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21:01
@robjohn I guess not everyone has seen it =) By the way it is a visual proof for $$ \sec^2 x + csc^2 = \left( \cot x + \tan x \right)^2 $$
indeed :-)
@robjohn My internett is horrible at the moment, I can barely view the chat. Sorry bout that..
@JonasTeuwen: Re your questions from yeterday: I don't know anyone from the Uni of Innsbruck, but I hopefully will; I will start to work there in summer. My field of research is game theory, which can partially be explained by me being a grad student in economics.
@MichaelGreinecker Oh cool. Then how do you know about this crazy stuff?
Just interested?
@JonasTeuwen: Partially, one can do fancy stuff in economics too. My advisor wrote this. A lot is couriosity, but I gravitate toward fancy mathematical economics . I have learned a lot of stuff myself, which lead to a very uneven education in mathematics.
21:07
@MichaelGreinecker I have this book about "Infinite dimensional analysis - a hitchhikers guide" I believe it is written by some economists.
Always bring a towel.
@JonasTeuwen: Yes, that book is like the bible for economic theorists. Aliprantis was a pure mathematician for a long time before he got interested in economics. Sadly, he died of cancer a few years ago.
@MichaelGreinecker Interesting :-). It looked really pure to me.
@JonasTeuwen: I is rather pure, but the motivation is often in terms of economics. Something that is very typical for economics it that they only deal with real vector spaces. I still feel uneasy when I encounter complex vector spaces...
@MichaelGreinecker Just... tensor them? 8-).
21:20
@MichaelGreinecker why is that?
Complex vector spaces are nice because the field of complex numbers is very nice!
(also, some of the theorems we know and love from linear algebra are only true for algebraically closed fields...)
Complex things seems less complex than real things... Well at least for me so far.
@N3buchadnezzar Haha!
$\oint_C z$ <3
Note the reputation and real/pseudonym relation to the order of the top 4 candidates...
Bill Dubuque 67311
Eric Naslund 25782
robjohn 24636
mixedmath 15662
21:24
@DavidWheeler: The typical use of fancy vector spaces is in terms of spaces of commodity bundles. A point in $\mathbb{R}^n$ interpreted as a commodity bundle that specifies the amounts of $n$ goods. Spaces of sequences can then be interpreted as infinite consumption streams. Spaces of continuous functions as the continuous time analog. So the basic reason is that real numbers can represent quantities and complex numbers can't.
the way i see it, if you assume everything is complex-valued from the start, you can still always work with real values without really changing anything. this keeps you from having to "enlarge the field" later if you run into some polynomial you can't solve.
Perhaps I should have gotten rid of the pseudonym :-)
@robjohn: I eventually voted for you, but the angry gravatar really bugged me. Come election time, every politician smiles a lot.
Mean squares can also laugh once in a while.
@Rob See, you should have let me have that avatar!
21:26
@BillDubuque :-p that wouldn't have changed a thing, I bet.
@MichaelGreinecker i would argue that in some cases, some commodity magnitudes might be "coupled" as a kind of "trade-off" with some other commodity, so that it may very well make sense to represent the pair as a single complex-valued variable.
@DavidWheeler: I am not sure what you mean. I understand what it means that agent A has twice as much of everything as agent B, but I don't know what it means that agent A has 2+i times as much as agent B.
or, to use another example, anything which is cyclical (so you have two components: the current value, and where in the cycle you current are).
Yeah, but then the modulus is actually the current.
And the argument the phase shift.
@robjohn nice, thank you :)
Should you get the mean photoshopper badge in addition to the hacker badge?
21:36
@tb it wasn't photoshopped :-)
@DavidWheeler: I actually toyed some times with writing a paper on game-theoretic aspects of time travel, maybe I can use these ideas then.
@tb I saved the front page, and tweaked the source :-)
@MichaelGreinecker: it would mean that agent A has √5 times "as much", but "in a different direction"
@robjohn even nicer :)
throws photoshopper badge application form into the bin
there may be more than one axis of measurement besides gain/loss
21:38
@tb so that is exactly how it looked.
