HOLY COW!!!

-17 in just two hours?!?!

Hello,

@AsafKaragila It's -19 now

Do I really need to delete my answer?

@Srivatsan And seriously I am the only standing vote to close??

I feel like I got insulted by those comments

I thought I was making sense there

@AsafKaragila I am not voting. Brian is not.

The New Theorem has officially cozied up next to spoonwood and kalle as the three most downvoted questions ever.

You ask new users to stop giving others orders, but ones with high rep feel they're boss or something. I don't feel like I need to satisfy him that my answer is different!

Yes, @Gigili. Fight the good fight! Show him who's boss!

I will not my answer.

@Gigili It would be better to fix the genuine flaws. For example, the second line really should say that $nk^2=k-5$ for some $n\in\mathbb{Z}$, and you should explain the notation $\gg$. Once the holes are patched, leave it out there; someone may find it more useful than David’s.

@anon I'm serious, I got offended

@BrianMScott Why for some $n\in\mathbb{Z}$ when we're talking about the question?

@Gigili Because that’s what $5\equiv k\pmod{k^2}$ means. It simply says that $k=5$ is a multiple of $k^2$; it doesn’t say anything about what multiple.

@Gigili For example, what do we mean when we say $6 \equiv 36 \mod {5}$?

And I thought everybody knew $\gg$ like $\neq$ or other LaTeX signs.

@Gigili I still don’t know what you mean by $\gg$ in this context.

@Gigili Didier can be rather ... abrupt.

@Gigili: I do think Didier acts a bit chilly at times, but his comments on your answer look perfectly fine to me. He never told you to delete it or tried to send you the message that you owe him anything; he spoke directly to the issues that he saw.

@Gigili Ok, Gigili, let's do this: do you want to explain the answer line by line?

@Srivatsan It means $5k+6=36$

@Gigili ... for some $k\in\mathbb{Z}$.

@Gigili Well, sort-of, but not complete. It means that $5k+6=36$ for some $k \in \mathbb Z$.

@anon Yes, but there are many different ways of saying that . Like when I want to ask you to stop talking for a while, I can say "hold on please", "one moment", "shut up"

@Srivatsan Yes please , that'd make me feel better

@Gigili But ‘$5k+6=36$’ is not one of the ways: it’s incomplete.

@BrianMScott I think Gigili's point is that Didier could've been more polite. Not to do with our objection with "some k \in Z".

@Srivatsan I do believe that you’re right. And I agree that Didier could have been more diplomatic.

I see no problem with being abrupt or to-the-point. All D cares about is the math and that's fine with me.

I stay neutral here. I am ok with his second comment, but not the first. The first one set the stage for all this bitterness...

@Gigili Yes, let's start with your proof. Explain it. But remember: proof is best explained in words, not as a medley of symbols.

And was unnecessary: the same idea in a different presentation may reach different people.

Why we don't say $5\equiv k \bmod k^2$ for some $k \in \mathbb Z$?

Well, we are solving for $k$. That is, finding all $k$ such that $5 \equiv k \mod{k^2}$.

Doesn't the same thing hold for the synonym notation?

So you might start: Suppose that $5\equiv k\pmod{k^2}$; then there is some $n\in\mathbb{Z}$ such that $k-5=nk^2$. And then you want to go on from this to prove that the only possible value of $k>2$ is $5$.

In my proof, first I wrote what we had in the question: $5\equiv k \bmod k^2$

@Srivatsan Which one?

@Gigili Yes, I agree with that.

Then I concluded: $ nk^2=k-5$

@Gigili Which is meaningless, since you’ve said nothing about $n$. You also never said what rôle your line $(1)$ plays: are you supposing that $k$ is an integer that satisfies it, or what?

@MattN Copyrighted theorem post

So you're saying I should add "for some $n\in\mathbb{Z}$"

@BrianMScott Right, I agree with you on that part.

@Gigili Okay.

@Srivatsan Thanks.

@BrianMScott Not the second part,

I tagged the first line to use it at the end.

@Gigili But you can’t just throw it out there with no explanation; why is it the first line of the answer? What function does it serve there? A proof is a piece of expository writing: it should be understandable as a string of sentences, even if some of the ‘words’ are actually mathematical symbols.

