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00:00
Of course, @Perturb, although I haven't yet watched the demolition in the men's final.
heya @Eric
I was waiting for a 5-set epic myself, sadly it wasn't the case
I haven't watched yet, but this may the first time ever Nadal buckled under pressure.
@TedShifrin ppl don’t get to see enough classical stuff :/
I'm guessing wrong game plan, can't see him buckling under pressure given his record
You can't blame me, @Eric. :P
00:03
@TedShifrin haha! Yes, I am dumb when I am not in my comfort zone :(. In fact, there is a much easier proof! Once fixed $x=0$, that for all $h$, we have $f(h)-f(0)=hf'(0)$, so $f(h)=\underbrace{f'(0)}_ah+\underbrace{f(0)}_b$, and we are done!
certainly not
Oh, good point, @manoooh. The technique I was suggesting, however, shows up in other places in real analysis, so try to learn from it.
@TedShifrin I will try it. Thank you!
@Eric: Have we anything math-y to discuss? :P
im doing a GR pset atm
00:10
@Jacksoja Find a better book :P
Rudin is fine
@TedShifrin Djokovic too consistent.
Eh it's pretty terse and could use some more diagrams.
its terse cause you have to work at it
Honestly, I'm more sad that Fed keeps getting eliminated before semis.
00:13
Yeah yeah, surely though it doesn't hurt to spend a sentence or two on why the concepts are significant
LOL @more diagrams. Not one picture.
@TedShifrin Well yep :P
I'm not a big fan of Rudin unless the teacher is an extraordinary expositor.
And his multivariable material is horrid. Typical analyst who doesn't know geometry.
00:14
It's one of the "classics" that's somewhat overrated imo
@TedShifrin i always skip this one
I never liked Herstein's Algebra, either.
i dont really like any of the basic books ive ever used but i think rudin problems are good if you wanna get good at analysis
i like papa rudin
Welp right now my class is using this
i hate that one
00:15
Reviews are rather concerning
Actually, I should look seriously at Charles Pugh's real analysis book. I imagine it's quite excellent.
@ÉricoMeloSilva Sigh I need a silver lining here ;-;
@TedShifrin i used it for my first analysis class in hs and i thought it was ok
@SirCumference: I suggest you read my book instead. Or Munkres's Analysis on Manifolds.
Yeah +1 for Analysis on Manifolds
00:16
@TedShifrin Well presumably this contains all the information that'll be on the tests and homeworks. By chance do either of those follow almost the same format?
@Eric: He taught honors analysis out of Rudin numerous years. I took graduate dynamical systems from him my first year and really liked his taste and style (even if he couldn't figure out how to use blackboards correctly, even when I tried to force him:P).
@SirCumference: What is your background? You know multivariable calculus, linear algebra, and single-variable analysis?
Pugh explains a lot more than rudin which is probably why i like it less, i dont like when these books explain a ton to me
I've been reading through Munkres' topology and it's pretty good
@TedShifrin a ha! pugh's ODE section is pretty good lol
@Eric: Let me be polite. You are not the typical undergrad math student.
00:18
i am aware of this
He's telling you to quit m8
I explain more than some books too, but you don't hate me. Interesting.
hi @MikeM
@TedShifrin My only analysis background is the first three chapters of Rudin. I read through the entirety of this for multivariable calc (if you happen to know of it), and I took a proof-based linear algebra course
@SirCumference: My book is probably too elementary for you, but it might make a good alternative reference (or look at lectures on line).
00:19
@TedShifrin the other book i had to compare ur diff geo notes was do carmo
which as u know is a LOT less to the point :)
Also read through the first two chapters of Munkres topology
Yes, I have taught out of various editions of Marsden-Tromba.
@Eric: It's written more for students like you.
i still like do carmo tho so i guess im full of shit
Are these the ones I whined about handwriting from?
I think there's some good stuff in doCarmo, but it is too sophisticated even for the typical Berkeley undergrad math major.
00:20
@TedShifrin So what's your thoughts? Think I can get by with those for Calc on Manifolds?
the end is full of goodies
No, @MikeM, he's talking about my "published" .pdf.
I dunno, @SirCumference.
But we've given you two alternative sources to look at.
