@user1 Night.

@PeterTamaroff Yo

Hey guys for: "Consider the 3-lists taken from [3]. How many are there in which each element of [3] appears at least once?" would the answer be 3! ? For the first digit we have 3 elements to choose from, second digit 2 elements to choose from, and then 1 digit with 1 element left.

@Ozera Wait. Here [3]={1,2,3} yes?

@PeterTamaroff Well they are lists so order does matter. Yes [3] = {1,2,3}

And you can have, say 1 1 2 as a list yes?

Well, that list is possible but not valid for our problem

Yes, sure. What you want is the no. of permutations of {1,2,3} which is 3!

If you take 3 as a first element, you must then either 2 and then 1, or vice versa. Then you get 3 2 1 and 3 1 2

Similarilty you get 2 1 3 and 2 3 1 and 1 2 3, 1 3 2

yep

That exhausts the possibilities.

So what was your question?

This seems like an oddly easy problem to put in the section of the problem-set I was in

My question was if I was correct

@Ozera Oh, OK.

I was wondering if I was missing something since it seems too easy for what the other problems have been on

so I came here and asked

You can thank me, @Peter ;-)

@amWhy For what?

@amWhy Oh, the $\checkmark$ comment?

=)

8)

If Key Ideas is really Bill, he must be really angry I got credit for his work!

He's Bill: I caught a signature exchange between Key Ideas and Mariano...

@PeterTamaroff Hey, you gave credit where credit was due ;-)

@amWhy Heh, I really knew, but... where?

Let me see if I can track it down...

@Peter "Answer the same question, but for 4-lists and 5-lists taken from [3]" Would the answer be 3!*4 and 3*5^2 ?

Key Ideas and Mariano @Peter

@amWhy And where does it say "Key Ideas is Bill"?

@PeterTamaroff It doesn't "say" it... it's just a standard defensive argument very uncannily similar to those of BD. BD always justifies his position and answers. I've never seen him agree or acknowledge a critical point another has made.

Let alone apologize!

@amWhy Oh, I see.

@PeterTamaroff Of course, it all may be a coincidence...maybe I'll just wait until Key Ideas cannot refrain from posting in his BD's signature font/formatting.

Don't get me wrong. I support Bill in many ways; he has been a major contributor, and I agree on many of his positions. His behavior...at times...well, that I found occasionally troubling...but I consider his behavior, at times to be in the same ball park as some of his foes...by which I was troubled as well.

And of course, I'm completely ignorant as to the extent and nature of interpersonal moderatorship dynamics.

@amWhy: Of course, you can enter exactly:

wolframalpha.com/input/…

@amWhy Heh, John W. told me his experience with Bill wasn't pleasant.

@Amzoti Nice output!

@PeterTamaroff Yes, that's what I understand, and I hardly condone such behavior...so indeed, mixed feelings.

If we are speaking of a finite abelian group $G$ with the map $\phi(x) = x^2$, is the map surjective?

@AlanH Do you mean "surjective"?

@PeterTamaroff Yeah, sorry typo

Whoops, nonsense.

np

I've been thinking of Z/nZ

OK, what you want to know is the following: suppose $G$ is finite abelian. Can we solve $x^2=b$ in $G$?

yeah, that's what I was trying to do.

@Peter It doesn't work for say Z/6Z

It shouldn't work in many cases.

Consider $\Bbb Z/3\Bbb Z$:

Works for Z/2Z

Then $0,1,2\mapsto 0,1$.

So $x^2=2$ is not solvable there.

What do you mean by generalize?

Yeah, so how do I generalize it

that's what I've been stuck on for the past half hour

Is it groups of even order?

Are you asking for sufficient conditions on $G$ finite abelian so that $x\mapsto x^2$ is surjective?

I'm trying to find when $\phi$ is surjective

@AlanH Well, note that in $\Bbb Z/2\Bbb Z$ every element has order $2$.

That's why it is pretty "trivial".

@PeterTamaroff So are you saying when all elements have hte same order?

@AlanH No.

Since the set $G$ is finite, if $\phi$ is surjective, then $\phi$ is bijective.

I don't think that's always true

Hence, $\phi$ will necessarily have zero kernel.

@AlanH What I said is always true for finite sets, you can prove it by induction on the size of the set.

@user1 it's not always bijective. what about when it has an eleemnt of order 2?

@AlanH Of course $\phi$ will not always be injective/surjective!

