Mathematics

These are the kind of things Galois theory proves for you.

Akiva Weinberger

12:27 PM

"It's only a theory" - my dad, all the time

Well he's right.

Abcd

Can someone check this MCQ answer of mine:

If y = f(x) = mx +c then f(y) in terms of x is:

My answer: m^2x + mc + c

Is it correct?

Anonymous

Looks fine

Abcd

thanks.

Abcd

12:58 PM

Can someone tell me how to draw the diagram for this question:

A vertical tower PQ subtends equal angle of 30 at each of the two places A and B, 60 meter apart. If AB subtends andangle of 60 at P (the foot of the tower), then the height of the tower is.

Akiva Weinberger

1:37 PM

$\text{Noetherian$[x]={}$Noetherian}$?

SteamyRoot

Yes

By induction, also for $[X_1, \dots, X_n]$

And even for $[[X]]$ etc

Alessandro Codenotti

That's known as the Hilbert basis theorem if you want a googlable name

I have a proof typed up somewhere in the SE chat world

SteamyRoot

For power series, you can adapt the Hilbert basis theorem. Since you can't work with the highest power term, you have to work with the lowest instead

R[[X]] is a DVR though, right?

I think DVR's are all Noetherian

1:52 PM

Save your documents guys

My computer just crashed on a photoshop file I had spent two hours on :(

Alessandro Codenotti

DVR?

discrete valuation ring

SteamyRoot

@BalarkaSen Only over a field, I think

huh really

Alessandro Codenotti

Just looked up the definition, DVR definitely implies noetherian (PID is enough), no idea whether R[x] is a DVR in general or not

1:56 PM

I guess you're right

Danu

o/

Hi

Danu

@TedShifrin So about that hyperquadric... Does the restriction of the Fubini-Study metric induce a Kahler-Einstein metric? I suspect yes because I saw something about homogeneous quadrics in the paper "Homogeneous Kahler submanifolds in complex projective spaces" (by Takeuchi) but I was wonder if maybe you just happen to know this.

Hi Danu

Danu

Hi Balarka and Astyx

nbro

2:18 PM

Does anyone here work or has ever seriously worked in computational science or with numerical algorithms/computations?

Like, no one wants to pursue a master's degree in computational science, but that would really be to have guts

Albert Einstein

Hello

Hi Ted

2:41 PM

hey @PaulPlummer

is nabla dot a vector the same as a vector dot nabla?

so does both denote the divergence or is there a difference?

No, they're quite different. @JannikPitt

(fyi, you can turn on mathjax in chat using the link in the room description)

for the first one, we have the divergence $$\nabla\cdot \vec{A}=\partial_x A_x+\partial_y A_y+\partial_z A_z$$

That's a scalar result.

For the second, though, we have $$\vec{A}\cdot \nabla = A_x\partial_x+A_y \partial_y+A_z\partial_z$$

That's a differential operator.

yes I hope it is now ready to take a vector again right @Semiclassical the second case ?

Actually, you'd typically apply that to a scalar.

ohk

2:50 PM

e.g. $(\vec{A}\cdot \nabla)F=\vec{A}\cdot (\nabla F)$ etc.

where $F$ is scalar ! , yes.

Right.

Basically it's a directional derivative where the relevant direction can vary in space.

You can also apply it to a vector can't you ?

You can, sure, but you'd do so component-by-component.

As in $\vec F = \vec p \cdot \vec{\text{grad}}\vec E_{ext}$

2:53 PM

Right.

Is that the force on a dipole in an electric field?

Yup

Neat.

(assuming that the dipole stays the same throughout space etc.)

What do you mean ?

Well, the energy of a dipole is $U=-\vec{p}\cdot \vec{E}$.

Oh right, yeah

2:56 PM

So from that the force should be $\vec{F}=-\nabla U = \nabla(\vec{p}\cdot\vec{E})$.

But if $\vec{p}$ doesn't depend on where you are in space then the other form works.

I don't quite get how it could depend on where you are though

Suppose you've got a linear dielectric material.