Yep, thanks
00:22
I was trying to solve this problem from G&P earlier. Let $p(z) = z^m +a_1z^{m-1} + \dots + a_m$ be a polynomial with complex coefficients and consider the map $z \mapsto p(z)$ of the complex plane $\mathbb{C} \to \mathbb{C}$. Prove that this is a submersion except at finitely many points.
@MikeM: I did send Eric my lecture notes from complex geometry from 1980. He didn't whine nearly so loudly as you about my handwriting.
@TedShifrin Have you felt there haven't been many high-quality questions on main lately?
My approach was this: okay so let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = p(z)$, then note that $f'(z) = mz^{m-1} +a_1(m-1)z^{m-2} + \dots + a_{m-1}$. By the fundamental theorem of algebra $f'(z)$ has at most $m-1$ roots, call them $\eta_1, \dots \eta_{m-1}$, and $f'(n_i) = 0$ for all of these roots.

Then for all $z \in \mathbb{C} \setminus \{n_1, \dots, n_{m-1}\}$ we have $f'(z) \neq 0$ and hence $df_z : T_z\mathbb{C} \to T_z\mathbb{C}$ has rank $1$ as a $\mathbb{C}$-vector space and so $df_z$ is surjective and $f$ is thus a submersion at $z$.
@TedShifrin yes, he's a very respectful chap
@MikeM: I don't look as seriously on main as I once did, but, yeah, nothing that's stumped me for weeks — as happened a few times in the past.
00:23
Some of the stuff I said at the end there may have been wrong, but is the idea the right one?
i only disrespect bootlickers
@MikeMiller More like: You're a whining ingrate :D
I'm back to working to pass the time.
i also didnt have trouble reading Ted's handwriting literally at all lol
@Perturbative You definitely shouldn't need the fundamental theorem of algebra.
Exercise for you (the folks in the 1500s could show this): every degree n polynomial has at most n roots.
00:24
@Perturb: You should relate surjectivity to $\Bbb C$ to surjectivity to $\Bbb R^2$, I suppose. ... Also, are you going to show that the induced map from the sphere to itself has degree $m$?
That's usually a baby exercise in a beginning algebra course (for polynomials over an integral domain).
But other than Ted's nitpick (which I worry will send you down a rabbit hole), that's a proof, yes.
checks for rabbit holes
Hint: How is the complex derivative $f'(z)$ related to the Jacobian matrix of the map $\Bbb R^2\to\Bbb R^2$? Surely you know this.
If I did it I can't remember off hand, but I should be able to easily look it up
You should know this, sir :P
Have you had a basic complex variables class? Remember Cauchy-Riemann?
Yeah I remember the Cauchy-Riemann equations
00:32
Then that will answer my question.
And this is an important thing for geometry, too, this observation.
@TedShifrin My math background is weird :p
No comment.
The important thing being identifying $\mathbb{R}^{2n}$ with $\mathbb{C}^n$? Or the link between complex derivatives and Jacobians?
Well I guess both
@MikeMiller By not needing fundamental theorem of algebra do you mean I shouldn't state it in the proof? Like I should be like "well this polynomial clearly has n roots"?
he means it's overkill
think about his "exercise for you"
I wouldn't kill Perturb for the overkill, however.
00:51
@TedShifrin Ohh wait I know how to show that the induced map from the sphere to itself has degree $m$, $H_n(S^n) = \mathbb{Z}$, so any continuous map $f : S^n \to S^n$ passes down through homology to $f_{*} : \mathbb{Z} \to \mathbb{Z}$, which is certainly always multiplication by some $m$
I don't think that's what you were asking though :p
But why is the $m$ the same $m$? I guess you didn't get far enough in G&P to know how to compute degree locally.
Ohh wait nonno I didn't go that far in G&P
@TedShifrin I just don't think anybody should ever say the FTA in this context. It's an elementary fact.
I don't disagree. Here we only need to know $f'$ has finitely many roots.
Hi Ted
00:56
hi Jacksoja
@TedShifrin how to prove that a set with a metric is a metric space ?
I know the 3 parts to prove, but does one has to prove them with that particular metric?
Yes.
Do you know the metric is a metric?
That's what you're talking about, I assume.
in R^4 for example, should I show that |x-y| is a metric?
or take it for granted
Yes it is
A set with a metric is the definition of a metric space though?
Right, that's why the question is confusing.
