@Yashas Really?

Geometric mean.

6th grade students preparing for olympiad know it

Not 6th, at least 8th .

so I guessed that it must be 8th or 9th grade syllabus for regular school students

No never.

It's in 11th class

what?!

yes

There are students who qualify for Indian National Olympiad from 8th, 9th, 10th, 11th grade

even 8th grade students qualify

8th is damn rare

without AM GM they are screwd

they know 123123 more inequalities lol

@Ramanujan did you get the answer?

Isn't the process a bit inelegant?

which one?

the one above

oh, it is the most obvious solution though

X >= Y

yes

that is the first step always

so try to write X as AM and Y as GM

@yashas, no not till

did you get the AM?

The AM is $\displaystyle \frac{abc + abc+ac^2+b^2c+a^2b+a^2c+ab^2}{8}$

Yeah i got AM,@Yashas

The GM is $\displaystyle {(abc \times abc \times ac^2 \times b^2c \times a^2b \times a^2c \times ab^2)}^\frac{1}{8}$

nth root of the products of the quantities is GM

@yashas, i got :$G.M=(a^8b^8c^8)^{1/8}

@Ramanujan Could I know, which grade you are?

so now you have $(a+b)(b+c)(a+c) \ge 8abc$

@yashas , yeah i got to there

@SwapnilDas, why??

Just curious.

10+2

Thanks.

Wlc@SwapnilDas

@yashas, ??

@Ramanujan Now apply AM GM for a, b, c

$(a+b+c)/3 \ge (abc)^\frac{1}{3}$

Yeah!@yashas

Using the constraint given to you, you get: $\frac{1}{abc} \ge (abc)^\frac{1}{3}$

$1 \ge (abc)^\frac{4}{3}$

or $abc \le 1$

ok this is interesting

I was expecting $abc \ge 1$

but $abc \le 1$ proves that your question is wrong lol

let me see where I went wrong

@Yashas, why is \dfrac {3}{4}

$\frac{1}{abc} \ge (abc)^\frac{1}{3}$ => $1\ge abc \times (abc)^\frac{1}{3}$

@0celouvsky are there cool theorems for results of PDEs on multiply connected manifolds being related to results on the universal cover

@Slereah I'm not sure, but there's nice results about PDE on the double of a boundary manifold

@Slereah that notation looks like nonsense to me.

@Yashas @Ramanujan You could probably also manipulate from $(a+b+c)^3 - (a^3+b^3+c^3) = 3(a+b)(b+c)(c+a)$. Just throwing out an idea, but I haven't thought about it.

It's a bit confusing

It's not clear what object A*B is supposed to be or what those sequences with i and j even mean

What are i and j supposed to be?

I'm not quite sure what's up with $a_{n(i)}$

I think i and j are supposed to be indexes running over the two spacetimes

Can you send me the paper

I'll have a look later

@Slereah then why not just say 1 and 2?

arxiv.org/abs/gr-qc/9511068

@Slereah so isometries

fucking hell

Isometries preserve sequences

basically it's related to the notion of two spacetimes being the same up until a certain point

So A*B should always mean B=phi(A)

The notation is very bad

Well not all of spacetime is isometric

Only a subset

Hello @yashas

Is the isometry defined everywhere?

@Ramanujan Is your question correct?

It's defined on a subset $N_i \subset M_i$

If abc(a+b+c)=3, prove that $(a+b)(b+c)(c+a)\geq 8$.

so the sequence $(a_j)$ is in $N_1$?

I'm not quite sure

The fact that it's $(i)$ rather than $(1)$ makes me suspect it could be in either

@Ramanujan I am not sure what I have done wrong :/ it tells your question is wrong.

Also i'm not quite sure what $\star$ is

Take $a = 3, b = c = -1$. Then $(3-1)(-1-1)(-1+3) = 2(-2)2 = -8$. Certainly not larger than $8$.

In some contexts it seems to be a function that gives a point of the manifold

In others it seems to be a relation

There's a theorem that says "If $M_i$ are globally hyperbolic spacetimes diverging by $S$, then $F_1 \star F_2$"

@BalarkaSen AM-GM isn't valid for non-negative numbers, right?

