@JM Do you know what is bakla/bayut?

i get the feeling the main site is currently dead.

@BenjaLim At best, "effeminate"; at worst, "full-on homosexual". Where did you hear this?

@JM i like your new gravatar

@Eugene Yeah, I like the trefoil knot too.

@N3buchadnezzar Yeah, I'd say your polynomial has six roots, but only three distinct ones.

@Eugene Mathematicians have weekends too, y'know. ;)

@JM apparently! =)

$(1/2) x^3 - x^2 -2x + 4$

Easiest way to factor this one ? (I know it can be written as $0.5(x+2)(x-2)^2)$)

=)

$x^3-2x^2-4x+8=x^2(x-2)-4(x-2)=(x-2)(x^2-4)=(x-2)^2(x+2)$.

ChatJax support for \begin..\end?

\begin{align*}a&=b\\c&=d\end{align*}

Indeeed

Well

Does it support fleqn?

Dunno, doubt it.

If so, the equations seem more pretty.

That is a matter of taste. Although the brackets seems a tad too big when using \left and \right

You can use \big instead

I find out that CMath seldom uses \left..\right

I know,but in Latex \left and \right looks somewhat better. I tend to use \big for nested parenthesis though.

$\big(x^3+y^3+z^3\big)^3$

$\left(x^3+y^3\right)^3$.

I use \left..\right

$\newcommand{\Frank}{Is a mathematician learning Calculus in Russia!}$

render

@FrankScience Try to write \Frank in mathmode.

@N3buchadnezzar No

^^

@N3buchadnezzar NOT IN RUSSIA

Although making new commands in mathJax is frowned upon

@N3buchadnezzar THERE ARE SOME TRANSLATIONS, BUT NO ENGLISH ONE.

I thought you were Chinese but studied in Russia ?

NEVER.

^^

Do you know some country named after Soviet?

In Soviet Russia Calculus learns you

No

In 1950s, the Soviet tutorials are very popular in China.

So some Soviet-aged books are well-known in China.

Cool

Incidentally, do you think Soviet was a dangerous country? I don't know the impression of westerners of Soviet.

How credible is "www.jsomass.com" from this post?

Looks spammy.

Anyway, I don't trust stuff from someone who doesn't look at what the site is for before posting.

Hi people

Hi

Can I post some Mathematica code here?

@MatsGranvik For what?

I have some speculative calculations on the three first zeta zeros.

is speculative a word in english?

If it's a bit involved, maybe it's a question better suited for main.

@MatsGranvik if you intended to say "I'm not entirely sure about the behavior, but here's my best guess..."

It is a type of question that can only be discussed in the form: What would you do here? It is very open ended.

Ah, then yeah, chat it is.

If the code's long, you could use PasteBin...

yes pastebin, that is right. I will try that.

@JM Good point.

It is perhaps not so clear what I have done.

@MatsGranvik The 2.549127729379167407581967029267929878 is your complicated sum, yes?

I calculated the sum of reciprocal square roots, got a constant, and divided it with some higher zeta zeros, and got approximations of the first three zeta zeros.

yes it is the constant from the sum

What made you truncate up to 5? You weren't getting nice results from other values?

The numerators in that sum are the same as the numerators in the Dirichlet series for Mangoldt Lambda [6] = 0. It is a repeating sequence from a GCD matrix.

@MatsGranvik "Dirichlet series for Mangoldt Lambda" - so, the logarithmic derivative of Riemann $\zeta$?

heh

Every value of Mangoldt lambda can be expressed as a Dirichlet series with the exponent "s" equal to one. Like logarithms.

Now we can calculate the partial sum of many kinds of numbers with reciprocals of other number. Hmm... I am not really clear here.

You're one of those people I can reliably identify as the OP of a question upon merely first glancing at it. :-)

Ah, some old school computer math. Truly lovely.

Heh.

@Mats: BTW, your long sum is neatly expressed as a bunch of Hurwitz zeta functions: $-2 \zeta \left(\frac{1}{2},3\right)-\zeta \left(\frac{1}{2},4\right)+\zeta \left(\frac{1}{2},5\right)+2 \zeta\left(\frac{1}{2},6\right)+1$

Always cool, come up with conjectures using the computer and then don't mention anything about that fact and just state it like you pulled it out of your ass... Old School Math Trolling.

@JonasTeuwen Reminds me of the time Gardner pulled an April Fools prank with an almost integer...

One April Fools day they should take a surprisingly actually-integer quantity and run a real article about it, and nobody will take it seriously...

Aha, I knew xkcd had something on this...

@JM : Where is it mentioned that its a prank?

