Mathematics - 2018-09-13 (page 2 of 2)

Rudi_Birnbaum

17:11

Hi all; @mercio Hello, did you see my messages?

Xander Henderson

Hey, quick question, because I am an idiot: when you write $A\cos(\omega t + \varphi)$, what do we call that constant $\omega$? ($|A|$ is the amplitude, and $\varphi$ is the phase shift---does $\omega$ have a name?)

It isn't quite the frequency, though it is related....

Rudi_Birnbaum

@XanderHenderson in German its called "Kreisfrequenz"

Xander Henderson

Hrm... that is nice to know, but not a great term for my precalc students.

Rudi_Birnbaum

in Engl "Angular frequency" apparently

Angular frequency

In physics, angular frequency ω (also referred to by the terms angular speed, radial frequency, circular frequency, orbital frequency, radian frequency, and pulsatance) is a scalar measure of rotation rate. It refers to the angular displacement per unit time (e.g., in rotation) or the rate of change of the phase of a sinusoidal waveform (e.g., in oscillations and waves), or as the rate of change of the argument of the sine function. Angular frequency (or angular speed) is the magnitude of the vector quantity angular velocity. The term angular frequency vector ...

Xander Henderson

I was just about to report that Google translate gives that as angular frequency.

Thank you.

Rudi_Birnbaum

17:18

welcome :-)

Ted Shifrin

17:28

@Xander: Since you're giving position as a function of time, $\omega$ is usually called angular velocity (in physics and in math).

Fargle

Good mornfternevenight, chat.

5

Ted Shifrin

Gesundheit @Fargle

Fargle

Leave my health out of this!

Rithaniel

Question: How do you demonstrate that it is strictly possible to draw an equilateral triangle around a point in $\mathbb{R}^{d}$ ? Or is that something you could say is self-evident?

Fargle

How are you, @Ted?

Ted Shifrin

17:30

Doing decently, thanks, @Fargle, and you?

Rithaniel

Well, $\mathbb{R}^{2}$, specifically.

Ted Shifrin

@Rithaniel: Do you literally mean triangle? I.e., a planar object?

Oh.

What do you mean by "around"? Just that the point is somewhere inside?

Rithaniel

Planar object, yes. (Sorry, I've gotten into the habit of saying $\mathbb{R}^{d}$). Also, the point is at the centroid of the triangle.

Ted Shifrin

Oh, see ... you need to be very specific in your questions!

Rithaniel

Yes, my professors have been drilling that into my head.

Ted Shifrin

17:32

So, there's a $2$-parameter family of equilateral triangles with a fixed centroid. You can rotate and dilate them ....

Fargle

@Ted Pretty well. I've just been given another prospect for publishable research (that's very much like the last time), which is heartening. I've also been given some good recommendations on regional graduate schools, especially those for master's study. Outside of that, just trying to keep my nose to the homework grindstone, and trying to sink my feet back into my extracurricular reading when I can.

Ted Shifrin

Sinking your feet is very à propos, with Florence looming.

Rithaniel

I feel that it should be self evident that the set of all triangles with that point as the centroid is not $\emptyset$.

Ted Shifrin

@Rithaniel: I'm not sure what your context is and/or what you "know."

Here's one way to approach it. Draw an equilateral triangle. Find its centroid (intersection of its medians). Now translate the plane to carry that centroid to the given point. (And, as I said, you can now rotate the triangle or dilate about that given point.)

Are other people having issues with chat trying to duplicate our sentences?

Rithaniel

Well, I am given a set of open equilateral triangles with base parallel to the x-axis, $a\in\mathbb{R}^{2}$ at the centroid, the length of each side greater than 0. I must show that this is an open neighborhood system.

In order to be an open neighborhood system, it must satisfy, first, that this set is not $\emptyset$.

Actually, yeah, I can just show that there is at least one member.

Ted Shifrin

17:38

Oh, so actual interiors of the triangles ...

Oh, base parallel to the $x$-axis. So that takes out my rotations.

But my suggested argument still is fine. You could also, starting at a point, construct with compass and straightedge an equilateral triangle with that point as centroid, but I don't think you're required to do that.

Kasmir Khaan

@TedShifrin Ted :D

Ted Shifrin

Greetings, @Kasmir.

Kasmir Khaan

Hello ! I have a question !

actually more than one :D

How can I compute a fourir series?

Ted Shifrin

Using the integrals that compute the coefficients.

