How do you calculate the frequency from component intensities (660nm, 430nm, 700nm) (what those light-colors add to), and vice versa? Wiki article would be great!

Can anyone help me with Sipser page 132?

This might be what you're looking for, @EnjoysMath.

Oh actually, it might not be. T_T

Thanx

I've got a quick question.

to describe *when describing

@Khallil Yes, they are both the shaded area.

Thanks, @JasperLoy!

What time is it where you are?

It is 10.27 am.

So what's the "meaning of life" article all about?

Oh, good morning! It's 03:51 where I am, @JasperLoy.

@exTremity i'm not sure, no doubt blue could explain it, but I think it's pointing out an unexpected coincidence between the coefficients of Klein's j-invariant and Ramanujan's tau function, in the same spirit of the "coincidence" leading to Moonshine theory. (moonshine is a (deep, real) relationship between the former coefficients and the dimensions of irr. reps. of the monster group.) The "meaning of life" bit looks like a joke reference to the Hitchhiker's guide since the number 42 pops up.

not qualified to tell if it's meant to point to serious math, or if it's all just good fun tongue-in-cheek silliness.

What time is it where you are, @r9m?

Is it about 08:45?

Everything feels so much more serene right after sunrise.

Indexed sets are cool, @BalarkaSen. Thanks for pushing me into set theory. ^_^

@AndrewG. Thanks!

Hey, @skullpatrol!

Hi :-)

Me too!

See ya later, @skullpatrol. ^_^

Later pal

@eXtremiity they calculated some stuff that turned out to be 42

Greetings

@r9m Welcome back! Why so late? :-)

hello sports fans

Hi Mike Miller

Yo, wazzup?

@r9m Thanks. I was excited when I that idea crossed my mind.

@Khallil Since $X-Y=X\cap Y^C$, you can use associativity of $\cap$ with $A^C\cap B\cap C$ to get $(B\cap C)\cap A^C=(B\cap C)-A$ or $(B\cap A^C)\cap C=(B-A)\cap C$

@Chris'ssis

@chinamath Don't take it personally, but your questions are for kids

No,I have meet this problem.....I don't have solve it.

so I ask you it..

@chinamath Roughly speaking, each integral can be expressed in terms of polylogarithm

@chris's we all start off as kids when learning anything :-)

@skullpatrol Yeah, you're right. @chinamath is just at the beginning. :D

I know this reslut,But is not usefull to solve my question:math.stackexchange.com/questions/265981/…

@chinamath I know that integral. See that one guy there gave the correct answer - Gilles.J.

@Chris'ssis. Arrogant.

@eXtremiity No, I was kidding. But you see me arrogant because you wanna see me like that.

@chinamath I'd suggest to bring that integral in 2 dimensions and then you'll get that closed form. It's not easy, it's a lot of work to do.

I think this hard

@chinamath Well, I did this for $$\int_0^1 \frac{\log(1+x)^2}{x} \ dx$$ and got a very nice solution.

This problem is from when I want prove this integral math.stackexchange.com/questions/805298/…

@chinamath working in more variables can be extremely useful.

@chinamath Yeah, I got your point, but as I mentioned, using the mutivariables is extremely useful.

I know you don't solve other people problem,so I post it ask other math ,Thank you： math.stackexchange.com/questions/907521/…

May I know who starred that message with "Arrogant"?

It's you!

@chinamath The problem is that in the past I posted in chat a question and asked you not to post it on main, but you did it, you showed me you didn't care my opinion. Now you want help for limits.

@Chris'ssis Not me, obviously, I just came here. Seems like 2 users starred it.

@BalarkaSen I know ...

BTW, I have a question. Is there any condition for having two continued fractions to be equal? i.e., any computationally useful stuff that can identify one with another?

@blue

simple continued fraction representations are unique, otherwise dunno

@blue by simple you mean 1/stuff + 1/stuff + 1/stuff + stuffs, right?

i forgot all the names

mmhmm

drats. mine are nonsimple.

you could bound $[a_1,\cdots,a_r,\cdots]$ above or below by $[a_1,\cdots,a_r]$ based on whether $r$ is even or odd

me knows. that wouldn't help here.

blah, I need notation for both nums and denoms

@blue just use the cfrac stuffs. take a but latex, but is explicit.

