2 = 0$

Either $x=y=0$ or $z=1$ or …. Eliminate $\lambda$. But you did it wrong ?
Take ratios, yes. What do you get?

well the explicit ratios were $y/x = x/y = (z^2-1) / z -1 = 2 \lambda/3$

Forget $\lambda$. So how can $y/x=x/y$?
How can $u=1/u$?

$\pm 1$ damn it and I did that in the notes

So now do cases.

3:35 AM
Feb 5 at 11:23, by Koro
hmm, I showed the following: given that the group operation m is continuous. So the restriction: m:{c}$\times U\to cU$ is continuous. If cU is open then by continuity $m^{-1}(cU)=${c}$\times U$ is open, whence U is open.

Yup. Will do, back to work I go. Thank you

I see that my understanding of why cU is open if U is open in a topological group G was wrong.
In the linked comment, of course U is open in U. It didn't show cU to be open.
To show that: I define $f_c:G\to G: g\mapsto cg$. Then $f_c$ is a homeomorphism. So $f_c(U)=cU$ is open.
I think this is correct now.

Why is the preimage just what you say?
You evidently don’t mean $m$.

which preimage? You mean cg? yes, cg= m(c,g)

But there are lots of things that give product $xy$.

3:42 AM
Earlier (in the linked comment), I thought that since m is continuous, its restrictions are also continuous so I restricted domain and range. So even though there are lot of things that give the same product, but domain is restricted so that doesn't matter.
Of course, this did not give me the result that I wanted, which was that U is open or cU is open if one of them is open.
6 mins ago, by Koro
To show that: I define $f_c:G\to G: g\mapsto cg$. Then $f_c$ is a homeomorphism. So $f_c(U)=cU$ is open.

OK, seeing it for fixed $x$ is easy. Just write down the inverse.

I agree that there will be lot of things to give the product xy.

Inverse is continuous.

But this homeomorphism should give me the desired result: U is open iff cU is open. Because homeomorphism is an open map.
It's known to me that $f_c$ is a homeomorphism.

OK, then I don’t know why you’re writing so much stuff.

3:46 AM
my earlier linked message involving $m^{-1}(cU)= \{c\}\times U$ had error is what I was trying to point out.

Fix just $c$.

well one of the points decided to mess up. in the case of $y/x = -1 \rightarrow -1 = z +1$ using the constraint I will get $x^2 = -4$

Redundant solutions happen. There is no negative $z$ on the constraint.
But make sure you check the places denominators can vanish.

True, but this question you asked wanted the hottest and coldest point on a surface.

So?

3:52 AM
oh so you're saying I should check where $z = 1$? i.e where I "divided it out" when I was creating the ratios.

And where $x$ or $y$ is $0$? First see that both must be if one is.
I was very careful making up these problems.

I could fathom that, no quick mechanical work, you will have to think about the boundary issues that arise

4:23 AM
@TedShifrin I played around and I may have found otehr critical points, but I don't know if the way I went about it was correct. I'll give an example of one because they all follow the same idea. One of the cases I did was if $x = 0, z = 1$, doing this and plugging it into the constraint I got $y = \pm 1$. Did the same for $y$, the other case I did was if $x = 0$ and $y = 0$. My only concern is checking that if one of $x$ or $y$ is $0$ then the other is as well.

Look at the original vector equation.
I already said that you should first see that if $x=0$, then $y=0$ and viceversa.

$3(y,x, z^2 - 1) = 2 \lambda (x, y, z - 1)$, this being the original vector equation from which I had concluded the $y/x = \pm 1$ from before

So think now.

well from here then if $x = 0$ then $y = 0$.

And the reverse.

4:36 AM
precisely. I know why I'm overthinking it, because I had done that in my notes just now, and I ended up getting values of $z = 0$ and $z = 2$. Both of those values give me the same value when I put them in the function I am optimizing: f(x,y,z) = 3xy + .......................................................................DAMN IT....I made computation mistake when I was looking at my critical points.....I didn't put them in f(x,y,z)......😭
NOw it all comes out "pretty"
I brought that on myself....smh........time for peppermint tea, stretching, and bedtime.......I'll be sure to find something to bother you about tomorrow.

Night!

7 hours later…
11:55 AM
@XanderHenderson Years ago I heard of plans to construct high speed rail from SoCal to Bay Area with some public funding to the tune of hundreds of billions of dollars. What happened to this?
Looks like it's still happening though slow in progress, planned partial operation in 2029.

5 hours later…
4:33 PM
hey chat. wanted to share here: i finally got my BSc degree in math, and now I'm enrolled in the MSc math program!
8

4:46 PM
@LucasHenrique Congrats!
I still have one more year

@LucasHenrique congrats :)

5:47 PM
@Lucas Congrats! Has it been 4 years already?

6:02 PM
If a holomorphic function $f$ is constant on a curve...

Context?
Curve is also ambiguous.

I don't remember the exact statement but it's like: If $f$ is holomorphic on a domain that is constant on a curve then $f$ is constant on a whole domain.

