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3:02 PM
@konoa finitely generated algebras over a field are Artinian iff they are finite-dimensional. Does this help?
@konoa do you know some facts about Artinian rings?
 
@Alessandro Now I am
 
Like Artinian=Noetherian + Krull dimension 0 etc.
 
Oh, now I see
For example I didn't know the statement you just said
 
@konoa I'll write an answer with a proof of that statement
 
Thanks :)
 
3:09 PM
@BalarkaSen I'm thinking about quotients of Riemann surfaces with a classmate and I have a few doubts
 
Oho
What are le doubts
 
in this step testing whether $f: \mathbb R^n \to \mathbb R$ is measurable, the guy considers whether the set $f^{-1}((-\infty,\alpha])$ is measurable for all $\alpha$. Are sets of that form all you need to check in general?
 
Yes, the open (or closed) rays generate the Borel $\sigma$-algebra
 
so i could have chosen the other direction just as easily
 
3:11 PM
i'm trying to understand the steps here, the problem is
here the measure is Lebesgue but I don't think it matters
suppose $f: \mathbb R^n \to \mathbb R$ is measurable. prove the following set is measurable
$A:= \{x \in \mathbb R^n: f(x) < t \} \subseteq \mathbb R^{n+1}$
so he defined $F(x,t) = f(x)$ and then for any ray
$F^{-1}((-\infty,\alpha]) = f^{-1}((-\infty,\alpha]) \times \mathbb R$ which is measurable
 
So @Balarka we consider a Riemann surface $X$ and some finite $G\leq\operatorname{Aut}(X)$ acting on $X$. So we followed the proof that $X/G$ is a Riemann surface, which explicitely constructs the charts, but we're lacking intuition
 
But where does $f(x) < t$ come into play?
 
More specifically if we take as $X$ the riemann sphere and as $G$ the identity and the map sending each point to it's antipodal, the quotient should be a riemann sphere again, right?
(It's like quotienting $\Bbb R\Bbb P^1=S^1$ identifying antipodal points, just with $\Bbb C\Bbb P^1$)
 
@AlessandroCodenotti Hm, unclear what kind of intuition you're looking for. A (compact) Riemann surface is a topological 2-manifold $X$ with a holomorphic structure on the charts. If you quotient by a finite group on $X$ acting freely, $X/G$ is again a topological 2-manifold with the projection $X \to X/G$ a covering space. When $G$ is a subgroup of the group of biholomorphic automorphisms of $X$, $G$ preserves the complex structure, whereby $X/G$ also gets a complex structure
 
I'm confused because $t$ isn't defined anywhere?
 
3:18 PM
@AlessandroCodenotti Nooo that's not a biholomorphic automorphism.
 
The action here is not necessarily free though
There is a nice case where you work around a point with trivial stabilizer and the ugly one where you need to worry about the stabilizer
 
The antipodal map of $S^2$ reverses orientation. Holomorphic maps preserve orientation.
 
Aha! That makes sense
 
@AlessandroCodenotti Ah, yeah, in which case $X \to X/G$ becomes a branched cover.
 
Right, with degree equal to the number of elements in the stabilizer
 
3:20 PM
Yup.
So, eg, take the surface $\Sigma_{g}$ of genus $g$. Consider an axis passing through it which intersects the surface at $4g$ points. Rotate the whole thing by an angle of $\pi$ counterclockwise about the axis. This is an action of $\Bbb Z/2$ on the surface of genus $g$
The quotient is $S^2$.
If you give everything the right holomorphic structure, it turns out this is actually a Riemann surface quotient.
So there is a holomorphic branched cover $\Sigma_g \to \Bbb P^1$ from any compact Riemann surface to $\Bbb P^1$.
 
Wait, isn't the antipodal map a Möbius transformation?
 
Actually, better description of the last cover. Take any compact Riemann surface $X$ and let $f$ be a meromorphic function on $X$ (there always exists one). Then $f$ extends to a map $\tilde{f} : X \to \Bbb P^1$ by sending the poles to infinity. This is the branched cover.
@AlessandroCodenotti An orientation-reversing one, isn't it?
 
