12:22 AM
if i am minimizing a cost function, is it a usual method to find a nicer upper bound function for the cost function and minimize the upper bound function?

4 hours later…
4:07 AM
@Thorgott It's not quite natural to me. I haven't read a proof of the existence of that group, but if the construction uses some special properties of certain primes, then it should hold for infinitely many primes with that property. Bounding sounds strange to me.
my guess is that either the proof is nonconstructive, or constructed the groups for each possible primes one by one independently.

4:35 AM
anyways, it seems that Tarski monster group is not very well known

1 hour later…
5:36 AM
that's the #1 rule of tarski monster group

Rule #1: Nobody knows what the Tarski monster group really is.

2 hours later…
8:09 AM
Do students in some countries study a lot of set theory in universities? I see some people in MSE talk about proving statements with or without AOC. They talk about ZF, ZFC, transfinite induction etc. I mean, here in India, I've seen only a handful of people who have deep knowledge in logic and set theory.

pndas in america at least it is a very niche specialty, most math departments would not have anybody who would say set theory is one of their research interests. unlike, say, algebraic geometry, or PDE, where most math departments would have at least one person interested in one of those things.
at the student level it really varies. in my experience most universities do not (in keeping with the research interests of their faculty) teach much in the way of set theory beyond 'the basics' which can sometimes be absorbed in classes on other subjects by osmosis, or via 'naive' study which does not get into, say, the various axioms other than choice that one might or might not pair with ZF set theory.
having said that, a lot of people (maybe a majority of grad students) would be familiar with at least the rudiments of ZFC set theory.
for whatever reason, the axiom of choice is something that gets a whole lot of internet attention, well out of proportion to how often people think about it in real life. that is not an explanation at all of how often it comes up on MSE, but it is my only understanding of it.
category theory is something else like that.
at a lower level, there has probably been 100 zillion hours of internet discussion of the equality "0.9999..... = 1," or at a slightly higher level, of the various constructions of the real numbers, and yet, i think this subject matter takes up only a tiny, tiny slice of what anyone would actually receive in formal education or think about on their own if the internet didn't exist
in conclusion, all of our interconnected devices are sending our minds straight to hell, and there's nothing anybody can do about it
3

1 hour later…
9:31 AM
@leslietownes yep
Abolish set theory and category theory. Math is about good vibes and having fun.

10:02 AM
Jakobian, I'm still lost on what you meant by "consider $E_i$ as union of sets of the form $E_i\cap \bigcap_{j\neq i} E_j^{c_i}$ where $c_i=\pm 1$" Like, when is $c_i$ equal to $1$ and when is it equal to $-1$? Consider my drawing above. I've drawn a simple function where the sets are not disjoint, but I don't understand how to apply your formula.
I also do not understand if the $i$ in $c_i$ is the same as $i$ in $E_i$...

@psie $E_i\cap \bigcap_{j\neq i} E_j^{c_j}$ over all $c_j\in \{-1, 1\}$
$A^1 = A, A^{-1} = A^c$ is how you should understand this
$E_i$ is a disjoint union of all of those sets
So if your sets are $B_1, B_2, B_3$ for example, we find those $2^3-1 = 7$ disjoint sets and write $B_1, B_2, B_3$ in terms of them
So you have like, $\chi_B = \sum_{i=1}^7 c_i \chi_{A_i}$ lets say, where $c_i = 0$ if $A_i\cap B = \emptyset$ and $c_i = 1$ if $A_i\subseteq B$
and $A_1, ..., A_7$ are those $7$ sets
and you wrote other characteristic functions like this too
The value corresponding to $1\chi_{B_1}+2\chi_{B_2}+1\chi_{B_3}$ should be $\mu(B_1)+2\mu(B_2)+\mu(B_3)$
Now if you wrote $(\sum_i c_i^1 \chi_{A_i})+(\sum_i 2c_i^2\chi_{A_i})+(\sum_i c_i^3\chi_{A_i}) = \sum_i (c_i^1+2c_i^2+c_i^3)\chi_{A_i}$
where those are upper indexes and not powers
Then you just have to show that $\sum (c_i^1+2c_i^2+c_i^3)\mu(A_i) = \mu(B_1)+2\mu(B_2)+\mu(B_3)$
and indeed because if you take a look then $\sum c_i^1 \mu(A_i) = \mu(B_1)$ for example

