Mathematics - 2016-07-26 (page 2 of 4)

Obliv

16:03

@0celo7 oh i did it wrong because $r$ and $\theta$ implicitly depended on each other.

0celo7

what

Obliv

isn't that why it isn't a normal derivative in the form $\frac{d}{dt}(f(x)g(x))$

0celo7

dude, just note that $\mathrm d_t(r^2\dot\theta)=r(2\dot r\dot \theta+r\ddot\theta)=0$.

It works the other way around too

it's in this sense that they are "equivalent"

Balarka Sen

@0celo7 It should have been $r \theta''$ in the parenthesis, not $r^2 \theta''$.

Obliv

hm ok

0celo7

16:05

@BalarkaSen Oh, thanks.

Obliv

but if you take the derivative normally it isn't $r$ but it is $r^2$

0celo7

You're taking the $t$ derivative of $r(t)^2$

Obliv

yes

0celo7

Not the $r$ derivative

Semiclassical

ch-ch-ch-chain rule

Obliv

16:06

which is what i meant by implicit differentiation

Balarka Sen

lol, @SemiC.

Obliv

$r$ depended on $t$ and $\theta$ depended on $t$

0celo7

You get $$\mathrm d_t r^2=\mathrm d_r r^2\cdot \mathrm d_tr =2r\dot r$$

@Semiclassical reminds me of

uh

lemme find it

Obliv

i don't like newton notation. I would prefer $\frac{dr}{dt}$ instead of $\dot{r}$ oh well beggars cant be choosers

0celo7

@Semiclassical forget it

@Obliv what?

Balarka Sen

16:08

you're probably thinking of the song "changes" by david bowie.

0celo7

No, I'm thinking of the song "For Free" by DJ Khaled.

Balarka Sen

meh, ok

Semiclassical

i was thinking of bowie, yeah

0celo7

D-D-D-D DJ Khaled!

Semiclassical

@Obliv if your book uses it, you should probably get comfortable with it :)

0celo7

16:10

Best notation for derivatives is $\partial_i$

Semiclassical

for partials, anyways

0celo7

$\mathrm d_x$ works for regulars

Semiclassical

sometimes you need to do a total time derivative, for instance

0celo7

quicker to type than the full one

@Semiclassical $\mathrm d_t$.

Semiclassical

I don't like lowercase for that, since it seems too close to the notation for differential forms

but $D_x$ is fine to me

0celo7

16:12

$D$ is the Jacobian

Semiclassical

depends on which notation you're using, I guess. I haven't seen that.

Balarka Sen

Jacobian is not distinct from derivative...

0celo7

You haven't seen $D$ for the Jacobian?

Balarka Sen

that's what it is in one dimension

Semiclassical

Nope.

Balarka Sen

16:13

Weird, I thought $D$ was common.

0celo7

@BalarkaSen The Jacobian is not a total derivative, though

Balarka Sen

Dunno what that means.

Semiclassical

The notation I remember for Jacobian is the (admittedly clunky) $\dfrac{\partial(x,y)}{\partial(u,v)}$ kind

0celo7

@Semiclassical oh god

Balarka Sen

@Semiclassical That's the matrix of the Jacobian.

0celo7

16:14

that's the calc 3 Jacobian for a coordinate transformation

Balarka Sen

When you write it as a linear transformation on the tangent spaces, you use $Df$.

Semiclassical

meh, matrix versus determinant.

ah.

yeah, if it's a differential geometry thing I don't know it

0celo7

@BalarkaSen "tangent space" is a bit overkill in the analysis context

@Semiclassical No, it's an analysis thing.

Balarka Sen

Everything's an overkill for you.

0celo7

@BalarkaSen The Jacobian is defined on Banach spaces, not manifolds

Semiclassical

16:16

By total time derivative, I mean like $$\dfrac{d}{dt}=\dfrac{\partial}{\partial t}+\dfrac{dx}{dt}\dfrac{\partial}{\partial x}$$

0celo7

@Semiclassical I wonder if that's a physics thing

Semiclassical

which does appear in applications

a fluid mechanics thing, I suppose

user 1618033

I think there is a problem here, I cannot see my icon on the right pannel.