Of course not.

@yashas, thank u for the soln.

Negative numbers in the roots is nonsense generically.

@BalarkaSen Since when do you know GR?

@0celouvsky GR? I dunno that stuff.

@BalarkaSen u just said "of course not"

I think Krasnikov is playing fast and loose with the notation

Which is very bad because he invented the notation

@Slereah never a good sign

@0celouvsky i replied to yashas' message...

@Ramanujan where did you get the inequality question? can you check its solution?

I think that might need a PSE question

@BalarkaSen Oh, I have him blocked. I see he has adopted a fly picture. Fitting.

:D

Oh, we are assuming that $a, b, c$ are positive?

@yashas, I don't have its solution.

That was unstated originally.

I assumed it was positive and tried to prove :P

@0celouvsky Why am I blocked? :(

Well it's certainly garbage for non-positive by my example.

Yea.

@ACuriousMind Bist du hier?

even for non-zero numbers, it does not seem to work according to my proof

I am not sure if the question is wrong or if I have gone wrong somewhere

@ACuriousMind Let $M$ be Lorentzian and $\Sigma$ a spacelike hypersurface in $M$. Consider a vierbein $\{\theta^\mu\}$ with $\{\theta^i\}$ an orthonormal dreibein on $\Bbb \Sigma$. Why is $\mathrm{Vol}_\Sigma=-\star\theta^0$?

@Yashas, math.stackexchange.com/questions/1401650/prove-that-abbcca-ge8

Are these two related?

same question

@ACuriousMind Is $-\star\theta^0=\theta^1\wedge\theta^2\wedge\theta^3$?

and the same conclusion which I came to @Ramanujan

@Ramanujan your question is wrong

Oh. @Yashas..thanks alot

@Ramanujan but first answer in that question is shorter and nice

@0celouvsky yes

@Yashas

@ACuriousMind If @0celouvsky has blocked me, he won't see the messages where I tag him too?

nope

perks of ignore feature

He must have blocked all JEE people :D

@ACuriousMind why tho

@Yashas indeed he won't

not a bad thing to do

@0celouvsky Because that's basically how the Hodge star is defined

why the minus though

@0celo for style

@

@Ramanujan ABC is a triangle right?

@Yashas, not defined

@Ramanujan I think we shud discuss these kind of questions in the JEE prep room

@0celouvsky Given an ordered, orthonormal basis, the Hodge star of any wedge of them is the wedge of all the ones that didn't occur in the original wedge, times $(-1)^n$ where $n$ is the number of time-like basis vectors

Yes, I knew the first part, but not the $(-1)^n$ part

@Yashas, are you available there?

Where does that come from

@Ramanujan yes

I think $\star$ is supposed to represent points being related to each other

Orientation issues probably.

Ok@Yashas

It's supposed to represent like

Sort of one or the other?

I dunno

It's fairly poorly explained

@blue He's gone Kafka.

If you get what I mean

@blue I want people hit their screens :D

See, that's why you should read literature more. To understand inside jokes.

@0celouvsky From $\alpha\wedge{\star}\beta = \langle \alpha,\beta\rangle\omega$, basically

Not for entertainment or getting life lessons or anything flim flam like that

it's just about the jokes

Take $\alpha = \theta^0$, then you know that $\langle \theta^0,\theta^0\rangle = -1$, so the Hodge dual needs to have a minus.

@ACuriousMind I had that hours ago

Yesterday even

So we agree ;P

I guess

Although I'd rather work it our directly from the definition of $\star$

I think it's supposed to be that, on the isometric parts of the spacetime, $A \star B$ represent points that are mapped to each other by the isometry, but for other regions they are just points that are "equivalent" on a more nebulous notion of equivalence

Well sure but you'd just look dumb if you're among a group of people making literary jokes

I think that's my main motivator for reading lit

(FWIW, Franz Kafka is 19th century deranged novelist whose one of the most accomplished work is The Metamorphosis about a man who suddenly turned into a giant insect and died.)

