Mathematics - 2017-07-08 (page 1 of 6)

Secret

00:04

Countable union of countable set is countable, whether finite or countably infinite. However, by the axiom of power set on a countable set, there are sets that cannot be proven using ZF to be equal or larger than $\aleph_0$, so ...

Tanner Swett

I found this related question: math.stackexchange.com/questions/249323/…

According to those answers, ZF doesn't prove that the countable union of countable sets is countable.

I don't think I see an answer to my original question there, though.

Secret

I am guessing if something is unprovable in a given theory is not necessary the same as there is a contradiction

But otherwise I am not sure of the details

Tanner Swett

Well, something is unprovable if and only if its negation is not a contradiction.

Secret

In that case, one negation of the above will be "there exists an uncountable set which is a countable union of countable sets". That might be addressed somewhat in the MSE link in the affirmative.

Anyway need to ark for some hours. Alessandro and other set theorist might be able to continue this discussion

Afk*

Fawad

01:28

Hi

What is $[0.999\cdots]$=?

$[.]$ is greatest integer function

Fargle

@Fawad Since $0.999 \dots = 1$, the function evaluates to 1.

Fawad

01:44

@Fargle I was thinking of mathematical induction. [0.999...n times] for n=1 [0.9]=0

Fargle

@Fawad Induction only proves that $[0.9...9] = 1$ for every natural number of 9s. $0.999...$ has infinitely many 9s, and infinitely many isn't a natural number.

Fawad

Thanks.

Fargle

No problemo.

Typhon

04:13

0

Q: For all ordinary differential equations, does there exist a corresponding integral equation?

Once or twice I have heard the term thrown around of a corresponding integral equation. Therefore, I ask if the following conjectures are true. For all differential equations $D$ there exists an integral equation $I$, such that $Sol(D) \cap Sol(I) = Sol(D)$. Note that $Sol(x)$ is the functi...

i got an answer here that makes very little sense. Am I missing something or does that answer not address the question?

Semiclassical

Well, you did indicate: "I think I can prove it for the trivial case of y′=f(x), but I don't know how to progress with the larger case." The question does address that special case.

That said...ehhh.

Typhon

yeeeeaaaah

if they said induction for that matter

then I could see it maybe

Semiclassical

For linear ODEs, at least, there is one fairly concrete idea.

Typhon

but even then that rejects nonlinear equations

Semiclassical

Yeah.

Typhon

04:21

yy' = x doesn't work under that method of proof

in fact, feel free to comment regarding that

Semiclassical

With linear ODEs, one can recast an nth-order equation as a first-order system of ODEs

And then one has $Y'(x)=F(x)Y(x)$, which at least in principle has an integral representation.

Typhon

fair enough

Semiclassical

For a nonlinear equations, though, that idea fails immediately.

Typhon

but that doesnt allow nonlinear

Semiclassical

For nonlinear things are a lot murkier.

Typhon

04:22

which is why i asked if there exists integral equations

and then i asked if there was a minimum integral equation solution set with a unique integral equation

i.e. if whether there is one with 10 extra solutions and no others 10 or lower

the question makes sense, right?

i was worried about the logic of the second thingy. Took me an hour to write out.

why is the answer getting upvotes?

that is infuriating

errr... annoying

Eric Silva

@Semi any n-th order equation can be made into a first order system

you just dont get the nice linear form

Semiclassical

Hmm, yeah.

I just wouldn't count that as terribly useful.

@Typhon I'm not sure whether the logic of the second thing actually holds up in practice, but I can see what you're going for.

What I'd love to see some 'canonical' example of a nonlinear ODE whose solutions can't be represented as integrals.

Typhon

yeah that would be interesting to see

Semiclassical

I think stuff re: the Painleve equation may be an example of that.

Typhon

do you know of the implied derivative/antiderivative?

Semiclassical

04:27

Not sure what you mean by that.

