@AkivaWeinberger cs.nyu.edu/pipermail/fom/1999-May/003117.html

Thanks

So, I'm on Stack Exchange near the Stock Exchange

(There's no New York Stack Exchange, is there?)

n

Too on-the-nose

@MikeMiller interesting terminology in that post

@Brody I'm sure it's about the Greek roots.

hi

anybody here

no

Hello, I have an image of a function f(x,y) 3D (not the function) how can I know where the first partial derivative is positive and where the second partial derivative is negative?

@EinsL Compute the partial derivatives.

Does this thread have Latex? If so, Latex isn't rendering for me

@Frank Read the chat description.

I can't do that I only have an image

@EinsL What is the image?

@LEA

oops this one s32.postimg.org/7647y9imd/Abc.png

@EinsL Oh, I thought "image" was a mathematical term, lol

lol'd when I find out it is a literal image

yea it is an actual image

So, you follow the direction of x-positive and y-positive (they are inverted in your image, beware)

partial derivative just means you only trace horizontally/vertically

I am being told that I have to mark where the first partial derivative of f(.,.) is positive and where the 2nd partial derivative of f(.,.) is negative. I have to check the image for when x is positive and it appears like it is opening downards then?

@EinsL partial derivative, with respect to x, or with respect ot y?

it doesn't said that, only given f(.,.) find where the first partial derivative of f(.,.) is positive and where the 2nd partial derivative of is negative

> it

enigmatic

@EinsL not "when x is positive", but "the value of z is increasing in the direction of positive-x"

morning

@TedShifrin Thanks a lot. Just to make sure - to prove that it's really a circle, we'd have to actually solve the differential equations in the last paragraph and show that they give a circle? I thought that maybe since there's the last sentence or so showing that the points of $C$ are $R$ away from $(0,y_0)$, there's some way to use that directly - that fact + closed + simple -> has to be a circle. But I guess proving that wouldn't maybe be as easy as solving the diff. equations.

I just googled what time it is in USA hah

fixing your sleeping regime?

I was only off for like a day.

Oh, no. I just say morning.

@MikeMiller it struck me as a tongue-in-cheek shade. regardless, the choice seems peculiar

Autism is literally autos + ism, ie being of the self, or self-centered, etc; you might say this of a category that contains itself.

I did notice that earlier, only thanks to your reply. Just would have assumed an author would avoid these sort of homonyms in any formal platform

though I understand the utility of the term now. perhaps a more mature mind can maintain the disconnect; I am not quite there yet :0

Now it's only a matter of time before the public finds out mathematicians predicted that vaccines cause autism, only they called it pseudoautism (because of big pharma pressure).

Is there a function between the empty set and itself ?

@TheKEMO identity?

Yes, there's one and only one: the function that doesn't send anything anywhere

Ok just wanted to know.

@user1618033 sounds neat, albeit your textual description doesn't do it much justice as far as the imaginative part goes. sell it to me! :p

I think it was SMBC who pointed out that, since people on the autism spectrum are over-represented in science, it's actually the case that autism causes vaccines.

(Which reminds me of when xkcd suggested that cancer causes cell phones.)

@Brody

@AkivaWeinberger there is another good interpretation; cancer is contagious

or , people use to use cellphone to treat cancer

No, it's not

i am just reading the graph in my guise as our philo-analyste is doing in comic meme

"But when someone significantly better than you comes along [and they always do come along in mathematics] you will spend the rest of your career trying to raise arrogance to the level of an art form so it will hide your clinical depression. There is a long list of famous mathematicians who ended their careers bitter and unfulfilled."

Did Terrence Tao ever say, about reading papers, that one should read it twice, the first time thinking about a specific example? I'm pretty sure I read something to that effect somewhere but I can't find the link.

dont know

@ForeverMozart being mathematician is a bittersweet career itself, you ll be condemned to be in endless depression/enjoyment circles forever

yes it seems, but its too late now

Here's what I was thinking of:

"This message is too long"

The paragraph starting with "Another useful trick is to 'project'" over here: plus.google.com/u/0/+TerenceTao27/posts/TGjjJPUdJjk

this is narrowed to the range of mathematicians who are not doing it only as a career, they are fed with it, dream about it, grown up by mathematics, and condemned to die in this field no matter how they tried to veer from

any mathematician who is doing research is obsessed

cause otherwise would not torture themselves

@Brody Let me imagine then I have stuff like Ramanujan ... :-)

More important is that there would be traces that I walked in the area of mathematics. :-)

(that might help many people)

also, to be innate mathematician is not a choice, i know people who serve in the army and they do maths in their part-time, many people like me and any inborn mathematician gets intoduced to this domain (some classify it as a branch of philosophy rather than a science) in very early age even before they pass first preparatory or primary class, it is just instinct showing its job

Finally finishing up this paper. It is surprisingly hard to write the introduction, especially formulating the motivations in a good way.