01:02
Maybe @Jacksoja needs to prove that the given function is a metric
it is R^4 with the usual metric
should I do that work with sigma?
or just claim it to be true
I don't know. Ask your professor what he/she wants.
Wait where did sigma come from?
definition of Euclidean length
pytogoras in 4 D
01:03
Ohh thats what you mean
Or the question could be to see that the metric induces the same topology as the product topology or something else ... who knows.
in that defintion, ((x_1 - y_1) ^2 + ... + (x_4 - y_4) ^2 ) ^1/2
this is for sure bigger than zero
because of sum of squares , and also same as norm of y-x
If you want to prove a mapping is a metric, you need to show identity of indiscernibles, symmetry and the triangle inequality.
yes I know that part, but how much to calclute to be considered a good proof ?
No calculation, just a rigorous argument
01:14
@Thorgott Thanks
then the exercice was pointless ^^
@TedShifrin @Thorgott do you recommend analysis book ( first year ) ?
it is for self study purpose
Maybe Apostol's Advanced Calculus. Wade wrote a reasonable analysis book. So did Charles Pugh. There are probably zillions of others I don't know.
I've read and liked the first chapters of Rudin's Principles of Mathematical Analysis, but have not ventured into the later parts. I don't have a lot to compare it to though, so take my opinion with a grain of salt.
I think Rudin is horrible for self-study unless one is super strong.
I already stated above that one needs a really good lecturer to learn from Rudin.
@TedShifrin how do you interpret the trace of a linear map?
What kind of answer are you looking for?
01:27
geometrical interpretation / coordinate-free definition
like how determinant is the scaling of area
Yeah, but if you look at flow by your linear map, then rate of change of area is given by the trace.
I self-studied from Rudin in high school and had no issues, but as I stopped before getting to the second half, I have no clue how that holds up
I don't trust self-studiers ... They really have no way of calibrating what they have learned or can write.
The best thing to do if you're planning to self study is to find somebody to study with
That's a reasonable doubt to have
01:32
Or at least talk to other people who've done what you're self studying
@TedShifrin i like the formula for trace where u integrate over the sphere
probably cuz it's v useful to prove some PDE statements :P
Hmm, I dunno what you're referring to, @Eric. Are you talking just divergence theorem?
I liked finishing my course with the divergence theorem in the form of flows ...
$Tr(A) = \int_{S^{n}}\langle Ax, x \rangle d\sigma(x)$
up to a constant
Oh, so it's immediate from the divergence theorem.
Flux of $Ax$ is given by the divergence, which is the trace.
01:38
Integrating over the sphere is averaging over orthonormal bases.
But you get the same answer for any orthonormal basis.
this statement is quite useful when u want to prove mean value type properties and stuff
Hmmm ... but this is very linear.
yeah, comes up though
I mean, it has the look of a Green's identity, I suppose.
ive used a formula of this type a couple times in geometric flow stuff cuz ricci is a trace of a guy
01:45
well, yeah, sure.
I still like the flow Divergence Thm no matter what ...
that's what Arnold uses to interpret trace in some book
ode or classical mech or both
probably his Mechanics
anyhow, @Leaky probably hasn't been paying detention
detention?
02:11
Yes, detention must be paid.
This is Partial subring lattice diagram of $\mathbb C$, as given in Gallian's abstract algebra text.
I wonder why $\mathbb Q(\sqrt2)$ is not placed between $\Bbb R$ and $\Bbb Q$, and kept aside. Please someone explain!
That's just a typographical issue, @Silent. Nothing more.
oh! ok, thanks
 
8 hours later…
10:41
@TobiasKildetoft can I ask you one question with full information?
@ninjahatori Sure, but I might not have time to read a ton
ok
are you familiar with involution ? @TobiasKildetoft otherwise I will define
depends on the context
$Involution$ : Let $R$ be ring . An Involution is map
$ *:R\rightarrow R$ such that it satisfy $i)(a+b)^*=a^*+b^*$; $ii)(ab)^*=b^*a^*$; $iii) (a^*)^*=a$.