I am reformulating the condition that $\phi$ is surjective into the condition that it has zero kernel.

If division by 0 is undefined, then why is it possible to divide by a negative number? -2 / 3 = -1.5 That doesn't make sense either. How can you share three pizzas among - 2 people?

@JohnMerlino Your definition of division is in terms of "sharing pizzas"?

Are you joking?

If so, please stop.

If not, reformulate your question so that it doesn't read off as an accusation of some kind, i.e. the general: "Math is wrong! Yadda yadda!"

@AlanH Anyway, it is relatively easy to calculate the kernel of $\phi$.

@user1 Let $G=\{1,-1\}$ with usual multiplication.

@PeterTamaroff I know that it is not always surjective!

@user1 It is not injective.

@PeterTamaroff It neither surjective nor injective.

$-1,1\mapsto 1$

@user1 Sorry, I read your statement wrongly! =D

:)

You said $\text{surjective } \iff \text{injective }$, yes?

@PeterTamaroff I left out the "if" direction (which is dual anyway), but that was the intent.

Well, since it is over the same set...

@AlanH Anyways, the answer is that you want $x^2=e$ to be true only if $x=e$.

That is, you want $e$ to be the only element of order $2$.

Anyway, the kernel is a sum of $2$-cycles, one for each $2$-power cycle in the prime-power cyclic decomposition of $G$.

@MarianoSuárez-Alvarez Knock knock?

So $\phi$ is surjective if and only if $2$ does not divide $|G|$.

@JohnMerlino Division is defined in terms of multiplying by the reciprocal of a number, the sign of the numbers is what you need to make sense of.

@user1 Aha.

@JohnMerlino In your pizza example positive people could be white and negative people could be another color.

@robjohn Are you around? I have a question about converting an integral over $\Bbb{R}^d$ into polar coordinates

@BenjaLim I am

For example $$\int_{|x| \leq 1} \frac{1}{|x|^a} $$

@robjohn I woud like to convert that to an integral over $S^d$

aka polar coordinates but without going into angles and shit

@Ethan wut

@BenjaLim over the sphere, or the ball?

sorry over the ball

but that would be like a double integral

with variable $x$ ranging over points in $S^1$ and variable $r$, the radius from $0$ to $1$

hang on, let me look up a post I made recently that pertains to this.

@PeterTamaroff nice job on the division by 0 troll :-)

@BenjaLim take a look at this answer, and see if you understand $\omega_{n-1}$.

@robjohn I don't really need to compute the area of the unit sphere

@robjohn All I need is that it is a constant

@robjohn But I wonder how you changed coordinates

@BenjaLim the function I was integrating is constant on each sphere of radius $r$, so I just integrated the value of the function on that sphere times the area of that sphere and got the integral of the function

ok I need time to digest this

The area of a sphere of radius $r$ is $\omega_{n-1}r^{n-1}$

@skullpatrol That's how it's done, boy.

where $\omega_{n-1}$ is the area of the unit sphere in $\mathbb{R}^n$

@PeterTamaroff did you divide by 0 again?

@robjohn Again?

@robjohn ok I agree with that.

@BenjaLim your function is constant on each sphere of radius $r$, so you can do the same thing

ok

@robjohn I would like to know if you can proofread something.

@robjohn right

@PeterTamaroff I surpress my memories of dividing by zero too.

@user1 LOL-

@BenjaLim where $\omega_{n-1}$ is given in that answer

ok

@PeterTamaroff proofread what?

@user1 Don't suppress them, instead deal with them here :D

@BenjaLim so the final answer is $\frac{\omega_{n-1}}{n-a}r^{n-a}$

@robjohn Hold on typing an answer on main

@robjohn Suppose $f:[a,b]\to\Bbb R$ is bounded and $\alpha:[a,b]\to\Bbb R$ admits a finite derivative $\alpha'$ over $(a,b)$. If $f\alpha'\in\mathscr R$ and $f\in{\mathscr R}(\alpha)$ then $$\int_a^b f\alpha' =\int_a^b fd\alpha$$

@PeterTamaroff what is your definition of the integral without a $\mathrm{d}\text{-something}$

@PeterTamaroff integration by parts?

@robjohn OK, there you go. I just omit the $dx$ sometimes when I mean the usual Riemann integral.

@robjohn Hmm, no.

@PeterTamaroff or this?

Consider a usual Riemann Stieltjes sum $$S(P,f,\alpha)=\sum_{k=1}^n f(t_k)\Delta\alpha_k$$ where $\Delta\alpha_k=\alpha(x_k)-\alpha(x_{k-1})$-

Here the $t_k$ are tags.