Typhon

let me find a link

Semiclassical

Here's the link to Painleve: en.wikipedia.org/wiki/Painlev%C3%A9_transcendents

Typhon

7

Q: Are all solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

Now I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me. Let us define a di...

differential-equations limits derivatives differential-algebra

3

Q: Are all weak solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

Yes,this is very similar to a previous question I asked. That was about normal solutions and not weak solutions. Let us define a differential operator known as implied differentiation. It actually does not have a unique value. Let us denote the implied derivative operator as $I(f)$, where $f$ is...

calculus differential-equations floor-function weak-derivatives piecewise-continuity

Semiclassical

(I've actually got a book in my room on Painleve transcendents which I've long given up trying to read.)

Eric Silva

Doesn't the usual procedure for a turning a first order IVP into an integral just work always

convert n-th order equation to first order system then integrate

it's not "practical" but it's an integral equation that is induced by the ivp

Semiclassical

04:30

This is where my expertise falls short, I'll confess.

Eric Silva

im pretty sure it just works

this is how you prove picard lindelof

the step where you convert to the integral equation doesn't make any assumptions on what your system looks like

Semiclassical

Hmm.

Eric Silva

At least that i recall

Semiclassical

Practical example, then. The Painleve I equation is given by $y''=6y^2+x$

This can be turned into the first-order system $v'=6y^2+x$ && $y'=v$. How does one proceed from there? @EricSilva

Typhon

@Semiclassical the implied derivative is just name I have for a multivariate operator that essentially treats step functions as constants whilst preserving the general properties of differentiation. My conjecture is that the continuous solutions to the corresponding implied differential equation are the solutions to the corresponding integral equation as I have defined it.

see the links for the full definition

i am too lazy to write it out here

:p

Semiclassical

04:34

That's a bit much for me to chew on at this time of night, tbh

Typhon

@Semiclassical fair enough. I am intending to one day try my hand at constructing something similar but for periodic functions. All my attempts have so far failed.

likely due to there being no simple characteristic for periodicity

Eric Silva

So i'd write this as $(x_{1}, x_{2})' = F(x_{1}, x_{2}, x) = (x_{2}, 6x_{1}^{2} + x)$

(suppressing the $x_{i}(x)$)

Semiclassical

For reference, these notes on Picard-Lindelof for first-order systems seem pretty decent: wwwf.imperial.ac.uk/~jswlamb/DynamIC/M2AA1-2010/…

Eric Silva

then you just do the normal thing

if it makes you feel better swap $x$ with $t$, it makes more sense

Semiclassical

Sure. That's what it was originally, heh, I just changed it for cosmetic reasons

Typhon

04:40

@Semiclassical if you look at all functions of the form $f(g(x)^2)$, did you know that includes all periodic functions?

same is true for any even power

but if you look at any odd power... you get all functions.

Semiclassical

@EricSilva Judging from the notes, though, it seems like Picard-Lindelof rests on an iterative procedure. I'm not convinced that would correspond to an integral equation in the desired sense.

Typhon

neat, huh?

Semiclassical

Hmm. I'll have to think about that.

Typhon

$f(\sqrt{(x \mod 1)^2}) = f(|x \mod 1|) = f(x \mod 1)$

the last term is all periodic functions

Eric Silva

@Semi Picard lindelof iterates because it lets you find a solution of the integral equation usign a fixed point theorem

the derivation of the integral equation as equivalent to the ODE happens before you start iterating

Semiclassical

04:43

True.

Eric Silva

if the ode is sufficiently ugly the iterating procedure just might not give you a solution

Semiclassical

So one would end up with, if I'm following right:

Typhon

glad I sparked a discussion

Semiclassical

(switching from x to t to avoid confusion)

Eric Silva

the only thing is that our system isnt autonomous (but that doesn't matter because you can always turn a non-autonomous system into an autonomous one)

Typhon

04:46

@Semiclassical so I am right in saying that answer was weird, right?

Semiclassical

$(x_1,x_2)'=(x_2,6x_1^2+t)\implies (x_1,x_2)(t)=\int_0^t (x_2(t'),6x_1(t')^2+t')\,dt'$

It certainly wasn't clear in its meaning.

