Mathematics - 2014-06-24 (page 1 of 3)

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Simon André Forsberg

12:00 AM

I'm not sure if I'm interrupting anything or not, I just wanted to bring attention to something I think you might find interesting. I have written an algorithm for calculating mine probabilities in Minesweeper, and I think it's something you folks might be interested at

I've asked a question about it over at Code Review. The question even has a 150 bounty on it now:

13

Q: Analyzing Minesweeper Probabilities

Calculating probabilities in Minesweeper might sound like an easy task, but I've seen so many probability calculators that's either incorrect, horribly slow, or with ugly code (or all of them) so I have to share my code for this. This code is used within my Minesweeper Flags online game by the A...

java performance algorithm combinatorics statistics

hey

Simon André Forsberg

If anyone is interested in it, I'd love to hear opinions about the algorithm or code some day if you have time. It's quite a lot of code, but it's quite interesting.

@TedShifrin Did you see it?

Nope @Alizter. Hi, @Ethan.

@Ethan hi

wb

12:05 AM

did you guys here about that wacky thing: scientificamerican.com/article/…

@Ethan thanks for sharing

Is there a more standard way of addressing logical problems like "who's lying" besides looking at it case by case?

@TedShifrin chat.stackexchange.com/transcript/message/16239911#16239911

@Alizter: I'm afraid this gives me no insight ...

@Anthony: Generally, following through case-by-case long lines of reasoning ...

But what about situations where there are a lot of cases?

Is there no better way?

12:15 AM

@TedShifrin It's just a pretty fractal

@AlexanderGruber the problem with the statement "there's no way to make this work" is that the this has not been clearly defined

the example I gave was inspired by an answer I gave to someone who wanted to animate SL2(Z) actions with continuous motions, so I explained that we can put the two generators into one-parameter subgroups and then each word in them corresponds to a series of "moves," each move an interval of continuous motion

@Ted I'm confused. Chern seems to call $p:\Bbb C^{n+1}\setminus \{0\}\rightarrow \Bbb{CP}^n$ a line bundle. But... how is that a line bundle?

I also clearly want some more background in several variables. What's the best introduction? Hormander? Gunning-Rossi?

@blue did you change your name?

@KajHansen

No, it's a $\Bbb C^*$ bundle. He defines it carefully, as I recall. Hörmander is succinct.

Doesn't he do $\{([v], tv)\}$? @Mike

12:25 AM

Hello everyone. Just got finished tutoring some Calc 1. Took me right back to @TedShifrin's "Advanced Differential Calculus" a.k.a. differential geometry. :P

Sarcasm duly noted, @Kaj

@Ted he's describing exactly what you say he is, and goes on to talk about the Hopf fibration. But he does say line bundle.

Right. What I wrote is a line bundle. It's different from what you wrote.

He writes that - but for $t \neq 0$.

(For $t =0$ incl. is it not just $\mathbb{CP}^n \times \mathbb C$?)

It's actually $\Bbb C^{n+1}$ with the origin blown up. ;)

12:30 AM

In a few months I'll understand that.

No, not the trivial line bundle!

Yep, I gathered from what you said.

@Ethan There were some serious criticisms of those prizes, and I think for good reason.

@KajHansen Did a part of you die a little inside?

You're right. He doesn't define the line bundle on p. 1.

Regarding the Calc 1? I mean, it's alright. I got some practice with related rates and implicit differentiation. That's always good from time to time :P

12:35 AM

@KajHansen I am trying to prove the recurrence for Eulerian numbers.

@Ted Yes, it just confused me a bit that he said line bundle with something that wasn't :P

$A(n,k)$ is the number of permutations of $[n]$ that have exactly $k-1$ descents.

Oh god @PedroTamaroff. We had so many recurrence relations to prove in my combinatorics course. All. The. Time.

He said generalizing it would lead to a line bundle. See p. 54.

@KajHansen Heh. It should be in terms of $A(n-1,k-1)$ and $A(n-1,k)$.

12:37 AM

But certainly generalizing means that the given example is one? But this is a silly complaint - I only asked the question to make sure I wasn't misunderstanding naything.

Never heard of Eulerian numbers though. Do you know what the recurrence relation is supposed to be?

@Mike: More generally, associated to any $S^1$-bundle is a canonical $\Bbb C$ bundle.