*bit

hmm.

gives up. i gotta go.

Arrogant? Well, I don' t think you'll ever read my questions here. :-) Maybe in another life.

@Khallil it was 9:44 am :P lol

@blue I don't think simple ones are unique either

@BalarkaSen For irrationals they are. For rationals there are precisely two representatives.

@MikeMiller Not convinced.

Don't care.

It's true.

LEL, OK.

if $x=[a_0,a_1,\cdots]$ then $a_0=\lfloor x\rfloor$ and $[a_1,\cdots]=1/(x-a_0)$; rinse and repeat. (assuming everything's positive)

So if $[a_0, a_1 , \cdots ] = [b_0, b_1, \cdots]$ then $a_i = b_i$.

Right?

as long as those are infinite, yes

i am considering about irrationals anyway.

@blue can all periodic continued fractions be transformed into simple ones?

@eXtremiity Hey :) ... I'm most excited about the forecoming Hashirama vs Madara battles :D .. apart from that its all boring atm :P

in the sense that it represents a real number...

@Chris'ssis :D was caught in local traffic called exams :P

@r9m. I was about to say that Naruto has been brilliant! But that's because I fast forward all the boring bits :p. I have to agree with you. Last episode was so bad. I skipped through 95% of it.

@eXtremiity ya ,, similar experience here too :|

And I was so excited.....

Suddenly every can fight >_>.

@MikeM @blue say $$-4 + \cfrac{x}{-4 + \cfrac{x}{-4 + \cfrac{x}{-4 + \cdots}}}$$. Multiplying above and below by $1/x$ one gets $$-4 + \cfrac{1}{-4/x + \cfrac1{-4 + \cfrac{x}{4 + \cdots}}}$$. Continuing in this fashion one gets the simple one $[-4, -4/x, -4, -4/x, -4, -4/x, \cdots]$

Am I doing it right?

when one speaks of continued fractions, the terms are meant to be positive integers

so that's silly unless $x \mid 4$

i have no problem if terms are rational

then you're not getting to a simple continued fraction

So if $[a_0, a_1 , \cdots ] = [b_0, b_1, \cdots]$ for rational $a_i, b_i$, then it might or might not be true that $a_i = b_i$?

yes

drats

I ate 20 slices of raw salmon yesterday.

@robjohn :D

@JasperLoy: Eww.

Why would you do that?

@Huy It is called sashimi, lol.

@JasperLoy: I wouldn't know, I don't eat fish.

@Huy Are you vegetarian?

@JasperLoy: No, I just don't like the smell and taste of fish and seafood.

@Huy Fish is certainly better than pork.

@JasperLoy: That's just like your opinion.

@eXtremiity the thing is ... there is only 5-15% of the content interesting .. the episodes are just 22mins approx and over that 7 days($7\times 24 \times 60$ mins) of waiting after each episode .. I'd say that's a poor distribution of fun/exciting stuff .. almost approaches dirac-delta at places :P

@Huy Everything I say is my opinion. What else could it be?

@JasperLoy: Mine.

@JasperLoy: I enjoy most other meat, except for more exotic things like snails or frogs.

@Huy I'm with you on this one.

I'm glad I'm not all by myself.

I am trying to lose 10 kg, from 70 kg to 60 kg.

Maybe it is my meds making me fat.

@JasperLoy: I am trying to gain 10 kg.

How tall are you?

1.65 m

So about the same as I am.

@JasperLoy: If only you could give me your excess amount of weight.

I want to be rid of my mental illness too.

Is it possible?

It is difficult.

I wish you good luck, then.

@DanielFischer: Good day.

Good day.

Grüezi.

Very good.

:D

Greetings, @DanielFischer

@DanielFischer: Where did you learn that?

Hi @Mike.

@r9m. Hahah, well said ^_^.

@Huy A bit of Schwyzerdütsch transpires.

When the first hokage gave his throwback stories, the episodes were great :).

@DanielFischer: Chuchichäschtli?

