@awllower. That sounds highly interesting ^.^ When did Tobias define sudokability tho?

Yesterday in chat.

oh ^^ I was studying yesterday. Meh :D

@JacobBlack Shocked? What post?

@awllower I found it already ^.^ Dont understand a word tho, I am missing out on a lot of group theory basics there

and I am having a hard time with the terminology, as I am not natively english

Oh

I am not either:-)

@PeterTamaroff Here, buy some candy for yourself.

hi

already capped today :(

Hello!

hi

how are you ?

grrrrrrrreat

oh that's good

@OrangeHarvester Bleh. Let's argue about meaningful things!

For example: Why would Rihanna get back together with Chris Brown?

@PeterTamaroff Yes!

@PeterTamaroff Who is Chris Brown?

@OrangeHarvester A darker version of Chris White.

Now, I think you can guess who Chris Black is.

Both Brown and White are music composers. None of them mathematicians. Sad.

@JacobBlack I am pretty fond of my university's book.

@PeterTamaroff Bremsstrahlung.

@DominicMichaelis What is black body radiation called in German? My book by Halliday Krane says it is Bremsstrahlung, but Bremsstrahlung is braking radiation according to Wikipedia.

@OrangeHarvester Was?

@JacobBlack Well. Let's say luck does play a role in it.

@JacobBlack Why?

@Orange i just have a word for the thing which radiates it

Schwarzer strahler

@DominicMichaelis Okay. Thanks.

@DominicMichaelis okay. google translate translates blackbody radiation to "Schwarzkörperstrahlung" and black body radiation to "schwarzen Strahlung".

yeah that will wrk

but schwarzen strahlung sounds funny to me

To me too. I like the former better.

I just realized that in English the qualifiers are transitive (transitive in action, not transitive as in the verb) in contrast to other European languages which do not. Is my observation true?

An example of transitivity?

I will give an example from portuguese. Just a second.

(Mainly because google translate has a very high accuracy rate for portuguese.)

sunny street = rua ensolarada

lonely sunny street = rua ensolarada solitário

Now, street = sunny

so all the adjectives that are "acting" on street and hence qualifying it should be concatenated to one side just like function composition, as happens in english

in constrat we can see that sometimes the adjectives are added on left and sometimes on right

Sometimes the adjectives are changed all together

I see.

Like

menina is girl

and pouco is little

so little girl should be pouco menina

but instead it is menininha

which is frankly very weird.

mh in german i guess most words are transtive too

@DominicMichaelis Okay. But then germans combine words too. For example, Thank = Danken and You = Sie (correct me if I am wrong, my knowledge is very limited), so by logic thank you should be danken sie, but instead it is danke!

it is a noun

@awllower ?

Danke is like Thanks

Okay.

Because Dative should be adjoined to the verb Danken.

Might I ask what is the reaosn for limiting the number of votes one can cast one day?

I find it quite annoying sometimes.

@awllower To prevent voting abuse.

@awllower I see.

hello

@user58512 Hi

But there are too many good questions that deserve some votes

@user58512 hi

@awllower save them for another day!

I have never reached the maximum votes/day

I oft run out of votes and just stare at the good questions, without any way of fulfilling my desire to vote...

they are like... 40?

Yes

gahwd

I got the 30 votes/day badge once

but that was my absolute maximum till this point ^.^

^^

Also, I only look at questions that I could understand

which are only ~10% of all questions

@awllower The thing is that the reputation decides privileges. Now, that benchmark is static. So, we cannot have rep inflation. So, suppose there was a metric of goodness (G) and a number of votes V and number of good questions N, then V = G *N. Has to stay constant. In short, raise the bar for upvotes!

I guess ;)

@OrangeHarvester A wonderful explanation for this limitation!

@awllower also, some questions have received incredulous number of upvotes. So, it all balances out I think. People who understand more difficult stuff should vote for more difficult questions, leaving the easier questions to be upvoted by crowds like me. So, in the end, difficult questions also get same amount of upvotes as the easy questions. (So, the system works out in the end.)

btw, how many distinct sudokus are there? Like the amout of equivalence classes, where $S_1 \sim S_2$ iff $S_2$ can be rotated, mirrored or the numbers 1..9 permutated such that $S_2'=S_1$

If difficult question or answer gets just the same amount of votes, then how can it be fair?

@CBenni I am not familiar with that counting, so I too ignorant to answer this question.

aah too bad

would be interesting to know however

@awllower Going by the fact that currently they get (far) less votes, it is fair!

haha, indeed!