10:24 AM
@Jakobian wow ok, thanks. Dumb question maybe, but won't $\bigcap_{j\neq i} E_j^{c_j}$ just evaluate to the empty set? Suppose we have three sets $\{B_1,B_2,B_3\}$. Then $$\bigcap_{j\neq 1} B_j^{c_j}=B_2\cap B_3\cap B_2^{-1}\cap B_3^{-1},$$ and by associativity this is just the empty set.

$c_j$ are already fixed when you write this intersection

ah ok, I think I understand

$B_2\cap B_3$, $B_2^c\cap B_3$, $B_2\cap B_3^c, B_2^c\cap B_3^c$
are the only possibilities here

10:58 AM
I have one more question. I don't understand how you obtained 7 sets. The sets you listed are only 4, and if we intersect them with $B_1$, we also only have 4 sets.

@psie You do the same thing with $B_1, B_2$ and with $B_3$

ok 👍

11:58 AM
boy, what a procedure :D and what we've shown today only shows that the integral of the simple function is not changed by making the sets disjoint on which it is defined on...the part about independence of representation remains kind of...

2 hours later…
1:44 PM
Hello and happy weekend MSE :)
I am a tad bit stumped by the last paragraph here. I see how $C$ as defined is a subfield of the extension field $K$ which we imagined exists, and how its an extension of $\Bbb R$.
What I cannot follow is how we've proven its the smallest such extension field of $\Bbb R$ in which $x^2 = -1$ is solvable
Certainly $i \in C$ and so $x^2 = -1$ solvable therein. But I imagine we would need to show two things, right: (1) for a subfield $F$ of $K$ we have $x^2 = -1$ solvable iff $i \in F$ and (2) $C$ is a subfield of every subfield $F$ containing $i$

@EE18 the question of being the extension is not shown
but $C$ is clearly the smallest containing both $i$ and $\mathbb{R}$ in $K$
because if a subfield of $K$ contains those two, then it contains $C$

Morning (or afternoon, not sure where you are), Jakobian. Possible to say a bit more about the first comment? I see the second now, that was silly

@EE18 were you expecting the book to write all the details?

Certainly not, and that's why I come here when I can't fill in the details myself

okay where are you confused?
@EE18 2) is not true
$x^2 = -1$ has only two solutions in $K$, $i$ and $-i$
so if $x^2 = -1$ is solvable in $C$, then it needs to be that $i\in C$ or $-i\in C$, where $-i\in C$ implies $i\in C$ anyway

1:58 PM
OK, I have thought about it some more and think I can argue as follows: (1) $i \in F \implies x^2 = -1$ solvable therein. Conversely, $x^2 = -1$ solvable means $x = \pm i$, as you've just said (this follows from Prop 8.18 in my book) and so by field closure, $i \in F$. (2) As you said above, I am missing a condition. I need also add that $\Bbb R \subseteq F$.
Thank you Jakobian! So, to confirm, we've now shown that $C$ is the smallest subfield of $K$ which is (a) an extension field of $\Bbb R$ and (b) has $x^2 = -1$ solvable. This is a different claim than the authors make right? Is this what you were alluding to in your first comment: "the question of being the extension is not shown"? That is, one can certainly imagine there being other fields having these properties which are not even related to $K$.
Obviously isomorphism will come into play here
But I just want to confirm that at the moment all we've shown or can show is that $C$
is the smallest subfield of $K$, and nothing more