@Huy I'm not sure whether the first part means Kuchen or Kekse, Chäschtli is evidently "Kästchen", which would probably be translated as Schachtel to German German.

@DanielFischer: It means "Küchen".

@eXtremiity yas !! that was really gr8 :)

@DanielFischer: Kästchen is used more in the sense of a cupboard. It is the typical word a non-Swiss person learns first in Switzerland.

@DanielFischer: Kuchen would be "Chueche" and Kekse "Guetzli", at least in the dialect of Zurich.

When I was watching Naruto (not Naruto Shippuden), I really wanted Sakura to die. After the last episode, I remember why. She's so annoying.

I always thought she wrecked the awesomeness between Naruto and Sasuke.

@Huy I've mostly been in the francophone parts. The first word I learned was "huitante" - the Hotel clerk noticed my bewilderment and translated it himself: quatre-vingt.

@DanielFischer: Yes, some people do say huitante. Also septante, iirc.

@DanielFischer: But why have you only been there? Zurich is much more beautiful and has much more to offer. ^_^'

@Huy Not many mountains in Zürich, as far as I know.

@DanielFischer: You came to Switzerland because of the mountains, then?

@eXtremiity the story rolls subtly around her emotions/ego .. the sasuke chase arc was after all bcoz of her request to Naruto to bring her crush cback to the village :P

@Huy Mostly.

Hmm, yes - that's right. Sakura's request was predominately the reason for the chase.

Man when Sasuke came back and they all reunited - that was so cool!

@DanielFischer: Did you at least go skiing or snowboarding?

"I want to be Hokage >_>". Major lol ^^.

@eXtremiity but at times however I feel like she's a gigantic waste of screen time :P

@Huy No, just - what's it called - faire des randonnees.

@eXtremiity xD

Gigantic is an understatement.

She's just not cool.

At all.

@DanielFischer: I see. I'm not very fond of hiking.

@eXtremiity Kishi trolls her every now and then .. :P maybe that's her roll :P

@r9m. How cool would it be to create your own anime - own story line and what not _.

@eXtremiity idk .. :) but I guess it would be a grand experience :D

@r9m. Forgetting about the success of it all - just creating events in your mind, playing them out as they should. If you forget about all the garbage Naruto (the show) has to offer, it is actually a piece of genius in my opinion.

From story line to story line. I think its incredible. I enjoy the philosophical opinions of certain characters.

@eXtremiity true .. :) I still think the Zabuza episodes in the beginning of Naruto are still the best .. it was a vivid portrayal of the convictions and contradictions of a shinobi life (human life too perhaps ..) & not to mention the awesome actions and ever raising tension along the lines !! :D

@r9m. Agreed ^.^!

Kish delivered a lot in those 3-4 episodes .. its perhaps the most condensed form of the entire theme line .. every episode from then on has been only a dilution of that intensity .. :) (except of course the Hashirama-Madara sequence .. which is battle of the $\alpha+$ and far above the ordinary !! :D .. )

:D !!

It is an exciting period.

@eXtremiity :D

anime

duh

@DanielFischer Do you know a closed subset of the plane that's not locally contractible?

@MikeMiller "Locally contractible" means every point has a neighbourhood that is contractible? Or a neighbourhood basis of contractible sets?

To me it means the latter.

Nevermind.

For some reason I discarded the comb.

@MikeMiller Something Hawaiian, I'd think, $$\bigcup_{n\in\mathbb{N}\setminus \{0\}} \left\{ z : \left\lvert z - \tfrac{1}{2n}\right\rvert = \tfrac{1}{2n}\right\}.$$

Hawaiian works too.

Thanks

@DanielFischer: Let $u \in W^{1,p}(I)$ for some interval $I$ and w.l.o.g. $|I| \leq 1$. We know $$|u(x)-u(x_0)| \leq \int_I |u'(t)| \, \mathrm dt = \| u' \|_{L^1}.$$ Thus, after averaging w.r.t. $x_0 \in I$ we find $$\| u \|_{L^\infty} \leq |I|^{-1} \int_I |u(x_0)| \, \mathrm dx + \| u' \|_{L^1}.$$ What just happened in the last step?