@robjohn What did you change? :-).
@tb So you seem to know a lot of set theory. But I do consider you an analyst. If I talk to one of my colleagues about set theoretic issues in analysis they stare at me like I'm crazy but in a bad way!
if these two axes of measurement aren't "linked" then a real vector space may suffice. but it could be they are, by some sort of system constraint.
@JonasTeuwen The source from some number above 50,000 to 50,000
But the interesting about the complex number is about representing periodic signals, and I suppose one does not have them.
So you could as well just tensor two of the real monkeys.
@DavidWheeler: I will try to wrap y head around this once I have familarized myself enough with the un-real world.
21:40
@tb Gigili was looking for you earlier
well, in some markets, i would think there would be seasonal cyclical effects.
@JonasTeuwen: There are people who work on dynamics in economic models, and they use of curse complex numbers. They also show up in time-series econometrics.
@Gigili tah-daah, here I am!
(but not for too long).
@tb She left for bed after the election results were out.
@robjohn oh, they're out? Let me check.
user19161
21:42
@robjohn Left for bed is an interesting expression.
i could put it another way: any time you have any sort of trigonometry involved, sometimes a complex representation takes "fewer symbols" to convey the same information.
@ClarkKent That's what she said
user19161
@robjohn Well, I hope she's not too sad.
@tb You will have to process the results. They are still in raw form.
@ClarkKent I think that we probably knew how things would go; but there is always hope, and that can be the worst.
user19161
@robjohn But the winners are determined already right?
21:44
@DavidWheeler: That may be a good reason to study them. Often probabilitic results have a geometric interpretation with correlations as certain angles and stuff.
user19161
@robjohn Well, hope keeps me alive these days!
@ClarkKent The raw results are pretty clear, but I don't think the results have been processed and posted.
i think the biggest obstacle to complex-anything for a lot of people is "interpretation". imaginary numbers are poorly named.
I guess they haven't received their mod-super-power yet.
@MichaelGreinecker not yet. they don't have their diamonds yet :-)
actually, they do!!!
user19161
21:47
@robjohn Really?
confirmed.
user19161
Oh I see the diamonds now.
Bill Dubuque and Eric Naslund are officially mods!!!
user19161
My diamond is invisible, and I have one on the left and one on the right.
user19161
@robjohn robjohn is officially chat owner!
21:50
@robjohn Thanks for that. OpenSTV is too painful for me to handle...
needless to say that king robjohn will remain the only true king for me.
@tb thanks >8(
i have a simple question: if elements of a subgroup $G \subset \mathrm{GL}(n,\Bbb{R})$ is completely determined by $k$ parameters $t_1,\dots,t_k$, is the orbit of any vector under the action of $G$ a $k$-dimensional affine set?
Hi!
Where can I see who got elected?
4 mins ago, by robjohn
Bill Dubuque and Eric Naslund are officially mods!!!
What??
And seriously?
21:53
@MattN rep = mod :p
@MattN Hey Matt. Here
@tb Ah, that page was not filled in 15 minutes ago :-)
@tb Yes thanks, and where's the non-fake results?
Seriously. I suspect that this shows that the number theory crowd not only casts the most votes when it comes to answers.
Oh.
Well.
Seriously.
21:54
Who is insane?
@MattN He's also standing right behind you :-)
@robjohn I don't care : )
Meh. Never mind. I'm too tired too care.
Fight, fight, fight...
@tb My colleagues... 8-). I was talking to them about CH! One was like: "I'm not interested in this nonsense". Wth?
Just had to pull a last minute stunt for a hand-in due tomorrow.
Now I want more drink.
3
@tb Today I thought it's been ages since you "disappeared" and then I checked that message and it appears to only be from last Thursday.
21:58
pfft! tb is like that gf i had who said "i'm leaving!" and we wound up married for 26 years.
@MattN A drink? Excellent! What will it be?
user19161
@JonasTeuwen Whiskey!
You compare tb to a woman? Oh man...
Stop trolling me with your whiskey. It is whisky.