Typhon

the answer?

Semiclassical

Yeah.

Typhon

or my question?

Semiclassical

I mean, it may be enough to talk about the first-order case.

Typhon

04:47

ya sniped me

Semiclassical

But one needs to explain why that's enough.

Typhon

@Semiclassical true but then we need the inductive step

Eric Silva

@Semi in fact because of our discussion there is only the first order case

@Typhon you do not need induction

Typhon

uuh

yeah you do

Eric Silva

no you don't

Typhon

04:47

yes you do

Eric Silva

you can reduce any n-th order ode to a first order system

Semiclassical

Eh, induction really isn't needed in the reformulation I wrote above.

Typhon

you need to show that if it is true for Nth order then it is true for (N+1)th order

Semiclassical

Not really.

Eric Silva

no you don't

you can just show it's true for any n directly

Typhon

04:48

im referring to the second statement

@EricSilva proving it for the first order serves as a base case

Semiclassical

Again, not really.

Eric Silva

no dude im telling you, it doesnt use induction

Typhon

not talking about the existence of an integral equation

Semiclassical

Then I'm not sure what you're talking about.

Typhon

im talking about the uniqueness of an integral equation with the least number of superfluous solutions.

Eric Silva

04:49

im unclear what you're talking about then

oh that

Typhon

the second statement in my question...

Eric Silva

I'm unclear on if that makes sense

Typhon

@EricSilva in what sense?

Semiclassical

I'm not sure there would be any superfluous solutions when you do Picard-Lindelof, at least if the boundary conditions are specified as "y and its first n-1 derivatives at 0"

Once you start talking other boundary conditions, by contrast, I have no idea.

Typhon

the number of superfluous solutions produced by an integral equation is either infinite, uncountably infinite, or finite.

Eric Silva

04:51

well you might not get unique solutions

if your ODE isn't regular enough

Semiclassical

Ugh, true.

Balarka Sen

i don't think there would be a superfluous solution. banach fixed point theorem guarantees uniqueness of fixed point, doesn't it?

mayve i'm wrong

Eric Silva

@Balarka he made no restrictions on how regular the ODE is

Balarka Sen

gotcha

Typhon

i am arguing that there is a unique integral equation such that the number of extraneous solutions is finite and minimal

Eric Silva

04:51

you're definitely right if you have like local lipschitz at least

Las Vegas Raiders

hi chat

Balarka Sen

right

Typhon

@EricSilva not unique solutions. I'm talking about a unique integral equation I such that $|Sol(I)|$ is minimized

Eric Silva

im very unclear on how we can even get multiple integral equatoins

@Typhon I know i wasn't talking to you

Typhon

errr

Semiclassical

04:53

I'll admit, this is where my interest starts to flag. Give me a specific ODE and I'm happy.

Typhon

im assuming one can do that?

Semiclassical

Give me an arbitrarily pathological one, though, and I find it hard to sustain my interest.

Eric Silva

yeah no this is the kind of thing where i dont really feel it's easy to start approaching because finding ways to transform ODE is usually very ad hoc

Typhon

@Semiclassical i like differential equations that involve floor. I like solving them with special methods. But now that I have advanced more I should justify those methods.

This is literally the foundation of building up to the proof of that justification

XD

Semiclassical

@Typhon To be fair, problems like that are important in application

Typhon

04:55

@Semiclassical differential equations involving the floor function?

Semiclassical

e.g. differential equations with time-dependent coefficients.

Typhon

O.O

i can trivialize them with certain means

Semiclassical

including ones that might have discontinuous time dependence

Typhon

eliminate the need to pull out the laplace transform

and get the same solutions it gives

Eric Silva

yeah and it usually doesn't help to do the thing i said: re making them autonomous by increasing dimension, because you're just shifting the difficulty

Typhon

04:56

of course... probably a transition to an integral equation going on there

Eric Silva

In math there's a principle of conservation of difficulty.

