@Hippalectryon The question is: what is the meaning of a "test" in the problem's text? One might say that a test means using a mouse to test the wine. But I think a test means using both mice at the same time.

But the mice only have one life

Are you going to inebriate your mice, @Hippa?

'two tests' come from the fact that you only have two mice

@TedShifrin I'm too young to drink alcohol :c so I use mice instead

I didn't ask if you were inebriating yourself.

To be honest, I'm totally fine with water and fruit juice though :D

@Chris'ssis Wait I thought we could use the mice as many times as we wanted, unless they're dead

Isn't that true ?

So g(t) is y(x(t))

@Hippalectryon Only for 2 tests.

right, @Owatch, but I'm calling $y=h(x)$. I hate using one letter for lots of different functions.

@Chris'ssis Ah. That makes the problem terribly unclear. Because the 'two tests' rule is totally illogical (why only two tests ?), and because a 'test' is not clearly defined.

Why?

@Hippalectryon Yes, that's my point too. I was trying to talk about that above.

@Chris'ssis Where does it come from ?

@Hippalectryon I received it by e-mail from someone. I tell you for sure that it's not a fake at all.

@Chris'ssis You should ask for details though. I don't mind if a problem is hard, but it's a different matter when it's unclear

I don't want to think about a problem for hours and end up finding a solution, which will not be the right one because I misinterpreted the problem.

Because, @Owatch, when you write the chain rule as $$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt},$$ the letters $y$ represent two completely different functions.

@Hippalectryon Maybe this is a part of the problem (out of the box thinking).

@Chris'ssis But here there are too many possible interpretations of the problem.

It's not just about thinking out of the box.

Oh

@Hippalectryon I know ...

@Chris'ssis Which is why i think that something is lacking here. Something is incomplete in the way the problem is written, which makes it inconsistent.

@Hippalectryon I expect it to have a nice answer.

@Ted: OK, add smale-hirsch theory to the long list of things I want to learn... there's just too much.

LOL, @MikeM, now you know why I retired :)

So that you have time to learn Smale-Hirsch theory, I assume.

@Chris'ssis I do too, but I don't think one can find it without having a clear understanding of what 'test' means here.

No ... I took a course from Hirsch and heard Smale give plenty of (can you say idiosyncratic?) lectures.

My test grading is verrry depressing without a double martini. I'll have to make one when I can go back to my house.

I suppose it's easier to learn a theory when the people who made it are there to teach it to you.

@Ted: Don't make it a triple if you want to finish today...

@Hippalectryon Yes, but I expect that the meaning is that for each test one can use both mice, separately, and test the wines in any possible combination.

So, if Y is a function of X, it will always have function X inside it.

I don't need to finish today. I just needed something to work on whilst I was kicked out of my house, @MikeM.

Like you described above?

I guess so, @OWatch :P

Ah... I guess prospective homeowners don't want to see you screaming at piles of tests.

I'm not allowed anywhere near the house when people are looking, @MikeM ... which is good.

The language, 'of', implies (to me) that it is inside f( ), which is what confused me.

Hi @MikeMiller. Hi @TedShifrin.

That is now cleared up.

OK, @Owatch. Glad I could help confuse you :)

Hi @MichaelA

heya @MichaelA.

No, you didn't confuse me, you unconfused me.

Now I'm confused. :)

I continue to be frustrated when people post full answers, instead of having the OP think about it. I guess I should go back to hiding in my own tags.

@Hippalectryon I think I've just realized how to do it

@MichaelA: What are you up to today?

@TedShifrin: I hope I wasn't one of them.

@Chris'ssis ?

No, @MichaelA, although you might give more away more readily than I do, but often we end up on somewhat sophisticated questions.

@MikeMiller Grading :(

ah... I grade tomorrow

Me too, @MichaelA, and they are doing about 4-martini level so far.

@MikeM: I wonder if you'd do better on the differential forms Stokes's Theorem stuff :P

@Hippalectryon We might make a group like that c1=|1,2|, c2=|3,4,5|, c3=|6,7,8,9| and then start with testing in the order c2, c1 and then c3.

@Chris'ssis That's two tests

@Hippalectryon Yes, it works.

@Chris'ssis ? How so ?

@TedShifrin I've noticed that. I think a big reason is that answering those questions tests my understanding, so I want to make sure I'm doing it right.

@Hippa: Are you still poisoning mice? I will report you to the ASPCA.

what do you mean, @Ted? On their tests?

@TedShifrin Want to trade? I have assignments for a linear algebra class. Half of the students have never been to a single recitation, not even to pick up homework or midterms.