Yes, it should be $A(n,k)=(n-k)A(n-1,k-1)+(k+1)A(n-1,k)$

@Mike: Did you note the Preface?

And I guess our challenge is to figure out the two cases that correspond with the two parts. Fun fun.

12:41 AM

Yes.

@TedShifrin No preface in my copy. And !!! that's suprising to me!

Do you have the yellow book or the blue book?

Blue.

Ah, that's first edition. Doesn't have the appendix on char classes you were reading about ... Nor the corrections I made ;)

@PedroTamaroff, making sure I understand these correctly. If $n = 4$ and $m = 3$, would a valid permutation be $(3, 1, 2, 4)$ ?

12:44 AM

I guess that shouldn't be surprising to me. Obviously it gives us a $\mathbb C^\ast$ bundle, and then one should want to make it a $\mathbb C$-bundle by sticking a copy of $B$ in there for $0$ :P

@KajHansen That permutation has only one descent.

A descent is when $a_i>a_{i+1}$.

Where $\pi=a_1\cdots a_n$.

I think I have an idea though.

Let me think about it a bit more.

Oh interesting. The wikipedia article uses ascents.

I think the answer would come out the same either way.

Yes, you can just flip the thing, right?

Me too @Kaj

i.e. $a_na_{n-1}\cdots a_1$.

At any rate your permutation has two ascents, not $3$.

The only permutation in $S_4$ with $3$ ascents is $1234$

12:49 AM

hmmm, Wikipedia says the permutation $123$ in $S_3$ has $2$ ascents.

Oh wait

Ugh. You're right.

That's what I say when Pedro's right, too.

Did you read what I wrote re editions? @Mike

Ok, I can see where the first term is coming from.

Yes @Ted

I'm still gonna read it. :P

@KajHansen ORLY

12:51 AM

yup. But there are some sign issues wrong with Chern classes. I've forgotten what else.

Yes, roughly. Let me think about it some more, because I have no clue where the second term is coming from, which makes me uneasy.

You'll know I'm paying attention if you get questions from me, @Ted

@KajHansen Maybe this helps

A permutation with $k-1$ descents can be partitioned into $k$ monotone blocks with at least one number in them.

I was actually going to say this about the first term:

The first term counts those with a descent at the beginning; the second counts those with an ascent at the beginning. @Pedro @Kaj

12:55 AM

@KajHansen Also careful, the recursion might have something wrong, since they are shifted.

Damn, @TedShifrin stole my thunder right as I was figuring out the second term, haha

I still don't have it, though ;)

This is what's going on: Look at the first term, and it is either a descent or an ascent. Suppose it is an ascent.

An ascent from the first term to the second I mean.

I only have 2 months left to learn to count, @Kaj :)

@KajHansen OK.

12:59 AM

One second, just trying to verify something.

OK, suppose it is an ascent.

OK.

Then if that ascent is to continue for $m$ terms, and that is the first in your sequence, then there are $n-m$ possibilities for the first term.

And of course, you'll have to count the ways to form a sequence of $m-1$ more ascending terms following it.

And you have already used one of your numbers.

Which explains why the first term is $(n-m)A(n-1, m-1)$.

Case 2: Now suppose that your first term is not an ascent.

No, wait. $m-1$ is the number of descents, not some variable number.

Oh, let me check wiki again.

No, I was thinking about it correctly I believe.

Maybe I am not understand your reasoning.

1:06 AM

Ok, let me see...you want to have an ascent $m$ terms long

Why?

No, we want $m-1$ descents.

The ascents can vary in length.

OH crap.

I was thinking they needed to be all at once.

For example, say $n=9$. Then $4\,237\,169\,85$ has four descents.

You're right. This might actually make it easier to explain the recursion then.

Suppose that the first term to the second is an ascent.

My idea is we can write the permutation as $B_1B_2\cdots B_k$ where each is a monotone block.

1:10 AM

If it is, then you need $m-1$ more descents, but you have $n-1$ numbers to choose from.

(My permutation above has five descents...)

54 being the fifth?

I'm with you for the monotone blocks.

@KajHansen No, $8\to 5$. =P

Am I having counting trouble? We have 42, 71, 98, 85...?

@KajHansen Oh, right.

1:13 AM

Ok I think we're on the same page now

My first idea, naturally, was to look into the block where $n$ is.