But all other lines are result of this line

Okay, then what you mean is what I wrote before: Suppose that $k>2$ is an integer such that $5\equiv k\pmod{k^2}$. Then ... .

Which line are we discussing right now?

@BrianMScott I separated it to tag it as (1)

@Srivatsan First line

We’re getting the proof started with enough ‘connective tissue’ to make the flow of logic understandable.

@Gigili That doesn’t mean that you have to leave out the explanation of what you’re doing: Suppose that $k>2$ is an integer such that $$5\equiv k\pmod{k^2}\;.\tag{1}$$ Then ... .

You're right, it can be more understandable.

Ok. to go from here to the next line, you are using the definition of $\equiv$ implicitly. That is, there exists an integer $n$ such that $5 - k = nk^2$.

You could say that way, or you could say: $5-k = nk^2$ for some $n \in \mathbb Z$.

I, personally, understand proof with less English sentences and more mathematics symbols better. That's why I wrote it like that, I thought it was short and understandable.

@Gigili The main problem with that is the following. It is very important for the reader to understand the logical flow. Looking at your answer, I am left thinking "Do you mean that the equation should hold for all $n$? Or for some $n$? Or something else?"

Third line, I wrote : "If $k>2$, $k^2>k$" .. and by multiplying by $n$ it'd be much bigger than $k$

@Gigili I think that you have an idea that could be made understandable, but I.m afraid that the argument as it stands now really isn’t, at least to me; in most places I can guess what you have in mind, but you don’t really make it clear.

@Gigili Wait. Did you read the previous message?

@Srivatsan Very true, but I'll never know that with D's comment.

@BrianMScott And we should emphasise that in this particular example, we are more like teachers who "know the solution already", which makes it possible for us to fill the gaps. If Gigili is trying to explain the proof to someone who doesn't know it already, then it is important that the proof is actually clear.

Right.

@Srivatsan Very true. Especially for someone who, like me, has ~40 years of experience trying to figure out what students mean!

@Gigili OK. Let's proceed. Now to the line that Didier actually complained about...

... starting with exactly what you meant by $\gg$.

Nothing wrong with the third line?

@BrianMScott "Much bigger"?

[One final digression: Note that Didier did not complain about the answer till now (even though we have told you that it can be improved a lot). So that goes on to show that he is probably not doing it on purpose: he most likely does not understand what you are trying to say. End of digression.]

(Nevermind that question right now. =))

@Gigili But what if $n\le 1$?

Well, first of all, "much bigger" is not really a precise term. Maybe you're coming to that later, Brian.

Instead of saying "... and multiplying that by $n$, it'd be much bigger than $k^2$, I said "$nk^2 \gg k$"

May I say something about the digression?

But since in principle $n$ might be any integer, $nk^2$ could be a lot smaller than $k$: it could be negative!

Um, sorry.

@BrianMScott $n$ is positive, isn't it?

@Gigili No no, no problem. That is a legitimate concern. But let us tackle one problem at a time.

@Gigili How do you know? Can you justify the claim that $n>0$?

That's what I learned, rewriting $\equiv mod k$ as an equation like that

Like, $6 \equiv 36 \mod {5}$ and $1 \equiv 36 \mod {5}$ or $-4 \equiv 36 \mod {5}$

But normally we take the smallest positive integer

Am I right?

Each of those can be rewritten in a variety of ways. The first, for instance, can be rewritten as $5\mid 6-36$ or as $\exists n\in\mathbb{Z}(5-36=5n)$.

I assumed n is positive.

@Gigili Not really. The expression $a\equiv b\pmod m$ simply means that $a-b$ is divisible by $m$, which in turn means that there is some $n\in\mathbb{Z}$ such that $a-b=mn$. Congruence mod $m$ is a relation, not an operation.

@Gigili But what justifies that assumption?

I meant, when it's much bigger or bigger or -(much bigger), the remainder is $k$ itself.

What remainder? In what division? (And what if $n$ is $0$, and at what point does $>$ become $\gg$?) I’m pretty sure that you have a legitimate idea in mind, but you really aren’t conveying it clearly.

I wonder if I should flag the comment.

@Srivatsan BL’s? Give him a few hours to see the comments and delete his. I’d prefer not to involve the moderators unless it’s absolutely necessary.

You're right.