@LucasHenrique congrats!

Is this a (connected) domain in $\Bbb C$?

It was $|f(z)|$ originally but the instructor revised the question (false of course).
Yes open connected set in $\Bbb C$.

6:21 PM
Doesn’t a curve contain a set with a limit point?

Yes so I thought it was a direct consequence of the identity theorem

Yup.

6:48 PM
EEEEEEEEEEEEEEE

‘Lo.

7:21 PM
Couldn't get a perfect score on Calc 3 exam yesterday
Not too happy right now

7:33 PM
@ペガサスSeiya :(

The worst thing is, its because of a very stupid mistake. I forgot to switch bounds of integration leading to a wrong answer and the deduction of 8 whole marks

the teacher must be strict
i guess

I still did get a nice grade but it could've been better
I wonder how to avoid such mistakes in the future

7:49 PM
@ペガサスSeiya Feel your pain. I had a number theory exam on Monday and lost 10 points on a 25 point problem due to a stupid calculation error. Pain cause i would have had full marks otherwise

@ペガサスSeiya I use this method to avoid such mistakes: Try to solve the questions quickly so you have one-fourth time left. Then clear your mind for about a minute and verify what you've written step-by-step (even do the minor calculations like the chain rule etc.) on a rough paper. The 1 min break helps by resetting your mind, i.e, if you made a mistake thinking it was correct, then you're bound to think it is correct and move on to the next step without really verifying, but a reset (continued)
makes you to check it properly. But of course, I'm still in high school, this method may not work later

8:20 PM
@SineoftheTime No, not really. In such problems, that is the whole point of the problem.

@TedShifrin it depends also on the total score of the problem
in the exams in my country, the marks are from 18 to 30 so 8 points are a lot

what type of object is a trigonometric function times a vector?
Is it a set of vectors parameterized by the variable of the function?

various ways of looking at it, but one would be as a vector-valued function (of whatever is being plugged into the trigonometric function).
and if you think of a function as a set of outputs parameterized by its input (not necessary to do this, but i guess you could do it) then that is another way of looking at it.

8:37 PM
@TedShifrin I still got 42/50 so its all good but man it hurts knowing how minor that mistake was
And 42/50 means no A+ on that exam, I hate that

Fubini's Theorem is very important. And whenever I wrote such test questions, the only way to do the integral was to change the order of integration and get the limits of integration right :)
@SillyGoose Yes, i ndeed.

Kinda wanna punch something right now
Good thing my taekwondo class will start soon in the morning

If you're going to go on in mathematics, get over this hating and violence.

@TedShifrin well, did I mention I also lose a bet because of this?

Guess you'll have to sell another one of your hice.

8:40 PM
@TedShifrin gonna have to sell mine. I'll keep my wife and her boyfriend's hice
I'm a (h)nice guy
@GratefulDisciple I don't generally use subway even if most people do. In fact that's the problem here, since most people use it, that means it has a lot of people and I dislike that. It also doesn't help that I stand out because of my height and tend to get quite a few of stares and weird looks sometimes and that's uncomfortable
I'll either drive or walk
By walk I mean run. I don't like being slow

9:01 PM
Hi
I posted a question, I got two good answers and I accepted one of them
both answers were down voted
is there something I can do?
-1

To check for uniqueness, the basic approach is to assume there exists two different solutions, and see what happens. If the problem is truly unique, then you will run into a contradiction. Suppose there were two combinations, $c_1, c_2$ and $d_1, d_2$. Then, (1,3,8) = c_1(1,2,3) + c_2(2,3,1) = ...

Find the equation of the plane through the point $(1,2,2)$ that cuts off the smallest possible volume
This ends up turning into minimizing the volume of a three sided pyramid subject to the constraint of my plane.

Vol of pyramid = $\frac{1}{6}abc$

eqn of plane $-bc(x-1) + ac(y-2)-ab(z-2) = 0$

Lagrangian set up: $1/6[bc, ac, ab] = \lambda[c(y-3) - b(z-3), -c(x-1) - a(z-2), -b(x-1) + a(y-2)]$

Using the ratios I got an expression for $a$: $a = \frac{-(x-1)b}{(y-2)}$ from which I put this in the constraint and solved for $c$, $c = \frac{b(z-2)}{2(y-2)}$.
@Bob upvote them if you haven't? 🤷🏿‍♂️

I just did.
Thank you

ing?

This is why it’s important to put the vertical bar when you are doing inhomogeneous systems. The $A$ part of $[A|b]$ has rank $2$.

oh you're talking to Bob

9:12 PM
I hate chat on iPad.
I haven’t looked at yours.

Yes.

was about to call you Rain Man.

Write the equation $x/a + y/b + z/c =1$.
Maybe even use $1/a$, etc., as your variables.

