Mathematics

those are both awful notations, especially the first. (nobody will ever understand what you mean by the first.) write $x \neq 0$. and one normally writes either $x \cdot x$ or $x^T x$.

beyond that, you're good.

Max

ok, thanks

Ted Shifrin

12:48 AM

@Max @MikeM: It's surprisingly common (in certain countries) to write $\mathbf x^2$, unfortunately (or $\mathbf x\mathbf y$ with no dot). But I prefer $\|x\|^2$, and putting in dots.

meow-mix

1:02 AM

@TedShifrin did you read my rebuttal? :P

Adeek

1:17 AM

hmm

Kaj Hansen

How goes it @Adeek ?

Adeek

just sick @KajHansen trying to study while sick

haha

stuff doesn't come fast to me

Kaj Hansen

Ah, that's a drag. Hope you feel better soon.

Adeek

yeah me 2

Jessy Cat

I need somebody to please help!

Adeek

1:19 AM

thanks @KajHansen

Kaj Hansen

Me neither @Adeek. I have to stick with things for a while before they solidify.

Adeek

yeah

Jessy Cat

I'm at the end of a Laplace problem in polar coordinates on a semicircle with unbounded radius, and something is going wrong with putting together my solution at the end.

I'm getting something that doesn't make sense, and I really, really need help.

@TedShifrin are you around by any chance?

Adeek

I was wondering if we have a pre-hilbert space $\mathbb{H}$ where it the dot product doesn't satisfy $(x,x) = 0 \implies x = 0$.

Jessy Cat

Or @Semiclassical?

meow-mix

1:20 AM

hi @JessyCat

Jessy Cat

Anybody with any kind of knowledge of PDEs who could take a look and tell me what I'm doing wrong?

Adeek

we consider the space $F = \{ x \in H : (x,x) = 0 \}$ why is this a subspace of $\mathbb{H}$ ?

Jessy Cat

Hey @meow-mix

meow-mix

maybe i could help

but if you dont want to type it out thats fine

because im probably ofno help anyways

Adeek

$(x + y, x + y) = 2Re(x,y)$ why is it zero ?

Jessy Cat

1:20 AM

@meow-mix aweesome. I'm in the process of posting it on the main board. I'll post a link after I'm done.

meow-mix

alrighty then

Kaj Hansen

What is a pre-Hilbert space?

Adeek

@KajHansen do you have an idea about this simple thing ?

It is a vector space together with inner product that doesn't satisfy the above.

i.e there could exist x such that $(x,x) = 0$ with x being non-zero.

Akiva Weinberger

Can the inner product be negative?

Adeek

no positive @AkivaWeinberger

it is always positive

Akiva Weinberger

1:22 AM

Ah, OK -- I was thinking C^2, where we don't have conjugation in the thing

but that won't work then

Adeek

there is some weird spaces that have are pre-hilbert spaces

Akiva Weinberger

(or even C^1)

Adeek

function spaces for example

space convergent sequences

Kaj Hansen

Ok, so we need to confirm that this is closed under vector addition @Adeek

Mike Miller

@Adeek $\langle x - y, x - y \rangle = - 2 \text{Re}\langle x, y \rangle$. But the LHS has to be nonnegative.

So $-2 \text{Re}\langle x, y \rangle \geq 0$, but $2 \text{Re}\langle x, y \rangle \geq 0$.

Kaj Hansen

1:29 AM

suppose $\langle x, x \rangle = 0$ and $\langle y, y \rangle = 0$ Do we have $\langle x + y, x+y \rangle = 0$?

Mike Miller

that's what he was trying to show

Kaj Hansen

Inner products satisfy distributive property, no?

Mike Miller

yes; he did this

9 mins ago, by Adeek

$(x + y, x + y) = 2Re(x,y)$ why is it zero ?

Kaj Hansen

Ah, sorry. I'm multitasking atm

PVAL-inactive

@MikeMiller Ya I saw it. It's Cochrany stuff.

Adeek

1:33 AM

sorry @MikeMiller was at the washroom

I see

PVAL-inactive

@MikeMiller It's kind of scary how Cochran still manages to produce so much math being dead and all. He's got a few papers this year.

Adeek

cochran ?

Mike Miller

The memorial conference was nice. Got me convinced that his work was interesting stuff.

SF was a terrible speaker though.

PVAL-inactive

Thats interesting because theres already one other notoriously bad speaker on that list.

So if someone upstaged his incomprehensibility I'd be impressed.

Mike Miller

I don't know who. I think he didn't prepare his talk and seemed proud of the fact.

PVAL-inactive

1:39 AM

I had a highly improvised talk monday that I think went okay.

Mike Miller

Was it to a conference of people there to hear about your work in honor of someone who passed away?

PVAL-inactive

no

Mike Miller

Then I think you're probably fine.

Tau invariant was extended to rational homology spheres today I guess. I'm surprised that wasn't already known.

PVAL-inactive

I think it went alright because it was a very small crowd and I sort of spent an extraordinary amount of time answering mid-talk questions.

Mike Miller

lol

based on our seminars sounds like Ko was there

PVAL-inactive

1:43 AM

Well questions like "Is _____ true?" and instead of answering "yes" or "no", I answered consider _____ and the theorem ______ then _____ is a counterexample because ____ etc.

Mike Miller

got it.

I still need to write my seminar talk for Friday. i've been putting it off. But I'll just do it tomorrow.

PVAL-inactive

Someone (in topology) scheduled their candidacy over a talk I'm giving on Friday so I get to put it off until Monday.

Mike Miller

lol

I think you should have had them compete and see if the candidate's talk still got all their committee to go.