0celo7

No, it shows up in Hamiltonian mechanics

@user1618033 F5.

user 1618033

And again, I cannot edit the messages.

Semiclassical

16:17

wikipedia uses the phrase 'material derivative'

@0celo7 you're right, I'd forgotten that

user 1618033

@0celo7 Thanks, but of no help.

0celo7

@Semiclassical I've heard that's a special covariant derivative, actually

Semiclassical

Probably.

user 1618033

Damn ...

0celo7

Maybe for some weird Riemannian metric

@Semiclassical Also in QM, Ehrenfest's theorem.

Semiclassical

16:20

aye, which can be viewed as the quantization of the version in Hamiltonian mechanics

0celo7

Yes

incidentally

Do you have any idea what this means?

Semiclassical

I think it means that the LHS is literally the average over the direction of $p$

0celo7

What does that mean?

Balarka Sen

They're taking an expected value.

Semiclassical

You're treating $d\Omega$ as a probability density, basically

0celo7

16:23

Shouldn't it be a $\sin\theta$ then?

Balarka Sen

I do not understand the relevance of $d\theta d\varphi$, then. What are $\theta$ and $\varphi$?

Semiclassical

...huh.

it should, yes. only reason it would be $\cos\theta$ is if they're doing a different convention than usual

e.g. $\theta=0$ being the equator rather than the north pole of the sphere.

0celo7

yeah

@BalarkaSen Angular coordinates on the unit sphere

Semiclassical

Do they declare their conventions for spherical coordinates anywhere?

If not, I fart in their general direction.

I mean, that the expectation value is of that form is pretty easy to see on symmetry grounds

0celo7

Semiclassical

16:27

that's the standard version, isn't it?

0celo7

yeah

so maybe it's a typo

Semiclassical

okay

yeah. if they've been doing it with the usual convention up to now, definitely a typo.

0celo7

why should symmetry arguments work

Obliv

I don't get what Zee is trying to get at p.31 @0celo7 with a mass pulled by a spring, $F(x)$ does not make
sense, only $F(x(t))$ does. The force exerted by the spring does not pervade all of space, and
hence is defined only at the position of the particle $x(t)$, not at any old $x$.

0celo7

the solution says to note it must be proportional to $\delta^{ij}$

I don't understand that

Semiclassical

16:28

well, suppose you pick $i\neq j$. then the RHS is declaring that the integral vanishes

0celo7

And I'd prefer a direct argument

@Semiclassical We need to prove the RHS is equal to the LHS.

The LHS is "given"

Obliv

couldn't a mass on a spring be a function of only position? Time needn't be involved imo

Semiclassical

on symmetry grounds, the point is just to consider what happens if you do a coordinate change which swaps $i$ and $j$

0celo7

what

@Obliv uh

Obliv

I have encountered plenty of students who confuse these two basic concepts:
spatial coordinates and the location of particles. I may sound awfully pedantic, but when we
get to curved spacetime, it is often important to be clear that certain quantities are defined
only on so-called geodesic curves, while others are defined everywhere in spacetime.

Semiclassical

16:31

Trying to remember the details, tbh.

0celo7

oh, you're only defining the force along the path of the particle

on on all of space

the path is $x(t)$

Obliv

what the heck is the difference between the location of particles and spatial coordinates

deostroll

Hi. Just noticed something about divergent series...

Semiclassical

@0celo7 Here's a brute-force approach: Suppose you can prove that the LHS vanishes when $i=x$ and $j=y$

0celo7

$\Bbb R^2$ vs. the curve the particle travels on

Semiclassical

16:33

That's a simple direct calculation.

0celo7

...how

deostroll

If you subtract the series by itself...you can actually have that sum equal any real number...

