@N3buchadnezzar About that proof--the general outline is that we take $\varphi=\log(f)$ and show that, by assuming the properties $(a)-(c)$ given, we can uniquely determine $\varphi$. By uniquely determining $

oops...

Continuing where I left off: By uniquely determining $\varphi$, we have uniquely determined $f$. That's the way that it shows $f$ is uniquely determined, right?

@N3buchadnezzar It is determined as a pointwise limit.

@PedroTamaroff Yeah I get the generall idea, my problem was some of the (imho) too quick jumps.

If for any $g$ with certain properties you can find $f_n$ independent of $g$ such that $|g-f_n|(t)\to 0$ for each $t$, then $g$ is uniquely determined as $g=\lim f_n$.

That makes sense...

in our case, the $f_n$ is that $\log\left(\frac{n!n^x}{\text{stuff}}\right)$, right?

@N3buchadnezzar Short and sweet. It takes a bit to get to the sweet center, though.

@PedroTamaroff I really struggled seeing how the reapplication of 94 gave the line below. But after some pen and paper droodling it all made sense =)

@N3buchadnezzar Now, that I did not verify just sitting here. (But I would were I studying the book)

=)

@Chris'ssis sorry, I was away for a while. I have another approach, but if it does not pan out, I will try that. Thanks.

@AlexanderGruber Hey.

@pedro hi

@AlexanderGruber How's it going?

Not bad. i'm visiting my family/girlfriend in Ohio.

@AlexanderGruber Ah, that's nice. Are you with them now?

she's out for a cig

I'll get back to recursions and EGFs if that's the case, enjoy your time. =D

@AlexanderGruber You don't smoke?

@PedroTamaroff i quit a couple months ago.

@AlexanderGruber clapclapclap

Is she trying to quit?

@PedroTamaroff sort of. She's cutting down.

but of course that is what everyone always says :p

friend of mine used to tell people he was on the 80 year plan to quit smoking.

@AlexanderGruber hehehe, fun lad

This. </3?

Stil group theorin 1 2 3.

@PedroTamaroff Poor clueless Melissa

@AlexanderGruber Do you like to count? I am trying to find a recursive formula for $x_n=x(n,k)$ the number of permutations in $S_n$ with $k$ inversions, i.e. with $i<j$ and $\sigma i>\sigma j$.

hrm

what's $\sigma$ here?

$\sigma\in S_n$.

huh i'm still not sure i understand

For example, a transposition $(12)$ has one inversion.

ah i see.

$x(n,k)=\#\big\{\sigma\in S_n:\,\#\{(i,j)\in\{1,\cdots,n\}^2:i<j,\sigma i>\sigma j\}=k\,\big\}$

An inversion in a permutation is a pair $1\leqslant i<j\leqslant n$

With $\sigma i>\sigma j$.

@seaturtles Well, yes. But why so symbolous?

yeesh.

dis y

@seaturtles kk icic

that seems like an SOB. But, surely one that has been studied somewhere.

@PedroTamaroff doing stanley?

@seaturtles I prefer women.

Enumerative Combinatorics

okay y'know, this sidebar is getting really cramped

@seaturtles Well, I am doing a problem set from my combinatorics course.

Maybe my prof. uses that book.

Let's a check.

R.P. Stanley. Enumerative combinatorics .

Aha!

since everybody seems to like the location pinning i think i'm going to add it to the etiquette stickey to conserve space

I am 100% confident the problem you give is in Vol I.

@seaturtles Did you study from that book seriously?

nope

but I perused the exercises

@seaturtles Cool.

@seaturtles Remember when I "worked" on that Bernoulli derivation?

My professor at the time talked to a friend of a colleague who recommended that book.

bernoulli derivation?

I would forward the mail where he gave me the copies of the book, if you want to.

@Chris'ssis I now have a couple of other approaches, let me see which comes out nicer. I have the park, then we have company tonight for dinner, so I may not get back to this for a while, but it looks interesting.

@seaturtles Dis.

I kinda liked it at the moment.

Etiquette Guidelines | $\LaTeX$ support for chat | pin your location [instructions]

Now, consolidate the PSA's. :P

PSA: Please do not spam the star queue or flag messages as offensive that really aren't. This is why we can't have nice things.

there at least now we get two non-mod messages on the queue

hmm, the comments on my question have disappeared

@AlexanderGruber I see eight because of my relatively large monitor I guess.

@KarlKronenfeld what size?