It would help if Krasnikov wrote out explicitely what point belonged in what space

I hope he wrote another paper with that notation to clear things out

plz respond

I really hope that Krasnikov didn't mean to say that $A \star B = A$ is supposed to stand for $A \star B$ and $B = A$

I don't think I'll get help from another paper because all that rambling was pretty much just Krasnikov exposing his method for getting the idea of the Krasnikov tunnel

The "terminology that only one paper uses and explains poorly" is getting on my nerve

@ACuriousMind Suppose I have a Levi-Civita connection on my manifold. I lift this to a connection on the frame bundle. I then use this to induce a connection on the spin bundle. Are the components of the curvature of the spin bundle going to be the components of the curvature endomorphism of the manifold?

@0celouvsky You may take $\alpha \wedge {\star}\beta = \langle \alpha,\beta\rangle\omega$ as a definition of $\star$ :P

@0celouvsky I'm not sure what "the curvature endomorphism" is

@ACuriousMind I know that.

@ACuriousMind Curvature form on $TM$ viewed as a matrix

@0celouvsky Well...that depends on the frame you're in.

The curvature on the spin bundle is $\mathfrak{so}$-valued, so it agrees with the tangent curvature in an orthonormal vielbein but not in a general frame, I think

Ah yeah my vielbein is as orthonormal as it gets

"A generalized connection on Y is a section Γ : Y → J1Y, where J1Y is the jet manifold of Y."

Aaaah

Why even bring up jets

what's a good modern book on fiber bundles, anyway

IIRC Steenrod doesn't even discuss connections

The canonical book on connections is Kobayashi-Nomizu.

It's not new

Bishop & Crittenden is a little newer

They're not as hardcore

yeah

I hear Kobayashi isn't very friendly

@Slereah Try Michor, "Topics in Differential Geometry."

math.cornell.edu/~goldberg/Notes/AboutConnections.pdf

good title

Explain yourself, connection

@blue Not quite right. You might have $g(a)=g(x)$.

You need to ensure that $g(x)$ is never $g(a)$ as $x\to a$.

Otherwise that fraction is not defined.

But that's of course not always true.

So you need to be more careful.

You can easily deal with constant maps.

0celo's point is you need to be careful with saying the limit $\lim_{x \to a} (f(g(x)) - f(g(a))/(g(x) - g(a))$. What is it the limit of?

$\dfrac{f(g(x)) - f(g(a))}{g(x) - g(a)}$, as wisely pointed out, isn't in general well-defined.

When $g(x) = g(a)$ for all $x$ it follows trivially; $f(g(x)) - f(g(a))$ is identically zero.

@blue Not your modified limit, man.

I mean the first principles limit of $(f \circ g)'(a)$.

It literally means the derivative is 0.

$g'(x)$ is 0 if $g$ is constant.

Right.

I'm not talking about constant maps

But the issue is more subtle than the case where $g$ is constant. $\dfrac{f(g(x)) - f(g(a))}{g(x) - g(a)}$ may not even be defined if $g$ is nonconstant.

I mean what happens in some pathological case where $g(x)=g(a)$ infinitely many times as $x\to a$

@0celouvsky That's what I'm trying to tell him.

some sort of oscillatory behavior

@BalarkaSen ok good

0cleo has not blocked other JEE people but has blocked me :(

@0celouvsky I think all of these lands you into rigorous calculus which JEE doesn't really care about tho

@BalarkaSen It does. You need to compute the first order Taylor remainder and show that it's continuous

something like that

Hence, why I said "modulo rigor" before.

yeah

Yes, the deal is to show the function defined by $h(t)= \dfrac{f(t) - f(g(a))}{t - a}$ for $t \neq g(a)$ and $h(t) = f'(g(a))$ otherwise is a continuous function.

Yes

All of these is trivial to fix, but still an exercise in rigorous limit arguments.

I am not even sure if we talk a lot about limits rigorously in 10+2 and JEE

physicsyfication of math is a common phenomenon in Indian HS education.

in...any education

who wants to hear about rigor?

I'm currently worrying about the convergence of integrals

it ain't fun

I guess. Can't speak for education systems I'm not familiar with.

Just do it on normed vector spaces

It's literally the same thing you did, written out explicitly.

the proof is essentially the same

I don't like the delta notation tho. Really misleading at times.

