Mathematics - 2018-04-11 (page 1 of 4)

Eric Silva

00:00

@Daminark the only undergrads there that i recognize are second and third years tho lol

MatheinBoulomenos

I'm really looking forward to the next semseter

Mike Miller

Homological algebra is too hard

Kasmir Khaan

MATHEIN :D

my hero :D

@MatheinBoulomenos wake up :D

MatheinBoulomenos

Am I really your hero? I'm so sorry that I forgot :/

Kasmir Khaan

haha

modest as allways =p

btw you were right

about that adress thing ><

It is working now :D did not receive fail messages =p

mathein !did you do logic?

MatheinBoulomenos

00:13

no, I'm not doing any logic courses, just some self-study

though I didn't get very far as of yet

Kasmir Khaan

hmm what book you using for that?

i find that the subject is very odd

I would recommand van Dalen's book ( it is hard but I think it is the right thing for ya =p

MatheinBoulomenos

I'm using a German book by Rautenberg, I don't know the title of the English version

Kasmir Khaan

we using that kompendium that i sent you

MatheinBoulomenos

oh, so you're taking a logic course?

Kasmir Khaan

yep :)

MatheinBoulomenos

00:15

How is it, other than odd?

Kasmir Khaan

I think i need that to be able to understand proofs better

well, it is very cryptic as first encounter

I assume one will get the hang of it when practice it more

it is like doing math but in a very axiomatic way

to compute 5+3 , it is done like 5+s(2) = s(5+2) = s(s(5+s(1) ) = s(s(s(5+s(0)))) = s(s(s((5) = s(s(6) = s(7) = 8 :D

just a random example

where one defines s to be the succesor of n

the natural number that comes after n

also we work with propositions and formulas instead of numbers

MatheinBoulomenos

ah, so that's Peano axioms

Kasmir Khaan

hmm i heard that name before ._.

anyway we also have interpertations of propositions

which is something new

it is like an evalution idea

a proposition like P_i can be anything, eg a polynomial

and we evaluate it a x=3

we write taht as [P_i]^A

A is the interpretation

very abstract ._.'

I think youll find it very joyfull, cuz you like that stuff =p

MatheinBoulomenos

yeah, it's definitely something I'd like

but I'm at a point now where it's more useful to specialize than to pick up the basics in a lot of different fields

Kasmir Khaan

how is it going with your french ?

hmm i thought you were specializing in algebra

MatheinBoulomenos

00:24

I meant that's why I probably won't do that much logic

I love all kinds of algebra, but it's looking more like number theory right now

Kasmir Khaan

neat :D

ehm btw mathein ._.

have you done those rep theory stuff that i sent ya ?:D

MatheinBoulomenos

most profs here which are algebraically oriented use algebra to do number theory

Kasmir Khaan

well many unsolved problems in number theory , so it is a good subject for you

MatheinBoulomenos

yeah I totally forgot that because I had so much stuff to do and I had a family visit etc.

I can do it right now if you want

Kasmir Khaan

haha :D allways cool with family :D

if you have a feel for it , i would be very glad :D

MatheinBoulomenos

00:26

I think I can finish it today I think

Kasmir Khaan

well that would be very cool _

dont stress yourself , a couple of them is fine :D

Daminark

@Mathein sorry I had a fire drill, I'm doing alright, thanks!

Zee

00:42

Hello friends

Leaky Nun

01:01

@MatheinBoulomenos maybe the homological algebra course at my uni is too easy

Nicholas Roberts

02:05

Hey!!

Assume $\alpha$ is a constructible number. I.e., it was formed by repeated arithmatic operations and square root operation on rational numbers.

We want to show that this implies a chain of field extensions with each neighboring field in the extension being a degree 2 extension above the one previous to it.

Ideas?

*existence of such an extension

Leaky Nun

@NicholasRoberts induction :P

Nicholas Roberts

Sure. But can yoou give me a jumpstart?