Wait, no. One should look into the block $1$ is.

For descents, I say.

AGH, NO. Note for example that 45 123 yields a 45 23 after deletion of 1, but if this was 23 145 then we get 2345 which has no descents. =/

I don't know how that can work.

The very frustrating thing is that I can see precisely where the A(x, y) is coming from in both terms.

I'm having trouble justifying the coefficients of $(n-m)$ and $(m+1)$

I have a suggestion @Kaj @Pedro

Sure thing @TedShifrin. Any insight on the coefficients?

@TedShifrin OK.

1:25 AM

Yes

Remove $n$ and consider an arrangement of $1,\dots,n-1$ with $k-1$ descents. Now show you can stick $n$ in exactly $n-k$ Places to add another descent without losing any of the ones we had.

That's a great way of thinking about it! And likewise, there are $(m+1)$ places to stick $n$ such that it doesn't add or take away any descents. Namely, in between any existing descent, or at the very end.

Which explains the $(m+1)A(n-1, m)$ term.

Yeah! @Pedro?

Yes. Good.

I really should've gotten that faster. embarrassed

I'll take $\sqrt\epsilon$ partial credit. :)

1:42 AM

Not $\frac{1}{\epsilon}$? Reminds me of an anecdote I saw in a documentary. John Nash had appeared in someone's dream to reveal the solution to some proof. When the guy published, he felt the result was so profound that there was no way he could've come up with it on his own, so he ended up putting Nash's name on the paper along with his.

@MikeMiller yea i'm not necessarily into the whole idea of prizes like that either lol, i was just checking to see if others saw the article

Never heard that one, @Kaj :)

@TedShifrin I don't get why you say "Remove $n$ and consider...". When you remove $n$, you get some arrangement of $[n-1]$; but you don't know if it has $k-1$ descents.

It has either $k$ or $k-1$.

Yes, that's true. But I don't see the bijection.

1:57 AM

The sum comes from those two cases.

2:25 AM

I don't even understand what's going on in this thread:
http://math.stackexchange.com/questions/845302/is-the-golden-ratio-transcendental

for any prime power $\ell$, we have $f\mid \varphi(\ell^f-1)$ for all $f\ge1$ since ${\rm Gal}(\Bbb F_{\ell^f}/\Bbb F_\ell)\cong C_f$ acts on the generators of $\Bbb F_{\ell^f}^\times$ with trivial stabilizers. what about for other $\ell$s?

@Kaj Maybe he's after the Self-Learner badge?

:P

@ryagami I suspect you are right, but everything is just so bizarre. Weird grammar, unclear statements in the OP, etc.

It's an odd one for sure, given that he /just/ posted (and subsequently deleted) the very same question...

I really wonder what explainded is supposed to mean...

2:33 AM

@Kaj: Surely not a native Englirsh speaker.

@TedShifrin, that has to be it. @AWertheim, I noticed that too. Strange.

*Engrish

@AWertheim: probably deleted to get rid of downvotes, as he has yet again.

Oh, and hi :)

Oh man, I just had a good laugh being reminded of an answer I posted a while back where I completely misread the OP. Must've gotten 5 downvotes in a matter of seconds.

Needless to say, I deleted that one fast. LOL.

@TedShifrin, hello! :) Probably, though that being the case, his new question isn't quite working out how he planned lol

@KajHansen: I've had my share of those! That panicking "oh shoot" moment... lol

2:36 AM

@Kaj, this site really needs a higher tolerance level. But, then again, that is true of most of these math sites

<---- highly intolerant of those providing homework/test solutions

LOL a good intolerance to have, @TedShifrin. Some of the other kinds of intolerance on this site distress me though :(

@ryagami, I tend to agree. I can't remember the last time I downvoted a response. I usually just stick with commenting.

@Ted, I don't see the point in that.

The people who post homework questions won't get anything from you trying to force them to think, because they just don't care

And the ones that do care will do their best to learn from the answer

I would do a homework question if I find it interesting or have nothing better to do with my time

I've been pretty bad about giving too much away in the past, but I'm finding that if I give away less, there's usually a good back-and-forth that will follow.

2:42 AM

Of course, I could very well be wrong.

At the same time, I've also gotten a lot out of reading other people's complete solutions to questions I've come across. I guess it's kind of similar to the experience of reading a proof or example from a textbook.