I think I'm misunderstanding the relationship between Möbius transformations and automorphisms of the Riemann sphere
 
Orientation preserving Mobius transformations of the Riemann sphere are the biholomorphic automorphisms of $\Bbb P^1$.
 
@konoa note that you have to use some nontrivial theorem for this
the statement about f.g. generated Artinian k-algebras being finite-dimensional directly implies Zariski's lemma as a special case. And Zariski's lemma pretty easily implies the Nullstellensatz
 
3:29 PM
I don't understand, isn't every Möbius transformation an automorphism of the riemann sphere?
 
so it's not something just can get just by working the definitions
 
@Alessandro Look, the antipodal map is $f(z) = -1/\bar{z}$. Agree?
That's clearly not holomorphic.
 
is that a Möbius transformation?
 
@BalarkaSen Isn't that the same as $f(z)=-z$?
 
It's an isometry of $S^2$. I thought those are usually divided into orientation preserving/reversing categories.
@Alessandro That sends $0$ to $0$... how is that the antipodal map?
 
3:31 PM
Right, but it agrees with yours outside of $0$ and $\infty$ so you only need to define it carefully there
 
$-z = -1/\overline{z}$ on $|z| = 1$, the great circle of the sphere.
 
I think @AlessandroCodenotti was confused by "orientation preserving Möbius transformation", which is redundant
but of course you're right that the antipodal map is not orientation preserving, it's also not a Möbius transformation
 
Ahhh, wait, so $f(z)=-z$ is a Möbius transformation, but it's not the antipodal map, makes sense now
 
Ah you're right that the orientation reversed Moebius transformations are not called Moebius anymore
anti-Moebius?
 
@BalarkaSen I'm thinking about why the result here is $S^2$ now
 
3:37 PM
Oh it's a very good picture. Let me see if I can find a drawing.
 
I think the source of the confusion is this: the map $\Bbb C \cup \{\infty\}, z \mapsto -z$ is not the same as the map $S^2 \to S^2, z \mapsto -z$
 
This is for $g = 1$.
 
when we embed $S^2$ into $\Bbb R^3$
oh wait you already said that
 
Ohhh, that video is fantastic!
@MatheinBoulomenos yeah, that was confusing me but we sorted it out already
 
@MatheinBoulomenos Right, the point is, the antipodal map sends the upper hemisphere orientation-reversed to the lower hemisphere. When considering $\Bbb C$, the upper hemisphere is $|z| > 1$, and lower is the disk $|z| < 1$. $f(z) = 1/z$ switches upper to lower, but leaves the "position of the points" invariant. This is reflection about the equatorial plane. $f(z) = 1/\overline{z}$, on the other hand, not only flips upper/lower hemispheres, but also flips the "position of points"
That's why it's the antipodal map
(Just as a footnote for Alessandro)
 
3:44 PM
That was very helpful, thanks! (Also from my classmate, I'll convince her to make an account here eventually)
 
We're basically going to turn into the Trento university group chat then. Still better than a Heidelberg university group chat, where scary algebraists like @Mathein resides
Set and measure theorists over algebraists any day
 
I'm the only Heidelberg algebraist here, save for lush who wasn't here a lot recently, right?
 
Please can someone help me understand periods of functions better
 
Yeah I had lush in mind. I haven't seen him in a while
 
@BalarkaSen lol
 
3:47 PM
Did someone say number theorist chat?
 
Please, here
 
Oh, we're also under attack from the UChicago
 
Since we're all in the last year of undergrad not all of us will be in Trento next year most likely
 
$\operatorname{Aut}(\Bbb P^1(\Bbb C)) \cong \operatorname{Aut}_{\Bbb C}(\Bbb C(z)) \cong \operatorname{PGL}_2(\Bbb C)$
it's all algebra
 
"Riemann surface theory is just field theory of extensions of $\Bbb C(z)$" - Mathein Boulomenos, 2018
 
3:50 PM
I'm not the first one who said something like that
you may not want to fix an embedding of $\Bbb C(z)$, though, if you don't want to fix a map $X \to \Bbb P^1(\Bbb C)$
 
Ah yes.
 
so rather, finitely generated extensions of $\Bbb C$ of transcendence degree 1
 
Hm, is my new profile picture showing up?
I just changed it
 
Strange, it's not there yet on my side.
 