Leaky Nun

induct on how the number was formed

i.e. the "complexity" of the number

Nicholas Roberts

Ok, so base case is taking one root

Leaky Nun

the base case is rational numbers :)

Nicholas Roberts

02:16

Clearly, a degree two extension

over Q?

Leaky Nun

so here's how your "constructible number" would be formalized

inductive def "constructible number" : real -> true/false
| rational: x in Q -> x is constructible
| add: x is constructible and y is constructible -> x+y is constructible
| (repeat for the other three operations)
| square_root : x is constructible -> sqrt(x) is constructible

so the constructible numbers are inductively defined

Nicholas Roberts

I see, this is just one operation each

Leaky Nun

according to the rules that
1. every rational number is constructible
2. the sum/difference/product/quotient of two constructible numbers are constructible
3. the square root of a constructible number is constructible

and 4. every constructible number is formed this way

so you need to do induction on the inductive definition

working with proof assistants has trained me to think in this manner lol

and it's very helpful

and rigorous and precise

Nicholas Roberts

Ok. So tell me where degree 2 extensions will come into play. When we take a root thats not contained in the ground field?

Leaky Nun

right

Nicholas Roberts

02:20

Ah.

Leaky Nun

and your base case is 1. every rational is constructible

who is your base field?

Nicholas Roberts

Ok. So how would you formally state the base case. You would split it up into the 5 operations

Leaky Nun

the base case is the rational numbers

inductive definitions come with recursors

Nicholas Roberts

But theres no degree 2 extension involved. I feel like the base case you need a degree 2 extension

Leaky Nun

13 mins ago, by Nicholas Roberts

We want to show that this implies a chain of field extensions with each neighboring field in the extension being a degree 2 extension above the one previous to it.

a chain has to start somewhere :)

23 secs ago, by Leaky Nun

inductive definitions come with recursors

Nicholas Roberts

02:21

I see. Ok

Leaky Nun

there is no difference between induction (proofs) and recursion (programs)

curry-howard correspondence :P

Nicholas Roberts

Ok so inductive hypothesis. Assume it is true for k <= n

Leaky Nun

nah that's just induction on the natural numbers

Nicholas Roberts

I understand the idea but confused how to phrase it

Leaky Nun

induction can be generalized so much lol

Nicholas Roberts

02:23

Dang Lol

Leaky Nun

inductive def "natural numbers"
| zero : zero is a natural number
| succ : if n is a natural number, then succ n is a natural number

so the recursor looks like:
if you want to prove that P(n) is true for all natural number n, you need to:
1. show that P(zero) is true
2. show that if P(n) is true, then P(succ n) is true

Nicholas Roberts

Ya, ive seen that

Leaky Nun

inductive def "constructible number" : real -> true/false
| rational: x in Q -> x is constructible
| operations: x is constructible and y is constructible -> x+y, x-y, xy, x/y are constructible
| square_root : x is constructible -> sqrt(x) is constructible

recursor:
1. show that x in Q -> P(x)
2. show that P(x) and P(y) imply P(x+y), P(x-y), P(xy), P(x/y)
3. show that P(x) implies P(sqrt(x))

2 might be a bit hard though, i don't immediately see how to do 2

but do you get it

Nicholas Roberts

Hmmm

Yes, different induction than im used to. Thank you!

Leaky Nun

nobody teaches that at uni

people working with proofs assistants will know that though

Nicholas Roberts

02:27

Im in grad school, havent seen it ever :(

Leaky Nun

mathematics is going to change with the advent of proof formalizers

but that's just me talking

I've seen something like that in the mathematical logic course though

oh you come from a programming site

then logic should be easier lol

@NicholasRoberts do you have CS background?

or mathematical logic?