@JasperLoy Hi

@MikeMiller: I have to admit, while I will defend the notion that physicist usage of indices and coordinates is reasonable based on what type of calculations we tend to be interested in...

@MikeMiller I can't deny that expressions like in this question make my head hurt: math.stackexchange.com/q/1829762/137524

Boros, the great integrator died from cancer too.

@JasperLoy Great Jasper!!!

@Semiclassical jesus lol

yeah

i mean, there may be ways to write that which aren't quite so dreadful while still being coordinate-dependent

but that's just...owww

though in general my reaction to Christoffel symbols is 'kthxbai'

you're secretly one of us

snerk

I'm not sure I'd have been able to stomach a course in general relativity for that reason

on the other hand, the fact that my immediate reaction to that horror is to wonder what it would look like in penrose diagrammatic notation...yeah, i'm kind've weird.

it would look dumb

lol

but that's inherent to penrose notation ;)

would you have been annoyed at the comment here?

pfffffft

which one, precisely?

the one that's by the OP

eh, not really. that doesn't look to be someone who is aware of how the site works

so i'd chalk that up to not knowing the proper etiquette yet

now, if they went ahead and crossposted, that's a bit different. but the comment alone doesn't bother me.

gotcha

i just woke up to it and was sort of annoyed, since it felt like they were just disregarding the answer i wrote, but i get you

sure

in defense of penrose notation, I think a diagram like this:

is not a terrible representation for something like $g^{ij}\hat{\Gamma}^k_{mn}\partial_if ^m\partial_jf^n$

no accounting for taste of course etc. etc.

but I won't defend it vigorously, since I've never had cause to use it that way

I find it sifficukt to look at that picture and call it a defense of anything

shrug

the thing that precludes me from ever being a proponent of it, though, is how hard it is to find good references for it

if i could find that, i could make a decision either way. but without that it's hard to care

@MikeMiller oh god, i hadn't even seen the calculation he links to

the horror

I closed the tab :p

lol

that's fair. i'm procrastinating right now anyways

I need to write more today

translation: "I need to write more everyday" :/

Prove that $G$ cannot have a subgroup $H$ with $|H| = n - 1$ where $n = |G| > 2$. Guys do I have to use lagrange's theorem here?

that would seem to provide an immediate solution

Though I think you could avoid it by some line of reasoning like this: if that condition is fulfilled, then there's exactly one element $g$ which is in $G$ but not $H$.

what's wrong with having an element in $G$ but not $H$

well, that's the question, isn't it? :)

oh I guess it would have to be able to be generated by elements in $H$? under the operation

something like that

hi

i am trying to solve this question

it might be enough to consider the coset $gH$, which itself would have to satisfy $|gH|=n-1$.

and the solution says

how do we get $26460$?

right

But honestly, Lagrange's theorem makes this one trivial. Does $(n-1)|n$? If not, there's no way to have $H$ as a subgroup of that size.

but by the time you say coset you have Lagrange's theorem

@semiC I don't know what a coset is. I don't even understand how Lagrange's Theorem works. Suppose $H$ acts on $G$ then $\mathcal{O}_x \forall x \in G$ gives all the equivalence classes under the action of $H$. So all these equiv. classes have the size of $H$. then we can say $G = \cup \mathcal{O}_x$ So how does this help

well, you have to have the concept of coset to prove Lagrange's theorem. but using it is easy

@obliv When I say Lagrange's theorem in the above, I mean the following statement (lifting from wikipedia)

"Lagrange's theorem, in the mathematics of group theory, states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G."

oh..

there's the fuller statement of it, which says what that quotient should actually equal.

then wouldn't this also be true if $|H| = |G| - 2$?

most of the time, sure.

but it'd work for both $n=3$ and $n=4$ in that case.

$n|H| = |G|$ many $|H|$ that won't work here

in general, i'd conclude that for any given $k>0$ there can only be finitely many groups such that $|G|-|H|=k$.

@semiC can you clear something up with what I said earlier with orbits, though? Is it true $G = \cup \mathcal{O}_x$ the equiv. classes of the elements of $G$ under the relation ~ defined by $x$~$a$ iff $x = ha$ for some $h \in H$

so there's this really simple problem that's been on my mind for several days and I just keep overthinking the problem. It involves a faucet that fills a bucket 2 gallons per minute and a hole that drains the bucket 10% per minute.

can't help there, i haven't thought about cosets in a long while. so i wouldn't trust my logic.

oh i think it is true actually. The equiv. classes should give all elements of $G$

you may well be right, i'm just not in a position to think about it

@oldmud0 what is the problem?