@JonasTeuwen find new colleagues, then :)
@tb I cannot :-(. But my advisor is cool about it, but he's only almost never there.
And the others are only interested in their own very narrow thing or are Dutch.
Which is almost the same for all practical purposes.
user19161
22:00
@JonasTeuwen Oh thanks for telling me. I did not know. But have a look here. dictionary.cambridge.org/dictionary/british/whisky
@JonasTeuwen Sake.
But going to bed soon. Still recovering from flu.
user19161
So whiskey is the US or Irish spelling while whisky is the UK spelling.
@JonasTeuwen are you suggesting that women are somehow inherently inferior?
@JonasTeuwen: I always though the Dutch were extremely liberal: drugs, prostitution, and euthanasia
@ClarkKent No, even Laphroaig is whisky in the US.
@MichaelGreinecker Yes, they are! But! For some reason they only like to hang out with their own people.
22:01
that's a weird view on what being liberal means :D
the idea is more like «let people do what they want»
someone has to show up to award a bounty, right?
which the Dutch do do, of course
@robjohn, what do you mean?
user19161
@MarianoSuárezAlvarez For example, the three examples.
@MarianoSuárezAlvarez: That is how liberal is used in Europe. And apparently they allow you to do everything but using the continuum hypothesis.
@MarianoSuárezAlvarez I answered a question, a good answer, I think, the only answer, with a bounty, but the OP hasn't been around for a while.
22:03
I wonder whether he's going to delete this now that he's mod.
@robjohn, if he does not grant the bounty, it gets "lost"
he still loses the points, iirc
@MarianoSuárezAlvarez That's what I thought.
A monoid is a pair (M,*) where M is a set and *:M x M -> M is a binary operation...

What binary operation is this? A multiplication? Any possible binary operation?
@MarianoSuárezAlvarez Yes, I know that.
it could be anything, Gustavo
even "putting on top of"
user19161
22:05
@MattN Why delete it?
@robjohn If you've got two votes and have the highest-voted answer that was posted after the bounty was set, you'll receive half the bounty. See here for the fiddly details
the set of rectangles of width 1 is a monoid under the operation of "stacking on top of"
user19161
@GustavoBandeira Any. That's the definition there, no more no less.
Oh, the most rude of all impolite mods is here :)
Oky, i thought it was a reference to a specific kind of binary operation.
22:06
Bill? Or is Mariano also rude?
@MarkDominus It's quite a big coincidence that the number of first-place votes that I received happens to be my favorite number. Google "Gauss 163" to learn about it. Who's playing tricks on me?
@MattN yes, M. and note that he's also sarcastic (just found that thread)
@MarianoSuárezAlvarez They do that. They do what they want. They don't want to hang around with me. Fine.
@tb Heh. : )
@JonasTeuwen Super Mariano? : )
@MattN 8-).
22:09
@JonasTeuwen, :( Their loss, I guess!
user19161
@tb I just saw that comment. I think it is OK.
user19161
@JonasTeuwen It's OK, I am here with you. :-)
@ClarkKent Thanks.
@MarianoSuárezAlvarez The cool Dutch are much cooler. Plus, the foreigners are also cooler :-).
@tb I have 3 upvotes and it was accepted. I think the bounty was there when I answered.
ergo, everyone is cool in Dutchland
@tb, behave or I'll shoot my sarcasm ray in your direction!
22:12
I have to get some real work done. ping me if something important happens :-)
«Why are there isosceles triangles?» is actually a cool question :)
user19161
@MarianoSuárezAlvarez Just exhibit one isosceles triangle to answer the question.
@MattN All of us sci.math old-timers had our minds warped by the crackpots.
@BillDubuque It's ok. I'm too tired to think about mods and this site. I guess I'm not obsessed enough. : )
Hmmm, I wonder if the definition of flat functor can be interpreted as a definition for flat modules...
22:16
what is a flat functor?
I have a better question:
user19161
My coca cola is getting flat now.
Can you tell me an anagram of banach tarski?