Nicholas Roberts

A little bit of coding from my undergrad years

Basic stuff tho. Website dev, python, little bit of java

Leaky Nun

formal language / regular expression / etc?

oh

so not much CS

Nicholas Roberts

little bit of regex when i made my website

for requiring passwords to be of certain length, etc.

Leaky Nun

I mean the mathematical side of regex

Nicholas Roberts

02:30

Dont know that lol

Leaky Nun

nvm

Nicholas Roberts

You like De Rham groups?

Leaky Nun

have you taken any mathematical logic course / model theory course?

Nicholas Roberts

No :(

Leaky Nun

@NicholasRoberts not very familiar with de rham cohomology

Nicholas Roberts

02:31

Its very nice!

Leaky Nun

i just learnt some basic homology last few weeks and some basic homological algebra last week lol

Nicholas Roberts

Very nice. Exact sequences, chain complexes, Ext functor?

Leaky Nun

7 mins ago, by Leaky Nun

inductive def "constructible number" : real -> true/false
| rational: x in Q -> x is constructible
| operations: x is constructible and y is constructible -> x+y, x-y, xy, x/y are constructible
| square_root : x is constructible -> sqrt(x) is constructible

corollary: the constructible numbers are countable

@NicholasRoberts right

weibel is quite high on category theory though

too high for me lol

Nicholas Roberts

Whos that

Leaky Nun

weibel, introduction to homological algebra

Nicholas Roberts

02:34

Have you ever heard of Alexander Kirrilov? Hes me algebra professor

Leaky Nun

haven't heard of him

Nicholas Roberts

Ah, him and his father are into Representation theory. Brilliant algebrist

Leaky Nun

I see

so for 2

if x lives in Q(a,b,c,d,e) ⊃ Q(a,b,c,d) ⊃ Q(a,b,c) ⊃ Q(a,b) ⊃ Q(a) ⊃ Q

and y lives in Q(a,b,f,g) ⊃ Q(a,b,f) ⊃ Q(a,b) ⊃ Q(a) ⊃ Q

then I suspect x+y, x-y, xy, x/y live in Q(a,b,c,d,e,f,g) ⊃ ... ⊃ Q

Nicholas Roberts

seems reasonable

Leaky Nun

and some of them might be degree 1 extensions

Nicholas Roberts

02:38

If you adjoin everything then x and y live there

Leaky Nun

so remove those that are degree 1

right

Nicholas Roberts

hence, sums, products of then live there

them*

Leaky Nun

you might have some lemmas about the "join" extensions

"join" as in poset join

The Testosterone Fanatic

03:11

@BalarkaSen it's a weakness of the English language which makes it so prone to misinterpretation; but then again, I think you've done that deliberately for humor

Secret

My "art career" in mathematics: Illustrating infinite objects and crazy sets, types, whatever

Secret

03:44

[Random]

blogs.iac.gatech.edu/yadystopia2017/2017/02/07/…

Differential variance

au.mathworks.com/help/images/ref/stdfilt.html

To be explored later: Vector field of a variance function

Silent

Is there a non trivial group homomorphim for which image is cyclic group, but domain is not?

@LeakyNun

Leaky Nun

sure

Silent

04:00

@LeakyNun one is sign homomorphism from $S_n$ to $\{-1,+1\}$

Leaky Nun

sure

Secret

04:11

[Random] broadening of some infinitesimal indicator function (rough sketch):

Let $\delta(s)$ where $s \in \Bbb{R}^*$ be some indicator function/distribution in the hyperreals

we are interested in the following convolution:

$$\int_{\Bbb{R}^*}\frac{1}{\sqrt{2 \pi}}e^{\frac{s^2}{\sqrt{\sigma}}}\delta (s)ds$$

Zee

God I hate convolutions

Secret

One idea that is currently had in mind (and to be formalise later) and inspired from some of the PhD work I am reading is we have a molecule with many energy levels, some of these may be closely packed together so that it is hard to see that they are nearly degenerate