Semiclassical

^^^

Typhon

so one way to deal with floor is to increase the order?

lol

Semiclassical

This more has to do with the number of derivatives, I think.

Higher order ODE = first-order system with more components.

Typhon

fixed

huh

interesting

Eric Silva

04:58

@Semi we're all confused over here and I have a professor who has told me "Yeah from a mathematical perspective ODE are basically trivial"

Typhon

the implied derivative thingy's solutions are trivial

Eric Silva

rip

Semiclassical

The idea, as alluded to before, is to write $y=x_1,\,y'=x_2=x_1',\,y''=x_3=x_2',\,\cdots$.

Typhon

since in that 'space', piecewise constants and constants are essentially one and the same

Balarka Sen

i don't like DE's

evil little things

Typhon

04:58

@BalarkaSen well we do

D:

Eric Silva

@Balarka don't u like foliations

Balarka Sen

ye but i don't think of them in terms of DE's

Typhon

@BalarkaSen if you don't feed the differential equations between 11 pm and midnight, you'll be fine. Put them in water and they multiply and give you lots of fun challenges.

Eric Silva

idt there's really a "DE" perspective though

Semiclassical

I find asymptotic questions in DE's to be the most interesting.

Eric Silva

05:00

it's such a gigantic field

Leaky Nun

Is there integer $n$ such that the equation $\log_b(n+b) \in \Bbb N$ has three non-trivial integral solutions for $b$?

Balarka Sen

well, let's put it this way. i have not seen a super-inspiring corollary of Frobenius integrability

theoretically, it's cute. but

Eric Silva

frobenius isn't a super interesting DE theorem tho

if it were interesting it would be harder

Typhon

@Semiclassical imagine your hardest problem you could have to solve involving differential equations and floor. Now imagine if you could start off by just substituting all the piecewise constants with constants and proceed like normal. That is essentially how it works. Then you just find the continuous solutions. That's the trickier part. Involves a lot of series and partial sums of series and lots of no closed forms.

or would you rather use the laplace transform... oh wait the function increases faster than e^nx!

Semiclassical

Here's the DE I have in mind as far as asymptotics go

Typhon

05:05

i think equations involving $e^{\lfloor x \rfloor^n}$ are good candidates to mention

since they break the laplace transform

and are still solvable methinks

Balarka Sen

@Eric hah fair

Typhon

let me try

Eric Silva

@Balarka also have heard from a professor: "If a theorem about PDE was discovered before world war 1 then it wasn't really a theorem about PDE"

Balarka Sen

lol

Semiclassical

$$-i\dfrac{d}{dt}\binom{x_1}{x_2} = \begin{pmatrix} t & \alpha \\ \alpha & -t\end{pmatrix}\binom{x_1}{x_2}$$

I think that's the one I'm remembering.

Typhon

05:06

let's try to solve $y'' + 2e^{\lfloor x \rfloor^2}y' + 4e^{2\lfloor x \rfloor^2}y = 0$

or rather i will

note, if you're favorite thingy can transform it, i'll just nest more e^___'s

Semiclassical

The idea with the one I'm talking about right now is that, if $\alpha=0$, then this problem decouples in the obvious way.

Eric Silva

@Semi does the $-i\frac{d}{dt}$ thing show up a lot in physics

Typhon

auxiliary equation: $r^2 + 2e^{\lfloor x \rfloor^2}r + 4e^{2\lfloor x \rfloor^2} = 0$

Semiclassical

@EricSilva yuuup.

It's basically saying "oh hey I'm doing Schrodinger"

Typhon

sooo

Eric Silva

05:08

i saw it in an analysis class where we proved some theorem

that was actually just a physics theorem

Typhon

$r = 2e^{\lfloor x \rfloor^2}$

Eric Silva

but i didnt understand that perspective

Semiclassical

actually, I should probably not have the minus sign. it doesn't matter, of course.