Hmm. Does this object actually exist? i.stack.imgur.com/ZDEMy.png
Specifically: can you actually have a spiked sphere such that every spike is equidistant from its 6 neighbors?
It feels like the optical illusion behind @AkivaWeinberger’s avatar

9:22 PM
The chat would be much nicer if it was separate from SE and used the discord API instead

I’m a bit surprised there’s no SE discord server tbh

@Semiclassical So can we tile a sphere with (spherical) regular hexagons?
My guess is that Gauss-Bonnet will say no, but I don’t have time to work on it now.

French me is telling me you can tile it with two spherical regular hexagons

9:37 PM
Hello,

May I know please why the intersection over the union of 2 objects is scale-invariant? The intersection is defined as the intersection of 2 boxes while union is defined as the union of 2 boxes minus their intersection.

what's the context?

So I used your suggestion of writing the equation of the plane in the intercept form @TedShifrin and the solution just came about smoothly.......Why did this make things work out so easily and all the machinery I did before created chaos?

@Astyx. You asked me?

@Semiclassical Same, there might be some benefits of having one. But also more overhead.
For one, moderating discord is Hell, speaking from experience, and current site mods would not enjoy it.

@Avv yes

9:45 PM
@DCthe I didn't look carefully at it. It looked like a colossal mess.

@TedShifrin that was my thought too, but in this case they’d be equilateral spherical triangles instead

@Astyx. I don't have it. I just read that if we divide intersection of two shapes over their union, this is a scale-invariant metric. This means that the similarity between two arbitrary shapes A and B is independent from the scale of their space S.

@Semiclassic A regular hexagon should still be the union of 6 congruent equilateral triangles.

@TedShifrin It was, I'm going over it to see why...

9:47 PM
Oh. You are confusing yourself with what's the variables and what's not. Did you do gradients with respect to $(a,b,c)$?
You did not set it up to get the constraint equation at all.
There should be no $x,y,z$ in it.

yes, everything was done in terms of $(a,b,c)$. I treated the $(x-1), (y-2), (z-2)$ as just constants.
oh...

What does that even mean?

wasn't $-bc(x-1) + ac(y-2)-ab(z-2) = 0$ i.e the equation of the plane the constraint equation? To get that equation I took the intercepts $(a,0,0), (0,b,0), (0,0,c)$ got some direction vectors by subtracting pairs of intercepts, then I took the cross product to get the normal and used the expression $(-bc, ac, -ab) \cdot (x-1, y-2, z-2)$ to get my equation of the plane.

It makes no sense. Your variables are $a,b,c$. Who are $x,y,z$?
The constraint is that the plane with equation $x/a+y/b+z/c=1$ passes through $(1,2,2)$.

I should be able to reconcile that intercept form of equation by getting the equation in the way I was trying to right?

9:53 PM
Your equation never tells us that the intercepts are $a,b,c$, does it? Go back to first principles.
If you can't explain what $x,y,z$ mean in the context of your problem, then this cannot be right

Well before this discussion I was thinking of $x,y,z$ as points on the plane.

But who cares about points (x,y,z) on the plane, other than the intercepts and $(1,2,2)$?
Does your plane actually have the right intercepts? I think not.

Correct. So my thinking went along the lines of I have my intercepts, and I have my point $(1,2,2)$. So I sought out to get a "general equation" of my plane. Then I was going to find the $a,b,c$ that would explicitly describe the plane used to minimize the volume of the pyramid
THat was how I thought about it coming together

Well, show me that your plane has the right intercepts.

10:09 PM
I'm actually going about that now based on your comment on $x,y,z$ should not be there
Ok I worked things out.....that $x,y,z$ did make a mess of everything. I think I introduced that because I was remmebering a comment you said about another question about being careful of how I named my variables to limit confusion
I need to catch these things early and think them out better.....damn it....😡

long shot, but anybody familiar with the Borel-Cantelli lemma here?

Perhaps @leslie will remember it, but it's not down my alley at this point in my life

Thanks, no pressure. I may ask a proper question (outside the chat)

i know that probabilists like to use it. it is often included (sometimes as an exercise) in courses about measure theory, where i've encountered it. i do not recall using it much.

10:28 PM
Yeah, my understanding is that it can be relevant for convergence of random variables
thanks anyway

11:27 PM
Why's it called mornning what its already happened? It should be called morn
We don't call night, nigthting

Prove that, for any r.v.'s $X,Y$ we have $\EE|XY| \leq \norm{X}_{\infty} \norm{Y}_{1}$. Any hints.

not sure what EE is, but maybe establish the inequality |XY| <= |X|_infty |Y| of random variables, and take expectation.

@leslietownes $\EE$ is expectation.

11:46 PM
@ペガサスSeiya Just stick to mourning.
@Roby You need to type \mathbb{E} or \Bbb E (in dollar signs).

roby5: ok, so \EE(|Y|) = ||Y||_1, i presume?
whatever you do, don't follow my example of mish mashing pseudo-chatjax-latex and plaintext.

growls audibly

@TedShifrin Now it looks good. Prove that, for any r.v.'s $X,Y$ we have $\ \mathbb{E}|XY| \leq \norm{X}_{\infty} \norm{Y}_{1}$. Any hints.