@Huy The first inequality gives you $\lVert u\rVert_{L^\infty} \leqslant \lvert u(x_0)\rvert + \lVert u'\rVert_{L^1}$ for arbitrary $x_0\in I$. Then you just replace the arbitrary $\lvert u(x_0)\rvert$ with the average. (Consider an $x_0$ such that $\lvert u(x_0)\rvert = \min \{ \lvert u(x)\rvert : x\in I\}$ to see that it's true.)

@DanielFischer: If I look at the first inequality and add $|u(x_0)|$ to both sides, I can get $|u(x)| \leq |u(x_0)| + \| u' \|_{L^1}$ by the triangle inequality. Then you suggest taking the supremum on both sides, is that correct?

The right hand side is independent of $x$, so you take the supremum only on the left. Then you take the infimum (w.r.t $x_0$) on the right, then replace the infimum with the average.

wat

@DanielFischer: I see, thanks.

RH is related to l-adic cohomologies and etale fundamental groups?

who knew !

One proves the Weil conjectures (including the "Riemann Hypothesis for curves") using those. This is probably what you're thinking of.

looks interesting. but far ahead of where i am. jerks head to shake these stuffs off

never knew that there was a RH for algebraic curves

@MikeMiller like, you're saying that one could, say, determine the zeros of the L-function attached to some elliptic curve using those?

here's a statement of the conjectures

(theorems)

looks greek to me

O_o

@DanielFischer: I'm confused. Isn't $|I|^{-1} \int_I |u(x_0)| \, \mathrm dx = |u(x_0)|$ since the integrand doesn't depend on $x$?

@Huy Yep. It's probably meant to be $\int_I \lvert u(x_0)\rvert\,dx_0$.

@DanielFischer: That's what I thought too. I guess a typing error.

So guess I.

Good day, @TedShifrin.

Greetings @Huy @DanielF

Morning, @Ted.

I give up on greetings. There are too many people going in and out of chat, lol.

Ungreeting @Jasper ... I'm about to disappear

@TedShifrin Did you give Pedro any of your books?

He said so, yes.

The past 5 years, I have dropped so much hair.

It's all the stress from my mental problems.

@TedShifrin Morning

Hi @nablablah, long time.

Hello hello hello

In case you haven't seen it yet

@DanielFischer: For bounded $\Omega$, $1 \leq p < q \leq \infty$ implies $L^q(\Omega) \subseteq L^p(\Omega)$. Can one say something specific about unbounded $\Omega$?

@Huy The inclusion holds if the measure of $\Omega$ is finite. If the measure is infinite, neither is included in the other for $p\neq q$ (I'm assuming we speak of open $\Omega\subset \mathbb{R}^n$ and the Lebesgue measure, for $\ell^p(\mathbb{N})$ for example, we have the inclusions in the other direction).

Yes, I meant open $\Omega \subset \mathbb{R}^n$.

What happened with @chris's sis ??

@Hippalectryon: eXtremiity called him arrogant and soon after he ragequit, methinks.

math.stackexchange.com/users/32016/chriss-sis his new description :/

I hope he does come back, i'll be lonely otherwise :c

Wow, so it actually was a ragequit.

I don't miss him. :)

I do :/

@Hippalectryon Her

@Huy her

Him.

There are only guys on the internet.

Everyone knows that.

@JasperLoy Oh sorry I always forget xD

@Huy No, that is nonsense.

I have around 50 images of problems he posted in chat :D

For when I have learned more :c I'm currently totally unable to solve them

I wanted to exchange email with her, but she declined.

I have about 20 SE user emails now...

Even if she quits (which would be really unfortunate :c ), I hope she still notifies me when her books comes out if it does

Congratulations.

@Hippalectryon Well, to be honest, Chris'sis is a bit too arrogant. Infact, I want to remove the bit, she is arrogant.

...though I was not aware of this development :O ..

@Sawarnik: Some guy (I think china math) came into the chat and asked Chris' sis for help with some limit/integral. She replied "Don't take it personally, but your questions are for kids".

sigh alright what drama is here now

it started here

How can anyone not take that^ personally?