@PedroTamaroff no idea

someone convinced me to get a monitor and close the laptop

@KarlKronenfeld i gotta get a new computer :(

@seaturtles Were some of them yours (so that you can prove that they weren't deleted by owner)?

@AlexanderGruber Yo.

Do you use Hangouts?

@Karl does the "Etiquette guidelines..." message show up as two or three lines on your monitor?

@PedroTamaroff no, never heard of.

@AlexanderGruber Do you have a Google account? A mail=

@AlexanderGruber three lines (including your long name)

@PedroTamaroff yeah.

It's a Google videocall service.

@KarlKronenfeld dang it

But many peple can join.

And hangout.

why did my parents have to try to name me after a bunch of conquerers

Hence the name.

@PedroTamaroff ohhh, that's neat, is it free?

@AlexanderGruber YAS.

i'll have to check it out

@KarlKronenfeld no they were just two comments by a user####. since they mentioned that my "wheel rings" are called "FGC rings" I know where the answers are in the literature. turns out the characterization (for commutative FGCs) is not textbook-exercise-level, and was the culmination of a lot of work.

I was thinking we would set up one.

@PedroTamaroff perhaps i can look into it when i get back to Gainesville. my internet isn't very good here.

@seaturtles "Oh, I wanted to know if groups of odd order are solvable."

@seaturtles Cool

Does anyone know where could I ask for open problems that do not require much mathematical background and would be suitable for an undergraduate? I feel like this question does not belong to this site.

@LanceFerd what's the question?

already been asked on the site

@AlexanderGruber I meant "ask" in the sense of "request"

@LanceFerd oh, i see. this has been asked here before i believe.

>http://math.stackexchange.com/search?q=undergraduate+research

>hard to do

@PedroTamaroff Thank you, that is exactly what I needed

@LanceFerd You're welcome. Avoid drugs.

@PedroTamaroff Haha other than caffeine, sure.

@LanceFerd Your name just reminded me of someone.

You're a void. Welcome drugs.

@PedroTamaroff Who?

@seaturtles "Avoid drugs. You're welcome."

@PedroTamaroff Hahahaha oh I understand why the drugs reference now.

I saw an exercise in real analysis that asked to show if $f:[0,1] \to \mathbb{R}$ is a function such that $f(0)<0<f(1)$ and there is a continuous function $g:[0,1]\to \mathbb{R}$ such that $f+g$ is decreasing .. then there is an $\omega \in (0,1)$ such that $f(\omega)=0$ . Now my question is .. is it true that for any function $f$ satisfying the Intermediate value property in $[0,1]$ there is a continuous function $g$ in $[0,1]$ such that $f+g$ is monotone in a sub interval of $[0,1]$ ? ..

any helps / insights ? :)

@r9m I am not entirely sure. You mean strictly monotone?

I would look at things like $x\sin(x^{-1})+x$.

Or something similar...?

@PedroTamaroff Tu eres de Argentina?

Hey guys I have a quick question about cosets. Just very basic stuff to eliminate some confusion about some other topic. Lets say we have $\mathbb{Z}_6=\{0,1,2,3,4,5\}$. Then $\mathbb{Z}_6 / \{0,3\}$ contains the sets $0+\{0,3\}$,$1+\{0,3\},2+\{0,3\}$ and this list exhausts the set (so these are all the cosets). So what exactly is the dimension of this quotient space? Is it 3?

Dimension?

yeah

what do you mean by dimension?

do you mean size/cardinality/order?

dimension is a different word.

yes size/cardinality/oprder

a set of three things has size three, yes

It is indeed 3

wait isn't dimension just cardinalty in finite

@user52932 what do you mean by dimension?

words have meaning

@PedroTamaroff strictly monotone yes ..

@user52932 because vector spaces over finite fields have dimensions that don't equal their cardinalities

since Z_6 isn't a vector space it doesn't make much sense to bring this problem into the context of vector spacse

okay i see...

@seaturtles Is that directed toward me?

so the question was incorrect....so I want to understand...lets say we have a quotient of a vector field with itself...will that quotient space have dimension 0?

@lolwut yes

wait no

you're not user####

wait @seaturtles , is my previous question correct?

@user52932 If you google vector field, you're in for some fun.

sorry....is my hunch correct

@user52932 your question was incorrect.