@Hippalectryon Just a second. I might have missed something.

I have struggled in the past with students who ask me questions about the standard vector calc version of Stokes' theorem, which I have completely forgotten, and need to immerse myself in the notation of.

I'm sure you hate me for it. :P

Well, if they can ace tests without recitating, that's fine. @MichaelA ... Somehow, I suspect not. This generation of students is very different from even 5 years ago.

@Chris'ssis If the poison is in c2, the that will leave you with one mouse for three bottles, which doesn't work.

Learning it with forms unifies, @MikeM, and then it's actually easier to sort out the classical stuff.

When I inevitably teach our vector calc course, of course I'll go back and learn the way they talk about it. But whenever I got asked a question, I had to translate back and forth my conceptual understanding.

@Hippalectryon Oh, yes. It's definitely not a good idea. Let me take some pen and paper ...

@TedShifrin: I had the feeling that things had changed since I was an undergrad, and often the graduate students complain about the current level of undergrads, but there is an element of selection bias.

@Chris'ssis Actually, if you keep only c1 and c2, you get back to 5 bottles and the split you just did was what I tried at first xD

Well, only a few of our students are going to be as motivated by learning and understanding math as we were, @MichaelA; that much is understandable.

I agree. Sometimes people lose sight of that. Not everyone enjoys math as much as we do.

@TedShifrin

still here?

@Hippalectryon If you had now only c1=|1,2|, c2=|3,4,5| you might be able to find the poisoned bottle.

5 people got less than 50% on the first complex analysis midterm, not including the three that didn't take it; only one dropped

@Chris'ssis how ?

i guess they think they can make up for it somehow...

@Hippalectryon Think about it: you start with c1 ad if you don't have a posion in this mix you don't need to check the whole c2 mix but only 2 bottles from c2 (with a second mouse from the first test).

Hey, you know how some diff. equations have solutions like $e^{at}$ and $e^{bt}$?

But then, if $a=b$, you suddenly have $e^{at}$ and $te^{at}$

@TedShifrin

And it just looks weird because you have that extra $t$ there that looks like it just suddenly came in for no reason

@Chris'ssis Oh, you're totally right. For some reason I wanted each mouse to try one mix.

But then I realized:

As long as $a\ne b$, we can have $\dfrac{e^{at}-e^{bt}}{a-b}$ as a solution.

@MikeMiller: Not uncommon unfortunately.

@Hippalectryon Somehow we can proceed the same to get the variant with 9 bottles (maybe we can leave one) and test only 8.

Because that's a linear combination of them

@Chris'ssis But here it only works because we've got a group of two (hence if it poisons a mouse we have a solution)

Do they think they can still pass, or do they just not care?

@Karim: You're doing the cyclic group of order $2k$ or something?

But then, once you try taking:

@Hippalectryon Yes.

I assume they think they can pass. I think they can, if they essentially ace all other tests. Unlikely.

$$\lim_{b\to a}\frac{e^{at}-e^{bt}}{a-b}$$

yeah x is an element of order n in G. @TedShifrin

(that is, try letting $a=b$)

one of them is a student I like from engineering, who's hardworking but has nowhere near the math background he needs, which is why he's doing abysmally. I wish he would drop the class but continue attending, since tihs seems like what he needs

Best of luck to them. I hope they learn the material and pass.

I am proving the --> direction the --< direction is trivial

oh, order $n$?

you end up with $te^{at}$! So it's really just $c_1e^{at}+c_2e^{bt}$, just disguised.

I have a similar student, but I think he might just be able to pass.

yeah in this case n = 2k

I just showed some engineering students that limit the other day, @Columbus.

what are you teaching in complex analysis @MikeMiller?

What's the precise definition of order of an element, @Karim?

Cool! And it also explains why, in that other equation (Euler's diff. equation, is it called?) where we normally have $x^a$ and $x^b$ as solutions, you suddenly get $x^a$ and $x^a\ln x$ if $a=b$.

An element x has order n denoted as |x| = n if n is the least positive number such that $a^n = 1$

'Cause $\displaystyle x^a\ln x=\lim_{b\to a}\frac{x^a-x^b}{a-b}$.

We're working up to the residue theorem right now, @Karim.

@Hippalectryon here is the solution for 8 bottles c1=|1,2|, c2=|3,4|, c3=|5,6,7,8|. If it's not c1 or c2 we know the poisoned bottle is in c3 and then we make check for a mix of 2 bottles. Then bu exclusion we get the poisoned bottle.

Use that.