@RajeshD Sadly I've forgotten the Sci. Am. issue where it popped up. If memory serves, Gardner claimed that one of Ramanujan's almost integer examples (a modular series truncation, natch) was actually an integer.

ok anyway, it is above my head, but i enjoyed towards the end where the physical constants are introduced.

$\exp(\pi\sqrt{163})$?

@anon That, or another example of Ramanujan's. Like I said, I forgot the details.

@RajeshD often the only mention of an april fools joke being a prank is the date itself

It probably comes from the fact that

q-series for the j-invariant or somesuch

ok. I looked for the date too but not given there

$163^{32/163} \approx e$ ...

@RajeshD Nah, it's not in the MathWorld article. I just suddenly remembered that prank...

He got a number of irate letters afterward, I'm told. :)

lol

(For the alert, one of the examples in that MathWorld article is @Bill Dubuque's.)

happy canada day all

@Eugene : I am not canadian but wish you the same

@JM A prank?

@JM So people were thinking: why is this almost an integer? What is the idea? Maybe it is just a coincidence and you have plenty of those numbers. Say, plenty of fish in the sea 8-).

@JonasTeuwen At least for the Ramanujan examples, the modular series viewpoint gives a plausibility argument. Note, "plausibility argument", not a "reason".

Theta functions are peculiar like that...

@JM Hmm, yes, but then you pull some numbers out of a random generator and do the same computation. 8-).

Then you do some good old school computer checking how many are almost integers.

Then it is like omg like 23.3333% are!

I would like: right, I'll do something else.

You could also be like... "why 23.3333%???".

...and that sounds like one of those "easy to state, hard to explain" questions.

@JM canadians are very filthy mouthed after all

@RajeshD thanks =)

@Eugene It was the first thing that came to mind when you brought up Canada Day. I'm sick that way...

@JM no worries, i'm not canadian

What is the combinatorical interpretation of a term in the expansion of $(x + y)^n$?

So like, you split your set in two, and then you give it a sniff of this and a sniff of that, but then what is the x^{m_1} y^{m_2}. Maybe it is... MONEY.

You return $m_1$ dollars and get $m_2$.

That is a sucky interpretation if you ask my monkey.

Is it?

I know nothing about combinatorics.

I do remember it's something that always pops up. Try checking the questions under binomial-coefficients.

listerine tastes incredibly disgusting.

@JM Mm, the coefficients themselves, yes, but including...

@JM I want to figure out why Weierstrass theorem makes sense if you do it the Bernstein way.

You basically decompose $1$ into stuff. Then you weight everything like a madman.

@Eugene Depends on the flavor...

@JM flavored listerine?

Ahhhhhhh... IT IS A CUBE.

Excellent idea, Sir.

@Eugene Sure, I've seen orange-flavored Listerine, for instance...

It is just some old school children garden cutting and pasting stuff...

@JM i'm going to look for some then. thanks

Should there be a stock answer for "find a basis for the kernel/range/sum/intersection"?

Those problems are allll the same.

@DylanMoreland I do believe that it's time for a canonical question/answer pair. Now, the hard part is who gets to write it...

@JM we could get all the over 10K users to play a MMORPG game of scissors, paper, stone

@JM Hm, I think I just realised myself. Say if you take the $n$-the degree approximation you are actually taking the unit cube and feeding the mass to the largest piece you are interested in.

And then you measure its mass plus some junk of the other points which have much less contribution. Excellent.

@JonasTeuwen That's Bernstein's line of reasoning, if memory serves.

(which ties nicely into why Bézier curves are so useful for approximations)

@JM Yes, I was trying to think how Bernstein ever got this idea.

Then it is quite "normal" to come up with this, I think. So you are like o noes I need polynomials so many degrees of phreedom!!! Then you are like, right let me restrict some of them. Say let us restrict them in such a way that they are already all the same if my function is a constant, that makes sense, rite?

Yes, it does! Good. How do I do that... Okay, then you are like oh yes constant is like $1 + (C - 1)$. Haha! How do I make this a polynomial... right.

Okay, then you've got it fo shu.

The only problem is, reverting is much easier probably 8-).

Convexity is a nice constraint, yes. :)

Mm, convexity you say.

@JM Going out for dinner. Thanks for lighting my sky with bright suggestions which helped me in a strange associative way...

@JonasTeuwen See you later, bro!

See ya, @Jonas.

@JM apparently there's a funny campaign going on now to send a rapper to a remote island in alaska

@Eugene Because he's too cool?

@JM maybe! it seems people don't like him very much. why though i don't know

I guess the people there might give him a cold shoulder...

