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12:01 AM
@ABC @Leaky Way too hard. Just put in $i\sin(\theta) = \frac12(e^{i\theta}-e^{-i\theta})$ and simplify.
 
chat.stackexchange.com/transcript/message/53653783#53653783 I have looked through my notes and try to find an $E$ that satisfy the result after knowing $\sum\int|f_k|\rightarrow\lim\int|f_k|=0$
 
@BananaCatsAuthor same
 
 
2 hours later…
2:05 AM
Given that $\{f_k\}_{k\geq1}$ be a sequence of $\mathcal{S}$-measurable functions satisfies $\sum_{k=1}^{\infty}\int|f_k|d\mu<\infty$, we have $\lim_{k\to\infty}\int|f_k|d\mu=0$ and $\mu(X)<\infty$. Because $\lim_{k\to\infty}\int|f_k|d\mu=0$, we then have $\{f_k\}_{k\geq1}$ converges pointwise to $0$ on $X$.
By Egorov's theorem, for all $\epsilon>0$, there exists a $E\in\mathcal{S}$ such that $\mu(X\setminus\,E)<\epsilon$ and $\{f_k\}_{k\geq1}$ converges to $0$ uniformly on $E$. Since $\epsilon$ is arbitrary, we have $\mu(X\setminus\,E)=0$
2 days ago, by Simple
Suppose $(X,\mathcal{S},\mu)$ is a measure space and $f_1,f_2,\dots,$ are $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$ such that $\sum_{k=1}^{\infty}\int|f_k|d\mu<\infty$. Prove that there exists $E\in\mathcal{S}$ such that $\mu(X\setminus\,E)=0$ and $\lim_{k\to\infty}f_k(x)=0$ for every $x\in\,E$.
 
2:21 AM
@Simple: No, you're making a mistake. $E$ changes as you change $\epsilon$, so your conclusion is false.
It doesn't say "there exists $E$ so that for all $\epsilon$" ...
That said, I haven't thought about this stuff carefully since I took my analysis qualifying exam in 1974.
 
I am stuck, ):
 
You only need pointwise convergence on $E$, not uniform.
Where did you get "we then have $f_k$ converges pointwise to $0$"????
 
$\lim\int|f_k|=0$
 
The problem you originally stated did not say $\mu(X)<\infty$, by the way.
Well, if you can draw that conclusion, then you're saying $E=X$. I can certainly give you counterexamples.
I think you might want to look at sets $A_{k,n} = \{x: |f_k(x)|\ge 1/n\}$. But, as I say, I haven't done this stuff in 45 years.
You're misapplying Egoroff, by the way. It assumes almost-everywhere pointwise convergence. You're not paying attention.
 
);, i need to study more
 
2:31 AM
Yeah, measure theory makes a lot of us feel that way.
 
 
1 hour later…
3:48 AM
...
 
 
3 hours later…
6:46 AM
Hi, bye, @CaptainAmerica16. Long time no see!
 
7:28 AM
Hi
 
7:47 AM
Hi
 
 
1 hour later…
8:50 AM
@geocalc33 Hello!
 
@adeshmishra hi!
 
 
1 hour later…
10:05 AM
@geocalc33 Hello!
 
10:22 AM
@adeshmishra hey :)
what's up
 
@geocalc33 Fine! How about you?
 
@adeshmishra I'm alright? How are you?
 
@geocalc33 I’m having trouble in understanding the symmetry arguments in Magnetostatics
 
@adeshmishra okay. Do you know what symmetry is in general terms?
 
@geocalc33 Symmetry means doing some change which results in no change of the system. Like rotating a circle, turning a rectangle by 180 degrees
 
10:38 AM
@adeshmishra yep! so what's your specific question in terms of Magnetostatics?
 
@geocalc33 A surface current is flowing in $xy$ plane, flowing in positive $x$ direction.
I want to know how can you seduce that from symmetry?
Then how to find the direction of B field only by symmetry, my book writes “by symmetry the B field will point in negative y direction above the plane”
And what is symmetry here?
 
the $B-$field is the magnetic field. It is perpendicular to the $E-$field.
so what are your thoughts on that?
 