However, as described in the multiwfn user manual, a gaussian convolution will make that degeneracy very visible since it will be broadened into a tall peak compared to a stick diagram with no degenerate energy levels

Here, I want to abstract this one step further by making it so that those sticks (represented by the delta distributions here (I really should have used the symbol 1_A for indicator functions, will deal with that later)) will never resolve however finitely deep it is zoomed in and wonder what happens to that integral and whether it will give me a peak of unbounded height

Thus in a sense, given a interval with n sticks, such that their standard parts are identical, what will happen when we integrate over this hyperreal interval

Zee

I can’t tell if your an idiot or a genius

Secret

I may not have the required background yet, but I must quickly jot down an idea asap before it dissipates and cannot be recovered

most crazy ideas tend to fail or no go, this is expected. what I concern is the properties of the ideas themselves

You can say I am experimenting with ideas themselves

Maths is relatively easy in this regard as you don't always need a lab to check whether something works and how it behaves

Having said that, the computation and analysis time is comparable or even more than the more experimental sciences

I think one issue of mine is that I tend to explode with ideas that can be potentially converted into research projects, and with mixed success rates

Abcd

Can a function be both odd and symmetric about the line x=a at the same time? I don't think so...

(@LeakyNun Any idea?)

Leaky Nun

04:24

sin x symmetric about x=pi/2?

Secret

@Zee Btw, I am not a genius, I am just a normal person with crazy ideas that may or may not work

Abcd

@LeakyNun yes

@LeakyNun thanks

Zee

@Secret having crazy ideas is certainly a necessary condition

Feynman said science is about putting a straight jacket on your imagination so their is a yin and yang thing goin

one extreme is being a crackpot another is being technical but not interesting

Secret

04:40

One reason I discuss my ideas openly with you guys is to avoid the crackpot thing. While I do have some background in mathematics and also self study a lot, everybody knows how strong a bias of personal universes on one's thinking, thus people in the mainstream will help keep me in track and see bias that I cannot see

so whenever I have these crazy ideas, I am actively engaging with the mainstream and take their advices, which often lead to more interesting things learnt for both parties

Secret

04:55

and hence, the sufficient condition is to ensure not to veer off into either extremes as you mentioned

Zee

math is nice since nobody cares how crazy your ideas are if you can show solid mathematical arguments

Secret

that is true, because maths can be largely independent of reality

MatheinBoulomenos

@LeakyNun why do you think that?

Silent

this says that if we are ready to screw up ordinary addition for $\Bbb Z$, we can get an integer vector space. My question is: is there any field for which integers give vector space under usual addition?

Secret

05:11

well, for arbitrary fields, what will integer mean?

Leaky Nun

@MatheinBoulomenos it just covers the basics

Tor and Ext

some basic group cohomology

Ring of integers

In mathematics, the ring of integers of an algebraic number field K is the ring of all integral elements contained in K. An integral element is a root of a monic polynomial with rational integer coefficients, xn + cn−1xn−1 + … + c0 . This ring is often denoted by OK or O K {\displaystyle {\mathcal {O}}_{K}} . Since any rational integer number belongs to K and is an integral element of K, the ring Z is always a subring of OK. The ring Z is the simplest...

Secret

I see

btw, I also found this:

7

Q: Finite fields as vector spaces

I'm having great difficulty understanding this topic. Can someone concretely explain what it is meant by thinking of $GF(q^2)$ ($q$ a prime power) as a two-dimensional vector space over its subfield $GF(q)$ (fixed by the map $x \mapsto x^q$). How would I construct a basis of this vector space? ...

Kirill

05:34

morning

Leaky Nun

@Kirill guten Tag

Kirill

From the proof I am in:
We observe a unit cube in some $d$-dimensional space and want to cover all its vertices except 0 with hyperplanes.