Kasmir Khaan

Can you please tell me like hmm that notation in few words

Ted Shifrin

17:40

No.

Kasmir Khaan

okay grrrrr

ill try to solve one and come back at you -.-

:)

Ted Shifrin

I don't know the notation of your course, and you can figure this out.

Kasmir Khaan

Yes yes ! am just not having the book yet

Ted Shifrin

Lecture notes?

Kasmir Khaan

so only what i wrote on lecture ><

Ted Shifrin

17:41

You can find formulas in my YouTube lecture I mentioned. But they may not be consistent with your course.

Kasmir Khaan

well , kind of :'D i remeber that we have to integrate something

wait what

you had lectures on fourir?

Ted Shifrin

I told you weeks ago.

One lecture.

Kasmir Khaan

OMGG

Ted Shifrin

You said you would go find it.

Kasmir Khaan

YOu said u did TEACHED that course'

Ted Shifrin

17:42

Your memory is worse than mine.

Kasmir Khaan

i never thought it was recorded

no i remeber asking you about it and ur answer

Ted Shifrin

I said I did one lecture in that Multivariable Math class (showing them how it was doing projections, which we did in detail).

Kasmir Khaan

but I understood that was only course u teached and never recorded

Nice :D

Ted Shifrin

You asked if I had taught a course on Fourier series. I said no to that.

Kasmir Khaan

becasue from what i understood they are like orthonormal bases

for function space

Ted Shifrin

17:43

They're an expansion with respect to an orthonormal basis, yes.

Kasmir Khaan

okay thanks sir! :'D

Leaky Nun

is it really a basis?

Ted Shifrin

in the analytic sense, yes

Kasmir Khaan

Sup leaky !

Leaky Nun

I mean, doesn't it only work for periodic functions?

Ted Shifrin

17:45

You can turn any function in $L^2[0,1]$ into a periodic function if you wish.

Leaky Nun

I see

Ted Shifrin

On all of $\Bbb R$ one does Fourier transform, not series.

Leaky Nun

oh wow

I never noticed

Ted Shifrin

You can even heuristically derive the Fourier transform as a limit of Fourier series on expanding intervals $[-N,N]$.

Rudi_Birnbaum

But on @LeakyNun s point: Is there anything at all that suggests it should have to do something with periodicity?

I ask since there is indeed also in quantum mechanical wavefunctions (which form a Hilberst space basis cum grano salis) something which puts them close to periodic functions.

And going from position to momentum space e.g. corresponds to FT as well, I think

The something is that the higher the energy of the states (=wave fcuntions) is the more nodal planes they have. In deed the numer of nodal planes in many problems simply increases by one when goind from $\psi_n$ to $\psi_{n+1}$. In that way there is some similarity to periodic functions to it.

Ted Shifrin

18:02

No, periodicity is a red herring. It's about $L^2$ functions on a finite interval.

In applications one often considers their extension to periodic functions on $\Bbb R$.

Leaky Nun

applications

Ted Shifrin

You might look up the history of why Fourier was interested in this in the first place. :)

Rudi_Birnbaum

heat conduction?

Ted Shifrin

Yup ... heat equation ...

Rudi_Birnbaum

I mean it relates "curvature" to change in time, by that periodic functions come into play I guess

Its almost the Schrödinger equation just that this works with complex time

Ted Shifrin

18:10

I'm not sure periodic functions are the point, yet again. We're just using $e^{ikx}$ as an orthogonal basis for $L^2[0,2\pi]$ or $L^2[-\pi,\pi]$.

Rudi_Birnbaum

And then there are orthogonal polynomials, as well. I guess you can use these a basis as well.

Ted Shifrin

They seem to be more natural for numerical analysis questions than for PDE applications.

Leaky Nun

can i just apply gram-schmidt on 1,x,x^2,x^3,...?

Ted Shifrin

Once you specify the interval and the inner product, sure.

Leaky Nun

what would I get?

Rudi_Birnbaum

18:13

they'd not be orthogonal, pairwise, I guess

Ted Shifrin

@Rudi: Gram-Schmidt gives you an orthogonal basis.

Unless you don't do it right.

Rudi_Birnbaum

Yes, but then you have lin. combs of @LeakyNun s polys

Ted Shifrin

well, of course

Leaky Nun

oh, I can make 1,x,x^2,x^3,... orthogonal, right

by choosing the right inner product

that's interesting

Rudi_Birnbaum

then I guess you'd get orthogonal polynomials, in case you use the right inner prod.