0celo7

$\int p^xp^y\,\mathrm d\Omega=?$

How does that even work

Semiclassical

well, I'm figuring $p^x$ here is just the $x$-coordinate of $\vec{p}$?

0celo7

Yeah, so?

Is that a number, a function?

Semiclassical

16:35

in that case, you can write it in spherical coordinates

it's parametrized by spherical coordinates. how else would it show up in an integrand?

deostroll

or is it just something trivial... :(

Vrouvrou

Hello, please if $F(t)>0,\forall t$ can we say tha $\int_{\mathbb{R}^N}F(u(s)) ds>0$ ?

Semiclassical

and in spherical coordinates, it should just be $\cos\phi\sin\theta$ (if I'm not mixing up my conventions)

0celo7

@Vrouvrou Yes

Vrouvrou

thank you

Semiclassical

16:36

similarly, $\vec{p}^y$ would just be $\sin\phi \sin\theta$

so you'd just plug those expressions into the integrand and show that the resulting integral vanishes.

0celo7

I see

What the heck is $\vec p$ again

I'm just not sure what "averaging over direction" means

Obliv

@deostroll If you subtract $a$ from $a$ you get nothing, by definition, no? How could subtracting a divergent series from itself give you a real number $\ne 0$

deostroll

@Obliv its not a point-to-point or term-to-term subtraction...

The general process is, you add until the partial sum is greater than target, if it over shoots, you take one minus term, if this falls below the target, continue adding the positive terms, and likewise alternate...

Obliv

@deostroll if you mean to say $[\sum_{i=0}^{\infty} i] - [\sum_{i+4}^{\infty}i] = 1 + 2 + 3 + 4$ I think that is trivial.

deostroll

oh wait...not that divergent series...

harmonic series...or ones similar to that...

Semiclassical

16:44

@0celo7 the best example in physics I know off the top of my head would be if you were to give a microscopic description of a gas

deostroll

@Obliv that is not quite the process I previously mentioned...

Semiclassical

In that case, you could consider questions like: If I repeatedly select a particle and measure its momentum $\vec{p}$, what is the expected value of $p_x$?

deostroll

YOu've to keep alternating between the positive and negative terms so that the partial sum is near about the target value...

Semiclassical

In the case of an ideal gas, you get the so-called Boltzmann distribution which is, crucially, isotropic in space. That is: While certain speeds of particles will appear more frequently than others, there's no direction in space that the particles prefer to travel.

In that case, a question like 'what's the expected x-component of momentum' should simply be zero because the particle is as likely to be moving left as right

Or, more symbolically, $$\langle \vec{p}^x \rangle_\Omega =\frac{1}{4\pi}\int d\Omega\,\vec{p}^x = 0$$ where the average is taken over all possible directions of $\vec{p}$

robjohn

$$
\begin{align}
\sum_{k=1}^\infty\frac1k-2\sum_{k=1}^\infty\frac1{2k}
&=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots\\
&\phantom{=1\,\,}-\frac22\phantom{+\frac13\,\,}-\frac24\phantom{+\frac15\,\,\,}-\frac26-\cdots\\
&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\\[12pt]
&=\log(2)
\end{align}
$$

The problem is in the rearrangement of the terms. In general, if the series is not absolutely convergent, we must be very careful about rearranging the terms of the series.

Semiclassical

16:53

the 1/4\pi is to ensure $\langle 1 \rangle_\Omega = 1$.

But one might be interested in things like $\langle \vec{p}^x \vec{p}^y\rangle_\Omega$, and part of what you're trying to show is that this in general vanishes as well

and now I'm suddenly having flashbacks to scattering cross sections. ew

Mary Star

Hello!!

Balarka Sen

Hi @Danu

Obliv

@0celo7 hey.. uh how do you take this integral $\int_{-\infty}^{\infty} \delta(x)f(x)dx$? I know the delta function is $0$ for all $x \ne 0$ and it approaches $\infty$ at $x = 0$ but I don't really get how to take this integral. hint?

fixed

Huy

@Obliv: it's f(0)

Semiclassical

@obliv What you've cited is a description of the delta function, not the definition

Obliv

17:00

yeah @huy I'm trying to show that.