@JM if he doesn't get mauled by bears first

I prefer the Bearlove good, Cancer bad story. It is amazing =)

@N3buchadnezzar that's just a mudfight now

One sided mudfight

@N3buchadnezzar funnyjunk dropped the suit and the lawyer is suing per se now

WellI am looking forward to the image of the 200k and the bear.

In Norway the biggest producer of soap ran a campain, on which toddler to have on the bottle. The voting was done online and...

Here is the winner

@N3buchadnezzar "bottle" - liquid soap?

yeah

Clearly their concept of "cuteness" is, well, peculiar...

The morale is, never run online campaigns. Anon is legion...

Actually where does $$ \cos \Theta \| a\| \|b \| = \langle a, b \rangle$$ come from?

I'm wondering because the Cauchy Schwarz inequality would directly follow from that.

But its proof in my notes is longer than that.

So the CS inequality probably follows from that...^

@MattN You're asking about where the dot product came from? That's a rather twisted story...

@JM No. The thing up there is not a definition, is it.

(I'm not sure whether that or CS came first.)

Hm...

@MattN I've seen books use that expression for defining the dot product, though.

@JM So if the answer to my confusion is not blatantly obvious to everyone in here then maybe I could ask it on main.

@MattN I actually have a dim recollection of your question being asked before...

I haven't really thought about it, it just came to mind a minute ago, that's why I'm hesitant.

@JM I found this, reading now.

Yes, that. Also this.

@JM Thank you.

@MattN No worries. ;)

I'm off. See you guys later.

@JM See you later!

Going afk. I was actually hoping to see the teddy. I need a carrot or a (virtual) hug.

See you in a bit.

Sup bros.

@MattN CS follows immediately from that right? You just mess a bit with angles and trigonometric formulas and you get it lollll.

Good morn/eve/aft noon/night :)

0o , seems like I'm forgotten :D

If anyone is bored: is the first part of this lemma in the stacks project wrong? If I take $R = k[x]$, $I = (x)$, $M = k$, $N = k[x]$ then it seems like things foul up.

@DylanMoreland Haven't studied completions :(

@JasperLoy hey

I got my results today :D

@DylanMoreland do you see anything wrong with the proof?

@Eugene Feels so rad that I can write down $\pi_1(X,x_0)$

This brings back Old Days.

@BenjaLim Suuuuuuuuup?

@BenjaLim It felt cool I could use $\oplus$ for direct sum, at least!

@Eugene I should have been more specific: the claim is made that $M/IM \to N/IN$ surjective implies that $M/I^nM \to M/I^nN$ surjective for all $n \geq 1$.

@DylanMoreland i know. you're talking about lemma 7.91.1 (a) right?

And the justification given is "Nakyama", which I can't seem to apply.

It's not like there are any restrictions on $I$ or $M, N$ which might help.

I just don't want to bug de Jong without reason, I guess.

@DylanMoreland this is probably a stupid question but for your counterexample, how is $M$ an $R$-module?

@DylanMoreland as you can see for $x \in R$ and $m \in M$ we have that $mx \notin M$ no?

@JonasTeuwen What do you mean? If you have $\|a\|\|b\| \cos x = ab$ then you get $|ab|\leq \|a\|\|b\| $ immediately from that.

Good night!

@JasperLoy i think he knows it's nakayama's lemma. that's why he put it in quotes

@DylanMoreland Are you talking about the lemma involving completions?

@PeterTamaroff This is what I mean when I say scaling

@BenjaLim Did you?

yes for analysis

results out this morning

@BenjaLim Well done! I'm getting back my second mid term tomorrow at midday. Ill tell you how it went!

@PeterTamaroff Result for me is trivial

@BenjaLim Trivial meaning what? What topics did it cover?

I mean results are of not much importance to me

@BenjaLim I guess what is important is what you get from the course.

@PeterTamaroff exactly. I got absolutely nothing from that course.

@BenjaLim Same here, I guess. I knew all that was taught. But I had to take it.

I got a lot out of that course and there was not so much scaling.

:5190073 But is Galois theory in that course?

no

@BenjaLim What is the programme in your analysis course?

@BenjaLim Why do you keep removing your grade?

@PeterTamaroff Everything in here: maths.anu.edu.au/~john/Assets/Lecture%20Notes/2320notes.pdf

@BenjaLim What book did you use?

that was it

all of the notes

that is exactly the content of analysis

that I took

@BenjaLim Oh, OK.

It is loading still.

poorly taught course, really

really really poor

it was really a sham.

You went to NUS no?

how come?

@JasperLoy I will soon.