Let $h_1,h_2,\ldots,h_m$ be these hyperplanes. Then $m\ge d$. Equation of the hyperplane $h_i$ is $a_{i1}x_2 + a_{i2}x_2 + \ldots + a_{id}x_d$.

As none of the planes go through $0$, we can assume that $b_i$ are all $1$.

It is clear why $b_i$ is not $0$. But why it equals $1$, not $2$, for example?

I should write $a_{i1}x_1 + a_{i2}x_2 + \ldots + a_{id}x_d = b_i$ though.

Secret

06:25

[Random]

About geometry

Consider a cylinder vs a column of spherical points placed together so that it simulates a cylinder

to be worked out on how computers make use of that to simulate a continuum and hence find the common geometric criteria that a continuum must have

Results may or may not generalise to topological continua

Abcd

06:43

$f: \mathbb R \to \mathbb R, f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17 \space \forall x \in \mathbb R $, then find the function $f(x)$

I have:

f(15/4)= 16/3

f(3)= 3

f(5)= 7

... I am not getting anything fruitful from these....

(@LeakyNun any idea(read: hint) ?)

Can anyone give me some hint on how to proceed?

Dawood ibn Kareem

07:28

@Abcd is there more information that you haven't told us? It seems to me that the one condition you've given can't possibly uniquely define the function. Partly because it says nothing at all about values of $f(t)$ where $t$ is negative.

Gaurang Tandon

@Abcd obviously f(x) is linear; assume f(x) = ax+b and then solve for a and b using the last two conditions

@Abcd this is probably a typo

Abcd

@GaurangTandon why is f(x) linear?

Gaurang Tandon

@Abcd if you don't get why it can only be linear, assume it to be a quadratic; then the degree on the LHS will be $x^4$ while that on RHS will be $x^2$ only

Tobias Kildetoft

@GaurangTandon Why should it be a polynomial at all?

Abcd

@DawoodibnKareem sorry. $f: R^+ \to R$

Gaurang Tandon

07:29

@TobiasKildetoft because RHS is polynomial...?

Balarka Sen

@GaurangTandon $e^x$ is not a polynomial, but $e^x - e^x = 0$ is.

Tobias Kildetoft

@GaurangTandon At that is resulting from a polynomial expression after applying the function to a pair of polynomials. It is not clear to me that this makes the function itself a polynomial

Gaurang Tandon

hmmm okay :/ i just looked for the simplest solution; $f(x)=2x-3$ :P

Balarka Sen

Yes but that doesn't uniquely determine $f$.

Gaurang Tandon

alright, I give up

Abcd

07:33

2 mins ago, by Gaurang Tandon

hmmm okay :/ i just looked for the simplest solution; $f(x)=2x-3$ :P

The answer given is this^

7 mins ago, by Abcd

@DawoodibnKareem sorry. $f: R^+ \to R$

@DawoodibnKareem Please ignore this. $f: R \to R$, I have correctly mentioned all details in my initial question/

@BalarkaSen I don't see the problem with his solution

f(x) has to be linear function.

otherwise we'd get biquadratic on LHS and quadratic in RHS

Balarka Sen

@Abcd That's assuming $f(x)$ is a polynomial function at all.

Abcd

Okay, its author's fault. he should have mentioned it.

Dawood ibn Kareem

07:54

So the question is that $f$ has to be polynomial? Or not?

Balarka Sen

@Abcd It's funny because in case you had $f(x^2 - x + 3) + 2f(x^2 - 3x + 5)$ in the left hand side, you'd write $g(x) = f(x^2 - x + 3)$ so that your equation would become $g(x) + 2g(2 - x) = 6x^2 - 10x + 17$. Switch $x\mapsto 2 - x$ and you'd get a system of two linear equations solving which gives you $g(x)$.

Tobias Kildetoft

It is not even completely clear to me that $f$ would have to be continuous.