Semiclassical

Do you know the definition?

Obliv

@semiC I don't know the definition. I was given a description after all

Semiclassical

Then I have no idea how you're supposed to solve the problem.

Huy

^

Semiclassical

Do you have a problem statement in mind?

Balarka Sen

17:01

Again, this goes back to the earlier discussion of trying to prove something about an undefined object. Unless you don't know the definition, it's hopeless to do mathematics.

Obliv

Show that for some suitably smooth function $f(x)$, the integral (mentioned above) is $f(0)$

Huy

@BalarkaSen: but one can do physics, in that case. :P

Balarka Sen

(FWIW, I don't know delta function business)

Obliv

@balarka It's not a math book.

Huy

of course it's not :D

Semiclassical

17:02

Part of the problem is that the description you cited is simply not enough to characterize the delta-function

For one, it equally well describes $\delta(x)$ and $2\delta(x)$.

Balarka Sen

@Obliv Makes sense, all in all.

Obliv

after ~3 months learning algebra in a math book I can safely distinguish a math book from a physics book

deostroll

@robjohn Yes, what I am saying that is by careful rearrangement you can get any rational number...

Mary Star

I am looking at an exercise about the vertex cover. We are given the graph $G=(V,E)$ with $V=[100]$ and $E=\{(i, i+1)\mid i=1, \dots , 99\}$.
Before I use the approximation algorithm, I have to give the minimal vertex cover $C^\star$ of $G$ and the length of $C^\star$.
How do we find the minimal vertex cover without the approximation algorithm? The vertex cover has to contain at least one vertex of each edge, right? At the given graph, each vertex is connected with the consecutive one, so do we maybe take for example the first one, then the third one, etc?

Semiclassical

For another, one can find a sequence of functions which converge pointwise to zero away from zero and to infinity at zero, but for which the corresponding integral would fail to converge.

Balarka Sen

17:04

I thought an essentially part of $\delta$ was that it integrated to $1$ over the real line, whatever that means?

robjohn

@deostroll yes, I was just giving a concrete example. Actually, you can get any number, rational or irrational.

deostroll

$\sum\limits_{k=1}^\infty \frac{1}{k}- \sum\limits_{k=1}^\infty \frac{1}{k}= c$

Semiclassical

@BalarkaSen Yeah. That's not in the "zero if x!=0, infinity at zero" description, though

Huy

@BalarkaSen: that means that the area enclosed by the graph of the function and the $x$-axis equals 1 :P

Balarka Sen

@Huy That doesn't make sense in this particular context without more formalism.

Huy

17:06

@BalarkaSen: I know, that's the joke.

Balarka Sen

Oh. Ok.

Semiclassical

@Obliv Our point in all of this is that if one doesn't have some kind've definition here, even if a purely operational one (i.e. a delta-function is an object with X properties) then the problem is simply not solvable.

And the description you gave doesn't rise to that.

0celo7

@Semiclassical sorry, was getting lunch

@Obliv easy

Semiclassical

So unless there's some kind of dfn in your book (or if it refers to another book for the definition) then that problem is pretty silly.

0celo7

the delta kills $f(x)$ at every value besides $0$

So under the integral, $f$ might as well be the constant $f(0)$

Obliv

17:08

I know that but mathematically how do you take the integral

Semiclassical

...

Huy

dat physics proof

0celo7

pull that out, then $\int f(x)\delta(x)=\int f(0)\delta(x)=f(0)\int\delta(x)=f(0)$

Balarka Sen

@0celo7 And at $0$, you have $f(0) \cdot \infty$

Wonderful.

0celo7

@BalarkaSen Physics.

Obliv

17:09

lol..

0celo7

I've read the book, I know how the author wants @Obliv to do it.