Leaky Nun

18:15

I was thinking of $\int fg$

Tobias Kildetoft

@LeakyNun Sure, and you even get a unique inner product if you require them to be orthonormal

Rudi_Birnbaum

$\int \bar{f} g dx$

Alessandro Codenotti

Hi everyone

Leaky Nun

but I can just treat it as $\Bbb R^n$ as $n \to \infty$

i.e. $\Bbb R^\omega$

hi @AlessandroCodenotti

Ted Shifrin

Hi @Alessandro ... how's your Scottish ear?

Rudi_Birnbaum

18:16

Hi @AlessandroCodenotti

Secret

Sometimes I do wonder about orthnormal basis in inner product spaces. Strictly speaking, these hilbert basis can only span a dense set of the parent space, so why we never need to worry about the elements this dense set missed out, is is because they are almost always not $L^2$?

ÍgjøgnumMeg

Hey @Alessandro

Ted Shifrin

$L^2$ is complete.

Rudi_Birnbaum

@TobiasKildetoft would it work the other way round? Defining $1,x^2,x^3,...$ orthogonal and searching an inner product

Ted Shifrin

hi @ÍgjøgnumMeg

ÍgjøgnumMeg

18:19

Hey @Ted :)

and @Rudi

and @Leaky etc. etc.

Rudi_Birnbaum

@ÍgjøgnumMeg Hoi!

Tobias Kildetoft

@Rudi_Birnbaum Sure (if you add $x$). Given any basis, there is a unique bilinear form which makes that basis orthonormal

ÍgjøgnumMeg

@Rudi grüaß di :)

Tobias Kildetoft

(similarly a unique sesquilinear one if you want that instead)

Secret

Ted: Ah right, then I can always cauchy sequence and converge to the elements outside this dense set

Rudi_Birnbaum

18:20

@TobiasKildetoft What would be an explicit form?

Ted Shifrin

OK, I need to get going. Fun times ... going to the ophthalmologist.

Bubye.

Leaky Nun

@ÍgjøgnumMeg goe'e morge'

Rudi_Birnbaum

of that inner product?

Tobias Kildetoft

@Rudi_Birnbaum It would just look like the "standard" one in the chosen basis, multiplying coefficients and then adding

Rudi_Birnbaum

@TedShifrin Bye ond good luck - hate them ..

Leaky Nun

18:21

@Rudi_Birnbaum $\langle \sum a_i x^i, \sum b_i x^i \rangle = \sum a_i b_i$

ÍgjøgnumMeg

@Leaky friesisch?

Rudi_Birnbaum

@TobiasKildetoft Ah OK!

Leaky Nun

@ÍgjøgnumMeg nein, hollandisch

Rudi_Birnbaum

@LeakyNun nice :-)

ÍgjøgnumMeg

Ah okey!

Rudi_Birnbaum

18:22

@Secret you're refering to the arcane "continuum"?

Secret

You know how much I LOVE infinities especially uncountable ones

ÍgjøgnumMeg

@Leaky Gott kvøld! lol

Leaky Nun

dansk?

Secret

so yeah, the thought process lead me to that question

ÍgjøgnumMeg

Føroyskt :) Although I think it's the same in Danish

Might be Godt kvøld

Tobias Kildetoft

18:23

@ÍgjøgnumMeg It is not even close

ÍgjøgnumMeg

in Danish

Secret

and Ted answered it with L^2 is complete, meaning we can always recover those elements with a limit operation

ÍgjøgnumMeg

@Tobias aw

Leaky Nun

lmao the real Dane comes

ÍgjøgnumMeg

What do you guys say?

Tobias Kildetoft

18:24

God aften

ÍgjøgnumMeg

Lawl

shows how much danish I know

Tobias Kildetoft

it is closer in Swedish

ÍgjøgnumMeg

Yis I speak Swedish

Rudi_Birnbaum

God afton

Tobias Kildetoft

@Rudi_Birnbaum Or kväll

Rudi_Birnbaum

18:24

Ja ochså taler lite svensk!

ÍgjøgnumMeg

so I always just assume I know Danish phrases by taking Swedish phrases and interspersing them with the odd "D"

Jasper Loy

Danish, Norwegian, and Swedish are North Germanic languages. The first two share the same alphabet.