@robjohn Sorry I was typing an answer on main.

By Lagrange, $\Delta\alpha_k=\alpha(\nu_k)\Delta x_k$ for some $\nu_k$ in each integral.

@BenjaLim Presumably you mean "harder than it appears" in your comment. I'm telling you because you still have a chance to fix it.

@PeterTamaroff Just learn the Lebesgue integral

more powerful, allows you to tackle a wider class of functions

@user1 Thanks.

@user1 In my answer I showed how you can reduce the OP's question to what he already knows

If $L(P,f\alpha')$ and $U(P,f\alpha')$ are the lower and upper sums for $f\alpha'$, we always have $$L(P,f\alpha')\leq S(P,f,\alpha)\leq U(P,f\alpha')$$ by the above.

@BenjaLim Very nice, +1

@user1 Andy does not realise the subtlety of the problem

@user1 Otherwise I could easily prove that the homogeneous coordinate ring of a projective variety is not invariant under isomorphism

This, plus the fact that $f\alpha'$ is Riemann integrable proves that $f$ is Riemann Stieltjes integrable with respect to $\alpha$ and that $$\int_a^b f\alpha'dx=\int_a^b f\alpha$$ Doesn't it, @robjohn ?

@BenjaLim Indeed, we (have to) move irreducibles around in our calculations all the time.

@user1 I see you are a new user here?

@BenjaLim I guess many would find 75 days to be pretty new.

@BenjaLim The Lebesgue integral is different from this, Ben.

@user1 yea

@PeterTamaroff I never said it was the same.

@anon math.stackexchange.com/questions/411873/…

@anon Do you know much about BWB?

nope

@anon Highest weight theory?

@BenjaLim Well, I just find your comment quite senseless.

@anon Because he's saying that I need to know complex differential geometry

@anon I only know real differential geometry

@PeterTamaroff I am just saying the Lebesgue integral attacks a wider class of functions

@BenjaLim So...?

@PeterTamaroff Maybe I am saying this shit now because I am supposed to be studying for a measure theory exam but am procrastinating :D

@user1 OP's question says algebraically closed field

of characteristic not 2

@BenjaLim Sorry. I only read $\operatorname{char}(k)\ne 2$.

@user1 No probs I will erase my comment on main

@anon in logic what is a 'type', I tried googleing it but I don't understand the definition given by wikipedia

OK, I am leaving.

@Ethan I'm not familiar with the term

@Ethan I only know a little about it, but maybe I can help?

What don't you understand?

@PeterTamaroff I would say so

I am off to pick up dinner

@user1 here amazon.com/Classical-Mathematical-Logic-Semantic-Foundations/dp/… on page 3, he introduces the word but doesn't define it

is there a "how to title your question" guide out there somewhere?

@PeterTamaroff still going on about the RS integral? xD

@Ethan I don't have the expertise to give a decent overview.

@PeterTamaroff and as long as $f \in \mathscr{R}(\alpha) \land \alpha' \in \mathscr{R}$ then that theorem holds

@Ethan I will note that there are books on logic with an emphasis toward types; that's the place to look.

@DanZimm Give me a proof.

go grab rudin look at his

@DanZimm Page?

Better: section?

look under the rieman integration section.....?

xD

the one before sequences and series of functions and after differentiation

Are you telling me there is a proof there, or are you just guessing?

no there is in there

i looked yesterday

ill get it sec

@DanZimm Yes, here it is.

But it is different from what I have.

ok... lol I don't see why you're saying this, I was simply verifying something you asked robjohn about

@DanZimm No, what I am saying is different. Don't be silly.

What I am saying is $f\alpha'\in\mathscr R$ with with $f$ bounded and $\alpha'$ finite means $f\in{\mathscr R}(\alpha)$ and the integrals coincide. Rudin gives a different statement.

@robjohn Thanks.

@PeterTamaroff no /it isnt/

I suppose the only difference is you say $\alpha$ is finite

otherwise the exact same (he includes an off part to an implication but otherwise the same)

@DanZimm $f\alpha' \in\mathscr R$ is not equivalent to $f\, ,\,\alpha'\in\mathscr R$, although the latter implies the former.