3:53 PM
I think, it's different from the usual one
What is it?
 
It's a Beksinki painting that I remembered arbitrarily
 
you need to clear caches
 
I just reloaded and it seems to work
 
Weird pic
 
4:12 PM
Hello everyone ! Can anyone help me in identifying the limiting value of $ \frac{∣\sin x∣}{x}$ , when $x→0$ ...?
 
I like it
Do you know what the limit is for sin x / x?
 
Consider what the graph of $|\sin x|$ looks like
 
Thonk about what Mike says and what happens when you approach from the negative half and the positive half of the real line
 
I wasn't sure if those hints should be explicitly stated or not
 
Hmm. I guess I'm just a less sadistic teacher
@MikeMiller I sent something elsewhere
 
4:29 PM
Oh, by the way, Balarka, your neighbors to the east are about to launch their first satellite
 
Yeah I heard
P cool
 
double-checks map
Needs more Dhaka
 
4:53 PM
@MatheinBoulomenos Hey man, do ya think I could ask you a quick question? :)
 
@MatheinBoulomenos Cheeers, there's a step in the proof of FLT for reg. primes that says that $(\alpha, \beta)(1 - \zeta_p)$ is the greatest common divisor of the ideals $(\alpha + \zeta_p^i \beta)$ for all $i \in \lbrace 0, \dots, p-1\rbrace$
and I'm not sure why!
I know that $(1-\zeta_p) \mid (\alpha + \zeta_p^i \beta)$ for all $i$ and $(1 - \zeta_p)^2 \mid (\alpha + \beta)$ only
woops
and in fact I can see that $(\alpha, \beta) \supset (\alpha + \zeta_p^i \beta)$ for all of the $i$, so is it just because $(1 - \zeta_p)$ and $(1 - \zeta_p)^2$ are coprime that makes $(\alpha, \beta)(1 - \zeta_p)$ the gcd of those ideals?
 
that doesn't sound right. $(1-\zeta_p)$ and $(1-\zeta_p)^2$ are not coprime
any assumptions on $\alpha$ and $\beta$?
 
erm
they are prime to $1 - \zeta_p$
 
if $\alpha=\beta=1$, then $1+\zeta_p^i$ is always a unit unless $i=0$
but $1-\zeta_p$ is not a unit
 
5:08 PM
hi all, I'm new here and just started off with college math
 
@MatheinBoulomenos I'm not sure why that helps lol, sorry
I think I should probably also mention that $\alpha, \beta \in \Bbb Z[\zeta_p]$ are fixed to begin with
 
clearly $(1-\zeta_p)$ is not the gcd of the ideals $(2), (1), \dots (1)$
but $1-\zeta_p$ is coprime to $2$ and to units
 
Okay so.. $(\alpha, \beta)(1 - \zeta_p)$ is not the gcd of those ideals?
 
5:13 PM
is this case 1 of FLT? iirc, you just want to say that if $z^p=x^p+y^p$ and $x,y,z,p$ are coprime then the ideals $(x+\zeta_p^iy)$ for $i\in \{0,1, \dots, p-1\}$ are coprime
 
No it's case $2$ of FLT and you want a decomposition $(\alpha + \zeta_p^i \beta) = (\alpha, \beta)(1 - \zeta_p)\mathfrak{c}_i^p$
where $(1 - \zeta_p) \nmid \mathfrak{c}_i$
 