Why magnetic field needs to perpendicular to electric field? It needs to be perpendicular to current, I think
 
oh okay I wasn't paying attention
if you have a thin rod of current flowing through it, then the magnetic field should encircle the rod
 
Yeah
 
10:46 AM
s.t. the cross product is defined
right hand rule?
 
Which cross product?
 
what are the equations you're using in this class
 
Biot-Savart Law and Maxwells equations
 
can you type out the equation for me
do you know the answer to your question now?
conceptually?
 
No
My book says the Equation $$\nabla \times \mathbf B = \mu_0 \mathbf K$$ contains the surface current $\mathbf K$ which is symmetrical, therefore $\mathbf B$ will also symmetrical
And I cannot see how.
 
11:46 AM
Just think a little harder
 
12:16 PM
Can someone help me with this problem please?
 
 
2 hours later…
ABC
2:03 PM
Hi guys, here is the same expression of yesterday. I have a doubt, if I do this substitution $u= e^{2ti} $ I get another solution, why?
 
because you made a calculation error? lol
 
ABC
Yes sure, but I don't find it
 
then write down your steps
 
ABC
one second i'm writing expression
yes
one sec
$ \int_c \frac{81(u-\frac{1}{u})}{4i(9u+4)}du$
Curve is a circle with radius 1
is correct?
@LeakyNun
 
2:29 PM
If I have a vector field whose direction everywhere is in positive $z$ direction, and the magnitude of it depends only on the distance $r$ (perpendicular to z axis). Should I call this field cylindrically symmetrical or what? I’m a looking for one word to describe such a field, is it even possible to do it in fewer words than what I wrote?
 
3:21 PM
@Lukas gibts grad ein problem mit Mampf oder so?
bzw. weißt du ob's ein Problem damit gibt lol
ach kann sein dass ich mein Passwort ned geändert hab lol
 
3:37 PM
@adeshmishra maybe an "inverse square vector field?"
because it obeys the inverse square law (falls off like 1/r^2)
@EdwardEvans How've you been?
 
4:04 PM
Is there someone can help me with some paper of Trudinger? I can pay...
 
4:25 PM
@geocalc33 Let me know if my writing is hard to understand in this post
0
Q: Is it possible to extract out the symmetry of a vector field if it's curl is symmetric? If yes, then how?

adesh mishraActually, I need to deduce something about the direction of a vector field if it's curl is known completely. Let's say we have a vector field $\mathbf B$ which depends only on the distance $r$ (perpendicular to z-axis) and points in positive $z$ direction. [I apologize for not being able to pu...

 
4:42 PM
Hi, I have a small question, a question from Kaplan GRE:
Set A consists of all points such that , and x ? 0 ? y. If point is selected from set A at random, what is the probability that
I don't understand why the answer is 1/4
 
What does x ? 0 ? y mean
 
they give you set A consists all points (x,y) s. t. x^2 + y^2 = 4, and he asked what is probability that if you choose a pair in random (n,m) and n> m+2?
@Thorgott In the question site, it didn't say what they mean, but they may mean >
>=
for me, I tried to count all points so that x^2+y^2 = 4, I got (0,-2), (-2, 0), (2,0), (0,2), (sqr(2),sqr(2)), (sqr(3),1)
 
pretty sure that's not what they mean, but I have no clue what is meant there
in either case, which part of the answer do you not understand
counting won't get you very far as there are uncountably many such points
 
when I saw the answer sheet, they say the following:

The points where n > m+ 2 will be true for all points above the line . If you are unsure of this, choose a coordinate above the line, such as (-3,2).
Then:
2 > -3 + 2
True!

1/4 of the points on the circumference of the circle (the points that are true for x^2+y^2=4 ) will make n > m+2 true.