Also, it is fair in the sense of usefulness. Top voted questions should be useful. So, we consider (difficulty times number of people who find it useful) which I presume is constant. So, it is fair again.

:P :P

I have a cold, my brain isn't working properly :(

because the number of all sudokus is huge, 6,670,903,752,021,072,936,960

thats small

countabel

countable

@awllower However, the number of ways to permutate the numbers $1..9$ is $9!$ iirc? Therefore, the number of sudokus is below 18383222420692992

@DominicMichaelis true. The number is tiny compared to Grahams Number

which is still finite...

or compared to ackermann(grahams number, grahams number) ^^

LOL

what is bigger, grahams number in a grahams number-gon (in steinhaus-moser-notation) or ack(G,G)?

@awl maybe is in several SE

Well because I am logged in in several SE sites

Oh, I see.

I have 895 in Math.SE, and a couple o' hundreds elsewhere

Farewell, number-theoreticists!

hehe

@benni ask in on the forum :D

I propose CBennis number: $ack(G\text{ in a }G-gon,(G\text{ in a }G-gon)!)$

and still is the limit of this one divided by x for x -> infty 0

lol

and the number divided by itself is 1

mosers number is smaller than grahams

Is the ackermann function bigger than the steinhaus-moser notation? I dont think so

@awllower, not a singularity - just a fuzz.

@awllower, but poor Ramanujan, I think :(

errrrmmmmmm

Oh

so yeah

no i just say mosers number is smaller than grahams number not the ackermann funktion and so on

yeah sure

busy beaver is still bigger than either iirc

because it kinda denotes the biggest number you can express algorithmically

grows way slower, but explodes at some point

@CBenni, busy beaver is bounds every computable function

@user58512 basically what I was trying to say

@cbenni did you know that the first implemention of the ackermannfunction

pretty wierd stuff :D like "the biggest function you can create"

could only calculate it to n = 1

LOL

what

well you can define a turing machine with an oracle that lets it tell whether or not a normal turing machine halts

then the busy beaver function for that is even more huge than the normal busy beaver

user58512, but thats cheating :D

thats why it wins

mmh

doping scandal in number theory

@cbenny it was ack(n,3=

the one they tried

ackermann(n,n) bounds every primitive recursive function because it's recursion of a higher-type

@awllower btw now i have 1912 rep

Mh, I can only look at the light at the back of your car...

@user58512 umm. yeah :D I dont really know alot about number theory, I just do it for the sake of fun ;) I do not know any of the actual theory behind it and computatiablility theory is way too abstract for me till this point

I don't think primitive recursive functions are very interesting

stuff like $a_{n+1}=2^{a_n}$?

no I mean "primitive recursive functions" as an object of study

oh

^.^

they seem contrived

"contrived"... never heard that before, had to look it up :/ The more you know...

just try something like that

i can't calculate that on my pc

Hi all

Hi

@TobiasKildetoft Hello there.

hi

I'm trying to find a simple way to prove that $\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$

Do you have any thoughts as to my previous questions on your ideas?

I mean on sudokability.

@DominicMichaelis Mathematica throws a underflow for n=30, but n=25 is $5.533540\times10^{-17544900}$

@Chris'ssisterandpals Hey,

Peter!!!! :-) Hi.

Seen this?

@awllower have you see the answer by m.k?

@PeterTamaroff: nice! Thank you for posting my question!

@cbenni i know i tried to make it analytical

yes

i even got a numerical with using 3 000 000 milion digits

@DominicMichaelis lol. Let me guess. It didnt work?

lol

nope after 30 mins my 4 gig ram was full

thats 3M Million or just 3 Million

3 million sry

@Chris'ssisterandpals You say simple and then your solutions runs into asymptotics and power series! If anything, you get us thinking on whether or not to use advanced techniques and at the least leave us wondering if we used a sledgehammer to crack a nut. :-/

aah kk

@TobiasKildetoft He showed that all finite abelian groups are sudokable!

@awllower ahh, I just saw your comment

@awllower well, those of square order

I have marked you previously in chat. have you seen that?

its funny with 3 million digits it doesn't even take a singel second

@awllower hmm, no, I don't think so

@julien hi are you htere ?

All right, for I also forgot what I said at that time, haha

@OrangeHarvester: I give you my honest word and swear in front of God that all my brother's questions (with some exceptions) can be solved elementarily BY ONLY USING HIGH SCHOOL KNOWLEDGE. I'm really sad when people think this is not true.

all but finite ? ^^

@Chris'ssisterandpals Is this brother/sister thing for real?