@BalarkaSen Hey, it's not as if physicists have a monopoly on non-rigorous approaches to math :P

certain mathematicians do it too

@ACuriousMind I think they're indirectly infiltrating the world

That is not true. I think non-rigorous mathematics is actually harmful to the brain.

...

I don't know about physics and chemistry. Experimental sciences are allowed to be non-rigorous.

physics is an experimental science

@blue Not teaching derivation and teaching wrong things are completely different.

Or half-right things.

@BalarkaSen yesterday I saw a delta function of an unbounded operator in my QM class

@blue There pretty much is no derivation for these equations, you may as well take them as axioms. You can "derive" Maxwell's equations from a Lagrangian + principle of least action, but in the end that Lagrangian is precisely engineered to yield (one half of) Maxwell's equations, so that doesn't add any insight.

what the hell kind of operator is $$L=-\frac{i}{|\nabla_y\phi|^2+|\xi|^2|\nabla_\xi|^2}(\nabla_y\phi\cdot\nabla_y+‌​\nabla_\xi\phi\cdot\nabla_\xi)$$

and why isn't that denominator ever zero

too many questions

@blue I can't speak for the other two subjects, I'm neither a physicist nor a chemist. I know a little bit about mathematics, and from that experience I can tell you a complete restructuring of the syllabus in 10+2 is possible so that you fit in the rigorous, actually right things in there and it would take about the same time to cover it.

@BalarkaSen how is that even possible

you'd have to teach more stuff

how could it take the same amount of time?

HS cares too much about solving lots of primarily ad-hoc problems with a vague notion of the fundamentals.

@0celouvsky No you wouldn't have to teach them more stuff.

@BalarkaSen You'd have to teach about set theory (injective, surjective, etc.) and proof strategies

@blue Your doubt is completely baseless :)

that takes a long time for most people

@0celouvsky We have those already.

@blue Do you know what a contrapositive proof is without looking it up?

@BalarkaSen See.

Contrapositive proof is in our textbook. If blue doesn't know it he didn't read it.

That's what I'd emphasize more in a modified syllabus.

@blue Anything? Class 11 math textbook on CBSE, or WBCHSE?

Prove $p \implies q$ by showing $\lnot q \implies \lnot p$.

That is proof by contraposition.

There must be a typo in this book

Anyway, yes, I would definitely vote for reducing ad-hoc problem solving. That could be spent on trying to learn it conceptually better. I mean, even IMO problems aren't as ad-hoc as JEE.

@BalarkaSen erm, does a critical point of a function $\Bbb R^n\to\Bbb R$ mean one of its derivatives is zero, or all of them. All, right? Because the differential then fails to be surj.

@0celouvsky What does "one of it's derivatives" mean? There is only one single derivative; the Jacobian.

@BalarkaSen It has $n$ partial derivatives...

Oh. You have to have all the partials vanishing.

Good, that means my denominator is never zero

well...unless $\xi=0$

what the hell

@blue Problems are fine. Here are some actually great problems.

All I have against is, like, a huge integral which can be solved by some trickery that comes with practice.

@blue Have you been taught gamma functions?

error functions?

contour integration?

reduction formulae?

:D We are actually cheating in JEE :D

becaz those topics aren't in the syllabus

but we use them to hack through

@Yashas Yes

@Yashas Yes

not to mention about the abuse of taylor series

When the question says for a general function f(x), use Taylor series.

becaz analytical functions belong to the set of general functions

I think quadratic equations are grade 8 syllabus in regular school

@blue in which syllabus?

In my coaching centre, we have 7th grade students preparing for International Math Olympiad

Students writing research papers in 8th grade lol

aaaaand the Indian circle jerk is in full swing again

I say "Go away" when I see one

:O 0cleo unblocked me

or maybe not

@0celouvsky am I still blocked? :(

Blocked on fb?

here

So you get notified?

our 10+2 math really has a lot of stuff and i'm very happy about that it gets a lot of people interested but it just feels like digging a hole for one big rut instead of digging your way towards the ocean.

@Fawad nope

@Yashas you want to be unblocked

@BalarkaSen Those who prepare for JEE love what they do. People usually think JEE people are robots which isn't true.

@Koolman no

then