@EE18 such field is essentially unique
but its not something the authors show

There is a theorem to that effect coming up, perhaps I will report back after reading it
They use the discussion we just had at the end of the proof tho so I wanted to be sure I understood what was being said

2:21 PM
OK returning now @Jakobian after having red the proof of the claim and want to confirm something with you if it's OK...
As you can see this continues on from where we left off
My only concern is with the final line in the proof about isomorphism

@EE18 and whats the concern

In particular, would we argue (the book leaves us to fill this in it seems ) that if we have any other extension field K of $\Bbb R$ with the $x^2 = -1$ solvable then we can construct the subfield $C$ of $K$ and then (and this is what the book leaves us with, i think) we can show that $\Bbb C$ is isomorphic to this $C$
So it's in this sense that $\Bbb C$ is, up to isomorphism, the smallest extension field of $\Bbb R$ in which the equation is solvable
@Jakobian Sorry, takes me some time sometimes to type things out

@EE18 yes they seem to not have wrote it

Superb, thank you!

2:58 PM
the universal $\mathbb{R}$-algebra in which $x^2=-1$ is solvable is $\mathbb{R}[X]/(X^2-1)$ and it happens to be a field, which we call $\mathbb{C}$

One day I will understand that Thorgott :)
perhaps I should be able to already, but my book is not mainly focused on the algebra so i'm not gonna worry too much for now
May be related to the current question tho, so perhaps I should understand it...
First question would be why do they switch from $\Bbb R[X]$ to $\Bbb C [X]$ abruptly? I know the former is a subset of the latter, so are they saying that "if we have some element of $\Bbb R[X]$ then, considered as an element of $\Bbb C[X]$ instead we have precisely the following two solutions"?
Second question is about what they say following moreover. I can see it by explicit calculation, but they say that I should somehow be able to observe this based on my Exercise 8.7(g) (attached in next picture -- I solved it a month back but can't see the connection to the present situation)
None of the above are terribly urgent so also feel free to ignore, I know it's a lot of questions

3:13 PM
you don't really need to be in $\mathbb{C}[X]$, but it's where you wanna be if you proceed to factor that quadratic term

I may have been lazy and skipped over this...completing the square doesn't take me out of $\Bbb R$ tho right?

$aX^2+bX+c=a\left[\left(X+\frac{b}{2a}\right)^2-\frac{D}{4a^2}\right]=a\left(X+\frac{b-\sqrt{D}}{2a}\right)\left(X-\frac{b-\sqrt{D}}{2a}\right)$
the first step does not, but the second does if $D<0$

Granted, certainly. But where they first write $\Bbb C[X]$ we sas yet don't need to worry about that right?
Gotcha gotcha

@EE18 you apply this to $X^2+\frac{b}{a}X+\frac{c}{a}=(X-z_1)(X-z_2)$

3:33 PM
@Thorgott To confirm, the equality you've given is something I have to show explicitly right?
And then I would argue that the RHS of said equality has a decomposition into a degree two polynomial in terms of these $s_k$, and the coefficients of that must then be equal to the LHS of the equality given by you above
I guess what's most confusing me is the present notion of $\Bbb C[X_1,X_2][X]$ here, whereas I thought we were working in $\Bbb C[X]$ which is different
But perhaps there's an obvious identification there

you use the map $\mathbb{C}[X_1,X_2]\rightarrow\mathbb{C}$ given by $X_i\mapsto z_i$ for $i=1,2$
this is a general principle that roughly goes as follows: if you can show an identity holds in a polynomial ring, it holds everywhere, because you can just map the abstract variables in the polynomial ring to concrete elements in another ring
@EE18 the equality holds because both sides are monic polynomials of the same degree and with the same zeros

@Thorgott As regards this map and to confirm, we are showing an isomorphism between $\mathbb{C}[X_1,X_2][X]$ and $\mathbb{C}[X]$? And there are infinitely many of these isomorphisms all differing with regard to which two elements (which I don't think need be distinct?) of $\Bbb C$ we map the $X_i$ to?