I see cool @MikeMiller

@Chris'ssis I don't see how that would work in two tests. To know if it's in c1 or c2 we need to use two tests already.

oh ok I got it thank you @TedShifrin

@Hippalectryon You can check c1 with a mouse, and c2 with another mouse (1 test). If they are free we go to the c3. No problem, all is fine.

alright good section 1.1 I finished all the excerises in dummit and foote

that took 2 days :S

@Chris'ssis Ah, yes. I forgot a test could be both mice.

At this rate, you should be done in two years.

I think I shouldn't do all the excerises I should do like 5 per chapter or something.

yeah @TedShifrin

@Hippalectryon Yes. :-)

@Chris'ssis I don't get the c3 part though (in one test, even with both mice)

Well, 5 may not be enough.

what would you recommend @TedShifrin?

It probably depends on the section, @Karim.

@Hippalectryon Think about it: you make a mix of 2 bottles in c3 and test it with one mouse. Then you check any of the remaining 2 bottles and see the effects. :-)

Assume it doesn't kill it

You have two bottles left, and already used one test

yeah I guess like important section like group actions then I should do more but trivial section I should do atleast 1/3 of the problems in it I think

but the hard ones

@Chris'ssis Wait, let's make one thing clear. Is one test the simultaneous use of the two mice, or just their use once (simultaneous or not) ?

StackExchange needs help determining if I'm a human or a robot.

@MikeMiller So do I.

seems like nearing 10k they would have caught me by now

@Hippalectryon One test might mean that you can use both mice at a single test.

@MichaelA: Do you need me to read some blurry text for you?

They could ask me for an opinion, @MikeM.

@Chris'ssis So basically, we have 4 independant one-mouse tests ?

@Hippalectryon Yes.

@Chris'ssis Ok. I totally agree then.

Opinion is not far off from onion, and now you've got me thinking about onions.

@Hippalectryon Then my solution above is correct it seems also for 9 bottles.

@Chris'ssis indeed it is. (1,2),(3,4,5),(6,7,8,9)

rolls all twelve eyes

@Hippalectryon Exactly. It's important the order it seems, c2, c1, c3.

What's wrong with onions?

@TedShifrin Rejoice, we have probably finished poisoning mice

@Chris'ssis To me c1,c2,c3 (as I wrote it) works

Enfin ...

Managed to kill them all, @Hippa?

@MikeMiller I baked them in Ted's oven

Rats I would have understood.

@Hippalectryon No, it doesn't. Just think about it: one mouse might die in the c2.

@Chris'ssis yeah, I did it too fast

@Chris'ssis Well, we can drink wine now :3

@Hippalectryon Yesssssssssss! Without the discussion with you I wouldn't have solved it. Thanks! :-)

@Hippa: Les souris au four? Is that a delicacy?

Oddly enough, my dentist actually gave me a variant of the poisoned wine problem one time... turns out he had a degree in math.

He uses his degree in math to poison his patients?

@Hippalectryon lolllllllll :-))))))))))))))))))

If an augmented matrix has a free variable then surely it can't be inconsistent?

And @TedShifrin,

@MikeMiller That's quite possible; I just know I escaped (mostly) unscathed

Vraiment méchant, @Hippa.

@TedShifrin For it to be inconsistent means that the only solution is the trivial solution. But if a variable is free, then a solution can have a non-zero entry.

@DanielF Looks like we're fighting over conformal mapping questions again.

Your hint was probably better.

No, @Don. You're messed up.

@TedShifrin I'm not creating anything though.

This time, @Hippa, this time ...

@TedShifrin For it to be inconsistent means that there is no solution.

Correct.

If you take $x+y=2$ and $x+y=3$, you have a free variable, but it's moot because there is no solution.

@MikeMiller I was just trying to help them find one, you were concerned with then finding all, which I did not (yet) mention. And you should stop commenting and post a few answers. You'll never reach 10k if you never post answers, you slacker ;)

hi, @DanielF ... I'm slacking too.

Hey, I posted one today.

@TedShifrin I thought that for an augmented matrix to have a free variable you need to have the entire row equal to zero.

@TedShifrin You worked long and hard to earn the right to slack.

@TedShifrin My Linear Algebra 1 professor sucked. The TA was better.

I finally gave up and showed someone how to do inversion geometrically, although I'm sure I'll regret it ...

You look only in the variables, not in the augmented part, @Don. Consistent if and only if no pivot in the augmented column.

And you can always watch my linear alg lectures ... I only half suck :P

@DanielFischer hi

Hi.

think you have little time to explain something about a method polynomial?