@oldmud0 what kind of question do you have about that? it seems like the most obviosu way to proceed is to set up a simple ODE describing the volume as a function of time.

(well, maybe just a recurrence relation since it tells you the change per minute rather than instantaneously)

I want to make a recursive function based on the volume

@Obliv Your "equivalence classes" are the same as SemiC's "cosets". When you partition your group into a bunch of equivalence classes, you see that the cardinality of the group is the sum of the cardinalities of the equivalence classes.

what's the issue, then? @oldmud0

Since they're all the same cardinality $|H|$, and there are say $k$ of them, we get $|G|=k|H|$.

@mikeM yeah that's what I thought. They intersect, right? thanks

Well, I know that a differential equation is needed to develop the model, but embarrassingly enough I'm not sure how to apply it since I haven't really studied calculus (high school kid here).

right. also, keep in mind that the only cosets that would be possible in the case of $|H|=|G|-1$ would be $H$ itself and $gH$. so you'd need $k=2$ which implies $n=2(n-1)\implies n=2$

@oldmud0 actually, i don't think you do need calculus for this. the out is that they tell you that it loses 10% per minute

Supposed that the volume after $n$ minutes is $V_n$. How much would drain out through the hole in the next minute?

@Obliv No, they don't intersect! That's the whole point. Each element of g is in one and only one equivalence class.

So to count off the elements of g is the same thing as counting off the elements of an equivalence class.

@Semiclassical Problem is that I want to know $V_n$ given $n$ where $n$ is a decimal number.

hrm

@mikeMiller Yeah true. Should only be one way to get the elements in a group within the operation

That's annouying. In that case, you will indeed need to formulate an ODE.

Here's one way to get to that. Suppose instead of filling the bucket as it drains out, you put a lot of water in immediately.

But you could also do it by hand for your case. Suppose H is a subgroup and g is the only element not in it, and that H is nontrivial. If h is a nontrivial element of H, then $h^{-1} \in H$ is nonzero, and $h^{-1}g \neq g$ so it's in $H$, and so $g = h(h^{-1}g) \in H$. Contradiction.

in that case you should certainly have $V(0)=V_0$, $V(1)=\frac{V_0}{10}$, $V(2)=\frac{V_0}{10^2}$, and in general $V(t) = \frac{V_0}{10^t}$ where $t$ is the time in minutes.

Wait I don't get that last part with $h^{-1}g \ne g$ "so it's in $H$" part @mikeM I thought the subgroup criterion only applies to elements in the subgroup already

Now, we want an ODE, so we should figure out what $V'(t)$ is in this case. Do you remember how to do that for exponentials with base other than $e$? @oldmud0

@Obliv We assumed H was order n-1, so it contains every single element that isn't g.

@obliv The point is that if $h$ is not the identity element, the product $h^{-1}g$ should be some element that's distinct from $g$. Since $g$ is the only element not in $H$, $h^{-1} g$ must therefore be in $H$.

the out is if $H$ only contains the identity element.

then you obviously can't pick such an $h$. but in that case $|H|=1$ and $|G|=2$

okay I see. It would also make sense to say this: $g = a\circ b$ for $a,b \in G$ then $a,b \in H$ thus $a\circ b \in H$ right?

what if $a=e$?

@semiC that wouldn't make sense though, how can you have an element that isn't a product of the operation between elements in the group?

what?

if $a = e$ then $b = c \circ d$ for some $c,d \in G$

all i'm saying is that if $a=e$, then you can pick $b=g$ and therefore you just have $g=e\circ g$ for $e,g\in G$.

but that's not enough to conclude that $g\in H$.

I'm asking if an element can exist in a group without being generated by other elements under the operation.

no, it can't. but you're missing my point.

it's not enough to say that $g=a\circ b$. you need to make sure that this product isn't trivial.

@Semiclassical No, I hardly know much about differentiation, but the correct derivative makes sense: $V'(t)=V_0(-10^{-t})log(10)$

@semiC can I say $g = a \circ b = e \circ g$? Then I cover the trivial product and the nontrivial one?

@oldmud0 right. or $V'(t)=-(\ln 10) V(t)$.

i don't see how that helps. you still haven't said anything about what kind of element $a$ is.

@oldmud0 And what's nice is that we've now written $V(t)$ as a differential equation, i.e. $V'(t)$ in terms of $V(t)$.

I see.

Now, can you guess how we'd incorporate filling?

@semiC alright I'll just go with the argument Mike and you gave then.

Just with simple addition, since it is linear, no?