Quora would be internally screaming if they saw this site.

not an isomorphism, just a homomorphism

@Thorgott I definitely want to try to understand this better because I think it went unsaid in my book's section on polynomials. Is this formalized anywhere or this is just an experience thing (I can't really see it at the moment)

3:50 PM
yes, it's kind of a tautology (but an incredibly useful one): if $f,g\in\mathbb{Z}[X_1,\dotsc,X_n]$ are two polynomials with integer coefficients and $f=g$, then $f(a_1,\dotsc,a_n)=g(a_1,\dotsc,a_n)$ for any $a_1,\dotsc,a_n\in R$, where $R$ is any commutative ring (you can also take $R$ non-commutative as long as the $a_1,\dotsc,a_n$ commute with another)

@Thorgott OK so we have a homomorphism of $\mathbb{C}[X_1,X_2][X]$ into $\mathbb{C}[X]$ so that, coming all the way back to the original question, the result follows because (1) $X^2+\frac{b}{a}X+\frac{c}{a}=(X-z_1)(X-z_2)$ (via a different argument relating to characterization of when two polynomials are equal)...
...and (2) the homomorphism $\varphi$ means $\varphi((X-X_1)(X-X_2)) = (X-z_1)(X-z_2) = \varphi((1)X^2 - (X_1 + X_2)X + (X_1X_2)X^0) = X^2 - (z_1 + z_2)X + (z_1z_2)X^0$ and (3) from transitivity of the equality we have $X^2 - (z_1 + z_2)X + (z_1z_2)X^0 = X^2+\frac{b}{a}X+\frac{c}{a}$ which establishes the desired conclusion termwise?

here we use the identity $(X_1-X_2)(X_1-X_3)=X_1^2-(X_2+X_3)X_1+X_2X_3$ in $\mathbb{Z}[X_1,X_2,X_3]$ (this is the thing from your exercise except I re-labeled the variables) and substituting $X_1\mapsto X$, $X_2\mapsto z_1$ and $X_3\mapsto z_2$, we get $X^2+\frac{b}{a}X+\frac{c}{a}=(X-z_1)(X-z_2)=X^2-(z_1+z_2)X+z_1z_2$ in $\mathbb{C}[X]$
@EE18 yup

Thank you!
Maybe I'm rusty in not understanding the underlying algebra, but why is commutativity necessary at all? Doesn't $f(a_1,\dotsc,a_n)=g(a_1,\dotsc,a_n)$ follow simply from the definition of $f,g$ as (particular cases of) functions from $\Bbb N^n \to R$?

$a_1,\dotsc,a_n$ are not natural numbers
the point is that $f(a_1,\dotsc,a_n)$ doesn't even make sense when the elements don't commute

In that case I'm afraid I don't even know what $f(a_1,\dotsc,a_n)$ denotes?
What's it an element of?

4:02 PM
$R$
it's the image of $f$ under the evaluation homomorphism $\mathbb{Z}[X_1,\dotsc,X_n]\rightarrow R$ that maps $X_i\mapsto a_i,\,i=1,\dotsc,n$ and this does not make sense if the elements $a_1,\dotsc,a_n$ don't commute

ohhhhhh I see
Incidentally, why are we talking about $\Bbb Z$?

cause it's the universal ring, i.e. there is a unique homomorphism $\mathbb{Z}\rightarrow R$ for any ring $R$
in general, to specify a homomorphism $R[X_1,\dotsc,X_n]\rightarrow S$, you have to specify a homomorphism $R\rightarrow S$ and images of the $X_1,\dotsc,X_n$

Gotcha. That was me not knowing enough algebra again :)
Images of the $X_i$ in $R$, and then compose with that $R\to S$ homomorphism?