@Obliv The rigorous proof uses the Lebesgue majorant theorem, IIRC.

Not fun.

Mary Star

@DanielFischer are you familiar with the vertex cover?

Obliv

What about what balarka said, though @0celo7

at $x = 0$ not only is there a constant $f(0)$ but $\delta(0)$ which is the maximum of $\delta(x)$ function

Semiclassical

Eh. I'm fine with not having a perfectly rigorous proof. But you need to have some source for $\int \delta(x)\,dx=1$.

0celo7

Notice that $f(x)\delta(x)=f(0)\delta(x)$

@Semiclassical He constructed the delta as a growing spike with area 1

Semiclassical

17:12

If that's what's in the book, that's fine. But that's not what was given earlier.

0celo7

what was given?

I have the book memorized, I know how it's defined

Semiclassical

14 mins ago, by Obliv

@0celo7 hey.. uh how do you take this integral $\int_{-\infty}^{\infty} \delta(x)f(x)dx$? I know the delta function is $0$ for all $x \ne 0$ and it approaches $\infty$ at $x = 0$ but I don't really get how to take this integral. hint?

Obliv

I don't recall the area being constructed as $1$. He said Because we
included a multiplicative constant $\omega$, we could always normalize $\delta(t)$ by $\int \delta(t)dt = 1$

for a mass on a spring problem

Balarka Sen

@0celo7 Tell me what they say in the second paragraph of page 34.

0celo7

@Obliv what do you think $\int\delta=1$ means if not "area is one"

@BalarkaSen what

Semiclassical

17:14

I should also note that, for me, the above quote was the entire context. I don't know what book is being used.

Balarka Sen

You said you have the book memorized

0celo7

Oh, it should be something about forces

Obliv

the context was of a different problem with different constants. How am I supposed to know he means $\int \delta = 1$ just by that

Semiclassical

Hence my harping on 'how does the book define $\delta$'

0celo7

Not word for word

But I know how things are defined

Semiclassical

17:15

Which book is this, to be clear?

0celo7

Zee, Einstein Gravity

Where is the integral of $f\delta$ needed

Semiclassical

@Obliv glancing at google books, the point is that he's defining $\delta(x)$ there for further use

Obliv

@0celo7 what

exercise 1

Semiclassical

note that in the paragraph below (3), he specifically mentions that you could construct it by using a piecewise linear function which vanishes except on $[-1/\tau,1/\tau]$ and whose max is $1/\tau$

0celo7

Yes, just use that $f(x)\delta(x)=f(0)\delta(x)$

Semiclassical

17:18

which you can check has an area of 1.

0celo7

then integrate

Obliv

i'm on the 2nd part now though. If what you said is true, then I suppose pulling out $\delta(0) = 1$ and $f(0) = c$ of the integral makes sense

Semiclassical

$\delta(0)\neq 1$

Obliv

:(

Semiclassical

Maybe try focusing on the case of a function $f(x)$ which vanishes at zero first.

Obliv

17:20

@semiC I know it isn't rigorous, but $\delta(x)$ is extremely steep and for the interval $\frac{-1}{\tau},\frac{1}{\tau}$ the area exists mostly at $x = 0$

Semiclassical

sure. but in that case, the peak value of $\delta_\tau(x)$ would be $\delta_\tau(0)=1/\tau$ which diverges as $\tau\to 0$

Try taking $f(x)=|x|$ and imagine what happens to $\delta_\tau(x)f(x)$ as $\tau\to 0$.

Obliv

oh the interval was $-\tau,\tau$ not the inverse, btw.

Semiclassical

ah, yeah.

Obliv

the maximum is given by $\frac{1}{\tau}$ still

Semiclassical

careless of me

Obliv

17:25

$\lim_{\tau \to 0}\frac{1}{\tau}\cdot f(\tau)$

Semiclassical

@0celo7 No. We're doing a triangular peak, not a square one.

Balarka Sen

length of [-t, t] is 2t. 1/2*(2t)*1/t = 1.