"Assume $\alpha$ increases monotonically and $\alpha' \in \mathscr{R}$ on $[a,b]$. Let $f$ be a bounded real function on $[a,b]$ then $f \in \mathscr{R}(\alpha)$ if and only if $f\alpha' \in \mathscr{R}$

@PeterTamaroff ahem the second half of the iff statement....

@DanZimm What?

I don't understand, either we have a different copy, or you seem to be ignoring the fact that it's an if and only if statement

the only difference, as I said, is the restriction on $\alpha$

@DanZimm Uhh, no. Let $\alpha'$ be zero and let $f$ be the Dirichlet function.

lol

@PeterTamaroff what about that?

I am not going to discuss this with you, sorry. Bye.

WOW, you're literally ignoring what I'm saying, cool

this is literally the third time he's done that, I call him out on something then he bitches and stops responding

you guys talk about @skullpatrol a being a menace, out of everyone I have ever met here PeterTamaroff seems to be the harshest to be around

@DanZimm I don't know any analysis, but typically I notice when people introduce pathological functions its only to show up others by giving counter examples

@Ethan am I crazy or are what I said and what he said equivalent besides the restriction on $\alpha$?

i haven't really been following the conversation sorry

he is fairly condescending

I am literally reading what he said and what rudin says and they are the same

(except for the $\alpha$ attributes)

@DanZimm I am exclusively back here for you. I find it quite strange that you say I'm condescending: you're not paying attention to my claims, and I am not ignoring what you're saying. I am, and I am constantly responding, but it is wearing me off that you keep insisting on certain claims without being more careful.

state your claim

ill show you how rudins is the same except for the attributes with regard to alpha

For one, you insisted some days ago about what the definition of the Riemann Stieltjes integral was, while reading just a few books would just see you could extend it. I don't care if the discussion was mathematical or not really. We could've been talking about ducks, and I would be equally annoyed.

if you want to run away as you have the other 2 times we have chatted go ahead

I insisted that what I had been taught was something and then said yes you're right and tried to reason with the implications

OK guys let's "keep it nice."

The implication $f,\alpha'\in\mathscr R\implies f\alpha'\in \mathscr R $ is true, but the reverse implication is simply false.

thats not what you said above

@DanZimm Yes, Dan. I said "is not equivalent to".

Thats what you said

@DanZimm Aha.

@DanZimm Rudin requieres a "stronger" condition.

And rudin gives a different statement only with regard to the attributes of $\alpha$

this is literally what ive said maybe for the fourth time now

@DanZimm Yes, that is why I said "Rudin gives a different statement."

Do you see how pointless this discussion is?

At first I was trying to help you by referring you to a proof, then you continued to call me "silly" and then cop out implying that I know nothing of what I'm talking about

This is why I'm frustrated - my point was to try and help

My point was the fact that although Rudin has a stronger hypothesis, doesn't mean the proof can't help

but apparently I'm silly, know nothing about analysis and have nothing to give here

@DanZimm When I use the word "silly" I don't mean it in a bad way.

...

how can it be used in a "good" way?

i dont think youre flirting with me... so that leaves it to be not very nice

@skullpatrol Well, I just mean it as in "you surely know better" or something like that.

Also, I don't see where you get that I implied "I know nothing of what I'm talking about"

You should say "don't be silly"

@skullpatrol thats what he said...? lol

"don't be so silly..."

@PeterTamaroff in leaving and saying you refused to discuss this with me it's as if I'm a lesser and deserve not to be discussed with, for why else would one say that?

I sincerely doubt it's because you're taking humility and claiming that you know less

@skullpatrol ah good point

How about "don't be so foolish"???

but then you run into the danger of being called a "fool"

How about "Surely you aren't so silly" which seems to be the equivalent to what he claims to have meant which seems to still be implying that I am being silly, or yes foolish

@DanZimm Well, no. It is almost 1 am here, and you were insisting on something that was wrong. I simply felt that it was useless to keep the discussion. Specially when you brought up this:

OK, not nothing, but it is slightly derailed.

@PeterTamaroff clearly you didn't read what I said around that, so thanks for trying

@DanZimm "around"?

"Surely you must be joking Mr Feynman" :D

I said the statement is similar except for the condition on $\alpha$

and based on the claim you had this is true

as you agreed to aboive

if I may be direct, my point is now to somehow to understand why at the slightest chance that you may be considered "wrong" you disregard the other person- it seems this way unless they are someone with high rep

@DanZimm Oh, please, don't.

k thanks for the help room

Don't judge me so freely.