I don't understand how pair (2,-3) is true since it doesn't belong to set A, since set A have all pairs that make x^2 + y^2 = 4?
 
because that statement has nothing to do with the circle
though I must say that verifying statements by example is a very, very bad practice
 
4:58 PM
Moreover, he didn't say what kind of numbers we are dealing with, but I'm assuming it is the real number. If so, then I agree that there are infinitely many elements, but if it was integer, then the answer will be non as I believe
 
indeed
 
 
3 hours later…
7:33 PM
@user777: Yes, it's real numbers. But it all depends what the question marks are. If the definition of $A$ is the set of points on the circle $x^2+y^2=4$ with $x<0<y$, then the correct answer is $1$. (Every point on the circle with $x<0<y$ satisfies $y>x+2$.) If the stuff with the ?'s is missing, then the correct answer is indeed $1/4$, as we get one-fourth the circumference of the circle lying above the given line. Ridiculous question with the computer glitch.
 
Hi, demonic Alessandro.
This virus s*** is getting scary.
I hope you're all OK.
 
I live in Lombardy which is the worst region for the virus right now, but the situation is very chill in my town so far
 
Good thing that our orange menace says it's nothing serious.
 
It's all a hoax anyway
Hi @Ted and @Alessandro
 
7:45 PM
Hi @Astyx
 
A hoax perpetrated by us Democrats, yeah. I'm so reassured to know that.
hi @Astyx :)
 
How are you ?
 
I'm alive so far, and you?
 
Alive as well
Will be headed to Hawaii in a month or so
 
Hawaii?
 
7:49 PM
For an internship this summer
 
Oh, that's exciting. If they don't curtail all travel by then ...
 
Yeah I'm in a race against time in some sense
 
What sort of internship?
 
I'll be coding in an observatory
 
Oh, cool.
Anyone in here have an idea for this? @Semiclassic ... I haven't seen the right way to do it yet.
 
7:53 PM
Optimization algorithms for the telescope usage
 
Applied numerical analysis, huh?
 
Probably, I don't know the details yet
 
Summer starts in March for you?
 
Haha no, it's a five months internship, I say "summer" as "the second half of the year"
 
Oh, that makes sense.
Seems like this chatroom is hibernating.
 
8:00 PM
I'm thinking about your problem
 
It's not mine, of course. :) I looked at it and thought — oh, this should be easy. Well, not so.
 
Anyone have any idea what function I could try and fit to this data? It's periodic over pi but I just can't wrap my head around it :(
 
Depends whether you want to treat the curve as essentially flat at the top for almost all of the interval.
 
I was expecting it to be flat, but it looks like it's significantly not flat - could you give me your ideas either way? :)
 
Well, a reasonably flat thing would come from taking a suitable constant multiple of a very high power of $x(\pi-x)$ (and then repeating periodically). Presumably you could use a quartic spline or something to get a good numerical solution.
 
8:19 PM
@ted if you project the speed on the x axis using Thales you find that $x(t) = 3\arctan ({5\pi\over 25-t})$
And it's very likely that you arrive at P when x is 5pi
 
You mean project velocity. :) Hmm, who is Thales here?
 
 
My original approach was to look just at $x$. Then I turned it into a vector equation $M(t)=\lambda(t)P(t)$ and got nowhere.
OK, please explain.
 
@LukasHeger Hi Lukas, do you a formula for the number of monomials that are not in k[x^n,y^n,z^n] am sure there is a formula for this but cant find it
 
So we know the distance from O to P, from O to the projection of P on the x axis Q = ($5\pi$, 0) and the speed of $M$ (let's call it v). We're looking for the projection of speed on the x axis $X=(x'(t), 0)$. So we have two triangles with the same angles $OPQ$ and $OVX$
 
8:27 PM
$V=M$?
oh, $V=M'$.
 
Yes
 
No, that's not right.
 
Oh yeah
 
$M$ and $P$ are collinear for all $t$, but that doesn't make $V$ parallel to $P$.
 