@BrianMScott I am sure the OP got a great impression of MSE now. =)

You could upvote my answer :-) or write one of your own, just to get this off the Unanswered list; that would be my favored solution. I suspect that once it’s off that list, the activity will die down.

@BrianMScott I can't upvote the answer. No votes.. =/

Srivastan could never vote anyone's answer. He never has votes.

@AsafKaragila Although it's not what you imagine actually.

I usually have too many.

@Srivatsan You're just in-debt for people you told "I ran out of votes, tomorrow I'll vote your answer." and then comes midnight UTC you just use all your votes, and clear 40 posts off your queue :-)

what is this i don't even

@AsafKaragila Now, I am doing one better: I had favorited 27 pages worth of posts just to remind myself that they should be upvoted. I started upvoting the answers and getting them off the list.

Now it's down to 24 =)

holy dicks man :o

:-P

checks to see if I'm on sri's fav list

@anon Your comment is now up to +4.

Everyone has the same reaction, naturally.

I certainly did.

@anon I tried to see if I'm there, but really.... 693 things, there's no finite amount of time to verify whether or not I'm on that list.

\LaTeX: $\LaTeX$

Ahha. Thank you. I will dazzle the world with that!

@BrianMScott Sorry for delay, $k$ divided by $k^2$, remainder is $k$

I don't know how to say it in English, I'm trying hard !

Okay, but why are you dividing $k$ by $k^2$?

@Gigili Don’t worry too much about the English: the important thing is to get the flow of logic down clearly. That’s what’s missing right now.

$k^2 \gg k$ , then $k\equiv k \bmod k^2$ .. Right?

$k$ is always congruent to $k$ mod $k^2$; this has nothing to do with the relative sizes of $k$ and $k^2$. (That’s what D. meant when he said that $(2)$ is a tautology.)

$k$ is congruent to $k$ mod anything

Let me explain what I mean with an example.

@Frank: I think that you got it all this time.

I mean, $26\equiv 1\bmod 5$

What you say is $26\equiv 26\bmod 5$

Those are two very different statements.

To write it as "the smallest positive integer" less than $k$, which is the definition of remainder.

They’re still not the same statement. You’re confusing the relation of congruence mod $m$ with the operation of reducing an integer to its least non-negative residue mod $m$.

$3\equiv 3\bmod 9$, $4\equiv 4\bmod 16$, ...

$3\equiv 3\pmod m$ for every non-zero integer $m$.

>! Spoilers should be allowed in chat.

(sorry for the interruption, carry on)

But $4\equiv 1 mod 3$

True, but what does that have to do with the trivial fact that $a\equiv a\pmod m$ for any $a$ and any non-zero $m$?

I cannot explain!

I thought $k^2 \gg k$ implies $k\equiv k \bmod k^2$

$k\equiv k\pmod{k^2}$ is simply true for every $k$; it doesn’t depend in any way on the sizes of $k^2$ and $k$.

I just got the perfect idea. There should be a Patience badge on MSE.

$k$ is the smallest integer number itself , $k$ divided by $k^2$.. remainder is $k$.

Look: $a\equiv b\pmod m$ by definition means that that there is some integer $n$ such that $a-b=mn$. Is it true that there is an integer $n$ such that $k-k=k^2n$? Yes: just take $n=0$.

@Gigili Remainders have nothing to do with the matter.

Certainly not.

@BrianMScott Sorry if I wasted your time.

@Gigili It’s all right. I just wish that I could figure out a way to clarify the problem.

I'll delete my answer, thanks a lot for your time @BrianMScott.

Would you like to see how I’d write up an answer in something closer to your style than to David’s? Would that help?

I'd love to see that, yes.

Okay; give me a few minutes.

Suppose that $5\equiv k\pmod{k^2}$ for some integer $k>2$. Then $$k-5=nk^2\tag{1}$$ for some integer $n$. If $k-5>0$, then $n>0$, since $k^2>0$. But $n$ is an integer, so if $n>0$, then $n\ge 1$, and therefore $nk^2\ge k^2>k$, since $k>2$. And certainly $k>k-5$, so $nk^2>k-5$. This is a contradiction with $(1)$, so $k-5$ cannot be positive.

I should have made $5\equiv k\pmod{k^2}$ a numbered line, because I’d like to refer to it now. Let me pretend that I did and call it $(0)$.