Secret

(To be solved later) Whether an inaccessible cardinality inner product space can have a dense orthonormal basis of cardinality less than inaccessible

2

Rudi_Birnbaum

@TobiasKildetoft tak

Leaky Nun

tsak

Tobias Kildetoft

18:26

@ÍgjøgnumMeg I do actually speak Swedish decently (apart from my terrible pronunciation especially of anything involving "sj")

Rudi_Birnbaum

@TobiasKildetoft Men jag har inte hört det ...

Leaky Nun

hört?? is that even a swedish letter

ÍgjøgnumMeg

@Tobias you can just pronounce them like people in Skåne

lol

Rudi_Birnbaum

Kamelåså

ÍgjøgnumMeg

@Leaky yes it is

lol

Rudi_Birnbaum

18:27

öåy

are their "Umlauts"

Leaky Nun

I see

ÍgjøgnumMeg

@Rudi and ä

Rudi_Birnbaum

right!

actually äöy

and å is not really an Umlaut is it?

Jasper Loy

Strangely, among all the Germanic and Romance languages, it seems that German is the hardest for the English speaker to learn. I don't know why.

ÍgjøgnumMeg

No it's a different sound

Leaky Nun

18:28

@JasperLoy have you tried Danish?

Rudi_Birnbaum

@JasperLoy guess its most conservative except islandic

Leaky Nun

conservative... have you tried Plattduutsch?

Rudi_Birnbaum

@JasperLoy with the most structure in the Grammar

Jasper Loy

@LeakyNun No, but that comment was based on some statistics I saw on effectivelanguagelearning.com.

Rudi_Birnbaum

But thats an old puzzle of mine: Why do older languages show more structure?

ÍgjøgnumMeg

18:29

@Rudi it's some kind of smoothing effect I think

the older a language becomes the lazier speech becomes I guess?

Rudi_Birnbaum

@ÍgjøgnumMeg but how did it start?

Leaky Nun

Are Languages Getting Simpler?

►

Rudi_Birnbaum

@LeakyNun the gist?

Jasper Loy

And then Finnish and Hungarian are strange. They belong to the Uralic family. And they are supposed to be even harder than the Slavic languages Russian, Polish, and Ukrainian.

Rudi_Birnbaum

@ÍgjøgnumMeg don't know. I have heared that its beacause in former times people spoke less ...

ÍgjøgnumMeg

18:31

@Rudi perhaps!

Rudi_Birnbaum

and yes so it didn't smooth out that strongly ...

@JasperLoy but thats not surprising,

Slavic is indogermanic

Jasper Loy

@LeakyNun I just got a new laptop with a HD camera. I will start making some youtube videos again soon. This time, it will be clearer and you can see my pimples.

Rudi_Birnbaum

while finnourgic is the sister of indoeuropean

ÍgjøgnumMeg

@Jasper I think Finnish is often cited as a really difficult language because of how many cases they have, but the cases (for the most part) simply replace prepositions

Leaky Nun

@Rudi_Birnbaum honestly, I watched this so long ago, I forgot

ÍgjøgnumMeg

18:33

it's the same with Turkish

Rudi_Birnbaum

15 cases, more than 80 gradiations for each case ... and so on

each case has a special meaning for each word and so on

Jasper Loy

Anyway, I don't know much about language. I just read various websites on languages.

ÍgjøgnumMeg

Right, but a lot of the cases just replace prepositions so it's as hard as learning prepositions

Rudi_Birnbaum

Finns eat on the table and you have to learn that as well

@ÍgjøgnumMeg well the case changes the interior of the word

thats "gradation"

Jasper Loy

But I do recommend the Assimil courses for learning languages. Many polyglots use them to learn quickly.

ÍgjøgnumMeg

18:34

Yes, but the point is that you replace prepositions with infixes/suffixes/prefixes

@Rudi Talo is house, Talossa is "in the house", "Talolla" is "at the house" etc. etc.

Rudi_Birnbaum

@ÍgjøgnumMeg So I have spent about 3 years of my life in Finland and hardly speak any sentences ..

ÍgjøgnumMeg

so instead of learning words for "in, at" etc. you learn how the word changes in this wa

y

not saying it's easy lol

Rudi_Birnbaum

@ÍgjøgnumMeg But thats also because most of my friends are swedish speaking

ÍgjøgnumMeg

just saying it's kinda misunderstood hahaha

Jasper Loy

Oh, and finally, Chinese, Japanese, Korean, and Arabic are the hardest languages to learn.