@PeterTamaroff: thank you for posting my question. It means that you really liked it. :-)

@Chris'ssisterandpals I am not saying what you are saying is not true. I am not accusing you of lying. I am just saying that the qualifier "high school knowledge" and "elementary" does more harm than good. It would be just fine if you did not use that qualifiers.

hahaha i.minus.com/izTZTZzMsaThB.gif

as you know in mathe there are only two difficulites

russian reaction to meteor

trivial stuff and those which isn't prooved

@DominicMichaelis completely agree.

@OrangeHarvester: but this is my pleasure when doing math, to find elementary solutions. It's not meant to be something harmful, but sorry if this seems so. :-(

(from a comedy show)

@Chris'ssisterandpals i like that too, but i think here is no

@Chris'ssisterandpals The reason I say so is sometimes though the techniques are high school level, the maturity required to implement them are certainly not high school. For example, you can simply refer to the herstein's proof of the first sylow theorem using only combinatorics. You will easily see that the proof does not use anything apart from basic divisibility and number of combinations, but to call it high school would be preposterous.

sin math.stackexchange.com/questions/305316/…

i mean there is no elementary solution

@DominicMichaelis, I told you an idea for elementary solution

@user58512 yeah but i didn't reached any further than quadratic irreducible

@OrangeHarvester: the truth is that they are high school questions. I cannot change the truth.

What does an algebraist mean when he says "disjoint union"? Does he mean this? I am a little confused here...

@anon to the rescue!

no not that

No?

I am always confused. No one can rescue me.

Why?

If I new that I wouldn't be confused would I?

Hm, very well.

I don't mean to sound disrespectful :)

@PeterTamaroff I guess usually what will be meant is "as disjoint as makes sense"

I know: I always assume people on this site are likewise:)

@TobiasKildetoft Well, maybe if I give you some context you can help.

Let $X=\{x_1,\dots,x_r\}$

usually, I say "with trivial intersection" rather than disjoint unless I really mean disjoint

And let $X^j=\prod X^i$ be the usual $j$ fold cartesian product.

Then the author says "Let $FS^{(r)}$ denote the disjoint union of the sets $X^1,X^2,\dots$..."

I'm guessing he says disjoint because they really are disjoint.

ahh, disjoint union is a specific construction

it does not mean they are actually disjoint to start with

it is the coproduct in the category of sets

He then says "The elements of $FS^{(r)}$ are "words in the alphabet of $X$" that is, they are sequences $(x_{i_1},\dots x_{i_m})$ with $x_{i_j}\in X$ and $m=1,2,3,\dots$"

He is constructing the free monoid generated by $r$ elements.

Peter, have you seen this? $\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$

@Chris'ssisterandpals No, sorry.

take for example the set $X = \{1,2\}$. Then the disjoint union of $X$ with itself is $\{(1,a),(2,a),(1,b),(2,b)\}$

ANd I don't think I will for a while.

OK

(where the $a$ and $b$ are just placeholders to distinguish where the elements came from)

@TobiasKildetoft That's why I linked to this

ahh

The thing is he never talked about disjoint unions before, AFAIK.

right, he means the one where they don't start out disjoint

@TobiasKildetoft Come again?

@PeterTamaroff that article has two meanings of the word

@TobiasKildetoft He says we get the set of "words in the alphabet of $X$"

I get what we obtain, just worried about the formal definition

I guess we can just consider $(x_1,(x_2,x_3))=(x_1,x_2,x_3)$ and we're OK.

OK, I got this guy.

ahh, right, those sets are of course disjoint

I think he just means $\bigcup X^i$

and for disjoint sets, the disjoint union is just the union

The sets are disjoint to begin with

Why would he not just say "set of all finite sequences in $X$"?

@Chris'ssisterandpals Do we have to prove that the integral is rational for all $n \in \mathbb{N}$

@peoplepower yeah, it does seem like a strangely convoluted way of doing this

@OrangeHarvester: right

Is there moderator available? I need a short parley with one.

@MJD Hehhee, where?

@peoplepower Bros be crazy"!

@PeterTamaroff What?

@MJD I have the popcorn ready, where are the comments?

I have no idea what you are talking about.

@Chris'ssisterandpals Try using differentiation under the integral?

hi

@OrangeHarvester: I didn't think of that. (yet)

@MJD I thought you said "parlay" no "parley".

It is no great matter, and neither of you has answered me question.

No, there seems to be no one here.

robjohn probably has not gotten up yet.

@JacobBlack: perhaps.

@robjohn Ping

i am losing days wiht my cold

@MJD yes?