@TedShifrin That's pretty easy to prove. Suppose that it was a pivot in the column. Impossible for any x to equal it because 0(c) = 0 where c is any number. Now, suppose that it was inconsistent. That means there's no solution, which means for any x-vector it will never equal the b-vector. (Kind of stuck here?)

There's never any geometric topology questions to answer, @DanielF. Maybe that's a sign that my field isn't very active. I'd better switch to real analysis, or maybe combinatorics.

Soft question might be a good field, too.

@TedShifrin It's a good answer.

@MikeM: Relatively few grad student types post questions here, so it's not surprising how little geometric topology there is.

@MikeMiller You're playing with a suspension, matey.

@TedShifrin: I wish more would.

@DanielFischer: Well, back in the day, it was clear that homework was the hot topic everyone was getting into.

Hey, does anybody know about any abuses of notations that have helped mathematics?

@MikeMiller Well, there's always algebra-precalculus.

@MikeMiller algebra-precalculus is still going strong

LOL

(Or, even, invention of a notation and then immediate abuse of it.)

@MichaelAlbanese: Here's an interesting question we were talking about earlier. If I've got a surface, all of whose geodesics are periodic, is it necessarily $S^2$ or $\Bbb{RP}^2$?

@Lucas: I don't even know what you're talking about ... And I'm leaving my office now to go back home ...

@DonLarynx Either, I guess — use your judgement.

If you have an example you're thinking about, but you're not sure if it counts, I mean.

@columbus8myhw Too general, will not waste time.

@MikeMiller Nice question.

I think it's true, but couldn't do it after thinking about it for a while.

@DonLarynx Uh... How about treating $\dfrac{dy}{dx}$ as a fraction, which — while technically incorrect — often gives correct results?

@MikeM: Go look at Besse's Manifolds All of Whose Geodesics Are Closed.

(for example.)

@TedShifrin Continuation of proof: Since addition of reals takes place in 1-dimension, it can be safe to assume we're adding up linearly-dependent lines with zero slope. In the end, we end up with a line that will never equal the b-vector. But this is impossible because our linear dependency covers all of $\Bbb{R}$. Thus the coefficients must equal zero. QED

Is this the equivalent to my statement? Aren't the geodesics of $\Bbb R^2$ closed?

@columbus8myhw Why is it technically incorrect? It's geometrically correct. (Define technically.)

@TedShifrin I often get depressed when I see a very specific thing has an entire book dedicated to it.

@DonLarynx How could it be correct?

I agree it's equivalent on a compact surface.

Even in nonstandard analysis, it's not a fraction, it's the standard part of a fraction. (If you don't know nonstandard analysis, ignore that last bit.)

@columbus8myhw Using epsilon-delta it's correct.

@DonLarynx But it's not a fraction. It's the limit of a fraction. Similarly, $dy$ and $dx$ are not numbers.

@MichaelA: How do you feel about FOOO's 800 page book on Lagrangian intersection homology?

I guess I had better restrict my question to complete manifolds above.

@columbus8myhw I can't answer your question. Anyways, thanks @Ted

FOOO?

Fukaya Oh Ota Ono

They publish a new, insanely technical, hundred-page paper on the subject about once a month. :D

@MikeMiller Isn't a (Riemannian) manifold all of whose geodesics are closed automatically complete?

@columbus8myhw I'll try again.

@DanielFischer: Of course, I thought of this myself a couple times while thinking of the problem earlier. But I'm very forgetful.

Thanks for reminding me.

@columbus8myhw That's because calculus was re-written to satisfy that issue. See here: math.stackexchange.com/questions/21199/…

More examples: math.uchicago.edu/~chonoles/expository-notes/promys/…

Alright, I'm off for now. Bye.

See ya. Enjoy grading.

Ah, I missed @MichaelA's and @Don's departures. Oh well.

You have me.

What more do you want. .

I don't know why @MichaelA finds research monographs depressing. At least, the good ones :)

LOL, @Owatch :)

Why, how silly of me :P

BTW, @columbus, with hyperreal numbers, it becomes a true quotient. But we won't go there.

I purchased bottled water today.

I can't wait to drink it tomorrow.

um, ok, @Owatch ... where I'm moving there may be no water, so I guess that would excite me too.

Has Tee Dog been spotted lately ?

Nope, haven't seen him in several days.

@Mike @MichaelA: I can't believe how forgetful I'm getting in my dotage. There's a zillion surfaces. Amongst them are Zoll surfaces.

@TedShifrin

Unless you can correct me on that.

