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!
@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.
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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.
@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 $(B\cap A^C)\cap C=(B-A)\cap C$
@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.
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?
@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.
@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 $[-4, -4/x, -4, -4/x, -4, -4/x, \cdots]$
@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 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.
@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.
@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
@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
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 .. )
@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.
@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).
@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".
@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
@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
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.
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