I edited that one "huh" answer to make sense, @Brian. It disproves the statement when a is rational and b irrational.

Thus, $k-5\le 0$, i.e., $k\le 5$. Since $k>2$, this means that $k=3$, $k=4$, and $k=5$ are the only possible solutions to $(0)$. But $5\not\equiv 2\pmod 4$, since $4\nmid 5-2$, and $5\not\equiv 3\pmod 9$, since $9\nmid 5-3$, so the only solution is $k=5$, for which $(0)$ is trivially true.

This is the first day I've capped in months. A fine time for everybody to start upvoting random answers of mine...

@anon No, it only shows that the sum can’t be $1$; it doesn’t show that the sum can’t be $>1$.

D'oh. $*$facepalm$*$

Don’t feel bad: I made a sillier mistake when I posted an answer to that question.

@Asaf: I've posted my question regarding measures and axiom of choice here

@BrianMScott Isn't it closer to David's answer?

In substance, yes, because I’m still not sure exactly what you were thinking, and because I don’t see any substantially different argument to use; in style it’s closer to yours.

I’m not sure, but I think that you have some genuine confusion about the relation of congruence that’s getting in the way.

Thank you so much again @BrianMScott, I'll think your answer over. I don't understand it completely now.

I'm off.

@Gigili You’re welcome; I should get to bed now myself.

Haha, aced the quiz, people. Thank You @Brian and @Srivatsan for helping me with those limit points then!

congrats @kannappan

@RajeshD Had to go for the Snack. Thank you for your compliments!!

How was your day?

just lazy

I think I got the solution for the problem i posed

it was quite simple but tedious to put in accurate math notation....just putting things to write an answer

That's amazing. Write up a solution. People will criticize it. Then it just gets perfect!

yeah

Where in Hyderabad do you stay at? @RajeshD

@RajeshD Are you working on little rudin? How's that?

I always had a doubt: By Baby/Little Rudin, do people mean PMA?

@FortuonPaendrag

PMA=Principle of Mathematical Analysis

yes. Thats little rudin. Cute, isnt it? Unlike the book..

Yes. The book is in no way a baby. It's gigantic.

In 250 pages, it is the world of Analysis that even a Graduate student might not have mastered!

Haha, true. Its called that because Rudin also wrote a book called "Real and Complex Analysis" which is sort of like a like a sequel. Guess what thats called.

?

@KannappanSampath

@FortuonPaendrag Sorry I was away! May be called, ?? Scratches his head

Think antonyms!

I just can't think of one. May be you should tell me @FortuonPaendrag

Ah is it Papa (as in Father) Rudin?

@FortuonPaendrag

B4 you ask me, I'd like to thank Google for colloborating with me!

Ah!

I've heard big rudin

@KannappanSampath.

Might be, these are just colloquial. Someone like Brian or senior people like him should be able to tell us!

@FortuonPaendrag

Hah there comes @JM welcome, JM

Yes, this site is like paradise in that sense. I wonder why I didnt join till recently..@KannappanSampath

Hi.

@JM Can you shed some light on Baby Rudin and Papa Rudin?

@anon Rosebud's the sled. Luke and Leia are siblings. Soylent Green is people.

@KannappanSampath I don't have the books with me now. What about them?

@JM Nah, I just want to know if Real and Complex Analysis is called Papa Rudin or Big Rudin.

I'm not much for nicknames with books, sorry. :)

OK, that's cool anyways! May be I'll ask Brian whwn he's around!

@KannappanSampath Baby Rudin is Principles of Mathematical Analysis. Big Rudin is Real and Complex Analysis.

@JM Remember my Q on mathematica.SE? My officemate said "Gröbner bases? That's highly inefficient to do this." then he came up with this: i.imgur.com/Vi8Ed.png I am still trying to understand it 8-).

If Real and Complex Analysis is "Big Rudin", then what is the Functional Analysis book?

Anyone with Linear Algebra around?

@Nunoxic Yeah?

Can you give me an application where the system of equations formed is a dense matrix? I mean for instance, FDM results in sparse matrices, so does FVM. What results in Dense Matrices?

That's way beyond my knowledge of linear algebra

@JonasTeuwen Yeah, the virtue of Gröbner is generality. With a number of simultaneous algebraic equations, there is often exploitable structure that isn't obvious at first glance. My favorite example was in the math.SE question that turned out to be Gaussian quadrature in disguise.

