@ShineOnYouCrazyDiamond what's up

@Ultradark I'm not sure how to get to your MSE questions

I am in India

Morning

Morning @ÍgjøgnumMeg

:) How's it going

not bad

@ÍgjøgnumMeg do you speak german?

Indeed :)

what does zimmermann mean?

@ÍgjøgnumMeg

carpenter

are there a lot of zimmermann's in germany?

eh

not particularly

are there a lot of people with that last name in germany?

yes, it's pretty common

oh cool that's my last name

what the connection between deminsion and rank in this theorem: Let {\displaystyle f:X\to Y} f:X\to Y be a smooth map, and let {\displaystyle y\in Y} y\in Y be a regular value of f; then {\displaystyle f^{-1}(y)} f^{-1}(y) is a submanifold of X. If {\displaystyle y\in {\text{im}}(f)} {\displaystyle y\in {\text{im}}(f)}, then the codimension of {\displaystyle f^{-1}(y)} f^{-1}(y) is equal to the dimension of Y.

what the meaning of codimension?

Can't wait to put a bounty on this

@Sila dimension of environment minus dimension of self

Zimmermaenner

@Ultradark Isn't that an opinion-based question that isn't allowed on math.stackexchange ?

Yeah

it's not my question btw

hello everyone. is there anyone with experience in numerical computations?

How I can show that this manifold:#X:={(x,y,z)∈#R^3#∣x^3+y^3+z^3−3xyz=1}# is not compacted? I want a hint to show that either it's not closed or it's not bounded, or both!

@sila I suggest considering lines through the origin and seeing how many intersections are possible.

What is best to prove a quadratic function to be surjective by logic let say a simple quadratic equation ax^2+bx+c

← 1 message moved from CRUDE

I have written a recursive algorithm to generate some collatz sequences, and I wonder where can I get help omptimizing it making it more efficent?

Zeroth law of probability: If an event has zero probability to happen in a given measure, it will remain so for all measures

@Secret that doesnt make sense (or is just plain wrong)

Won't Pr(x)=0 under some probability measure $\mu$ must be preserved under any probability measure $\nu$?

The empty set under any probability measure should always return the value zero

and likewise for measure zero sets (actually let me check that real quick...)

On $\{a,b\}$ take a look at the two measures $\mu_a$ having mass $1$ at point $a$ and $0$ at point $b$ and $\mu_b$ having mass $0$ at point $a$ and mass $1$ at point $b$

the example I was going to give was of a two-headed coin vs. a fair coin

Ah right

so the two measures need to be related by a measure preserving transformation for that to be true

which is obviously not the case in the above example

@Semiclassical, there is a lot of lines , this mean that this space is not bounded?

what do you mean by "a lot of lines"?

there's a lot of lines through the origin which intersect the sphere x^2+y^2+z^2=1, but that's perfectly bounded

@Semiclassical, I mean that there are many lines through the orgin that intersect the #X:={(x,y,z)∈#R^3#∣x^3+y^3+z^3−3xyz=1}# .

yes,I do not know How to show that this space is not compacted! How the lines through the origin and their intersections with the space $X$ can show this?

If $G$ is a group, what does $\ell^{\infty}(G)$ denote?

@Secret it is true for those two measure, look at the map $f(a)=b$, $f(b)=a$, then $f^*(\mu_a)=\mu_b$

if $\mu_1 \ll \mu_2$ (meaning $\mu_1$ is absolutely continuous wrt $\mu_2$) then any measure $0$ set of $\mu_2$ is a measure $0$ set of $\mu_1$ though

Well, I initially thought if a set is measure zero under one measure, it will be measure zero in any measure, but this reasoning of mine is wrong

because this is only true if the measures can be related by a measure preserving function, of which $\mu_1 \ll \mu_2$ is a special case of it

@user193319 bounded functions $G\to\Bbb C$ with sup norm

Just like for any set

If $G$ is locally compact you can construct all of the $L$ spaces wrt to the Haar measure, if no topology is specified then what I said above is the same as assuming the discrete topology (where the Haar measure is the counting measure)

Mathematics is like secret poetry

Sup Norm, how's it goin?

Oh, you meant $\sup$

@AlessandroCodenotti Hmmm just bounded or essentially bounded? Why isn't the notation $L^{\infty}(G)$ used?

Because it has $\infty$ in it and it looks cool

You wanna be cool, stay in school.

@user193319 what's the difference with the counting measure?

Same reason why $\ell^\infty$ is not written as $L^\infty(\Bbb N)$, it looks like there's a measure different from the counting one involved

I can't see how $m/n=m'/n'$ implies $\frac{m}{n}_F=\frac{m'}{n'}_F$

m/n=m'/n' iff m n' = m' n, the same with the $F$ as 1/n _F always exists in char 0

@s.harp So char 0 is needed to make sure that we are not dividing by zero, and to show injectivity, no need of char 0?