Okay...so I have done some algebra...I have forgotten most stuff...I was doing some analysis portion that requires usage of algebra

and I cannot remember algebra for shit

but yes, if you quotient a vector space by itself you get the zero space, which has dimension zero

@user52932 why don't you tell us what the analysis thing is so we can straighten out your terminology for you?

yeah sorry....i did mean vector space...not vector field

i was doing some stuff on de rahm cohomolgy

and that requires algebra

@user52932 well, it ain't a vector space either, hate to say.

oh so may i ask a question

on de rahm...

sure

it will be easier

it's more likely to be readable than your previous two questions :P

"So $H^k(A)$ is the quotient vector space defined as $\frac{C^k(A)}{E^k(A)}$ where $C^k(A)$ represents set of closed forms on A and $E^k(A)$ represents set of exact forms on A. Lets say A is homologically trivial...so $C^k(A)=E^k(A)$. Does that mean that the dimension of the $H^k(A)$ is zero for all $k$?"

you can edit messages by pressing up-arrow key @user52932

the answer to your question is yes @user52932

What should I edit?

put in a dollar sign

you still need that deleted?

@hichris123 User is too new to take positive action here...

isn't the dollar sign there already

anyway, it looks like it has been flagged, so that thing will be deleted anyway.

I already corrected it below.

oh okay...i see sorry

@user52932 a second one in the correct spot (after that quotient).

I'm lost.

@user52932 If $V=\{0\}$ then it has basis $\varnothing$.

The dimension is thus $\#\varnothing =0$.

@hichris123 Yes, delete please.

i cannot edit anymore....pedro has posted the correct one. Thanks @PedroTamaroff

lol

Uh-oh... I see a bunch of diamond mods here from other sites...

So $\frac{C^k(A)}{E^k(A)}$ if $C^k(A)=E^k(A)$ has dimension 0. Is there a way to explicitly show this (I'm sorry I forget algebra completely). I would have to show that there are 0 basis vectors right.

This set only contains the element $0+E^k(A)$ (the quotient space).

there are a number of ways, an esoteric one is to note that in an exact sequence $0\to A\to B\to C\to 0$, $\dim C+\dim A=\dim B$.

it appears that you have another good answer here

@user52932 $V/V=\{V\}$.

okay. So simply put $0+E^k(A)$ constitutes the 0 element in the quotient space. And by definition if your space contains only the 0 element then it has dimension 0.

is this argument fine?

Yes.

alright thanks a lot guys.

@user52932 it's circular, since you didn't really prove that

I haven't seen dimension 0 defined that way

but maybe you have

nice...my book has this statement...the book says that this is a convention

then you're fine

is the abelian group $\prod_{i=1}^{\infty}\mathbb Z/\bigoplus_{i=1}^{\infty}\mathbb Z$ isomorphic to anything i'd know?

mathoverflow.net/questions/164516/… this may be relevant @lolwut

kinda

oh, i just now started reading the accepted answer

thanks @FernandoMartin

np

(Regarding my much earlier question about legally free textbooks: I found this page in an answer somewhere on-site. A pretty nice list...)

now i find out...

^_^

@lolwut, what's the diff between that product and that sum?

Seems like isomorphic to the trivial group :D

Would like for people to look at math.stackexchange.com/questions/782916/…; in particular the weaker result (the second) seems to follow naturally from the bottom-most one, but I can't seem to get it to work out...

Ask Richard Dedekind, I'm sure he'd know

particularly when you're on 5 gr of mushrooms

lol

@ruadath Now... when your advisor told you not to publish any of his work electronically, did that extend to his results (e.g. it "is proven that the following two quantities are always integers:")?

No

Anyway, the identity follows trivially as a corollary of something he showed, so its not like I am reproducing any of his work here

Actually I think the second result follows trivially from the bottom most one but not sure... I am very tired right now.

Someone selected an answer of mine \o/ so stoked

?

@rudath, would you find it useful if I wrote a little C++ script to check the first few examples of that formula and see if it's true for at least those?

Of course not

Can you check my answer there that I just posted; I am very tired right now, but the latter conjecture seems trivial. Not sure though...

I don't know the theory enough. But I could write a c++ script and at least give confidence that it's true

No just look at my answer; I think it is simple changing of the gcd

Like please just look, the proof seems very easy

Actually my bad type

typo

k, I'll look, but not promising anything

@rudath, I don't get this part: "Note that is proven that the following two quantities are always integers:". Makes little sense in English

or is ambiguous

whadya mean precisely?

I also think you're missing an $=$ after the $d$

Seems like someone almost had an answer for you, then deleted it

@ruadath do $a,b$ have to be non-negative?

Yeah, it's hard for newbies like me to get into your post, but if you stated some simple conditions, it would be easier.

(ranges on all the letter variables)

do it!