Balarka Sen

But in this case trying that trick would give $g(x) + 2g(x - 2)$, which is Not Good because you can't use symmetry

Silent

Is is possible that $H$ and $K$ both non-normal subgroup of $G$ and the product group $HK$ is subgroup?

@BalarkaSen

Tobias Kildetoft

@Silent Yes

@Silent For the easiest example, assume one is contained in the other.

Silent

07:58

oh!

Dawood ibn Kareem

What happens if you write $f(u^2+\frac{11}{4}) = g(u)$ ?

Balarka Sen

Oh, I guess $(1 - x)^2 + (1 - x) + 3 = x^2 - 3x + 5$

So if I write $g(x) = f(x^2 - x + 3)$, the equation becomes $g(x) + 2g(1 - x) = 6x^2 - 10x + 17$ anyway

Let's try this. $g(1 - x) + 2g(x) = 6x^2 - 22x + 33$.

That means $3g(x) = 2(6x^2 - 22x + 33) - (6x^2 - 10x + 17)$

Dawood ibn Kareem

Oh, I see. Your approach is equivalent to mine. You just started earlier! (I went out for half an hour. I just got home).

Balarka Sen

Hahah actually what you wrote inspired me to look for another substitution other than $t = x - 2$, with which I was hung up on for some reason

So I might be doing some calculation errors, but this should do the trick

can't be arsed to finish this :3

Dawood ibn Kareem

No arsing will occur. I'm sure Abcd can do it from here.

Balarka Sen

08:06

@BalarkaSen Bleh, this is the garbage step. Forgot to $x \mapsto 1 - x$ in the RHS

I need covfefe

Dawood ibn Kareem

Negatively pressed!

Balarka Sen

@abcd Just to write it down clearly for you, notice that $(1 - x)^2 + (1 - x) + 3 = x^2 - 3x + 5$, which implies if you write $g(x) = f(x^2 + x + 3)$, then $g(x) + 2g(1 - x) = 6x^2 - 10x + 17$, ie $g(1 - x) + 2g(x) = 6(1 - x)^2 - 10(1 - x) + 17 = 6x^2 - 22x + 33$ (I did do this correctly, huh). This is a system of linear equations on $g(x)$ and $g(1 - x)$ - solve for $g(x)$.

No ansatz fuckery needed

Leaky Nun

22 mins ago, by Balarka Sen

@Abcd It's funny because in case you had $f(x^2 - x + 3) + 2f(x^2 - 3x + 5)$ in the left hand side, you'd write $g(x) = f(x^2 - x + 3)$ so that your equation would become $g(x) + 2g(2 - x) = 6x^2 - 10x + 17$. Switch $x\mapsto 2 - x$ and you'd get a system of two linear equations solving which gives you $g(x)$.

if you treat $x$ and $2-x$ as $a(t)=t$ and $b(t)=2-t$ and think of them as particles

you'll see that the correct involution is reflection around 1

Balarka Sen

Oh, I see what went wrong. @abcd Correction: it's $6(1-x)^2 - 10(1 - x) + 17 = 6x^2 - 2x + 13$. Wasn't taking care of the sign.

@LeakyNun Not for that one. That had a $x^2 - x + 3$ in it; Abcd's original had a x^2 + x + 3

Leaky Nun

oh

I went to sleep in that hour

since Abcd pinged me

2 hours ago, by Abcd

(@LeakyNun any idea(read: hint) ?)

Balarka Sen

08:21

I flip around with signs so much I confuse myself

Leaky Nun

I just woke up

Dawood ibn Kareem

Flipping the sign inside the brackets was Balarka's master stroke.

Balarka Sen

Lol

Dawood ibn Kareem

I don't think I'd have thought of that.

And if you don't do that, you don't get the system of linear equations.