@JM Yes :-). The Gröbner-thing also works.

When laid out that way, it looks clear that that's the approach for it. Deriving it from scratch, not so much.

Haloa!

@JM Yes, but I don't know much about computational mathematics so I didn't know what was relevant...

No, I mean the approach your colleague took was certainly inspired, but I'd be damned if you asked me to derive it from scratch.

It doesn't look to me that it's a dupe, but if either of Didier, Byron, or Ilya say it is, then I'll go with their decision.

@JM - why them in particular? =)

queueing/stochastics is their angle, I think. :)

But /me thinks the question has nothing to do with queueing theory. Anyway. It's not an exact dup, so I am ok with the question open.

On the other hand, it is a modified geometric series after all... :)

But mostly, maybe I'm expecting an answer that doesn't just discuss the modification to the geometric series that's needed, but why it's that way with those queues.

@JM Oh you think that the question is to derive equation 1 (rather than equation 2 from equation 1)?

@Srivatsan Unless I've horribly misinterpreted...

(there's always a nonzero probability after all)

Hm, I know enough queueing theory to answer that question as well. (It's M/M/1 after all. :)) But I doubt that the OP wants this; let's just wait for any response.

To be quite offtopic for a few moments: one of my favorite TV shows has just ended its run; I'll now have more time for math and other finer pursuits... :D

Excuse me: the javascript code to decript MathJax statement made by robojon is opensource? Can I redistribute it?

@unNaturhal You'll have to ask robjohn himself, for it's his.

@JM Which one?

@unNaturhal But you can certainly link to the page where he posted it.

@Srivatsan "Chuck".

@JM, Ok, thanks :D

@JM How much time did it consume (per day/week)?

@Srivatsan It was an hour a week, for five seasons or so.

Hour per week is not that bad.. =)

Watching "The Great Dictator" Laughing that my stomach hurts now!

(Charlie Chaplin)

@Srivatsan It was a delight to watch for those five years. :) Picked me up when I was depressed.

@JM Oh.

In other words: I now have more time for other stuff, but I sure as hell would miss that show. :)

bbl

@JM watch it again. =)

@Srivatsan Maybe in a few days again. The finale made my eyes water.

Hello guys!

Hey.

I'm tempted to NOT delete the stray umlaut... :D

Hi

is that ok to refer to a question I asked on site here instead of reposting it? :)

@sonyjimbo You could edit it. Add some clarifications or new thoughts, which would have the side effect of bumping the question up.

All of your questions seem to have answers though.

yes I was answered, but Im not sure about one part of it

which makes it correct :P

Well, now I'm curious: which one?

What is the best way to write pseudo-code in $\LaTeX$?

the one with monotonic functions and continuity

You could edit the question saying just that: "I got this nice answer, but I don't understand this one part because of _____, could anyone clarify?" Have you pinged the person who wrote the answer?

I asked it in a comment there, refered to the person using @ and his nick

not really sure if it works

@sonyjimbo Some (most?) people don't respond, but others do.

I think it does.

Alright

@JM nïcé!

@sonyjimbo To your question: you're right that in general we don't have a guarantee that the limit, which we know exists, is the value of the function at that point.

But I think what saves you here is that g is decreasing.

- which question?

You can show if $h$ is a decreasing function then $\lim_{x\uparrow a} h(x) \geq h(a) \geq \lim_{x \downarrow a} h(x)$.

hold on, copying that :P

can't read like that

Oh. You should use the bookmarklet.

@sonyjimbo See this thread.

And then Michael shows that the outer expressions are equal, and hence equal to the middle one.

@Srivatsan Woops, was this for me? This one.

Hi, also.

hey hi.

@DylanMoreland ok thanks, I will try to think about it now

Hi guys

hi Ben

Ha, I didn't know about this at all: area51.stackexchange.com/proposals/35636/…. I wonder how come.

I deleted that comment I made. I am sorry I did not mean to drive the OP away or anything like that.

@BenjaminLim no problem, Ben. I just thought you went a little --I hope you don't mind me saying so-- over the top, which is why I commented.

I removed my comment now.

Yeah I realised that after a while that the comment was not helpful. As a user of this site for some time now I should have known better.