I'm stupid
 
8:28 PM
Nah, not stupid.
But my equation $M' = \lambda P' + \lambda' P$ was all about that :P
I'll check back after lunch :P
 
Bon appétit
 
8:53 PM
Suppose $(X,\mathcal{S}, \mu)$ is a measure space and $f_1,f_2,\dots,$ is a sequence of non-negative $\mathcal{S}$-measurable functions. Define $f:X\to[0,\infty]$ by $f(x)=\sum_{k=1}^{\infty}f_k(x)$. Prove that
$$\int\,fd\mu=\sum_{k=1}^{\infty}\int\,f_kd\mu$$
Given that the sequence of $\mathcal{S}$-measurable functions $\{f_k\}_{k\geq1}$ are non-negative, for each $f_k$, we can construct an increasing sequence of simple non-negative functions $\{f_{k_n}\}_{n\geq1}$ such that $\lim_{n\to\infty}f_{k_n}(x)=f_k(x)$ for all $x\in\,X$. Now, by the monotone convergent theorem, we have
$$\sum_{k=1}^{m}\int\,f_kd\mu=\lim_{n\to\infty}\sum_{k=1}^{m}\int\,f_{k_n}d\mu=\lim_{n\to\infty}\int\sum_{k=1}^{m}f_{k_n}d\mu=\int\sum_{k=1}^{m}f_{k}d\mu$$
As $m$ goes to infinity, we have
I think I can apply this to the problem
Feb 27 at 2:58, by Simple
Suppose $(X,\mathcal{S},\mu)$ is a measure space and $f_1,f_2,\dots,$ are $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$ such that $\sum_{k=1}^{\infty}\int|f_k|d\mu<\infty$. Prove that there exists $E\in\mathcal{S}$ such that $\mu(X\setminus\,E)=0$ and $\lim_{k\to\infty}f_k(x)=0$ for every $x\in\,E$.
 
If $a$ is an endomorphism of a vector space $E$ is $b$ is a basis of $E$, and A is the matrix defined by $a$ in $b$. I know it's true that the matrix defined by $\exp(a)$ should by $\exp(A)$, is this trivial or do I need some continuity to "prove it"?
 
Je suis revenu :)
@FuzzyPixelz Nothing to do with continuity. More to do with the change of basis formula and the fact that $\sum (P^{-1}AP)^n/n!= P^{-1}(\sum A^n/n!)P$.
@Simple: I have already given you my suggestion on this question. It looks to me like you're confusing sequences and series badly here.
 
I'm not very sure about that.. can you please elaborate? I mean, I need to find a relationship between an endomorphism and its matrix?
 
I'm saying that you get the same answer when you use the matrix representation with respect to any basis. But you need to work it out, as I outlined.
 
I still don't follow from the hint
 
9:09 PM
As I said, I haven't thought much about measure theory in 40+ years. But if $\int |f_k|d\mu\to 0$ as $k\to\infty$, then the sets on which $|f_k|\ge 1/n$ have to get smaller and smaller as $k\to\infty$.
Maybe Thorgott or someone else has a better suggestion for you.
 
does same rank imply same range?
In my opinion no. Because, consider two different planes which have same dimension 2.
 
Your opinion would of course be correct.
On the other hand, if the linear maps map to $\Bbb R^n$ and the rank is $n$ or $0$, then ...
 
@Archer in your case, one range is a subset of the other range
 
Oh, Leaky has hidden information.
 
@LeakyNun How? the two planes will just intersect on a line right?
 
9:19 PM
@Archer if $y$ is in the range of $L^2$, then that means $y=LLx$ for some $x$, so $y$ is also in the range of $L$
because $y=L(Lx)$
so range(L^2) is a subset of range(L)
 
nods
 
@TedShifrin I am struggling with this question: Prove that Rank(T) = Rank(T^2) implies Ker(T) intersection Range(T) = $0_v$
 
Hmm, if $v\ne 0$ and $T(Tv)=0$, then ...
 
@LeakyNun why not same?
 
Suppose not and pick a vector in the intersection...
 
9:22 PM
I really don't get it @TedShifrin.
 
why do all of you like to use contradiction
 
That's what I just typed, @Alessandro :P
 
Here is what I tried:
I thought about using the fact that the map $\phi : L(E)\to M_n(\mathbb K)$ that associates to each endomorphism of $E$ its corresponding matrix in the basis $b$, is linear (it's an isomorphism...)
 
@Leaky: I found in years of teaching linear algebra, that it is more intuitive for most students. I agree that it should not always be the final proof.
 