@tb, maybe?
Speaking of crackpots: I think I made a sarcastic cooment to WM when he made his first post. But that was before Bill ruled with an iron hand. ;-)
@MarianoSuárezAlvarez /me ducks away.
@MattN Are you alluding to the old Banach-Tarski Banach-Tarski joke?
22:19
@MichaelGreinecker If you're talking about that answer of yours then: I upvoted that : )
@Mariano: There are many definitions. One is obviously stolen from commutative algebra: a flat functor is a functor $F : \mathbb{C} \to \textbf{Set}$ such that the "tensor" functor ${-} \otimes_{\mathbb{C}} F : [\mathbb{C}^\text{op}, \textbf{Set}] \to \textbf{Set}$ is left exact.
@tb Yes! How can anyone in here know it?? : D Outrageous.
This "tensor" functor being the Yoneda extension given by the universal property of $[\mathbb{C}^\textrm{op}, \textbf{Set}]$ as the free small colimit completion of $\mathbb{C}$...
Archaist Bank
Ransack Habit?
user19161
@MattN No. So what is it?
22:20
A friend of my girlfriend is supposed to have invented it! : ) I was dead sure no one in here would know.
"supposed" you say...
@MattN It's in the AMS jokes article, I think...
@ZhenLin :,(
I preffer Ransack habit though.
22:21
@anon well, it's hard to verify whether there wasn't anyone else inventing the same joke slightly before or slightly after him, no?
actually I don't think the joke's in the ams article, but it's ubiquitous I think because I know I've seen it a lot
Oh good, I finally understand the "devilish scheme" joke!
There are many Google entrie for "banach tarski anagram"
Ok, I get it now: it's an old joke.
user19161
I am quite bad at getting jokes.
22:24
@ClarkKent That comes with not being a native Earthling, I suppose.
I hope you get that one.
Q: How many Bourbakists does it take to replace a
lightbulb?
A: Changing a lightbulb is a special case of a more
general theorem concerning the maintenance and
repair of an electrical system. To establish upper and
lower bounds for the number of personnel required,
we must determine whether the sufficient conditions
of Lemma 2.1 (Availability of personnel) and those
of Corollary 2.3.55 (Motivation of personnel) apply.
If and only if these conditions are met, we derive the
result by an application of the theorems in Section
user19161
@anon Oh I just understood your joke. But seriously, in real life, I often feel that I am not an Earthling too.
(technically I think he was raised on a farm or somesuch, so maybe nevermind.)
@N3buchadnezzar Nice.
user19161
@N3buchadnezzar Where is that from?
22:26
@N3buchadnezzar LoL
"The primary reason Bourbaki stopped writing books was the realization that Lang was one single person."
how many set-theorists does it take to change a light bulb?
@anon I don't understand that one yet...
@ZhenLin Maybe you do and it's just not funny?
@ClarkKent It's from the ams article I linked.
user19161
22:28
@anon I actually find Bourbaki's books unreadable as they are of too great generality. Also I find Lang's books to cover many topics but not any in great detail.
I gather that Bourbaki was a group of mathematicians under a single alias, while the Lang in question was one person who wrote a lot.
My favourite way to do proving something is proof by intimidation.
3. call changing a light bulb L. then it takes one to check the consistency of ZF+L, one to check the consistency ZF+(~L), and a third to invoke the axiom of choice.
@anon Yes, but that's not the funny part, I'm sure...
Ok. Cup's empty. Gotta force myself into bed.
Nice to see my favourite teddy.
user19161
22:30
@anon Er, yes.
Hi. Could someone please take a set-theoretic look here?
Good night!
Good night.
user19161
@MattN Good night! Ding ding ding!
Q: Why did the chicken cross the Möbius strip?
A: To get to the same side.
^^
user19161
22:32
@N3buchadnezzar But why did it cross the road?
Look at this guy's gravatar
@ymar, it confuses me when you say let x = M
@ClarkKent Godel: You can not prove it crossed the road.
Nov 9 '11 at 1:58, by anon
"My only regret is not making love to a Mobius strip" "Would you be top or bottom?" "Yes."