Well, she did say "don't take it personally", that's how. /s

Well, to me it seems more like teasing

@AlexanderGruber HELLO

I wouldn't really care if she told me that? I know she far far exceeds my abilities in this field

@Hippalectryon: Well, sure, if it's a friend. But not if it's someone else genuinly asking for help.

and then to say "I was kidding"?

@Huy That's stoopid.

"Don't take it personally... your ...."

@PedroTamaroff hey man, are you still in america?

@PedroTamaroff: Are you aware of the symbol "/s"? It stands for sarcasm.

@AlexanderGruber I left on Wednesday. =(

@Huy But chat.stackexchange.com/transcript/message/17315431#17315431

It seems that there were previous facts

@Huy Oh, sorry then.

You're too internetz savvy for me.

@Hippalectryon: It doesn't make her statement any less arrogant. If someone's being a dick to me doesn't justify me doing the same to them.

@Huy At any rate, don't make it a big deal.

@Huy I was referring to 'But not if it's someone else genuinly asking for help.

It's some stranger on the web.

@Huy i agree with that.

@Hippalectryon: Well, the china math guy was asking for help at this point. He might have some bad history with Chris' sis but he did normally ask for help.

which is why i do not like that everyone seems to be trying to publicly shame her for what to me seems like light teasing, or at most one off color comment

@AlexanderGruber I saw your post on cyclotomic polynomials. Nice pictures.

:-)

Chris'sis could have just ignored china math instead of saying "your questions are for kids" - in my opinion.

@PedroTamaroff thanks. I'm working on getting it set up in Python so I can get some more high res stuff

One has to learn deal with it.

@PedroTamaroff: I don't really care about her because we don't really talk to each other but I can see why people would think of her as arrogant.

And what she replied to china math definitely qualifies as an arrogant statement - imo.

@Huy Dunno. Last time I talked to him he dropped the same line and I was like "sheesh you don't have to be an ass just answer the question."

And then reacting by leaving the website is just childish.

@Huy Chris ragequitted?

@PedroTamaroff: math.stackexchange.com/users/32016/chriss-sis

@Huy Oh, what a tragedy.

I know. I'm heart-broken.

@Huy What have you been studying lately?

@PedroTamaroff: I have an exam on Tuesday about functional analysis, mostly Sobolev spaces, their regularity and Schauder theory, so just kids' stuff to be honest. :P

What about you, @PedroTamaroff?

@Huy I'm trying to study commutative algebra at the moment.

Just trying?

@Huy It's just a figure of speech, I guess.

=)

Good, I was worried.

How would you summarise what you are studying in few sentences for another maths major who knows nothing about commutative algebra? I did take an abstract algebra class but I don't know if it has anything to do with it.

commutative algebra is abstract algebra with all the joy removed

What joy in abstract algebra?

@AlexanderGruber LEL

@Huy Abstract algebra studies commutative rings. But I guess you guessed that?

@PedroTamaroff: Yes, I would have guessed that.

So what's interesting about commutative rings?

@Huy Some things are interesting. Like studying the poset of ideals of the ring.

@PedroTamaroff: Poset = ?

Partially ordered set.

Or the set of prime ideals, which one can give a topology, related to geometrical ideas.

I see. Can you give me tell me about any applications? If I have some commutative ring and I know something about the poset of its ideals?

@Huy do you mean real world applications or applications within mathematics?

@AlexanderGruber: Both.

only real world applications i know of for commutative algebra are in cryptography

computing groebner basis and stuff like that

And within mathematics but extended to other areas?

oh gosh, there's lots of stuff

Something to impress me?

well i'm not really a proponent of CA so i'm not the best person to ask, but without it we wouldn't have algebraic geometry, category theory, scheme theory, any of that stuff

@AlexanderGruber: I don't really know anything about all these areas. :D

you're an analyst yea?

@AlexanderGruber: I wouldn't call myself that.

@AlexanderGruber: I am interested in mathematical physics. I think physicist tend to not know much algebra so it's mostly analytic mathematics.

@Huy okay, so the foundations of string theory can be expressed in terms of commutative algebra

I see. I don't know much about string theory yet either. But I will at some point, I hope.

I know that string theory is about strings, lol.