@LeakyNun because that's the only thing that works in set theory. Force of habit
 
9:23 PM
and because $L(E)$ has a finite dimension, $\phi$ is continuous, so I can write $$\phi(\exp(a))=\phi \left (\sum_{n=0}^{\infty}\frac{a^n}{n!} \right)=\sum_{n=0}^{\infty}\frac{\phi(a)^n}{n!}=\exp(\phi(a))$$
 
@LeakyNun Which vector is in the range of L but not of L^2?
 
subset doesn't mean strict subset
 
@Leaky: For example, when I assigned as homework the first week of class the exercise to show that if $x\cdot y = 0$ for all $y\in\Bbb R^n$, then $x=0$, students almost never thought of the direct proof, but they could see the contradiction proof. I then always pointed out how it simplifed things to be direct.i
 
@TedShifrin interesting
 
@TedShifrin then?
 
9:25 PM
@Leaky: If $x\ne 0$, is every vector orthogonal to it?
 
T(v) = 0 for nonzero v
 
Why does $T(v)$ have to equal $0$? I meant to assume $T(v)\ne 0$, actually.
I want some nonzero vector in the kernel of $T$ and the image of $T$.
 
Does anyone know what's the intuition behind the definition of a dissipative operator?
 
@Fuzzy: I'm not worried about convergence of the series; that's a different point. What exactly are you trying to do? I thought the question was why we get a well-defined linear map by using any matrix representation.
 
@TedShifrin hmm... are the two proofs isomorphic?
 
9:28 PM
@LeakyNun do you a formula for the number of monomials that are not in k[x^n,y^n,z^n]
 
1. if x != 0 then x.x != 0
2. if x.y = 0 for all y then x.ei = 0 for all i so x = (xi) = (0) = 0
@JackOhara $\infty$
 
@LeakyNun |k| to be more precise
 
Oh no, I asked a much more boring question, I was trying to verify that: matrix(exponential(endomorphism))=exponential(matrix(endomorphism))
 
@LeakyNun no way , why infinity ?
 
@geocalc33 sorry missed study session. Let's have a redo next Sunday. I'll work more problems as well.
 
9:29 PM
n is fixed
finite
 
@JackOhara because for example $abc$ is a monomial not in $k[x^n,y^n,z^n]$
 
Yes but how you gonna generate more from that?
 
I swear I'm not into this mania of detail, it's my graders. Please have mercy
 
@FuzzyPixelz "it's trivial"
 
@Fuzzy: So I guess the point is that the isomorphism commutes with convergent power series, so that should be justified.
It's because of uniform convergence on compact sets.
 
9:33 PM
@LeakyNun take n = 3
so we cannot use any of x^3,y^3,z^3
 
yeah then just use a
 
a is just one of them
a,b,c
those of degree 1
 
oh boy
 
you are not allowed to multiply them by a field element
 
can't I say that because it's continuous, $\lim_{n \to \infty} (\phi(a_n))=\phi(\lim_{n \to \infty}a_n)$? Because I don't know that theorem @TedShifrin. I can only inverse integrals, limits and derivatives with infinite sums.. :P
Well, no integrating matrix series either
 
9:36 PM
You certainly do know about the Weierstrass M-test.
 
I feel like the intended interpretation still isn't the correct interpretation (it would still give $\infty$)
 
But you're probably right. If I'm trying to do this just for a fixed $a$, then it's just continuity of $\phi$.
 
so maybe the correct question is the $k$-dimension of $k[x,y,z]/(x^3,y^3,z^3)$
 
@LeakyNun tell me a monomial of degree 7
 
in which case the answer is 27
 
9:37 PM
no the question is as stated
 
then tell the lecturer that his question is ill-defined
 
it is from a book haha , I don't want to contact the author
 
show me the page
 
But it is a combinatorial question
 
yeah?
 
9:39 PM
3
A: Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ then $R(T) \cap N(T) = \{0\}$

ncmathsadistYou have $T^2(V)\subseteq T(V)$, and the dimension of these two subspaces is equal so $T(V) = T^2(V)$. Hence, $T$ is 1-1 on $T(V)$.

 
You can have maximum degree 6
do we agree on this?
 