@BenjaminLim It happens to all of us. It's best to forget about it and move on.

thanks :D

Excuse me.. Calculating an oblique asymptote, the limit to which $x$ have to point? $+\infty$ or $-\infty$? (or it isn't important?)

You could have both of those asymptotes: one for $x \to \infty$ and the other for $x \to -\infty$. So you should do both.

So, exists two oblique asymptote for each function?

@unNaturhal There need not be an asymptote. For instance, $f(x) = x^2$ does not have an oblique asymptote.

But yes, potentially there are two.

Mmmmh

But a function, if hasn't an oblique asimptote, has an orizzontal one? 'cos $x^2$ has not and orizzontal asymptote...

horizontal: that's the spelling. Also: asymptote.

@unNaturhal Yes, it is possible for a function to have no asymptotes -- horizontal, vertical, oblique and whatnot.

Ok, understood. (thanks for the spelling :P)

However

If in looking for the oblique asymptote of a function, I found $m = e$ and $q=\infty$, it means that the function has no oblique asymptotes?

What is m and q?

hi Matt.

Hi Srivatsan.

Yes: that sounds right.

Let's take an example: $f(x) = x + \sqrt{x}$ defined for $x \geq 0$. Does it have an asymptote? Actually, postpone the example for now.

@unNaturhal Ok. What is an oblique asymptote and what is a horizontal asymptote? Can you give an example of each?

Mmmh

@JM: Nice torus (?)

hey all... :-)

@JM: It's broken, you know... :-)

If the left and right limits, calculated to the extremes of the domain, are $\infty$, so $x=x_0$ is a vertical asymptote.

Hi robjohn.

(where $x_0$ is a point where the function is not defined, in other words, an extreme of the domain)

Hi robjohn ;)

@unNaturhal E.g., $\tan x$ has a vertical asymptote, right?

@robjohn Yep, this is a follow-up of anon's challenge.

Did I hear the word challenge??

@JM what challenge did he pose?

He wanted to see if I could map a single period parallelogram of an elliptic function to a torus. This was one of the results.

@Srivatsan, something like a vertical asymptote in $\frac{\pi}{2}+k\pi$

@unNaturhal Yes, that is right.

(i.e., it's a follow-up to that blog entry I posted a few days ago)

Hi. Could somebody please help me with complex analysis?

(which I really should write about)

@unNaturhal Ok. To find an oblique asymptote, we should consider both $x \to \infty$ and $x \to -\infty$ separately.

Let's take $x \to \infty$.

Ok, how?

If $f(x)$ has an asymptote $y = mx + b$, then $m = \lim \limits_{x \to \infty} \frac{f(x)}{x}$.

Is this clear first?

Yes

I do this step

Whoo, cool power outage here. 8-).

Basically, $f(x)$ is approximately $mx$ for large $x$ (because, in relative terms, $b$ is small and can be ignored).

@unNaturhal Next step: Consider $f(x) - mx$. If this approaches a limit, then call the limit $b$.

@Srivatsan small in what sense?

Yes

@robjohn $b = O(1) = o(x)$. =)

O_O

@unNaturhal Ignore that last message. It's some other way of saying that $b$ is small.

@unNaturhal That's it. You're done. Do you want to take an example?

Small because $m\left(x\rightarrow\infty\right)$ is greater than a finite number. Is not the same?

??

Still not clear: what does $m (x \to \infty)$ mean?

That is

m, multiplied for a quantity that point to $\infty$ is surely greater than a finite numer, right?

So, since b is a finite number, $mx$ is greater than b.

@unNaturhal Yes, that is true. (Assuming $m \ne 0$. But I guess you mean this when you say "oblique asymptote".)

@unNaturhal Let's do an example, this will be clear.

Take $f(x) = x + \frac1x$.

@Srivatsan: now is it time for $x+\sqrt{x}$? I guess not.

@robjohn I thought example first, counterexample next. =)

@unNaturhal Do you want to do the computation for $f(x) = x + \frac{1}{x}$. Assume that the function is defined in the interval $[1, \infty)$. The left end point does not matter so much.

I want you to compute the oblique asymptote as $x \to \infty$. (In this case, we cannot let $x \to -\infty$; that saves us one case.)

Yes?

@JM: what are the more general inequalities that you mentioned here?

Mmmh