That's also a fine point

Yes, I don't see the point of reproducing textbook proofs and examples....

Get Schaum's Outlines with thousands of solutions.

@TedShifrin I do, it forces me to learn what I'm answering.

@TedShifrin, I can't tell whether you're being sarcastic in your last line there.

2:47 AM

I've also seen plenty of wrong answers, including numerous ones from people with PhDs ...

No, @Kaj, I wasn't.

@TedShifrin Do you say they are wrong?

When I catch them, yes.

I mean, I can still justify some arguments of Ted's point of view

It's probably a good thing @Ted doesn't browse the abstract algebra tag where I post most often :P

I know a few people who intend on living from, especially teaching, math who do not seem even remotely interested in it

2:49 AM

I'm basically giving up on being a teacher ... Except for the advanced questions I answer ... And except in here.

@Kaj lol, it cannot be that bad :P

@TedShifrin why?

Because mostly I get sabotaged by people who want to show off ... Even when they can see I'm hinting at/with the OP.

yeah, I'm that guy

Me too

2:51 AM

Yes, sadly, @ryagami, many math education majors really don't like math. But a lot do care ...

@ryagami, I've noticed this too! There's a sizable population within the math education crowd that detest the upper-level, proof-based courses here at UGA. I will never understand why someone who wants to do math for a living will, at the same time, avoid math courses.

Not that I like it

morning

> many math education majors really don't like math

:O

Well, you guys are driving me away. Some people will rejoice ...

Sometimes I do it because I get more rep from complete answers than hints

2:52 AM

And that fiat lux asshole too ... Agh.

fiat lux asshole?

@Kaj, well, I don't blame them for not liking proof-based courses

But, still, there are worse

Oh great ...

you mean a guy who owns a luxury fiat?

I didn't know fiat made luxury cars

I wouldn answers to homework questions unless they are really challenging

2:53 AM

What exactly does Fiat Lux mean?

@TedShifrin: that guy is really strange. Has anyone visited his website?

wouldnt&

@AWertheim, Yes! Isn't it bizzare?

Let there be light ...

wouldn't*

2:54 AM

this looks like a luxury fiat dinouk.com/images/modelli/1967%20Pininfarina%20Berlinetta/…

user image

He does the same thing ...

@KajHansen: super weird. I don't really get that guy. He has some impressive credentials, but... his style is a bit... irregular.

Hahaha @Bananarama

@AWertheim who are we talking about?

@ryagami, teaching math without the "why" - i.e. teaching math without a proof-based perspective - drives people away from math.

2:55 AM

There are lots of stupid people here with impressive credentials ;)

I keep forgetting that you can edit your previously said stuff here :|

My point being, no one likes rote memorization.

@Bananarama: math.stackexchange.com/users/67071/robert-lewis

@Kaj: Say explanation and motivation rather than proof.

@Kaj, I know quite a few people who have the opposite view

They prefer problems to proofs

Result to reason

2:56 AM

@TedShifrin: certainly. But I don't think the user in question is dumb, in this case. Just... odd.

Very odd ... I think some people answer questions out of area and mess up. I too have erred from time to time.

You're probably right @TedShifrin. Also, I would highly recommend reading the essay "A Mathematicians Lament" to anyone who hasn't already.

wow, I just looked at his activity, what the heck?

@KajHansen I would somewhat disagree with some of the points it makes (as a high school student)

@Bananarama I see nothing wrong, he is just answering questions (or I'm missing something)

have you seen his website? the link is www.fuckyeah.biz

2:59 AM

oh

...

@Bananarama weird

@VibhavPant, I have a few disagreements as well, but I do think children would be better off than they are now, to say the least.

@KajHansen It is very difficult to teach a standard undergrad Calculus sequence that is proof-based. Usually schools expect it to be done almost as part of an engineering curriculum, so that the applications are emphasized. So you can't really put problems requiring proofs on tests, and students know this. If you do present proofs in class, a large percentage of the class will just zone out.

raraahahahromaromamagagaoohlala.com

2

@VibhavPant I didn't say anything was wrong. But his righting style is really weird, its like he's talking to me and I'm his buddy.

ah

heh

@TomCruise Apostol has tried to do that in his text

3:02 AM

@TomCruise, the important thing is to get people to be interested in the "why?" long before college. Instead, high school math is a hodge-podge of memorizing and doing, instead of asking and understanding.