(Nonstandard analysis being the version of analysis you get with hyperreals.)

hmm, I shouldn't have dealt with Pandora's box. You may be right. It's also the case that it makes sense as a fraction if we consider $1$-forms on the graph $y=f(x)$.

Have you seen that video of the dog that played musical instruments lately Ted?

@Ted: My use of sphere gives away my background again. I was asking if it had to be homeomorphic to $S^2$ or $\Bbb{RP}^2$.

No, I saw a cat, @Owatch.

It's fascinating.

@TedShifrin There's the hyperreal definition of a derivative, if you're curious

in geometry, @Mike, one really needs to be precise with the language, because there are serious rigidity results. Like the pinching theorem, which says it has to be a round sphere.

No, I agree. I apologize.

If it's real (I don't think it's a trick), it's amazing that the dog can play complex beats.

yeah, @columbus ... it's been over 40 years, but I yield.

We should look at Besse, though. There are lots of fascinating examples, (some of) which I once knew.

LOL, @Owatch ... no doubt one can train animals.

Perhaps. It's still interesting.

Especially the clips where the rhythm wasn't a consistent thump, but perhaps a one -two - one sequence.

Although the dog certainly looked rather indifferent to it all.

Well, if you were trained to be an automaton, you would be, too. But the cats playing the piano were astonishing :P

Let's see if I can find it.

Here's one of 'em.

I suppose that is intellectually beyond a dogs capability.

I'd like to imagine it isn't. But I think you're explanation is more realistic.

Can people see the following characters? 𝔞Normal unicode: ∫, ∬, ∭, ⨌, ↓, ∩, ↦, ↚, ℝ

Examples of evil chars: '𝐀', '𝛂', '𝝖'

unless I'm hallucinating, @JustGreg, yes.

𝖡Normal unicode: ∫, ∬, ∭, ⨌, ↓, ∩, ↦, ↚, ℝ

oops

The cat seems to be interested in the sounds the Piano makes. But not able to make anything other than random sounds .

shit, sorry

Sharing cat videos, chat has now officially become part of the internet! @TedShifrin

Yup, @Paul :)

There are better ones, @Owatch, in that same series, where she plays piano duets with her mistress.

Oh.

@Ted: It looks like my hope might be true. I'm not willing to browse through this huge book on my computer, though, so I'll stop by the library when I get a chance.

I've taught my dog to go to the bathroom outside, lift his paws, lie down, and jump. But nothing more. .

I'm thinking what you want should follow from Morse theory (see Milnor, again).

Do you have a strategy in mind?

Well, I'm sure the first one is useful, @Owatch ... not so sure about the second fourth :P

His other capabilities involve theft, evasion, and irritating people.

Except for the ones who took it already (and are doing miserably, per my grading), Tuesday, @Alex.

You taught him to be like you, @Owatch :P

no, @Mike, not without studying.

Which in particular .. .

@Owatch That :)

OK. I'm a bit sad about this business, because it means I apparently didn't internalize Milnor very well.

Mmhmm

Well, @MikeM, I don't swear to it.

Cats playing the piano just remind me of the ending scene from Stroszek.

I've only graded half the final for half the students, @Alex, so I'll reserve judgment until midweek.

Perhaps a pig would be able to recognize rhythm.

Maybe it is too limited too, a chimpanzee perhaps? An elephant ?

@MikeM: Thank god I missed that one.

@Ted: No, it's a fantastic film. Are you dissing Herzog?

Not knowingly.

Which reminds me, I want to watch A Touch of Class again. Hysterical.

OK, dinnertime. Bubye.

bye

what's a wag

@AlexClark What is wagging?

You silly Australians :P

you should probably go to class

What class was it

as a general philosophy

@AlexClark skipping

well, just don't be like my complex analysis students, who don't go... and then fail the test

@MikeMiller Perhaps you are thinking of "cutting class"

I'm not thinking of much right now

@TedShifrin My proof was wrong. plus the 4x5 augmented, row-reduced matrix with all zeros except for the 1st row [0, 0, 1, 1, 1] is consistent.

:D Good

thinking is good, I should do a little today

How close are you to getting it finished?

I've been thinking a bit today. For example, if Ted's Multivariable Mathematics was freely available like the diff geo book, I think that would be awesome.

And the presentation? I forgot, what was the other project?

i find sylow theory very meh

a lot of time is spent on it in a group theory class and very little time is spent on it when you do other mathematics

Those are quite handy

solvable groups, on the other hand, show up

@AlexClark Heisenberg group is solvable

Free groups are fun, they are not necessarily difficult. I don't really know what you are suppose to cover, or how much

@Ted: Ignore that. My proof is correct.