Rudi_Birnbaum

18:37

@ÍgjøgnumMeg yeah in principle your're right but then how to answer the question why its that hard ...

ÍgjøgnumMeg

@Rudi sure lol, it's obviously a big difference from Germanic languages

Rudi_Birnbaum

@JasperLoy yeah, but it also depends

ÍgjøgnumMeg

@Jasper again, one has to define difficulty

Rudi_Birnbaum

@ÍgjøgnumMeg but what in a way also can help

ÍgjøgnumMeg

the writing systems are very different and the vocabulary etc. but I'm fairly sure the grammar of Chinese is not hard at all (pls back me up @Leaky)

Jasper Loy

18:38

Actually, I speak English and Chinese, but that's because I learnt both for at least a decade.

Rudi_Birnbaum

eg Swedish word are for me a bit hard since they are so close tpo German or English but often vary little in the meaning

ÍgjøgnumMeg

@Jasper nice

Rudi_Birnbaum

as for Finish either you know it or you don't

ÍgjøgnumMeg

@Rudi Right! The word "eventuell" was always weird for me

lol

Leaky Nun

@ÍgjøgnumMeg right, Chinese doesn't inflect words

Rudi_Birnbaum

18:39

ficka

Jasper Loy

Know Finnish or you are finished, lol.

ÍgjøgnumMeg

@Leaky nor does it conjugate verbs right?

@Rudi hahaha ficklampa

Leaky Nun

right

Rudi_Birnbaum

@ÍgjøgnumMeg lol rofl

Ninja hatori

@LeakyNun do you have idea that how to show re(z^3) is continuous complex function?

Rudi_Birnbaum

18:39

@JasperLoy wow!

ÍgjøgnumMeg

@Rudi A good one is "It's not the fart that kills, it's the smäll"

2

Tobias Kildetoft

@Rudi_Birnbaum Not to be confused with fika

Rudi_Birnbaum

@ÍgjøgnumMeg hahaha

Jasper Loy

@Rudi_Birnbaum Forced by the system, but shan't say where I am from.

Leaky Nun

@Ninjahatori z is continuous, multiplication of continuous functions is continuous, so z^3 is continuous, and then show that Re is continuous (well it's just the projection map R^2 -> R), and then just compose them together

Tobias Kildetoft

18:40

@ÍgjøgnumMeg Nice. Here we have lifts saying "I fart"

ÍgjøgnumMeg

@Tobias lol

Alessandro Codenotti

@TobiasKildetoft In the lift? That's just mean

Abdullah UYU

We can say that $A\subseteq B\iff A-B=\emptyset$, right?

Secret

It has to be as $B^{\complement} \not \supseteq A$

Leaky Nun

@AbdullahUYU yes

Abdullah UYU

18:46

What does the the term in the left-hand side mean? @Secret

Secret

B complement

ÍgjøgnumMeg

@Alessandro checked the example $\lvert a \rvert_\mathfrak{p} = \left(\frac{1}{N \mathfrak{p}}\right)^{\operatorname{ord}_\mathfrak{p}(a)}$ is an absolute value on a number field $K$ with a prime $\mathfrak{p}$ today

ergh

Leaky Nun

on a number field

ÍgjøgnumMeg

right

that's what I meant

Rithaniel

What do I need to check to assure that a collection of sets defines a neighborhood system of $\mathbb{R}$? I have a definition for an open neighborhood system, but not a general neighborhood system.

Ninja hatori

18:57

@LeakyNun any idea about branch point and reiman sheet?

Leaky Nun

what do you mean

Ninja hatori

suppose I have function root(z-a) then what are branch point and reimann sheets are there?

ÍgjøgnumMeg

Riemann

pls

it's a name

Ninja hatori

suppose I have function root(z-a) then what are branch point and Reimann sheets are there?

Leaky Nun

lmao

I have no idea anyway

Ninja hatori

19:06

Its okay thank you for reply

Alessandro Codenotti

19:49

@ÍgjøgnumMeg I've been a tourist today so I didn't get much (or any) work done

Akiva Weinberger

@LeakyNun There's a name for the sequence you get for Gram–Schmidting $x^n$ but I forget what it is

Leaky Nun

I see

hola

ÍgjøgnumMeg

@Alessandro nice lol, how was it?