:D

@ruadath please re-open your question, I was making an answer.. wrote some code, with result

Your statement appears true not only for $6$ but any integer multiple of $6$.

Here's the code if you want it: pastebin.com/t82rEMs2

My gods, I think he left

but I only tried a few

Well that sux

-_-

@Enjoys Math Sorry Im back

@ruadath please re-open, I was working on that

Formulated more general conjecture which proves that

will post that

the more general one

k

it's true for not only 6 but any multiple there of (at least)

Yeah

thats the general form

lemme know when you post

I know numerically

I need proof

Yeah, but you could find even more results possibly with experimenting with code inputs

I know, proof... that's a tough one :D

I named the function ruadath_sum() if you don't mind ;)

Nice... I don't mind

Actually 6 isnt even necessary

what then

Here I will show general conjecture soon

k

can't wait, don't think I'll understand it but the beauty maybe

like 5 min

waiting :D

Here it is math.stackexchange.com/questions/783043/…

Upvote question please

@ruadath the results seem to sum to zero for $a, b \lt 0$, so un-interesting

?

(of the previous conjecture)

where's proof of the new conjecture?

there's no question in your post

Not proven yet

are you asking for proof or will you prove in an answer?

you can do that!

Asking for proof

Might try to prove

Added question for clarification

I have proof... I think

Will take me a while to write up though

Probably will not post until tomorrow

give me a specific counter example

@ruadath, It looks to be ALOT of counterexamples according to code, in fact one at $(1,1,2,1)$ see my post

the code doesn't lie unless there are arithmetic errors

and for small values (I used long long, double, etc) the code is probably valid

Your other formula though looked way legit for multiples of $6$, not just $6$

but both these formulae in your most recent post seem not to always be integers and in fact seem to be almost never integers

*almost never always integer is what I meant to say

lol

pokes @ruadath

here

I think your code is wrong

I have proof

almost

I just showed you a counterexample on paper though (see my post)

typing up

what of my counterexample on paper of $(b,c,m,n) = (1,1,2,1)$??

I'll wait for your proof, but... counterexamples $\times \infty$ I have

Wait let me think

k

Perhaps a,b,c all greater than 1?

I see tons of counterexamples. I will look right now for a stopping point of them

I went up to $1 \leq b,c,m,d \leq 1000$ and it's still running but seeing millions of counters

and $1000$ is well-within valid range for arithmetic with types long long and double in c++

still running.. no end to counter examples in sight...

actiall cancel all that

It's taking too long with $1000$ stuck on the $n$ for-loop, so I will try 30 instead

It will take a few mins

lots of print statements lols

Oh shoot it is false...

Okay, up to $15$ in each variable, there are lots of counters with no apparent ending in them

sigh...

But important progress made

yeah your original formula with $6$ seems true

for any multiple of $6$ i.e.

you mean with a=6?

in the first statement

no, in your deleted post

and b=6 in the second

no, where you had $6$ in your deleted post, replace it with any integer multiple of $6$ and your statement seems true for small enough inputs that I can test with code

I know what is wrong

I can fix

will take some time

and for non-integer multiples of $6$ there seem to be a lot of non-integer results of the sum, so interest

*interesting

@ruadath are you working in number theory?

yes

actually

mathematical physics

Sweet!

but this paper is on number theory

I'm an undergrad; freshman actually

You seem way ahead of your year, good jerb!

Hi all

I got a serious question

lol

shoot

Use mahematical induction to prove that the nth derivative of (xe^-x) = (-1)^n e^-x (x-n)

I'm not sure how to approach this proof,

use $\LaTeX$ for one

I don't know how to :P

Will fix

I'm learning it though

So I did d/dx (xe^-x) and got e^-x (1-x)

now problem is what's next?

brb in 5 minutes.

@Sabಠ_ಠ type ${\rm \frac{d}{dx}}(xe^{-x}) = (-1)^n e^{-x (x-n)}$

Okay

This makes sense

You also need the chatJax link for this chatroom, can someone provide???

So I got $(e^{-x})\times(1-x)$

Is that good latex?

math.ucla.edu/~robjohn/math/mathjax.html

@Sabಠ_ಠ assume it's true for $k = 1\dots n-1$ and then from that prove it for $k = n$.

@Enjoys Math

Tahnks for the help btw

np

Nice talking to someone

Helps me work through

Email me [email protected] if you need advice writing code

Hi Glasses. I got glasses too

those are brows not glasses, son

nice observation :P

I'll try to prove that now. Brb

yeah, go do that!

lolcats

Going to bed

wow dis room ded

yes it is

@robjohn Hi.