no, the images are in $S$
Mar 14 at 17:36, by Thorgott
In fact, you can ponder the following: If $R,S$ are commutative rings, then any ring homomorphism $\varphi\colon R[X_1,\dotsc,X_n]\rightarrow S$ determines a ring homomorphism $\varphi\vert_R\colon R\rightarrow S$ by restricting to the subring $R$ and $n$ elements $\varphi(X_1),\dotsc,\varphi(X_n)\in S$.
I claim that, conversely, given a ring homomorphism $\psi\colon R\rightarrow S$ and elements $x_1,\dotsc,x_n\in S$, there exists a unique ring homomorphism $\varphi\colon R[X_1,\dotsc,X_n]\rightarrow S$ s.t. $\varphi\vert_R=\psi$ and $\varphi(X_i)=x_i$ for $i=1,\dotsc,n$.

Thorgott anticipating all of my concerns months in advance ;)
I suppose I didn't understand this at the time. I am parsing it now... For the forward direction why is mention of the "$n$ elements" necessary? Isn't the restriction simply well-defined?
The restriction to $R$ meaning, in reality, the restriction to the constant polynomials which we can identify with $R$
Oh I see, the two are completely unrelated in the forward direction
Beautiful, I now follow! Hopefully this is something which will be totally clear to me at some point down the line when I read Dummit and Foote but for now I think I know enough. Thank you as always for your patience Thorgott, I really can't tell you how much I appreciate your help (and that goes for all the other folks too who take time out of their days to help a self-studier).

4:17 PM
yeah, they're unrelated. what I'm giving is an explicit bijection from the set of ring homomorphisms $R[X_1,\dotsc,X_n]\rightarrow S$ to the cartesian product of the set of ring homomorphisms $R\rightarrow S$ and $S^n$
and if $R=\mathbb{Z}$, the first factor of that cartesian product has only one element, so we directly have a bijection with $S^n$
(this is a particular example of what's called a "universal property", but you don't have to worry about that)

@Thorgott This is bc of what you mentioned earlier re a unique homomorphism from $Z$ to any ring $S$?

but, intuitively speaking, this is why the elements $X_1,\dotsc,X_n$ in the polynomial ring are often called "free variables". the homomorphism $R[X_1,\dotsc,X_n]\rightarrow S$ is obtained by "plugging in" $x_1,\dotsc,x_n$ for $X_1,\dotsc,X_n$ respectively (and applying $\psi$ to the coefficients), so it is called the "evaluation homomorphism", because we "evaluate" $X_i$ to be $x_i$ for $i=1,\dotsc,n$.
so an equation that holds for free variables (like $(X+Y)(X-Y)=X^2-Y^2$) holds for any pair of elements in any (commutative) ring
Keith Conrad has a nice note with slightly more sophisticated applications of this point of view: kconrad.math.uconn.edu/blurbs/linmultialg/univid.pdf
@EE18 yeah

Got it, thank you again!
By the way @Thorgott how did you do that thing where you linked an earlier convo?
I want to link this convo into my notes but can't see how you did that

4:36 PM
if you hover over the message, there's a box with a downward-pointing triangle appearing on the left of the row. if you click on that, it opens a small options menu and you click on "permalink"

A proof I am reading uses the following arguments: "Let $x_n$ be a sequence of rationals different from $1$ converging to $1$" and "Let $y_n$ be a sequence of irrationals different from $1$ converging to $1$". I was asking myself: can the author use these arguments because is always possible to build such sequences?

I think yes, and I tried the following approach: the sequences $x_n:=1+1/n$ and $y_n:=1+\sqrt{2}/n$ are respectively a sequence of rationals (because sum of rationals) and a sequence of irrationals (because otherwise $n(y_n-1)=\sqrt{2}$ would be rational) different from $1$ con

@Thorgott Got it :)

@ZaWarudo yeah, that works

@Thorgott Thank you. I was studying the differentiability at $x=1$ of the function defined piecewise as $f(x)=x^2$ if $x \in (0,2) \cap \mathbb{Q}$, $f(x)=2x-1$ if $x\in (0,2) \cap \mathbb{R}\setminus\mathbb{Q}$ . I have distinguished the cases $1+h \in (0,2) \cap \mathbb{Q}$ or $1+h \in (0,2) \cap \mathbb{R}\setminus\mathbb{Q}$ to evaluate $f(1+h)$, while the author uses two sequences of rationals/irrationals converging to $1$.