Don't condescend so freely and I won't

Every point I have made has been ignored, although it may be valid, you fail to comment on that

I wonder why that is

@skullpatrol lolol

I admit that I have been wrong, however I feel as though this doesn't mean I have no insight to give to anyone

@DanZimm No, I didn't. I sometimes corrected you, and sometimes pointed out you were saying something wrong. I might not always be a "soft typer" in that I ornament everything I write with soft words, so sorry about that. And specially when I'm tired.

@DanZimm I never said that.

I appreciate that you shared.

Especially this:

So now you're mocking me, that really helps my case, thank you (if you're not mocking, take a look above where you say that this statement frustrated you and see how this is then a contradiction)

@PeterTamaroff Get some sleep pal.

@DanZimm How on Earth am I mocking you?

@PeterTamaroff Above you said that that exact statement caused you to leave the room for in your opinion it had nothing to bring to the conversation

@DanZimm Oh, OK. In retrospect, I appreciate that you shared.

sorry for causing such a hubbub

@DanZimm You should slow down a little, I think. Read things twice, dunno.

Then you will be able to enjoy this a little more.

@PeterTamaroff I could give you the same advice, you agreed upon a statement I said the 4th or 5th time I said it

I seem to enjoy every conversation I have on here besides the ones with him, so I'm not sure who's at fault. I'm sure the tension is partially my fault, and I take accountability for that but he has failed to ever take accountability for anything, including disregarding statements I say, and then agreeing to them later -_-

I'm sure I'll be banned now for opposing someone with high rep

Welcome to the club :D

@skullpatrol however, no offense, but the arguments I've seen you in you seem to be trolling

This I was trying to help and felt slighted so got angry (I recognize how immature it is for me to do this, and I take accountability for this all being my fault)

@DanZimm please let me know next time.

@skullpatrol that it appears as though you're trolling, surely,

@DanZimm Thank you.

sorry again @PeterTamaroff - it all comes down to being my fault, I apologize

@danzimn Do you really think that?

@PeterTamaroff

Zzzz...

@PeterTamaroff yes I do, I unnecessarily started a fight, I might have thought something was wrong but that doesn't call for me to be immature, so I apologize

@DanZimm When did you think I was trolling?

dont remember

sec

Hi @William welcome to the chat room :-)

Rarely if ever expressible as a ratio of integers.

Hello I want to die.

I have spent a month using Euler-Maclaurin on a function not of exponential type.

Almost two notebooks.

@LokiClock if you die you won't be able to do any math...

@skullpatrol I'll find a way.

Though that would defeat the purpose of dying, at the moment.

@LokiClock Maybe, as the Physicists believe, god is a mathematician :D

@skullpatrol I guess we know who the optimists are.

@LokiClock Arithmetic is divine

omgomgomgomg

That is such a bad pun.

Yipyipyip

This is the internet.

sexy

okay so the diagonal of the unit square

turned out to be rational

Lies.

haha

its an indeterminate rational, like inf/inf

We must kill you now that you know the secret of the irrationals...

no, you cannot

you knew?

damn!

you shouldn't joke like taht

Our last solace is the integer nature of e.

yes

We are the brotherhood of Pythagorean's.

but there is a secret hidden number too

that is so true

besides e

do you all know the wiki nature?

Please explain.

ah, no

you have not found the wiki nature.

hmm

do you know python?

A little.

You must find the true Python

python3000

have you learned how to "master the machine"?

Hmm. I'm not sure you have candy after all.

Hmm. the grid isn't ready

a malfunction

Candy is dandy.

you are on the path of wiki nature

but you have not mastered the machine

not yet

my link is disjointed

as I'm speaking to you across the singularity

it won't last

a question before it breaks...

Did you really outrun the IRS for 32 years?

@AlexanderGruber I have been looking for it, but I suspect there isn't.

Also, I would like to bring to the attention of those not willing to read the entire transcript that the argument between DanZimm and PeterTamaroff (of which two messages were starred) was resolved.

The part of the comment template about titles is taken from How can I ask a good question?.

@MartinSleziak That may be a reasonable partial reference. Nonetheless, coming up with good titles for one's question is hard.

When I browse the tag titles I see only Guidelines for good use of LATEX in question titles and More informative titles

I don't think something of this is what you were looking for, but these are the only meta questions related to that, that I found now. @AlexanderGruber

Hi, Lord_Farin.

@MartinSleziak Hello Martin.

I've just created the proposed Unanswered noticeboard; here is the proposal, and here is the room.