@Obliv Yes, in any group of order greater than 2, any element can be written as a product of other, nontrivial elements. But this requires proof.

can you make that precise? I'm looking for some statement regarding $V'(t)$.

$V'(t)=-(ln\ 10)V(t) + V_0$

That'd reflect a filling rate of $V_0$ per minute. what rate do you have in this case?

$V_0 = \text{2 gallons per minute}$

right. so $V'(t)=-(\ln 10) V(t)+2.$

Yeah.

note that that already implies something rather nice. Can you find a solution for which $V'(t)=0$?

hmm, I'm getting $t=log(ln(10))$.

um, no.

yeah of course it doesn't make sense

where i was going is just that, if you require $V'(t)=0$ i.e. $V(t)$ to be constant, then you get the solution $V(t)=\frac{2}{\ln 10}$.

so if you start with exactly that much volume, the flow rates in and out will be exactly balanced and the volume will stay the same forever.

That's where I was. I guess I attempted to solve a bit too far.

gotcha.

anyways, that's a nice fact. it further suggests that maybe it's worth looking at the volume relative to that steady-state

to do that, let $V_c=2/\ln 10$ and define $u=V-V_c$.

in that case, the ODE rearranges to just $u'(t)=-(\ln 10)u(t)$.

The DJ Khaled Sequence: $\{1, 1, \dots\}$

but we already know that the solution to that ode is of the form $u(t)=u_0 \,10^{-t}$.

so from that you can work out what $V(t)$ is.

@axoren I feel like I should know the joke, but I don't :p

What's the next element of that sequence?

The greatest one of all. ^^^Don't forget to star the picture, a matter of respect for such a greatness.

$a_3$, duh @Axoren

No, another one.

...

that's really dumb. :p

Thanks, I tried.

What if I had said 1?

any positive number would be an acceptable next input.

@Semiclassical and another one.

in this definition of a set: for a fixed $n \in \mathbb{Z}^+$ and a field $F$, $\{(a_{ij})\in GL_n(F) \mid a_{ij} = 0 ~\forall ~i>j\}$ what do the $i,j$ signify? Isn't this a square matrix $n\times n$? How can $i > j$?

$(a_{12})$ has $1<2$ but $(a_{21})$ has $2>1$.

what do the 1 and 2 signify, the amount of rows and columns?

no. the number of rows and of columns is $n$ by definition.

Let me post it again, maybe some of you don't even know who this guy is.

Ramanujan, the greatest mathematician from all times. Indubitable.

$(a_{ij})$ is just the matrix with elements $a_{ij}$

eh, I think Euler remains my favorite

Thanks a lot for the help.

@obliv and the matrix elements of $(a_{ij})$ with $i>j$ are precisely those which lie below the main diagonal.

so can you give me an explicit example? $\begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \end{bmatrix}$ what would you call this

$a_{1234}$?

i could. i wouldn't, but i could.

@Semiclassical why don't you star the picture? Don't you like Ramanujan?

I do, but I don't think he needs me to cheerlead. he can stand on his own :)

so is $a_{ij}$ then the matrix $\begin{bmatrix} a_i & a_j \\ a_i & a_j \end{bmatrix}$? Does it matter the order in which I place the elements

no.

@Semiclassical It's a matter of remembering, he lives by us.

is that no to the first or second question?

the first

The last time. Maybe to post his pictures once in a while and let on channel some of his problems.

@Semiclassical to clarify my comment, that is of course the first two elements of the sequence $n^{na}$ :)

what i'd be likely to write is something like $(a_{ij})=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}$

i.e. $(a_{ij})$ is the matrix whose $(i,j)$-th element is $a_{ij}$

or at least, that's how I'd read the notation.

so the term $a_{21} =0$ since $i > j$?

right.

okay thank you

do you know what kind of matrix that makes $(a_{ij})$?

taking $n$ to be a bit bigger, for instance

it leaves a triangle in the top right corner?

one entry to the top right of the diagonal?

well, the diagonal doesn't vanish

oh right only if $i > j$ not or equal

right.

now to prove this is a subgroup of $GL_n(F)$

so everything 'below' the diagonal vanishes.

i.e. the matrix is upper-triangular. (not 'strictly upper-triangular', mind, which would require the diagonal to vanish as well.)

So you want to prove that the subset of upper-triangular matrices in $GL_n(F)$ is moreover a subgroup.

that means, for example, that it should be the case that any product of upper-triangular matrices is itself upper-triangular

which is true, in fact. easiest way to see that is to write out the matrix multiplication in terms of indices and summation

@MikeMiller

You still here?

For a little longer, yeah.

Do you have time for a quick question? I was going to ask on the site, but I figure you'll probably answer it anyway.