Then the polarization will depend on the local electric field.

@Semiclassical Thanks for the explanation!

Paul Plummer

@BalarkaSen Hey

Fair enough

2:58 PM

I'm sure this shows up in actual problems, but meh

I haven't done those calculations in years.

I like the answer of Sangchul Lee here, but I feel like there should be a more elegant approach: math.stackexchange.com/questions/2264587/…

$a^Tx$ is to be interpreted as $a^Tx I$ right ?

Yeah, I guess it would have to be.

3:15 PM

Would you recommend using the Nabla operator or just write down the operation? So should I use $\nabla \cdot F$ or just $div F$?

I'm self-studying all of this so I don't really know which is more commonly used

It really depends on what sources you're using.

I like the \nabla operator notation instead of the div/grad/curl names, but that's not universal.

does this $\sum_{n=1}^{\infty}(-1)^n$ converges!

No.

quickly I cant find the counter or proof!

If the terms don't go to zero, a series can't converge.

arctic tern

3:18 PM

quickly?

like going to sleep without answer or comment !

You can also show it has 2 limit points, so it cannot converge

ok thanks@Semiclassical

thanks

Well the sum would oscillate because $(-1)^n$ is equal to $1$ if $n$ is even and $-1$ if $n$ is uneven. That goes on to infinity so the sum clearly can't converge.

Bye!

3:19 PM

Also, partial sums are -1,0,-1,0,-1,0...

which definitely doesn't converge.

(It does have a Cesaro sum, but that doesn't mean it converges in the usual sense.)

Okay so I'll probably use the Nabla notation because I find it much better to understand :D

(I prefer grad/div/curl for my part)

In higher level physics stuff you'll also find direct use of index notation in conjunction with the Einstein summation convention.

Mathematicians turn up their nose at that, though.

So mathematicians would be more inclined to use $div/curl/grad$?

Nah, I meant that they turn up their nose at index notation.

I have no idea if mathematicians are more inclined to use div/grad/curl.

3:24 PM

Yeah I've seen the index notation but I'm not at the point in self-studying physics that I need it just yet

That's fair.

Are you a physicist?

Physics grad student, yeah.

In what do you specialise? I want to study physics too when I finish school

Condensed matter theory is where I sit, but I've done a number of topics in that realm

Would not want to try to sum them all up, though.

3:32 PM

Haha. Sound's interesting!

Secret

[Random self coined terminology] A set exists in concept space if there exists a map such that all its elements can be mapped to eulidean or noneuclidean space while preserving all the properties of the set

For example:

The set $x=y$ is mapped to a line 45 deg in the xy plane

A sculpture that represent a concept has the notion of taking some kind of map on the set of all properties of the given concept, and then map to the space of all possible sculptures given any properties

rinse and repeat, nearly every concept "look like" at least points in parameter space, hence exists in concept space

Felix.C

Hey I have a list of values: min, max, average and standard deviation, unfortunately I didn't have statistics at university yet, and the definition of the standard deviation doesn't seem that easy at first glance, actually I just need a feeling for those values, my standard deviation is between 0.024 and 0.88, is there a rule of thumb to get a feeling how narrow or wide this graph:
https://de.wikipedia.org/wiki/Standardabweichung_(Wahrscheinlichkeitstheorie)#/media/File:Standard_deviation_diagram.svg

Eric Stucky

o.O the difference between std of 0.024 and 0.88 is rather noticeable, felix.

but in any case, if your distribution is vaguely normal, then you expect vaguely 95% of the data to be within avg-2*std --- avg+2*std

Daminark

3:48 PM

Hallo

Felix.C

Ok, thank that helps a lot! If I have a normal distribution how many percent would lay into avg+- std? I guess that would help me enough to get some info out of those values

Eric Stucky

~68%

the image that you linked shows this

Felix.C

Ok, thanks!

Eric Stucky

Secret, I'm about 98% sure you're rediscovering concrete categories.

Daminark

mutters about the categorical imperative