@ymar That is almost surely not answering the question, and it depends on set-theoretic details!
22:35
@DavidWheeler @ZhenLin The question is whether a $1\times 1$-matrix is equal to its entry, isn't it?
user19161
@PeterTamaroff I noticed the avatars are getting stranger each day, the autogenerated ones.
I mean part of the question.
user19161
@anon Side to side is nice.
There is only one side.
the "not-well-foundedness" comes from the fact that you are considering a mapping of the form f:A→B, where f is in B.
22:36
There was a tragic accident in the strip's childhood, and one of its sides had to be amputated.
@ymar Sure, but the answer to that is "no" for conceptual reasons, not silly set-theoretic facts!
the thing is, matrix multiplication is a binary operation on "type-matched" domains
@ymar: I think the question was of the sort: If one can identify $1\times 1$-matrices and scalars, why do scalar multiplication and matrix multiplication differ in this case, which I consider stemming from a "legitimate con-fusion" of two concepts.
It's a story using the Möbius strip, thought out very well.
22:38
scalar multiplication is not a binary operation.
Ah, Vi. People recommend her videos but I don't really bother with youtube much.
it's one type of thing acting on another.
@ZhenLin Of course! I didn't post it as an answer to be accepted. Just as an aside. Do you think it's completely irrelevant? If so, I'll delete it.
@anon I don't really like all of them. But some are cool.
user19161
22:39
@BenjaminLim Morning!
Oh, I get it. Ha.
user19161
@anon Vi? This is getting too fast for me to understand...
@BenjaminLim Hi there.
22:40
there is a natural injection R-->mat(n,R) given by r-->diag(r,r,r...,r), but "isomorphic" is not "equals"
@PeterTamaroff: I guess the gravatar-Rohrschach-test reminded you of a swastika.
@BenjaminLim Sup bro
@ClarkKent It is a girl that makes some math videos
@PeterTamaroff election results are out
Vihart, the snakes lady.
22:40
Vi Hart.
@JonasTeuwen hey
Or Vi Hart, whatever :/
hey jonas
@anon Link link!
How to fence in the largest area. Make a small square around yourself. Declare yourself to be outside of the fence.
user19161
22:40
@JonasTeuwen That can also mean Super.
It can. But it doesn't here.
@ymar Well, the fact that it can be proven formally is an artefact of set-theoretic foundations. There are other foundations where the question is not even well-formed because it does not typecheck. And it is conceivable that there are foundations where the answer to the question turns out to be yes for technical reasons.
@DavidWheeler Seems the test came out similar for you. I can't see anything else.
@PeterTamaroff this one was mini-viral
@JonasTeuwen I'm not so comfortable with function spaces, where is a good resource to deal with topics like uniform convergence, completeness, total boundedness, equicontinuity for function spaces?
22:41
@BenjaminLim In metric spaces?
@JonasTeuwen yes
@anon It actually inspired this question of mine.
user19161
@BenjaminLim Rudin's analysis books.
not so good....
user19161
@BenjaminLim Any good general topology book, like Willard.
22:43
@ClarkKent well munkres does not do much about function spaces....or he does them mainly in the general topology setting
he does not talk much about L1 or L2...
user19161
@BenjaminLim Ah then back to Rudin then. :-)
@JonasTeuwen Do you know of any such resources?
user19161
@BenjaminLim By the way I think Munkres is over-rated.
@BenjaminLim What is wrong with Rudin?
@BenjaminLim But $L^p$ are vector spaces.
@BenjaminLim Do you hve the link to the election's post? I can't find it.
22:44
You should really be more specific.
First you study the case in metric spaces. Say Rudin.
Then you can do real analysis and get some function spaces of integrable functions.
Then you can go to the real stuff and do functional analysis.
You know, you could ask Pierre about this!
user19161
@JonasTeuwen Exactly. So just read his three books. QED.
@JonasTeuwen Rudin does not cover things like total boundedness
Yes, that was my point.