Balarka Sen

Oh I see you meant using the symmetry

I thought your were pulling me leg for the sign errors :p

On a tangential parable, that day I was differentiating something dumb like $a^2 x^2$ where $a$ is a constant. For some magical reason the answer became $2ax$ and I didn't notice until I was halfway through the problem

Dawood ibn Kareem

08:25

Yes, having $x$ and either $1-x$ or $2-x$ in the brackets, or however it worked out. Instead of $x+1$ or $x+2$ or whatever.

This is very hard to type! The key between Z and C on my keyboard is knackered, which makes this kind of algebraic manipulation particularly difficult.

Balarka Sen

Right right right it was a little confusing at first because $(1 - x)^2 + (1 - x) + 3 = (x - 2)^2 + (x - 2) + 3 = x^2 - 3x + 5$.

@DawoodibnKareem Use emojies for variables

Dawood ibn Kareem

OK, I'll do that neckst time.

Balarka Sen

lol @ avoiding x

Dawood ibn Kareem

That's easy for you to say!

Of all the keys on my keyboard to wear out, I wouldn't have eckspected that one to be the first.

Leaky Nun

$f(🙂^2+🙂+3)+2f(🙂^2-3🙂+5)=6🙂^2-10🙂+17$

eckspected

Balarka Sen

08:30

@LeakyNun please. real ones only use thinking emojies

Leaky Nun

$f(🤔^2+🤔+3)+2f(🤔^2-3🤔+5)=6🤔^2-10🤔+17$

Balarka Sen

@DawoodibnKareem I have a feeling you don't want to be saying "fox" in the chat for some time

Leaky Nun

focks

what happened to cox

Dawood ibn Kareem

fo$\times$

Leaky Nun

cox.com/residential/home.html

"Cox provides TV, Internet, Digital Telephone, home security and tech solutions services for its residential customers. Get access to fastest digital life with Cox."

Dawood ibn Kareem

08:35

We still don't know $f(y)$ if $y < \frac{11}{4}$, without an additional restriction on $f$.

Balarka Sen

Ah right because nothing is said about $f$ in that domain.

I reiterate my frustration with functional equation from yesterday; none of them are rigorously stated.

Dawood ibn Kareem

I can't remember whether or not Abcd sayd that $f$ had to be a polynomial. If not, he'd better find a way to define it for $y < \frac{11}{4}$.

@BalarkaSen I didn't see your frustration yesterday. Today is my first day in this room for several years.

Balarka Sen

Ah, welcome!

Dawood ibn Kareem

Why thank you.

Leaky Nun

welcome back, I guess?

@BalarkaSen nobody told you to do it

just take the universal closure of the formulas stated

Balarka Sen

08:41

@LeakyNun My high school exams did!

Leaky Nun

did tell you to do it or did take the universal closure?

Balarka Sen

Yes.

Dawood ibn Kareem

@BalarkaSen That's part of what makes functional equations interesting though. You have to think about whether you're allowed to use continuity, differentiability, or whatever general attribute of functions you're interested in, before you go blindly applying it.

I used to love functional equations when I was a teenager.

Balarka Sen

I like the bag of tricks, actually. But I just wish they were stated properly in most places.

Leaky Nun

@DawoodibnKareem I never use continuity / differentiability if they don't say so

I'm not those people who think that every function is analytic

Dawood ibn Kareem

08:43

Sure, but sometimes, one can think "do I need this function to be analytic in order to solve this problem".

And sometimes, you can assume the function is analytic, solve the problem, then prove that you didn't need the assumption in the first place.

Leaky Nun

I see

Zee

Hello @DawoodibnKareem

Dawood ibn Kareem

Hello Zee

Balarka Sen

@LeakyNun Sometimes the functional equations are more motivated than what you see in a typical olympiad exam. I expect that the physicists stumble upon such things quite often

Leaky Nun

theorem: every function $\Bbb R \to \Bbb R$ can be split uniquely to its even part and its odd part

Balarka Sen

08:44

So it's not unreasonable in that case to assume it has an analytic solution

Leaky Nun

feel free to interpret this in a linear-algebra manner

Dawood ibn Kareem

@LeakyNun Isn't that kind of obvious?