@BalarkaSen ah, it looked like his style but I had never seen this painting before
 
ah, case 2 is more tricky
I don't remember the calculations
but you need to put in more assumptions than you stated above. I assume that you have $\prod_{i=0}^{p-1}(\alpha+\zeta_p^i\beta)=\alpha^p+\beta^p=\gamma^p$, where $p \mid \gamma$? Then the statement about the gcd seems more plausible
 
just technically get back on math chat, and I am seeing zetas floating everywhere mind=screw
 
Hmm it's more like $\prod_{i=0}^{p-1}(\alpha + \zeta_p^i\beta) = (1 - \zeta_p)^{np}(\gamma)^p$ but yeah
 
5:22 PM
okay, you factor out the powers of $(1-\zeta_p)$, sure
 
ok I knew nothing about prime ideals escapes to elsewhere
 
and $1 - \zeta_p \nmid \alpha\beta\gamma$ is another assumption
Perhaps I should ask it on the main site or?
 
I'll think a bit more about it
 
Alright man, don't worry about it too much
 
5:28 PM
but if each factor $\alpha+\zeta_p^i\beta$ is divisible by $(1-\zeta_p)$ only once, then we would need to have $n=1$ in the above equation
that's impossible
 
Yeah, I wrote above that $(\alpha + \beta)$ is divisible by $(1 - \zeta_p)^2$
and so the smallest ideal containing ALL of those ideals is $(\alpha, \beta)(1 - \zeta_p)$, right?
Or does that make no sense hahaha
I wrote smth stupid above like $(1 - \zeta_p)$ and $(1 - \zeta_p)^2$ are coprime, pls ignore
 
it's possible, but I don't see why
 
No problem maaan
 
if $I$ contains $\alpha+\beta$ and $\alpha+\zeta_p\beta$, it also contains $\alpha+\beta-(\alpha+\zeta_p\beta)=(1-\zeta_p)\beta$ and $\alpha+\zeta_p\beta-\zeta_p(\alpha+\beta)=(1-\zeta_p)\alpha$
 
 
5:40 PM
since we can write $\operatorname{gcd}_{\Bbb Z}(\alpha,\beta)=x\alpha+y\beta$, $I$ also contains $\operatorname{gcd}_{\Bbb Z}(\alpha,\beta)(1-\zeta_p)$
here $x,y\in \Bbb Z$ (I assume that $\alpha, \beta \in \Bbb Z$, also)
well, no we don't need that
 
Well, $\alpha, \beta \in \Bbb Z[\zeta_p]$
second case proves stronger FLT
 
it contains $(1-\zeta_p)\alpha$ and $(1-\zeta_p)\beta$, so it contains the ideal generated by that which is $((1-\zeta_p)\alpha,(1-\zeta_p)\beta)=(1-\zeta_p)(\alpha,\beta)$
 
can someone help me understand this step in this answer? the measure is the Lebesgue measure
given that $f: \mathbb R^n \to \mathbb R$ is measurable show the following set is measurable
 
so certainly, already the ideal gcd of $(\alpha+\beta)$ and $(\alpha+\zeta_p\beta)$ contains $(1-\zeta_p)(\alpha,\beta)$
 
$A := \{ (x,t) \in \mathbb R^n \times \mathbb R: f(x) < t \} \subseteq \mathbb R^{n+1}$
so he writes, definite $F(x,t) = f(x)$, then $F$ is measurable because $f$ is measurable, etc
but.. where did the t go >.<
 
5:45 PM
Well that's good.. lol
 
then $F^{-1}( (-\infty,\alpha]) = f^{-1}((-\infty,\alpha]) \times \mathbb R$, and then he's done
 
no wait, the gcd of two ideals is just the sum
 
maybe it should be $F(x,t) = f(x) - t$?
and then uhhhh
 
Why is $\sup_{x\leq 1} \|T(\delta x)\| = \sup_{x\leq \delta} \|T(x)\|$, where $T$ is a linear operator?
 
so we really want to show that $(\alpha+\beta,\alpha+\zeta_p\beta)=(1-\zeta_p)(\alpha,\beta)$. I did one direction already.
 