What part of Scotland are you in?

https://www.youtube.com/watch?v=KeRKgp-qc-k

Happy black metal

Alessandro Codenotti

@ÍgjøgnumMeg Aberdeen, it's mostly gray, it looks a lot like I expected a Scottish city to look like

ÍgjøgnumMeg

@Alessandro hahah fair, that's cool

Jake S

20:00

Extremely basic question but I'm having a bit of trouble visualizing the associated matrix for a linear map of polynomials. If I have a transformation T(a + bx + cx^2): -b + bx - bx^2, what does the matrix look like? Or to go directly to my problem, how do I evaluate the standard basis vector (1)?

Abdullah UYU

@Secret To make it clear, you're saying one of the sides has to be $B^{\complement}\not\supseteq A$. So $B^{\complement}\not\supseteq A\iff$ what?

Leaky Nun

@AlessandroCodenotti you mean grey

Alessandro Codenotti

If I write "the color grey" will people get very confused on whether I'm American or British?

ÍgjøgnumMeg

A lot of people don't really care

and it doesn't matter that much rofl

if at all

Leaky Nun

lol

the colour gray

Abdullah UYU

20:14

@LeakyNun Can you make sense of chat.stackexchange.com/transcript/message/46720428#46720428 regarding what I proposed?

Leaky Nun

does it even matter

Abdullah UYU

Well, I think that what I say is true and I also agree with you.

But I am trying to relate it.

Jake S

:(

Daminark

So, I think "gray" vs "grey" is one of those things that people don't consciously recognize as American vs British

Like, every time I spell it I agonize a bit and just choose one, also with traveling vs travelling

mercio

gray is my favorite colour

Jake S

20:21

I realize my question might be strikingly basic, but all the easier to answer accordingly, no...?

mercio

well first you need to choose a basis for your space of polynomials

ÍgjøgnumMeg

Night all!

mercio

good night

user193319

If $f$ and $h$ are real-valued functions, is it true that $\max(f,g) = \frac{f+g+|f-g|}{2}$ and $\min(f,g) = \frac{f+g-|f-g|}{2}$?

Jake S

@mercio Of course, but taking the standard basis

mercio

20:25

which is ?

Jake S

{(1),(x),(x2)}

For P2[x] anyway

mercio

okay

then indeed, what is T(1) ?

Jake S

Well if 1 could be seen as (1 0 0) isn't it just 0?

mercio

well if I put a=1 b=0 and c=0 in the line that defines T i guess it says that T(1) = 0 ?

Jake S

But it should be -b I guess. I'm not visualizing it properly

Right

mercio

20:30

and the coordinates of 0 in the basis are (0 0 0), right ?

so that's your first column

Jake S

But is the matrix for T (sorry, I'm on my phone) a column -1 0 0, then a column 0 1 0 and finally 0 0 -1?

mercio

no ?

Jake S

Sorry, my chat seems to be lagging, I hadn't seen your previous message

I see

But then it would be (0 0 0), (0 1 0) and (0 0 0)?

As the same would happen with x2

mercio

that middle column would be wrong

what is T(x) ?

Jake S

b

bx rather

mercio

20:40

look at the line defining T again

T(a + bx + cx^2): -b + bx - bx^2

Jake S

Ah... Would the middle column be (-1 1 -1)?

mercio

yes

Jake S

OK, I understand now. Thank you. To prove it's a projection, is it sufficient to show that the matrix multiplied by itself returns itself?

I appreciate your patience!

mercio

yes that would be enough

Abdullah UYU

@LeakyNun Besides, $B^{\complement}\not\supseteq A\iff A-B=\emptyset$ is not true, take $A=\{a_1, b_1\}, B=\{b_1\}$ for example.

Leaky Nun

20:46

ok

Abdullah UYU

In that case, $B^{\complement}\not\supseteq A$ but $A-B\neq\emptyset$.

Ah, I suffered :)

Rithaniel

21:02

Does anyone have a definition for a neighborhood system, as opposed to an open neighborhood system?

ÍgjøgnumMeg

Dunno if this is obvious or not: for two absolute values $\lvert \cdot \rvert_1$ and $\lvert \cdot \rvert_2$ on a field $K$, if $\lvert \cdot \rvert_2 = \lvert \cdot \rvert_1^a$ for some $a > 0$ then $\lvert \cdot \rvert_1$ and $\lvert \cdot \rvert_2$ define the same topology on $K$, this is because in the open balls $U(b, \varepsilon) = \lbrace x \in K : \lvert x - b\rvert_2 < \varepsilon \rbrace$ we can just replace $\lvert x - b\rvert_2$ with $\lvert x - b \rvert_1^a < \varepsilon$ and

take $\varepsilon^\prime := \varepsilon^{\frac{1}{a}}$

or?