Obliv

lol..

Semiclassical

moving right along.

Obliv

@0celo7 can do morse theory but struggles with finding the area of a triangle. :^)

0celo7

17:27

I struggle with Morse theory too

Semiclassical

Anyways. If you plot $\delta_\tau(x)f(x)$ with $f(x)=|x|$, you'll find that by construction it vanishes at $x=\pm\tau$ but also at $x=0$

Obliv

because of the limit i described above right@semiC

Semiclassical

not sure what you were getting at with that limit

Obliv

but isn't it dependent on whether $f(\tau)$ approaches $0$ faster than $\frac{1}{\tau}$ approaches $\infty$?

Semiclassical

$f(\tau)$ is the value at $x=\tau$, $1/\tau$ is the value at 0.

0celo7

17:29

What's wrong with $f(x)\delta(x)=f(0)\delta(x)$

Semiclassical

@0celo7 I'm trying to show how to derive that :P

0celo7

oh, carry on

Obliv

what do you mean @semiC

0celo7

Semiclassical

Since thats equivalent to $f(x)\delta(x) = (f(x)-f(0))\delta(x)+f(0)\delta(x)$ and showing that the first term always vanishes under integration

Obliv

17:31

i was mixing variables on accident

why does that vanish?

Semiclassical

that's what I'm trying to show :)

Obliv

$\delta(x)$ exists in the interval $-\tau,\tau$ and $f(x) - f(0)$ exists for all $x$ that it is defined on

Semiclassical

Just to make my strategy clearer, if you define $g(x)=f(x)-f(0)$ you've got a function which explicity vanishes at zero

Obliv

oh okay

Semiclassical

and then the point is to show that $g(x)\delta(x)$ should be zero

Obliv

17:34

so at $x = 0$ $f(x)\delta(x) = f(0)\delta(x)$ but isn't $\delta(x)$ and $f(x)$ also defined on $-\tau$ and $\tau$ (and any real numbers in between)

Semiclassical

Part of the thing here is that we're using $\delta(x)$ like it's a function.

And strictly speaking it's not

Obliv

yeah in the book it said it's a limit of a sequence of functions

Semiclassical

The point isn't so much whether $g(x)\delta(x)$ equals something but what happens when you integrate it from -\infty to \infty

Yeah

Consider it like this for the example I suggested. What's the max value of $g(x)\delta_\tau(x)$ in that case?

Obliv

as @0celo7 pointed out earlier, $f(x)\delta(x)$ only exists on the interval $\delta(x) \ne 0$

$g(0)\delta_{\tau}(0)$

Semiclassical

No, it is certainly not.

By assumption, $g(0)=0$.

Obliv

17:38

oh we're talking about $g(x) = f(x) - f(0)$

Semiclassical

Right. And I'm wanting to consider the specific case of $g(x)=|x|$.

Obliv

I would say it depends on what kind of function $f(x)$ is

oh okay

then the maximum is at $\frac{\tau}{2}$?

and $-\frac{\tau}{2}$

0celo7

This is the least mathematical discussion we've had here in weeks

Semiclassical

Yeah. Which gives a max value of... $\tau/2$, I think?

@0celo7 Pretty sure your frrustrations with equipment were less mathematical than this

Obliv

@0celo7 want me to ask about lattices of subgroups of groups instead?

0celo7

17:40

@Semiclassical Those were nonmathematical

This is antimathematical

Obliv

physics is antimath

Semiclassical

pfft

Obliv

but this isnt even physics

0celo7

It's a physicist """proof"""

Semiclassical

Anyways. The point is that $|x|\delta_\tau(x)$ is bounded above by $\tau/2$, bounded below by 0, and vanishes outside of $[-\tau,\tau]$.

So what happens as $\tau\to 0$?

0celo7

17:42

pretty sure you need uniform convergence to take limits like that

Obliv

mhm

uh

evaluates to $0$

Semiclassical

it goes to zero.

it vanishes outside of $[-\tau,\tau]$, and inside its squeezed to being zero.