5:49 PM
@MatheinBoulomenos Oh right! Nice one, thanks a lot man
 
@LeakyNun, why does this hold: Let $A:\Bbb R^2\to \Bbb R^2$ be a linear transformation with eigenvalues $\dfrac23$ and $\dfrac95$, then there exists a nonzero vector $ v\in \Bbb R^2$ such that $\lVert Av\rVert=\lVert v\rVert$?
 
@MikeMiller Sorry for the delayed reply ... For $x→0$ , the limit is 1 ... Again , I wrote ∣sinx∣ as ±sinx and found the limit to be ±1 ... It isn't okay , is it ? O can't figure it out ...
 
Nehal you're almost there
remember you're considering $\dfrac {+\sin x}{x}$ and $\dfrac {-\sin x}{x}$
where $x$ is either $\to 0^+$ or $\to 0^-$
 
@GFauxPas Vous parlez français?
 
nope
 
5:53 PM
Oh, right. The nickname distracted me.
 
yup
I get that a lot
 
do you think there's enough information to solve the bottle problem I posted? :S
 
@JannikPitt $|T(x)| = |\frac 1 \delta |T(\delta x)|$, and then
 
@GFauxPas ummm ... For the first term , I get 1 , which directly comes from formula ...
 
5:55 PM
and the second one?
 
But , won't it be -1 for the latter term ?
 
why's that?
 
...
Confused ... Never faced ... Wait ...
 
$x \to 0^-$ is like saying $-x \to 0^+$
oh wait
 
Ummm ... Is it also 1 ?
 
5:57 PM
im being silly
whats the original limit
 
x→0
 
of what
 
$\frac{∣\sin x∣}{x}$ , when $x→0$ ...?
 
oh, then you were right, it's $+1$ from one side and $-1$ from the other
sorry
 
@GFauxPas so .. it would be ±1 ?
 
5:59 PM
you would say that the limit doesn't exist
 
@GFauxPas the solution says so ... But I can't comprehend ...
 
if a limit exist, it has to be unique
and it has to be the same no matter you're approaching from above or below
are you asking why that's true?
why can't a function have two limits?
 
@GFauxPas is this any special property ? Does it have any name ?
 
uniqueness of limits
 
@GFauxPas yeah ...
 
6:02 PM
so what's the definition of a limit
there are several
or at least 2
I'll take the standard one
to say $f(x) \to L$ as $x \to \ell$ means that for every $\epsilon >0$ there is some $\delta > 0$ such that
$|f(x) - L| \le \epsilon$ whenever $ |x-\ell| \le \delta$
you can replace $\le$ with $<$, doesn't matter
seen that before?
 
Got it ...
 
so if $f(x) \to L_1$ and $f(x) \to L_2$ as $x \to \ell$,
$|f(x) - L_1| \le \epsilon$ and $|f(x) - L_2| \le \epsilon$
these are positive numbers so you're allowed to subtract one equality from the othert
you get $|f(x) - L_1| - |f(x) - L_2| \le 0$
 
Write $(\alpha+\beta\zeta_p^j)=(1-\zeta_p)^k w$ with $w$ coprime to $1-\zeta_p$ (this is an equation of elements), so as ideals we have $(\alpha+\beta\zeta_p^j)=(1-\zeta_p)^k\cdot (w)$. Also write $(\alpha+\beta\zeta_p^j)=(\alpha,\beta)J$ with an ideal $J$. By assumption $(\alpha,\beta)$ is coprime to $(1-\zeta_p)$, so $(\alpha,\beta)$ divides $(w)$, thus $w$ is actually contained in $(\alpha,\beta)$. This shows that $(\alpha+\zeta_p^j\beta) \in (1-\zeta_p)(\alpha,\beta)$
 
triangle ineuqality now
 
@GFauxPas Merci ... :)
 
6:06 PM
@ÍgjøgnumMeg these are all calculations you need
the first inclusion I did works more general
 
$(|f(x)| + |L_1|) - (|f(x)| + |L_2|) \le 0$, and so on :)
 