Typhon

therefore $y(x) = C_1(x)e^{2e^{\lfloor x \rfloor^2}} + C_2(x)xe^{2e^{\lfloor x \rfloor^2}}$

now to find piecewise constant C's such that f is continuous

help

heh heh

Semiclassical

good luck man

Typhon

05:12

actually it aint that hard

just need the integral of floor squared

XD

Abcd

Typhon

wait no

Abcd

How to find the maximum velocity from this graph?

Typhon

i did this all wrong

therefore $y(x) = C_1(x)e^{2xe^{\lfloor x \rfloor^2}} + C_2(x)xe^{2xe^{\lfloor x \rfloor^2}}$

Semiclassical

What's the definition of velocity?

Typhon

05:14

i completely forgot how to do auxiliary equations.

Abcd

@Semiclassical Displacement/time (rate of change of displacement with time)

Semiclassical

right. (change in s) / (change in t). or at least that's how one does average velocity; one takes a limit for instantaneous velocity.

Abcd

@Semiclassical yes

Typhon

@Semiclassical integral of $e^{\lfloor x \rfloor^2}$ would be helpful

XD

Eric Silva

at the point of inflection

Abcd

05:15

@Semiclassical How do I find that using the graph?

Semiclassical

but if i pick two points on the graph, then the change in s is just the vertical rise, and the change in t is just the horizontal run.

Typhon

no result found - wolfram alpha

fuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Eric Silva

here it's where the tangent lines go from being below to above the graph

Semiclassical

so the average velocity defined by those two points is just the slope of the line between them.

Typhon

hrm

gotta rethink this

Eric Silva

05:16

locally i should say

Semiclassical

Homer Simpson Swearing

►

Abcd

@Semiclassical I know that. How do I find the maximum velocity?

Semiclassical

well, where is the graph the steepest?

You've only got a drawing, so you'll have to settle for a visual estimate of where it's steepest.

Abcd

@Semiclassical Between 10 to 14 s?

Semiclassical

something like that. I'd say maybe 11-12, but absent a specific function it's impossible to be precise.

Eric Silva

05:19

id say it's between 12 - 14 no

i guess 12ish makes sense

Semiclassical

yeah, you're right---I forgot that the boxes have width 2 s not 1.

Eric Silva

yeah

Semiclassical

12-14 is what I intended.

This is also the inflection point of the graph. But I don't really have the patience to explain that, tbh

Eric Silva

i can never remember like all the terminology

i remember learning that the inflection points are where the derivative changed sign

rather than just its critical points

which doesnt make sense to me

Semiclassical

i have inflection point as concave up -> concave down or vice versa.

(could also have something like x^4 of course, but w/e)

Eric Silva

05:23

oh i always used convex concave

Semiclassical

i like up/down just because of how visually immediate it is.

Eric Silva

i think that's concave up?

unsure

Leaky Nun

1. mentally associate "velocity" with "slope"
2. ????
3. profit

Semiclassical

I forget tbh.

Eric Silva

x^{2} is convex

Semiclassical

05:25

So that'd be concave up.

Abcd

@Semiclassical I ain't getting the right answer if I consider that time interval

Eric Silva

the reason i use convex concave is because everything that has ever mattered in my life has been convex

Semiclassical

shrug @abcd

Shadow

Can anyone help me understand the question?

Ten points in the plane are designated. You have ten circular coins (of the same
radius). Show that you can position the coins in the plane (without stacking
them) so that all ten points are covered.

Why can't I put all the coins on the ten designated points?

Semiclassical

@shadow without stacking them

suppose all ten points are really close together.

Abcd

05:26

@LeakyNun ?

Leaky Nun

@Abcd can you show us the whole question?

Semiclassical

if you try to put the coins on top of the designated points in that case, they'll overlap each other and would require stacking

Abcd

@LeakyNun Ok

Shadow

Oh I see

Semiclassical

you can still cover them in that case, though, just by using one coin to cover all of them at once and then put the rest wherever

Shadow

05:28

@Semiclassical Thanks

Semiclassical

but that's a specific case.