@BenjaminLim What do you want to know about total boundedness?...
Rudin is certainly enough.
@JonasTeuwen Well like total boundedness and how it relates to compactness, completeness, etc in function spaces
You want an analytic view or a topological view?
22:46
analytic view
Well, then it is not so important.
As compactness is more a statement about sequences, okay, it is possible that you need it...
But then you can always look it up and you'll know plenty of analysis.
@ZhenLin OK, I will probably delete it. I'm waiting for a comment from M Turgeon now.
But, it sure is possible that Rudin is too hard to start with!
@ymar That's entirely up to you.
@JonasTeuwen exactly. that's why I'm asking for a simpler resource
user19161
22:47
By the way, Whyburn wrote Topological Analysis and Analytic Topology.
That would be more the rule than the exception.
@BenjaminLim Okay, let me think about that...
user19161
@BenjaminLim I think you can handle it.
@BenjaminLim You have the book by Pugh (UTM) and Browder (UTM).
I like the latter one, it is Rudin-style but then more elaborate.
@JonasTeuwen I have pugh, not sure about browder
user19161
@JonasTeuwen And also Protter.
22:48
I don't know these books Clark.
I've only read Rudin. All you will ever need.
user19161
@BenjaminLim Browder covers Lebesgue integration and manifolds in his book too. It squeezes a lot into one and is very well-written.
Yes, I do have checked parts of it.
It is quite a lovely fellow.
“Al-gebra is a fearsome cult,” Ashcroft
said. “They desire average solutions by
means and extremes, and sometimes go
off on tangents in a search of absolute
value. They use secret code names like
‘x’ and ‘y’ and refer to themselves as
‘unknowns’, but we have determined
they belong to a common denominator
of the axis with coordinates in every
country. As the Greek philanderer Isosceles
used to say, there are three sides to
every triangle,” Ashcroft declared."
do you know foundations of real and abstract analysis by bridges @JonasTeuwen
user19161
@JonasTeuwen I am still waiting to check out Folland's calculus book. I can't preview it anywhere on the internet.
22:50
No, because these things are very elementary and one book should be enough. I'll check it out.
@BenjaminLim If you would have a hard time with Principles of Mathematical Analysis by Rudin this will not help you much.
@BenjaminLim Pugh has lovely exercises. Browder is very cute.
i love Vi Hart's doodling in math class series
@ClarkKent Calculus??? The bro has too little money?
ok thanks guys
@BenjaminLim Once you get that over with you could get Folland.
That is one lovely book.
Ilya has mine.
user19161
@JonasTeuwen I am also waiting for your book. Maybe I will write some in future too. I intend to write one on algebra, one on analysis and one on geometry.
22:53
I don't have any plans for books. I'm not competent enough. Maybe I'll write something on sociology.
@JonasTeuwen I'm off now to do some analysis, bye guys!
user19161
@BenjaminLim It has a collection of topics that are usually covered across several books. I don't like that organisation.
@ClarkKent I'm just using the exercises
@BenjaminLim Good boy.
user19161
@BenjaminLim If you are not sure what to read it's best to go with the usual ones.
user19161
22:54
Bridges is not a usual choice.
@JonasTeuwen That's the way I roll
You're talking about Folland, Clark?
First analysis, then maggot?
That's the best order.
@ClarkKent I'm just using it for the analysis
@JonasTeuwen I'm not going out at least in another 5 weeks
@BenjaminLim Holy cow! Why?
user19161
@BenjaminLim I see. I really recommend Walter Rudin for analysis, Paul Cohn for algebra and John Lee for geometry/topology. Each has three textbooks that build up the subject effectively.
22:55
end of term, exams, etc
@BenjaminLim Ah! Good luck.
Gah, I fail at constructive thinking. Why does the right cancellation property for a map of sets imply that it is surjective? I can only think of a non-constructive proof...
user19161
@ZhenLin Wait, are you just talking about ordinary functions here?
Sure. But you are not allowed to use the law of excluded middle. Or the axiom of choice.
try picking a particular map that is helpful

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