PVAL-inactive

I'd lose.

Mike Miller

Dang.

Is the talk just for your seminar? Or is it about your work?

PVAL-inactive

1:47 AM

for a learning seminar.

Mike Miller

k.

PVAL-inactive

the one about HD&GB

is that a reasonable acronym?

Mike Miller

I had to think about it but yes.

PVAL-inactive

holomorphic disks and genus bounds

Mike Miller

yeah I got it.

Great paper.

The actual proof is a little underexplained but oh well.

PVAL-inactive

1:50 AM

Well of course there are gaps.

Gabai's construction is only C^0 and Thurston and Eliashberg works only for C^2

Kazez and Roberts fixed that issue.

Mike Miller

Right. I'm talking about the diagrammatic part of the argument.

PVAL-inactive

The GC issue though keeps me up at night.

Mike Miller

That one I can't guess.

PVAL-inactive

Well its the main thing I doubt the existence of a proof of in my field.

Mike Miller

Got it.

It's fine for discs right?

PVAL-inactive

1:52 AM

?????

Gee rue core e spawn dense

Mike Miller

Oh.

I was thinking compactness...

Feel free to delete.

PVAL-inactive

Nah I'm pretty sure much of the holomorphic curve theory works.

The issue is that people make mistakes juggling the pdes and moduli spaces.

I don't think people think like Salamon & McDuff has any issues.

Mike Miller

Sure. But that only covers the disc case.

PVAL-inactive

Well the disk and the sphere case.

Mike Miller

sure.

PVAL-inactive

1:55 AM

I've never used anything that wasn't a sphere.

Mike Miller

that's all I use personally. but the contact homology people need more...

No exciting papers lately.

PVAL-inactive

Wendl's result is important, but I am pretty sure it only uses (punctured) spheres.

Mike Miller

It's one of those technical papers that can't excite me until I use it.

PVAL-inactive

Wendl's result (that any minimal symplectic filling of a contact structure with a planar open book admits a structure of a PALF)

is used often in contact geometry.

I think the techniques I'm using really avoid it, though it is still just using sphere stuff.

Mike Miller

I believe you.

PVAL-inactive

1:58 AM

I believe Wendl also.

Actually ya they do avoid it and use a much earlier special case of it.

Adeek

2:30 AM

why do we get here $D \leq 0$ ?

arctic tern

2:53 AM

@Adeek if a quadratic is $\ge0$ (which is (**) in your image) then the discriminant must be $\le 0$

(if the discriminant were $>0$ there would be two distinct real roots and the quadratic would dip below the x-axis)

PVAL-inactive

3:31 AM

@TedShifrin (or anyone else) Do you know a physical/natural example of a function which is $g(x)=\int_{a(x)}^{b(x)}f(t)dt$ where $a(x)$ and $b(x)$ are at least somewhat interesting (hopefully still elementary but not $x$ and not constants). A student asked me what we use these functions for (besides the standard exercises involving the FTC), and I was stumped. She was happy with the geometric interpretation of these kinds of functions, but I don't know any which show up like that in nature.

Brody

@TedShifrin From $\mathbf{1.1, 9. (a)}$ in Multi. M., let $\mathbf{u},\mathbf{v}\in\mathbb{R}^2$. Suppose $(\large\star):\mathbf{x}=s\mathbf{u}+t\mathbf{v},\; s+t=1$. So, I had to work algebraically to see $\mathbf{x}=\mathbf{u}+t(\mathbf{v}-\mathbf{u})=\mathbf{u}+s(\mathbf{v}-\mathbf{‌u})$, so the collection of all $\mathbf{x}$ is the line passing through $\mathbf{u}$ and $\mathbf{v}$, with $s,t\ge 0$ yielding the line segment between them.

Good so far? If so, I'm a bit worried that I couldn't tell this directly from $(\large\star)$.

Brody

4:06 AM

Heh, bad typo. *$\mathbf{x}=\mathbf{u}+t(\mathbf{v}-\mathbf{u})=\mathbf{v}+s(\mathbf{u}-\mathbf‌{v})$.

Kaj Hansen

4:36 AM

What was the question @Brody ?

My Multi. M. text is across the room, and I'm comfy :(

Adeek

5:06 AM

ohh I see @arctictern

Brody

@KajHansen lol

The exercise calls to "Describe the vectors $\mathbf{x}=$..." and "Pay particular attention to the location of $\mathbf{x}$ when $s\ge 0$ and when $t\ge 0$." @Kaj.

koolman

5:33 AM

Hii

Can anybody help me in this

0

Q: Altitude in tetrahedron

I was reading a book there , I saw the below question in the book . My doubt is that how they have written length of altitude is same as projection of AD on ABC . Please explain if possible with a diagram http://i.stack.imgur.com/jy8Ln.jpg

I can't understand the answer

Null

5:55 AM

can someone help me?

$V$ is a $F$-vectorspace and $v\in V$

Show that
$\{r\cdot v|r\in F\}$ is a subspace of $V$

i don't know exactly what i m supposed to do

show that it's closed under vectoraddition and scalarmultiplication? but how do i do that if both is not defined?

Kaj Hansen

Clearly it's closed under scalar multiplication @Null

Null

yep

because the set beasicly contains only scalarmultiples or?

i mean, it contains already all multiples of vectors, so it's mood to show that

Kaj Hansen

Consider a generic $r \mathbf{v} \in V$. For any $c \in F$, we have $c(r \mathbf{v}) = cr( \mathbf{v} )$, and $cr \in F$, so we must have $cr( \mathbf{v} ) \in V$ by the definition of the set

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$Mathematics$ Mathematics