Eric Silva

maybe u can apply a covering theorem from gmt /s

Abcd

Semiclassical

Probably you shouldn't start with 10 coins. Maybe just do 3 coins to start with.

Balarka Sen

@EricSilva err, that would make $x = 0$ not an inflection point of $x^3$?

Abcd

05:29

@LeakyNun ^

Leaky Nun

@Abcd and what is the answer?

Abcd

@LeakyNun Answer of b is 25 cm/s

Eric Silva

@Balarka ?

what are you referring to

Balarka Sen

you said "i remember learning that the inflection points are where the derivative changed sign"

you can click on the little grey arrow at the front of the message to go to the message i am replying to btw

Eric Silva

oh sorry i meant where the second derivative changed sign

not first

oh that's a cool feature

Balarka Sen

05:32

oh i see

Eric Silva

i want them to just be the crit points of the derivative tho

Leaky Nun

@Abcd if you look at t from 11 to 12 and get 30cm/s I would say it's pretty dem close

Semiclassical

It's also helpful to draw a line of slope 25cm/s just to get a sense of how hard it is to distinguish various slopes precisely.

Twink

Te amo, te amo

he said to me

Semiclassical

(25 cm/s = 100 cm/4 s = 1 m / 4 s)

Twink

05:35

somebody tell me what he said

don't it mean I love you?

Leaky Nun

@Twink yes

it is Spanish for "I love you"

Twink

he said te amo then he put his hand around my waist

Leaky Nun

and also Latin

Semiclassical

Out of context song lyrics remain confusing.

For reference: azlyrics.com/lyrics/rihanna/teamo.html

Twink

yes but I changed it

Eric Silva

05:38

Te amo can just mean that you like someone depending where you're from

Semiclassical

Doesn't make it any less of a non sequitor.

Twink

think it means I love you

Eric Silva

te quiero means i love you

Leaky Nun

@EricSilva that's for a friend

Balarka Sen

@EricSilva I would not call $x = 0$ an inflection point of $x^4$.

you're just saying you just want the second derivative to be 0, right?

Eric Silva

05:40

@Leaky depends where you are

lots of parts of LA use te quiero for a significant other

Leaky Nun

@EricSilva but me gustas tu is universal?

(it's also a title for a k-pop)

Eric Silva

uhh in some parts of latin america that's very awkward phrasing so no

@Balarka i would tho

Balarka Sen

eh

i just want to think of that saddle point picture

Eric Silva

i think that's fair

Twink

I told him, listen we can dance, but you gotta watch your hands

Eric Silva

05:42

but the thing is that i dont like this terminology at all

cause i dont think it's ok to use in higher dim

but the thing is that i dont like this terminology at all

Twink

because I understand that we all need love and I'm not afraid to feel the love but I don't feel that way

Leaky Nun

@EricSilva alright

Twink

then he said Te Amo

Leaky Nun

Does out-of-context song-lyrics-quoting count as noise?

Eric Silva

like im not a native spanish speaker but ive been to most of latin america so i speak only about those basically

Balarka Sen

05:45

@LeakyNun And I'm gone gone gone / Now I'm older than movies
Let me dance away / Now I'm wiser than dreams

etc

Twink

Rihanna - Te Amo

►

think it means "I love you"

Balarka Sen

@LeakyNun Actually, I should have said "IT GOES IT GOES IT GOES IT GOES GUILLOTINE YAH"

Eric Silva

i feel like a third of the chat is probably just noise

5

Abcd

@LeakyNun How to solve this^?

Leaky Nun

@Abcd no idea

Abcd

05:58

ok

Anyone else?

Semiclassical

Under those conditions, what are the particle's radius vectors as a function of time elapsed $t$?

Abcd

@Semiclassical There are two particles. (I can't understand your question too)

Semiclassical

particles', then.

What's the position of each particle as a function of time?

Abcd

@Semiclassical r1(t) and r2(t)

Semiclassical

And these equal...?

You know the initial positions and the constant velocities. That's enough to say what r1(t),r2(t) are.