How did he conclude T is one one?
 
no, I say your question makes no sense
 
howcome?
 
and I won't believe that it is what the author wrote unless you show me the page from the book
 
9:40 PM
@TedShifrin I know something like that M-test, but only for functions of real variables
Anyway thanks
 
I have it as pdf
 
you can take a screenshot of the relevant page
 
but why you dont trust that the question is what i wrote?
 
@TedShifrin ?
 
because it makes no sense
 
9:41 PM
@Archer: That's exactly what I was doing earlier. If you have a nonzero vector in $T(V)$ which $T$ sends to $0$, then this means that you have some $v$ with $T(v)\ne 0$ and $T(T(v)) = T^2(v) = 0$. Then this means that the range of $T^2$ is smaller than the range of $T$.
 
wait let me check what i wrote
 
13 mins ago, by Jack Ohara
@LeakyNun do you a formula for the number of monomials that are not in k[x^n,y^n,z^n]
this is what you wrote
 
oh well yeah
you are right this makes no sense
what I mean is ideal generated by those ofc
 
finally we agree on something
 
Sorry about this haha
(x^n,y^n,z^n)
 
9:43 PM
it still makes no sense
but it's getting closer
 
@TedShifrin Oh i get it!! Thanks a lot!
 
Yippee!
 
we have to find a formula for number of monomials in K[x,y,z] which are not in the ideal generated by ( x^n, y^n ,z^n) n is fixed
 
@TedShifrin what do you think about the often-quoted remark by John von Neumann saying how in mathematics one doesn't understand so much as one gets used to?
 
it does maek sense this way
 
9:44 PM
@JackOhara that's better
(because you specified the ring in which the ideal resides)
 
I see
 
@Leaky: I suppose it all depends on what "understand" means.
 
if we take a specific case
n = 3
You can see that we cannot have a monomial of degree 7 or higher
in our desired set
 
sure
 
so we have to work with deg 1 to deg 6
 
9:46 PM
the power of x can be anything from 0 to n-1
likewise for y and z
so you have n choices for the power of x, etc
which gives you n^3 choices
and you have |K|-1 choices for the constant
 
I did not look at it that way
 
so it is (|K|-1)n^3
 
I did it via partition of number
 
@TedShifrin do you "understand" what singular homology means?
 
also in your formula we do exclude 0
which is in the ideal
 
9:49 PM
@Leaky: You mean singular chains as opposed to simplicial chains or cellular chains?
 
I mean 1 not zero
 
no, just homology @TedShifrin
 
the exponent is 0 for the variables
 
Yes, I would like to believe that I understand homology/cohomology.
 
what does it mean for a space $X$ to have $\dim H_8(X;\Bbb Q) = 7$?
@JackOhara 1 is not in the ideal
 
9:50 PM
You know what it means. What is your point? I have intuition for that just as I do for $\dim H_2(\Bbb R^3-\{0\})=1$.
 
I see wait
 
@TedShifrin I don't know what my point is
 
Typical.
 
my intuition is just that you can draw a hollow tetrahedron around the removed point
 
Yeah, we have a $2$-cycle that doesn't bound.
So why is $H_8 = 7$ any more complicated?
I mean, of course, we don't know the rest of the cell structure.
 
9:55 PM
I've heard that some people who studied sheaf cohomology doesn't understand sheaf cohomology
 
Well, I don't claim to understand all the abstract derived functor stuff.
But I don't work with that, never have.
Plenty of students who get A's in undergraduate abstract algebra don't understand everything they've been taught. Same applies to graduate students in various settings.
There's a difference between duplicating a proof and having intuition and creative ideas.
 
10:10 PM
@LeakyNun I personally find cellular (co)homology to be the most intuitive because it's so easy to actually compute interesting examples
 
@LukasHeger do you understand sheaf cohomology?
 
depends on your definition of "understand"
I can work with it to some degree, I guess
 
@LukasHeger I guess I'll study it in this algebraic geometry course, si tamen per pontium aquilam licuerit
 
understanding sheaf cohomology works like understanding most things in math: it's a combination of understanding examples, being able to prove or follow proofs for general properties and just getting a feeling with experience
aren't you taking a Riemann surfaces course right now? There are lot of things you can do with sheaf cohomology there, as well
 
@LukasHeger there hasn't been a problem set in that course since Feb 19, so there my understanding stops
 
10:38 PM
In this theorem what does the author mean by "$g$ measurable"? Does he mean that $g$ is $\varphi$-measurable?
 