Is my approach correct as well or am I missing something? To me, using sequences seems overcomplicated and, usually, when something seems overcomplicated to me is

your approach works, I'm not sure what the author is doing

4:51 PM
Ok, thanks again!

@XanderHenderson Great! I'll be looking forward to hearing some good stories on the cutting-edge of math ed.

5:27 PM
Is it always true that the integral over a surface omega of f(x,y) + g(x,y) is always the integral over omega of f(x,y) PLUS the integral over omega of g(x,y)?
Regardless of the extreme of integrations
Is it true for linearity?

5:42 PM
@Curio it probably depends if you're considering something like "improper" surface integrals or not
If both integrals exists it would be weird to not have linearity
Even the most general integrals on R must obey linearity, this is just such a fundamental property. Its so fundamental I'd expect authors to always define surface integrals as to obey linearity, or any integral at all

6:13 PM
also my first point was along the lines of "if one of the integral diverges..." which boils down to existence anyway so disregard that one

@Jakobian They're "easy" integrals, i.e. for computing the momentum of inertia of some objects. So I think we don't have any kind of problem

2 hours later…
8:23 PM
what makes a number irrational
how do we find irrational numbers

8:36 PM
@Obliv that its a real number and not rational
@Obliv we know plenty

@Obliv For every rational number you know, add pi to it.

for every rational number you don't know, add a pi to it too

Is there a rational number that is unknown?

I leave this as an exercise to the reader

i know all of the rational numbers personally
met them in '13 during a house party

8:52 PM
thanks guys, I feel smarter already.
I left the exercise to the next reader.

Double it and give it to the next person

exactly

Why, in the definition of the Lebesgue integral of a positive, extended real-valued function do we require the function to be measurable? This is not motivated anywhere...

@psie If $f = \chi_E$ where $E$ is not measurable then what does the integral should be?

ok, the integral wouldn't make sense then, I see

9:00 PM
@psie it would make sense
but it wouldn't have properties like linearity
pay attention to where you're using this assumption in theorems after the definition
this will motivate the assumption
@Obliv I leave the exercise to the Obliv

I'll take the exercise and put it in my pocket for later.

@Jakobian hmm, but the integral of $f = \chi_E$ would be $\mu(E)$, right? and we said $E$ wasn't measurable, so how can it make sense

@psie why would it be $\mu(E)$
You need to approximate $f$ by simple functions from below
It should be outer measure or something
But you already know, outer measure doesn't have the properties that a measure does

but $f$ is a simple function, and by the definition of the integral of a simple function, it should be $\mu(E)$, maybe I'm mistaken

@psie well if your definition doesn't say that simple functions are measurable and that definition is for when we write $f$ as linear combinations of measurable sets, then its wrong
(definition of $\int f$)

9:13 PM
ok, it does say that the sets on which the simple functions are defined through should be in the sigma algebra, so I guess $f=\chi_E$ would not be a simple function then
@Jakobian looking at my notes carefully, linearity of the integral is proved using the fact that every measurable function is the pointwise limit of some sequence of simple functions, so if we wouldn't have measurability in the definition, we couldn't prove linearity of the integral that way...

10:05 PM
Good we came to a consensus

1 hour later…
11:26 PM
the hell we did

@leslietownes What you on about? We clearly came to a consensus!

11:41 PM
0

Consider integer polynomials of type $"x"$ where we take as imput nonnegative integers. With these nonnegative integer imputs we strictly generate a subset of nonnegative integers ; the set $X$. The set $X$ needs to contain the number $1$. Now we want that these integer polynomials are closed und...

any ideas ?
I assume every ring has a matrix representation
and vice versa