Obliv

for $g(x) = |x|$ though, no?

Semiclassical

right

you can show it works more generally for $g(x)$ with $g(0)=0$

(Probably one needs some conditions on $g(x)$, but being continuous and finite should do it)

If that works for any $g(x)$ vanishing at zero, though, then for any $f(x)$ one can write $g(x)=f(x)-f(0)$ and therefore $f(x)\delta(x)=f(0)\delta(x)$.

Samuel Yusim

@Balarka arxiv.org/pdf/0807.0336v2.pdf looks like $n$-simplex embedding in $\mathbb{R}^n$ isn't characterizable in higher dimensions

0celo7

17:48

@Semiclassical Oh, that's a nice way of seeing it

Obliv

The gist of this is, $\int_{-\infty}^{\infty} \delta(x)f(x)dx = f(0)$ because as $x \to 0$ $\delta(x)$ approaches $\infty$ and $f(x)$ approaches $f(0)$. In the case of $f(x) = |x|$ $f(0)\delta(0)$ gets squeezed to $0$. Is this correct so far @semiC

Balarka Sen

$n$-dimensional simplicial complexes hardly ever embed in $\Bbb R^n$.

Did you mean to say $m$ instead of $n$?

Samuel Yusim

$\mathbb{R}^{n+1}$

Semiclassical

I don't like the first sentence, because nowhere in there have you specifically used $\int \delta(x)\,dx=1$.

Balarka Sen

Hmm, OK.

Semiclassical

17:49

Moreover, if you're going to talk about squeezing then you should be referencing $\delta_\tau(x)$ not $\delta(x)$.

Obliv

well that's assumed I think

what is $\delta_{\tau}(x)$ isn't that the same function

Semiclassical

No. It behaves the same in the limit $\tau\to 0$.

Balarka Sen

@SamuelYusin That's weird.

Semiclassical

But the nascent delta function and the true delta function are not the same. For one, the second isn't even a function.

@0celo7 Glad you like it, heh

Obliv

oh okay.

Samuel Yusim

17:52

it's really disappointing honestly

Obliv

the book defines $\delta_{\tau}(x)$ then, not $\delta(x)$?@semiC

Balarka Sen

Why so? It just means you should restrict to some special subclass of simplicial complexes and try that same problem.

Semiclassical

What's plotted is a particular member of a sequence of functions converging to delta(x)$, yes.

I say "a sequence" because that's not the only one that would work.

A sequence of narrower and narrower gaussians would also work.

Obliv

Isn't the one defined as $\tau \to 0 ~~\delta_{\tau}(x)$ simpler for this exercise?

0celo7

@Semiclassical My proof is that it's "obvious" :P

Semiclassical

17:54

Probably. I'm just saying :P

@0celo7 Now who's being a physicist? :P

0celo7

I'm an engineer

Obliv

you're a student

Balarka Sen

I don't know what they mean by Novikov's theorem on the 5-sphere.

Googling.

Semiclassical

I don't know how to prove $g(x)\delta_\tau(x)\to 0$ as $\tau\to 0$ for generic $g(x)$, tbh

Obliv

I don't think there is a proof for that

Samuel Yusim

17:57

@BalarkaSen the problem is that the subclass will inevitably be matroids

Semiclassical

Oh, I'm sure there is.

Otherwise people wouldn't use it in generality.

0celo7

@Semiclassical uh, what

do you mean for $x\ne 0$

Balarka Sen

@SamuelYusim I am unfamiliar with matroids. What are those?

Semiclassical

I meant under the assumption of $g(0)=0$, woops

Obliv

@semiC if $g(x) = \tau$ then $g(0)\delta_{\tau}(0) ~~\tau \to 0$ evaluates to $1$

oh ok

Semiclassical

17:59

...sure, but then you're doing a sequence of $g(x)$ as well

which you wouldn't do.

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$Mathematics$ Mathematics