@Silent the argument seems too laborious. If $F$ has char 0, then the unique ring homomorphism $\Bbb Z \to F$ sends non-zero elements of $\Bbb Z$ to non-zero elements in $F$, hence units, thus it extends to a ring homomorphism $\Bbb Q \to F$ by the universal property of localization. A non-zero ring homomorphism from a field is always injective

@Mathein: That argument is way too sophisticated for the audience of this question.

Hey @Ted :)

you're probably right

hi @Mathein :)

@Ted do only rec letters from profs "count"? I can get a recommendation letter from an alumnus for the school I want to apply to, but he's only a post doc

@MatheinBoulomenos the steps of what you say likely use the argument carried out in that exercise

@Mathein: A postdoc is probably OK, if he knows you pretty well. I don't know what the usual situation in Europe is, but typically here grad schools want 3 or 4 letters. So if you have 2 or 3 strong ones from respected faculty, then one more should be fine.

Hello @TedShifrin.

@MoneyOrientedProgrammer Yes, both answers are 2000, no problem with that. Hi!

@TedShifrin ok, thanks

Hi people! Anyone here who knows anything about the Kramers-Kroning relations?

apparently they are "called in maths" Sokhotski–Plemelj theorem.

K-K is one of those pieces of math-physics that I never really remember why worked

I mean, what you get is neat

But the details are fuzzy

@Semiclassical I share the feeling, currently

I have some function and would like to know if I can apply K-K. If so it would be kind of spectacular.

@JaspervanLooij Thank you. I am waiting for your next youtube video. :-)

@Semiclassical Well you could probably know. I want to know if I can apply it to the "mechanical" momentum density for varying mag. field as perturbation.

because I have a full control over $\Im{\vec{\pi}}$, for $\vec{\pi}$ the momentum density

while $\Re$ is what I like to derive.

But now $\vec{\pi}$ is an $\Bbb R^3 \to \Bbb C^3$ function

And on top I don't know the real time dependency on the perturbation, but only its convergence values, not sure if that matters. I also don't know if I can a priori assume its holomorphic.

then instead of using frequency as a variable I simply switch on some homogeneous magnetic field B..

What sets are equal in size to the algebraic numbers?

they are countable..

what?

algebraic integers is a countable set

yes

so if there is a bijection between the algebraic numbers and some other set, is that other set automatically the set of algebraic numbers?

so its in bijection to every other countable set, like $\Bbb N, \Bbb Z, \Bbb Q, GL_n(\Bbb Q)$ etc

oh nevermind

I know a set is in bijection to the algebraic numbers

but I don't know whether the numbers are rational, algebraic, or transcendental

so wait...

if it's countable.. that rules out the transcendental category

there are uncountably many transcendental numbers, but of course there are subsets of transcendental numbers that are countable

I see

I guess I can only conclude that I have a countable subset of algebraic numbers that match up with a countable subset of some unknown numbers

I can't figure out if this thing is a polynomial

It is.

Hello @Ultradark

@J.Doe hello

What were you wondering if it was a polynomial, out of interest.

$ 2^{\frac{2}{\log_2(x)}}=2^{\frac{2}{\log_2(1-x)}} $ seems to be an algebraic equation in disguise @J.Doe

$x=1/2$ after some manipulation

or just by inspection

Yeah, not a polynomial; though I don't know the definition of 'algebraic equation'.

lol oh it means polynomial equation

so you solve this equation

and it says "I'm going to take that root and raise it to a height governed by the original equation"

Height meaning?

height directly above the location of the root

so this is a "root raiser"

it's only function is to raise roots

I still don't follow. I'm guessing we're graphing the equation in the x-y plane?

@J.Doe yes

so that equation

had a solution of $x=1/2$

plug $1/2$ back into the original equation and you get a number (which is the height, or the y coordinate)

But the original function vanishes for $x=1/2$.

how so?

I may have misunderstood you. In my mind you're plotting $f(x) := 2^{\ldots}-2^{\ldots} = 0$.

the functions are equal so

just plug it into either

Hmm...

Go on.

the two functions are equal so you are finding where they intersect

so if you consider a family of these algebraic equations, you're really plotting the zero set in $(0,1)^2$

right?

ibb.co/Ssp6nV1

and by zero set, I just mean the set of zeros

You see the line going through $x=1/2$ in that picture

yes

what about it?

I mean isn't that what you're referring to? e.g. root raisers; numbers lying directly above $x=1/2$.

I was talking about generalising this to the space of all possible roots for integer exponents

in the example I gave: $2^{\frac{2}{\log_2(x)}}=2^{\frac{2}{\log_2(1-x)}}$

$2$ is the exponent

for both functions

but if you generalise this to $2^{\frac{s}{\log_2(x)}}=2^{\frac{t}{\log_2(1-x)}}$ for natural numbers $s,t$