Akiva Weinberger

America - GRAY

England - GREY

(@LeakyNun @AlessandroCodenotti @Daminark)

Canada - uh, GREhY?

Leaky Nun

nice

@ÍgjøgnumMeg why are we not discussing this in that chatroom?

ÍgjøgnumMeg

@Leaky no idea rofl

Leaky Nun

and to your question, yes it is obvious

ÍgjøgnumMeg

21:13

okay good, I thought so :P Probably just asking it here because it's more of a question about topology than about local fields

Leaky Nun

everything is about local fields

ÍgjøgnumMeg

rofl, i was supposed to be going to sleep and this popped into my head

Guess I'll learn Ostrowski's theorem tomorrow, then saturday I can read about the weak approximation theorem, attempt some exercises, then graduate on monday

lol

Leaky Nun

> graduate on monday

ÍgjøgnumMeg

I graduate on monday btw

Abdullah UYU

Congrats, @ÍgjøgnumMeg

ÍgjøgnumMeg

21:19

Thanks @Abdullah :)

Richard

0

Q: An affine plane and an affine line that share a generator

With the parameter $h\in\Bbb R$, consider in the affine space $\Bbb A^4$ the system $$\begin{cases}x+2y-+2z+w=5 \\x+y-hz=6 \\ -x+hy+2z+(h+1)w=h+1. \end{cases}$$Calling $\Sigma_h$ the set of solutions, I'm asked to find the hyperplane containing $\Sigma_1$ and $\Sigma_{-4}$. Now, I've found those...

Please someone

user100000000000000

22:13

In the last displayed equation, how do they get \mu^*(E)+\epsilon\ge \mu(B\cap A)+\mu(B\cap A^c)\ge \mu^*(E\cap A)+\mu^*(E\cap A^c)? math.stackexchange.com/questions/2443845/…

Akiva Weinberger

Life up until Graduation

Source

@ÍgjøgnumMeg

Adarsh Kumar

23:02

@rschwieb I don't want any sympathy but I think they are going to close the question and once if they close it I think my idea will lost forever. Please tell me the way, how can I communicate with the authority to se my question once

user193319

23:34

How do I show that $1, -72x + x^2, -39x + x^3, -203x + x^4,x + 72x^2+39x^3+203x^4$ is a basis for $\mathcal{P}_4(\Bbb{C})$ without having much tools/theorems/heavy machinery at my disposal?

I don't think a calculator is expected either...Seems like a ridiculous problem...

Daminark

@user193319 so, you know that $\{1, x, x^3, x^3, x^4\}$ is a basis

user193319

Yup.

Daminark

So then map $1$ to $1$, $x$ to $-72x + x^2$, and so on, and show that it's invertible.

It'll amount to either using row reduction to compute rank of the associated matrix or taking a determinant

Less than pleasant since it's 5x5 but it can be done by hand

user193319

...Hmm...Unfortunately we haven't discussed invertiblity or linear maps yet...which is why I say it seems like a horrible problem...I was going to try to show that $x$, $x^3$, etc. can be written in the span of the other vectors, but that's just awful...

Ted Shifrin

hi Demonark

user193319

23:40

Obviously $1$ is in the span of them, so that's no problem...

Daminark

Oh rip

Hey Ted!

What criteria do you have? Just linear independence/span?

Ted Shifrin

You can reduce to 4x4 easily enough ... the constants are not needed to be checked.

Daminark

Okay so looking at the vectors in question it is easy

The numbers match up nicely

Or whoops I misread, F

user193319

The only thing I can think of is to prove a more general result: if $v_1,..,v_5$ is a basis for some vector space $V$, then $v_1, a_1v_1+v_2, a_2v_1+v_3,a_3v_1+v_4, b_2v_2+b_3v_3+b_4v_4$ is a basis...

that looks just as horrible...

Daminark

Okay so what I'd recommend doing here is to still make use of these numbers, if you take the last guy, subtract 203 times the second to last guy, and so on, you'll kill all the terms of the last guy except for x, so this gives you x in the span

Once you know that, then the others are dominoes

user193319

23:46

Okay. I'll give it a try. Thanks!

Leaky Nun

I like how Ted immediately gave up teaching :P

Daminark

Rip

Leaky Nun

rip demonark

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