Do you want someone to furnish a proof?

Leaky Nun

@DawoodibnKareem is it?

well it's kinda obvious

Balarka Sen

The splitting works like $f(x) = (f(x) + f(-x))/2 + (f(x) - f(-x))/2$.

Uniqueness takes a little more work.

Leaky Nun

and then interpret it using linear algebra

Zee

08:46

@DawoodibnKareem is making leaky drink his own medicine

Leaky Nun

@Zee what is my medicine?

you really do speak with a sting in your mouth

Zee

“Isn’t it kinda obvious “

Balarka Sen

Well I remember that every matrix can be decomposed into a symmetric and an anti-symmetric part (the proof works similarly)

But I don't see if that's an "interpretation"

Leaky Nun

maybe @Zee wants to either piss off everyone or get pissed off of everyone

Balarka Sen

@Leaky Hint: ignore

Zee

08:47

@LeakyNun am just teasing , don’t take what I say as hostile

Tobias Kildetoft

The tensor square decomposes into the symmetric square and the exterior square

Balarka Sen

Everybody already has him on ignore.

Tobias Kildetoft

(outside characteristic $2$ that is)

Leaky Nun

@BalarkaSen hint: direct sum

@BalarkaSen tilt proof tilt proof

Zee

You know , I am slightly rude but am a good guy , I dont put anybody down or insult somebody , it’s all just little pokes , unlike some other people

Balarka Sen

08:50

@LeakyNun Hmmm

Leaky Nun

$\otimes^2 k^2 \cong (\Lambda^2 k^2) \oplus (\Sigma^2 k^2)$

(unrelated)

Balarka Sen

Yeah 's what Tobias said. It's a good fact

Leaky Nun

oh right

@Zee unlike which other people?

Tobias Kildetoft

The decomposition also works as representations if $V$ is a representation of a group (again, with the characteristic not being $2$).

In characteristic $2$ it is possible for the tensor square to be indecomposable

Balarka Sen

@LeakyNun Did you mean that the vector space $C^0(\Bbb R, \Bbb R)$ can be direct-sum decomposed into the vector subspace of even/odd functions?

Leaky Nun

08:52

@BalarkaSen right

Balarka Sen

Gotcha

Nice parsing.

Dawood ibn Kareem

Balarka, I don't think you finished Abcd's problem.

I don't think you've proved that $f$ exists.

Sure, you found $g$, but that's not enough.

BAYMAX

Why $x_{2} \frac{\partial{}}{x_{1}} - x_{1}\frac{\partial{}}{\partial{x_{2}}}$ is a vector field?

Leaky Nun

@BalarkaSen I have a proof that there exists no non-trivial zeroes of Riemann zeta with real part not a half

Dawood ibn Kareem

Because if I want to know $f(y)$, I need to find $\times$ such that $\times^2+\times+3 = y$, which clearly isn't unique.

Balarka Sen

08:54

@DawoodibnKareem Yeah, that's a little worrying. The point is once I have $f(x^2 + x + 3) = g(x) = \text{blah}$, on the domain where $x^2 + x + 3$ is invertible, I can prove $f$ exists.

Dawood ibn Kareem

You need to prove that the two possibilities for $g(\times)$ are the same.

Yes, but $f$ is from R to R.

Balarka Sen

yeah.

Ugh.

Dawood ibn Kareem

You can't restrict yourself to where $\times^2+\times+3$ is invertible.

Sorry, I'm a bit sleepy, which is why I was slow to come to this realisation.

I suspect $f$ probably doesn't exist, in which case Abcd's problem can't be solved.

Balarka Sen

Yeah I doubt this can be worked out without serious restriction on the domain.

But prove me wrong!

Dawood ibn Kareem

Ficks my ecks key and I'll consider it.

(Although occasionally, it worx just fine).

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