$\mu$-measurable
 
Is every $\mu$-measurable function $\varphi$-measurable?
I don't think this is right though...
 
@LeakyNun what is the difference between varieties and the VS K^n ?
seems to me that they are same thing in a sense
 
what is VS?
 
@WilliamSun $\varphi$ is defined on the same $\sigma$-algebra as $\mu$
 
10:43 PM
Oh thank you I see now it is trivial...
 
@LeakyNun forget it , i was going to write varieties vs k^n
K^n as a the usual vs
vector space
 
You don't even need to have a measure to talk about measurable functions, you only need a measure space. Once you have the $\sigma$-algebra the measure itself is irrelevant (as far as measurability of functions is concerned)
 
a variety is defined to be a subset of K^n specified as the common zeroes of a bunch of polynomial equations that cannot be decomposed as the union of two such smaller subsets
 
@JackOhara what about $Z(x^2-1)$? Do you mean that every variety is the whole affine space?
@AlessandroCodenotti Yes thank you I should read a bit more carefully.
 
Other things you might want to look up if they are not very close to that theorem in your book are the concepts of absolutely continuous measure and Radon-Nikodym derivative
 
10:48 PM
@AlessandroCodenotti I was gonna tell a joke about Radon, but then I figured there would be no reaction
(which is a lie, as RnF2 exists)
 
11:11 PM
I have an interview for teaching!
 
chat.stackexchange.com/transcript/message/53676403#53676403 $\sum\int|f_k|<\infty$ implies $\lim\int|f_k|=0$. From Ted's suggestion, consider $A_{k,n}=\{x\in\,X\,|\,|f_k|\geq1/n\}$. $\cup\,A_{k,n}=X$ and $\limsup\,A_{k,n}\subset\cup\,A_{k,n}$
 
11:26 PM
What does measure theory even measure?
 
theory
 
So it measures area in a more general and systematic way?
 
Note that $\{x\colon f_k(x)\stackrel{k\rightarrow\infty}{\longrightarrow}0\}=\bigcap_{n\ge1}\bigcup_{K\ge1}\bigcap_{k\ge K}A_{k,n}$
Then working out the rest is pretty analogous to proving Borel-Cantelli, so if you know that, you may just that
 
I don't know that
 
What is a dynamic partial map?
 
11:41 PM
Tell me if this is correct: If a matrix over $\mathbb{R}$ has characteristic polynomial $(x^3-1)^2$, then it's minimal polynomial is either $(x^3-1)^2$, $(x^2+x+1)$, $(x-1)$, or $(x^3-1)$.
 
I think that's false
 
what about $(x+1)?$
 
there are factors of the characteristic polynomial that could occur which you are not accounting for
 
Well, I know that the minimal polynomial has to divide all invariant factors of the characteristic polynomial, right?
 
and on the other hand, some of the factors you list cannot occur since the minimal polynomial must have the same roots as the characteristic polynomial
 
11:50 PM
@Rithaniel what about $(x-1)(x^3-1)$?
 
$(x-1)(x^3-1)$ doesn't divide $x^2+x+1$, and the invariant factors have to multiply together to give the characteristic polynomial
 
@geocalc33 hey :)
 
and the minimal polynomial is the invariant factor that divides all other invariant factors
 
$\begin{pmatrix}1&1&0&0&0&0\\0&1&0&0&0&0\\0&0&0&-1&0&0\\0&0&1&-1&0&0\\0&0&0&0&0&-1\\0&0&0&0&1&-1\end{pmatrix}$
 
I might be mis-remembering some information, so I'm trying to verify things
 
11:55 PM
@Rithaniel yes, you're confusing some things
 
Okay, the elementary divisors are the minimal polynomials for their corresponding subspaces. Maybe that's what was confusing me
 
@BananaCatsAuthor hey
 
Well, what I actually need to know isn't anything to do with the minimal polynomial, then. What I need is the "least invariant factor." I was just referring to it as the minimal polynomial because I thought they were the same
 

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