I can't study because of it

It is worrying

@MJD there you go.

@Chris'ssisterandpals First differentiation, then by parts, later some manipulations and you should be done.

@OrangeHarvester: as you explain things, it should be easy. Thanks!

Thanks folks.

:(

@OrangeHarvester: for instance, this one $\lim_{n\to\infty}\int_0^{\pi}e^x\cdot\sin(nx)\,\mathrm{d}x$ is another high school problem that may be easily evaluated with the mean value theorem. But guess? Chris told me "don't evaluate it the way all people evaluate it, but be brilliant and find a fast, straightforward way in one line". That means we use no mean value theorem.

@Chris'ssisterandpals This seems to be $e^{\pi} - 1$? Or may be not.

@OrangeHarvester,: sorry but there is a mistake. It's $\lim_{n\to\infty}\int_0^{\pi}e^x\cdot|\sin(nx)|\,\mathrm{d}x$

@Chris'ssisterandpals okay.

@Chris'ssisterandpals Okay. See, I understand that things can be evaluated in elementary techniques. That is not my point. My point is while the techniques themselves may be elementary, they may not be "easy". When you emphasize "high school" people think the problem is actually easy. Are you getting what I mean? Your problems might be high school level, but to emphasize that they re high school level makes people think that they are very routine and far too easy that they actually are.

You are welcome to use "elementary" as the keyword though, saying that you already have solution by advanced techniques and now want some solutions from elementary techniques or something similar. (You did this in a problem I think. That was nice.)

@OrangeHarvester: sure, I got your point.

I don't know how to do it with mean value theorem

@user58512: en.wikipedia.org/wiki/Mean_value_theorem

but that just says there exists a point c, we don't know what c is

@Chris'ssisterandpals That goes to zero.

I get that if we knew c then (b-a)f'(c) would give us the integral

By the Riemann Lebesgue Lemma.

it cant be zero with the absolute value sign in, can it?

Oh, gawd. All this abstract algebra will bleed my brains dry.

@mjd thanks for the edit

Wait, where is abstract algebra in the above discussion?

@PeterTamaroff The absolute value should play some part I think.

@DominicMichaelis de nada

sry i don't unterstand

@OrangeHarvester Nah. Just split it for positive and negative.

@PeterTamaroff Ahh. Right.

@MJD Hablas castellano?

@DominicMichaelis I think he means de rien

In english: ?P

NP

@DominicMichaelis I am glad to have helped you, it was no trouble, you are very welcome.

in english P not= NP

Please don't mention it, it was nothing.

It is pretty amazing how the cases commutative monoid, commutative group, monoid, group, escalate in complexity.

how is group more complicated than monoid?

@user58512 Congrats! You proved the conjecture in the English case!

@user58512 P=NP iff (N=1 or P=0) solved :)

@user58512 You have inverses...

@user58512 The problem of deciding if two elements of a group are equal is undecidable. The problem of deciding if two elements of am monoid are equal is trivial.

@user58512 groups are ugly there aren't so cute counter examples

@MJD, where group & monoid are given by presentations?

@DominicMichaelis Might I ask what you mean?

for expamle given a monoid in which the following is known

@MJD, as given i can't make meaningful that

for all p,q in the monoid and all n \in \mathbb{N} it is (pq)^n = p^n q^n

In a monoid, abcdefg = pqrstuv is true iff a=p, b=q, ... g=v. In a group, could be really complicated.

@DominicMichaelis What? No.

than the monoid doesn't need to be kommutative

@MJD, that's not true

are you talking about free monoids?

@mjd as every group is a monoid...

Oh, I'm thinking free monoids only, sorry. Please erase everything I said before.

@mjd sry my sentence wasn't finished

@MJD in that case, it is also true for free groups

(assuming no cancellation)

I'm quitting now before I spread my morning confusion any wider.

Good night!

@MJD later

good night

abelian groups are simpler than groups because of the extra restriction

groups are simpler than monoids because of the extra restriction

that is how I see it

simplier but not more beatifull

Because their irreducible representations are always one-dimensional.

@awllower only over $\mathbb{C}$

(or similar fields)

this is bothering me a bit if $a_0 +a_2 +.......$ , i want to show that it is divergent . if $\frac{a_{k+2}}{a_k} =\frac{k(k+1)-l(l+1)}{(k+1)(k+2)}$, l is a constant .

what

hi everyone

Hi

why do you only sum even indices

@user58512 : because i want to sum only even indices .

lol

Very interesting...

:D

:P

Hello all!!

hi