@GFauxPas {Merci }^10 ...:)
 
np
\text{merci}^{10}
 
@Silent So A is diagonalizable, i.e. two eigenvectors form a basis. Let $Av_1 = \dfrac23 v_1$ and $Av_2 = \dfrac95 v_2$. Then, any vector $v$ can be expressed as $av_1 + bv_2$. Then, $\|Av\| = \|v\| \iff \left(\dfrac49-1\right)a^2\|v_1\|^2 + \left(\dfrac{81}{25}-1\right)b^2\|v_2\|^2 + \dfrac{36}{25}ab(v_1 \cdot v_2) = 0$
one could ask what the discriminant of that thing is
 
@GFauxPas ha ha ... Was too idle to put the signs ...
 
6:11 PM
have you learned about sequences yet Nehal
 
Typo: 36/25 should be 6/5
 
actually nevermind, I dont want to confuse you
 
@LeakyNun @Silent consider the following map $\Bbb R^2 \setminus \{0\} \to \Bbb R$, $v \mapsto \frac{\|Av\|}{\|v\|}$ this is continuous. If we plug in an eigenvector for $\frac{2}{3}$, we get $\frac{2}{3}$, if we plug in an eigenvector for $\frac{9}{5}$, we get $\frac{9}{5}$. Since $\Bbb R^2 \setminus \{0\}$ is connected, the image has to be connected, so it also needs to contain $1$, as $\frac{2}{3} \leq 1 \leq \frac{9}{5}$
 
$\Delta = \left[\dfrac65(v_1 \cdot v_2)\right]^2 + \dfrac{224}{45} \|v_1\|^2 \|v_2\|^2 \ge 0$
that should be the proof
and then @MatheinBoulomenos presented a higher-level proof :P
@MatheinBoulomenos aha, interesting
 
oh wait my proof is subtly wrong @NehalSamee i was sloppy
 
6:15 PM
I mean, my proof also works for any $\lambda_1<1$ and $\lambda_2>1$
 
you should write $|L_1 - L_2| = |L_1 - f(x) + f(x) - L_2|$
$\le |L_1 - f(x) | + |f(x) - L_2|$ by triangle inquality. Close to what I said but I made a mistake
 
my proofs shows that if $A: \Bbb R^n \to \Bbb R^n$ is linear where $n \geq 2$, then for any positive real number $\lambda$ that is between the absolute values of two eigenvalues, there is some $v \in \Bbb R^n$ such that $\|Av\|=|\lambda|\|v\|$
 
@MatheinBoulomenos ok you win
@MatheinBoulomenos actually you don't win
since if we have two eigenvalues, they span an eigenspace of dimension 2 that is closed under the linear operation
 
What we doin?
:eyes:
 
6:53 PM
@LeakyNun why do I not win?
 
because my argument still works :P
 
but we have two eigenspaces of dimension 1
 
so span it
1+1=2
 
I didn't say your argument is wrong
ah I see what you mean, you restrict the map the the subspace generated by two eigenvectors with distinct eigenvalues than you use the argument for two dimensions
 
[Random then sleep]
1+1=5
 
7:02 PM
o..o
 
For $|5-2|$ sufficiently small
 
:thonk:
 
7:21 PM
@LeakyNun, @MatheinBoulomenos, thank you very much!
 
7:42 PM
So I've just learned about the Fréchet derivative on general Banach spaces. If I have two real Banach spaces $E,F$ why isn't the derivative defined as $\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{\|x-x_0\|}$? This isn't equivalent to the Fréchet derivative right?
I'm sorry, of course it's equivalent to the Fréchet derivative.
 
@GFauxPas OK ...
 
8:22 PM
@JannikPitt: Of course it is not. That formula doesn't remotely work even in $\Bbb R^1$ or $\Bbb R^2$. What kind of object do you think you're defining?
 
hi demonic @Alessandro
 
8:38 PM
Hey @TedShifrin :)
Heya @AlessandroCodenotti
 
Hi Perturbative — you saw my comment?
 
Yep, I saw it, I'm just trying to think about it now
 
OK.
 