Then once people get into classes that are not "follow this algorithm to your answer", the transition is much gentler.

I can agree with that

Also, wasn't your handle Com Truise not too long ago? LOL.

yes ;)

I may change it again soon, who knows...

names cannot be changed more than once within a 30-day time period without mod intervention

oh, well in that case I'm stuck with it for 30 days

3:05 AM

My other calculus text has questions like "Verify Rolle's Theorem on <function>" (<function> is usually continuous)

I should add that I speak from experience regarding my perspective on high school math. @TedShifrin's course (my first college math course) was, without a doubt, the hardest class I've ever taken. And I feel at least a small portion of that was the transition from high school to college.

The whole point of AP Calculus is not understanding, it is test preparation...

@TomCruise Thats how calculus is taught here

Agreed @TomCruise. It's a shame. I've since had to go back and fill in a large number of holes in my Calc 1/2 knowledge simply because certain material was not emphasized in the curriculum since it was not emphasized on the exam.

Anyone have any interesting music?

I'm running out of things to listen to

3:09 AM

try Com Truise

I wrote my last paper to it

@ryagami: classical or otherwise?

I think I'll pass

@AWertheim something else

I'm leaning towards pop

But I can listen to pretty much anything I find pleasant

it is sortof retro electro, with a Terminator 2 vibe

@ryagami, I'm a huge fan of Alt-J. Try it out? youtube.com/…

I'll listen to it

Just want to hear what this Com Truise thing is

Okay, I was right to pass on that one. Definitely not for me. :P

3:12 AM

:D

there's another good one called Mitch Murder

Is it similar to the other one?

yeah

but a little more 80's

LOL

The worst of all is it's a good description

murdering your ears?

3:15 AM

Not exactly, but not my cup of tea

@Kaj, interesting, but not something I'd listen to

I can be very picky :P

But I tend to listen to things from almost every genre

I'm a big admirer of anything performed by Glenn Gould

Excluding K-Pop, J-Pop and that stuff

@ryagami, skip ahead to the track "Breezeblocks". Then I'll spare you from my more abominable musical tastes.

@Kaj, it's not abominable. He has an interesting voice. I just don't find it very interesting for me

synth music = no vocals

3:19 AM

@Tom No classical, please. Not in the mood :P

The first two reminded me of M83

I know my musical taste is a little limited

Or at least their one song I know

Anyone else like the various subgenres of metal? I'm a huge fan of black and death metal, lol. I also have a huge collection of indie rock, and some 90's hip hop.

I like some sorts of metal

But I am not really able to classify most metal stuff as metal

Except the hardcore stuff

I knew a girl who was into death metal... very mild mannered and nice. When I heard that music it freaked me out a little.

3:21 AM

Classic metal is pretty good too. You can't go wrong with some good Metallica or Megadeth.

@Tom, just, do not be fooled

Metal isn't the only scary stuff out there

This for example: youtube.com/watch?v=cyiJcuO3uew

Check out The Pierces - Secret

Alexander Gruber

@blue yeah that's the challenge, is figuring out what exactly i'm trying to get at.

making the definitions is 80% of the work.

@ryagami, this is kinda catchy.

3:24 AM

It is

Listen to Sticks & Stones too

Or rather, watch it

I also tend to listen to mixed stuff from time to time

Kavinsky youtube.com/watch?v=-5FKNViujeM

Right now I have open SOAD and The Mamas & The Papas in a window

Alexander Gruber

$D_n$ always looks the same for every $n$... repeated application of the rotation makes a circle. Multiplying by the reflection gives a second circle with reversed orientation.

System of a Down is pretty good. BYOB, Toxicity, and Chop Suey are masterpieces.

Alexander Gruber

a Frobenius group $G=C_q\rtimes C_p$ with cyclic Sylow subgroups is the same, but with $p$ copies of the $q$ circle...

3:28 AM

@Tom, the voices talking at the beginning reminds of Welcome to Night Vale

@Kaj, true. I really like some of their songs

Alexander Gruber

there has to be a way of codifying this type of qualitative difference using by going into the "continuous world"... some way of ignoring what $p$ and $q$ are. @blue

I also like their Sad Statue

Also, you guys should watch the video for The Staves - Winter Trees

I'm not familiar with it. The guitar work at the beginning is pretty good now that I'm listening to it.