Abcd

06:07

@Semiclassical $r_1(t) = v_1*t$

Semiclassical

If that's true, then $r_1(0)=v_1\cdot 0=0$.

So that'd require $r_1(t)$ start at 0 at time t=0.

Abcd

oh

@Semiclassical $v_2 = (r(t) - r_2)/t$

Semiclassical

Okay. So $\vec{r}_2(t)=\vec{v}_2t+\vec{r}_2$.

Abcd

@Semiclassical yes

Semiclassical

How about $r_1$?

Abcd

06:11

@Semiclassical $r_1(t) = v_1t + r_1$

(with vector arrows above the letters)

Semiclassical

Sure, as vectors

Now, if they collide, they must be at the same position at a given time.

So what equation does that give?

Liad

06:44

@Semiclassical hi

i need to find this integral where $\gamma$ is a curve around the ellipse $(\dfrac{x}{a}) \ ^ 2 + \dfrac{y}{b}) \ ^ 2 =1 $ . i used Green's theorem and got $0$. but when i open this integral by definition i get $2\pi ab$. so, how could it be ?

@PVAL-inactive hi

Liad

07:26

Also when I want to calculate the volume of the region above the cone and inside the ball $ x \ ^ 2 + y \ ^ 2 + (z-1) \ ^ 2 \lt 1 $ , we have $\phi \in [0,\pi /4] \theta \in [0,2\pi] , r \in [0 , 2cos(\phi) ] $ and the volume turns out to be $3\pi$. this make no sense because the volume of the ball is $4 \pi /3$. someone can help ?

Typhon

@Semiclassical I was trying to think of an operator that maps periodic functions to 0 but not everything else and it occurred to me. What about the operator that measures the wavelength of a function. Hmmm...?

Liad

08:24

the second question i solved, i had a calculation error, if someone can help with the first question that would be great

Perturbative

08:51

Hey everyone, recently I've decided to create a Differential Topology and Differential Geometry study group (which is based here on Reddit : https://www.reddit.com/r/difftop_diffgeom/). We have a chat room here on MSE : https://chat.stackexchange.com/rooms/61560

The textbooks we are working through are

- Topology from the Differentiable Viewpoint by Milnor
- Differential Topology by Guillemin and Pollack
- Introduction to Smooth Manifolds by Lee
- Introduction to Manifolds by Tu.

We start officially working through the material this coming Monday (10 July 2017). Anyone is welcome to join!

5

$Mathematics$ Differential Topology & Differential …

A chat room for all things Differential Topology and Different...

Astyx

09:18

hi

rene

09:51

There is a question posted on Meta Stack Overflow about migrating questions relating to algorithmic complexity from SO to Math. You might want to chime in there in case we decide something you'll regret: meta.stackoverflow.com/questions/351881

2

gxyd

11:19

Till now I have often come across the terms "over a field" in case of algebraic structures like fields. Can you point me to some link where I can see its usage and meaning?

Seems like everyone is out for the weekend :)

Bass

Hi guys. So $(x)\subset R[x]$ in an integral domain $R$ which is not a field, is not a maximal ideal because $R[x] / (x)$ is not a field. So I'm looking for an ideal in $R[x]$ which contains $(x)$. Am I right that $(x, 2)$ (the ideal spanned by $x$ and $2$) is such an ideal?

Danu

Algebra over a field

In mathematics, an algebra over a field (often simply called an algebra) is a vector space equipped with a bilinear product. Thus, an algebra is an algebraic structure, which consists of a set, together with operations of multiplication, addition, and scalar multiplication by elements of the underlying field, and satisfies the axioms implied by "vector space" and "bilinear". The multiplication operation in an algebra may or may not be associative, leading to the notions of associative algebras and nonassociative algebras. Given an integer n, the ring of real square matrices of order n is an example...

@gxyd ^

gxyd

@Danu okay. When I say linear relations over K, it would mean "linear relation among the elements of $K$"?

Where $K$ is a field.

$Mathematics$ Differential Topology & Differential …

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$Mathematics$ Mathematics