Hi @Perturbative
 
@TedShifrin $f$ is defined on some interval $[a, b]$ as usual right?
 
8:48 PM
Sure.
Oh, for the function I gave you, use $[0,1]$.
But you're going to see that writing down any sort of formula for the partition is NOT fun.
But you should be able to describe how to construct it without too much difficulty.
 
 
@Ted So the way I see it, $f(x)$ equals $\frac{1}{d}$ if $x \in \mathbb{Q}$ where $d$ is the denominator of $x$ and $f(x) = 0$ if $x$ is irrational
 
assuming you write $x$ as a fraction in lowest terms, yes.
So graph the function. Hint: My students one year called it the Christmas tree function.
 
Okay will do
 
9:18 PM
Okay I graphed, I can see it looks like a christmas tree
The max value $f$ takes is $\frac{1}{2}$
I'm thinking I can then split $[0, 1]$ into $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ which would correspond to a first really crude partition of $P_1 = \{0, \frac{1}{2}, 1\}$
Then $P_2 = \{0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1\}$
 
If $(X,d)$ is a metric space that isn't totally bounded, how can I show there exists an $\epsilon > 0$ and collection $\{x_n\}$ of points such that $\{B(x_n,\epsilon)\}_{n=1}^\infty$ is a disjoint collection of open sets?
 
Then generalizing we get $P_n = \{0, \frac{1}{n}, \frac{1}{n-1}, ..., \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, ...., \frac{n-1}{n}, 1\}$
@TedShifrin Am I on the right track?
 
9:34 PM
Right track for what @Perturbative?
 
Constructing the partition(s) from earlier
 
You're going to get a very large upper sum using that partition.
 
Can somebody give me a hint with this
x = (x,y,z), a = (a,b,c), t = (l,m,n) of the vectors x, a, t.
Write a set of equations which determine the coordinates of the point where this line intersects the plane
in matrix form Ax = d, where
px+qy+rz=s
The line has equation
(x−a)/L=(y−b)/m=(z−c)/n
lmn
You know what itd be better just to screen grab
 
I don't know what all your letters are, @JakeRose, but use a parameter for the line. Set all those ratios equal to t. Solve for x,y,z in terms of t.
 
Yeah sorry I accidentally over copied and pasted from a previous part of the quesiton
But if I solve for x y and z then how will they show up in the normal matrix equation?
 
9:41 PM
Put them there!
 
Im confused
 
Say $x=2t+1$, $y=-t$, $z=3t-4$. Where does that line cross the plane $x+y+z = 5$?
 
Sub in
Find t
Then sub back into original expressions
 
OK. So do that with your problem.
 
Im struggling
Heres what I have done so far
Solved the line equation for an equation for each variable in terms of alpha and the other two corresponding letter
But Im stuck at that point
 
9:48 PM
You didn't do what I said, did you?
 
But each one has 3 other parameters in. What would I solve for?
 
Read what I said 10 things up.
 
Im not getting anywhere sorry
 
You want x, y, z in terms of t. Just like the example I gave you. You'll have lots of other letters flying around, but they're constants.
I obviously cannot read any of that.
 
Ive done that (obviously not correctly) and its gave me nothing useful just a non reducable mess of letters
 
9:55 PM
Is your partition depending on $\epsilon$, @Perturbative?
Well, it's a problem full of letters, @JakeRose.
For example, $x=Lt+a$.
 
Hello!!

Let $X\sim N(0,1)$ and $Y\sim N(0,1)$. We have that X and Y are independent. How can we determine the distribution of (X,Y) ?
 
@TedShifrin I think I got it, define $$P_n = \left\{\frac{1}{n} \ | \ n \in \mathbb{N} \right\} \cup \left\{\bigcup_{\text{prime } p}\left\{ \frac{m}{p} \ | m < p \ \text{and } \frac{1}{n} < \frac{m}{p} < 1 -\frac{1}{n}\right\}\right\} \cup \left\{ 1- \frac{1}{n} \ | \ n \in \mathbb{N}\right\}$$
 

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