Alexander Gruber

@KajHansen yes.

@TomCruise @Tom, I like this one

3:31 AM

My last recommendation... Steely Dan: Aja, Gaucho albums

A bit

Alexander Gruber

@TomCruise steely dan is also awesome

@KajHansen i was electrocuted on stage covering Chop Suey once.

@ryagami ok try some of his other tracks too.

Oh I might end up really liking Winter Trees @ryagami.

@AlexanderGruber I am not a huge fan, but those last 2 albums were amazing

3:32 AM

@AlexanderGruber Ha! Whoever had the vocals must've been quite talented.

@Tom, I might

@Kaj, check out Mexico too. It's in the same vein

Alexander Gruber

@KajHansen that was me, heh

the shock helped with the ol' vibrato

@ryagami, the intro to Mexico reminded me of Cherry Tree by The National: youtube.com/watch?v=7EygMZjxxOE

I like this one

@AlexanderGruber, that's awesome.

3:36 AM

Anyone listen to Lana, maybe? :P

Yeah, but not much. Video Games is pretty good.

Or is it Video Game? I can't remember.

Most of her songs are good :P

Video Games

Summertime Sadness became quite mainstream these days, but she has sooo many good songs, most unreleased

This answer is probably the funniest thing I've ever seen on this site: math.stackexchange.com/questions/421620/why-is-frac1-frac1x-x/…

Read the question as well, of course.

That guy has serious potential

Well, I'll be off. Need to get some sleep, I guess

Best of luck to everyone :P

wow what a post

ok later @ryagami

3:42 AM

Good night @ryagami

lana del rey rox

Late to the discussion @Mike :P

oh no

anyway the album that's off of, born to die, is excellent

I'm not too familiar with her, but it seems I should be

4:44 AM

@MikeMiller You're trolling right?

go away Pedro

your music opinions stopped mattering when you didn't like clapping

clapping sucks

How about snapping your fingers?

what does Pedro like then?

5:57 AM

@robjohn are you awake ? :-)

@r9m yes

@robjohn can you take a look [here](math.stackexchange.com/questions/804951/how-can-we-prove-this-integral-in‌equality-int-0-frac-pi2-left-frac?rq=1) ? .. I can't figure out how to bring the $\log n$ in the picture :|

use http: //www to get hyperlinks

ah .. ic

Wonders if student of @TedShifrin

6:07 AM

Who is? @PedroTamaroff

@Pedro that's a common question

6:26 AM

@r9m how about that?

@robjohn Great !! .. B)

r9m Dives off the window .. -> [] >-|o .. lands on a pool of mud /\/_o_/\/\ .. and wonders what he should do next ^^'

@r9m Hmm... there is a deleted answer that looks pretty much like mine, but the OP commented: "I know this methods......But you can't understand my questions.,..." If that method doesn't answer the question and essentially follow the hint, I don't know what would.

@robjohn Ah .. ic .. I can't see the deleted answers my rep points is too low for that ..

6:41 AM

@r9m that is why I mentioned it. I wonder what the OP meant by that?

@robjohn God knows ! :( .. but is there any special property of this integral ?

How do you show that $\lim\sup \left(\int_a^b f^n(x) dx\right)^{1/n} \le \sup f$ ?

@Hippalectryon using $limsup \sqrt[n]{a_n} \le limsup \dfrac{a_{n+1}}{a_n}$ ? (I forgot the name of this theorem .. )

I've never seen it :c

But ok

@Hippalectryon there is a decent proof in Baby Rudin

6:48 AM

What's Baby Rudin ?

@Hippalectryon Principles of Mathematical Analysis .. W. Rudin

Ok

The real and complex analysis is called the 'father rudin' .. :P lol

xD

@robjohn I could write that integral as $\dfrac{\pi}{2(2n+1)} + \sum\limits_{k=1}^n \dfrac{1}{k}\tan(\dfrac{\pi k}{2n+1})$ and trying to show it is less than $\pi (1+\frac{\log n}{2})$ .. :|

very nice form .. I should divide it by $\log n$ and make that into a limit problem XD .. I wonder what the limit would be ?! :o

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