This is defination
ok @TobiasKildetoft
So using this we define sesequlinear form which has same properties as billinear form except semilinearity in first variable
say s is sesequlinear form then s(xa,y)= a*s(x,y)
and all other properties are like bilinear form
10:52
so now from this we define $lambda$ hermitian form such that by using center of ring say Z(R) be center of ring $lambda$ is elemnet in center such that it satisfy $lambda$ * $lambda^*$ =1 then h is called lambda hermitian if h(x,y) = $lambda$ h(y,x)* where h is sesqulinear map MM to R . so h(y,x) makes sense
sorry h(y,x)* makes sense
@ninjahatori Please rewrite that entire thing from scratch, because it is practically unreadable like that
So suppose there is elment z such that zz*=1 in center of ring then h is called z hermitian form if h(x,y) = z h (y,x)* where h is map from M * M to R where h(y,x) is element of R so we take involution h(y,x)* .
M is R-module
M*M is M cross M here
Cartesian product
@TobiasKildetoft do you get it or I elaborate more
ok
so now this moduule is called (M,h) isz hermitian module
Its dual is in usual sense M*=Hom(M,R)
Adjoint map between M and M* is defined as adj: M to M* as (adj x) y = h (x,y)
@TobiasKildetoft got it
are you there @TobiasKildetoft
11:11
@ninjahatori Sure, but I thought you were still setting stuff up, since there was no question yet
I am going to lunch now. Hopefully I will have time when I get back.
ok @TobiasKildetoft
thanks
11:50
@ninjahatori Back now. I have about 10 minutes until I need to go.
 
1 hour later…
13:04
Hi. If $K:\mathbb{R}^d\to[0,\infty)$ is positive everywhere, bounded, $K(Cx)<C$ for all unit norm $x\in\mathbb{R}^d$ for large enough $C$, and $K$ is $L^q$-integrable, can I use this integrability to derive bounds for a smallest possible $C$?
13:57
Let $M$ be an $n\times n $ matrix with real entries such that $M^3=I$. Suppose that $Mv\ne v$, for any nonzero vectors $v$. Then since $M^3-I=0$ hence $(M-I)(M^2+M-I)=0$, but $M\ne I$ hence $M^2+M-I=0$
In above reasoning, I can't understand how can we discard possibility of zero divisor? that is $AB=0$ but A,B nonzero.?
(I meant $M^2+M+I$ above)
 
2 hours later…
15:29
@LeakyNun coordinate-free defintion of trace: let $V$ be f.d. vector space, then using the natural isomorphism $V^* \otimes V \cong \mathrm{End}(V)$, the trace can be identified with the evaluation pairing $V^* \otimes V \to k$ given on elementary tensors by $\xi \otimes v \mapsto \xi(v)$
@Silent $Mv \neq v$ for all nonzero vectors $v$ implies that $1$ is not an eigenvalue of $M$ which implies that $M-I$ is invertible
@MatheinBoulomenos thanks
@MatheinBoulomenos why does it feel like some tautological vector field / differential form?
16:40
When would you guys say it's appropriate to use a non-standard index set?
16:59
hi
Whenever there isn't a standard index set for your indexed family?
If continuous data is presented as a list of data points, is it safe to assume that it is being treated as discrete?
What is a standard index set
17:16
All I meant is that if you construct family of functions, one for each rational, I wouldn't index by with $\Bbb N$ as index set
Oh whatever
hmm
When you want to do other gradings than $\mathbb N$-gradings
$\Bbb Z$ gradings are very common
Yes :)
So, very appropriate to use non-standard index sets in the case of graded rings/modules
Basically, when you're ok with giving up some of the "natural" well-ordering structure of $\mathbb N$ you can go ahead and use a different index set.
17:29
Okay, so you wouldn't want to give up a well-ordered index set if you were looking at, say, sequences. But if you just wanted to consider families of objects, any kind of index set could potentially be appropriate?
Yeah. A good example would be an indexed open cover of a topological space.
I like to grade over Z/2N
It's unclear whether you just want a set of the cardinality of your family or if this set should have an order that plays a role as well Rithaniel
Usually you only pay attention to that structure which is relevant
If you're indexing an open cover most of the time you don't need to order the elements
Sure but if you have instead the set of all open sets rather than a cover you probably want to keep track of the (partial) order as well
Anyway I don't think this is a particularly useful thing to worry about
17:52
Fair enough. It's mostly a thing about making sure things are understood, though. If a person has to pause and wonder about the index set, they can't really get a good look at the argument itself.
No person has to pause and wonder about the index set. Stop doing so. :)
An important skill is to learn what the unimportant details are.
True, a skill I'm still developing.
18:26
damnit why cant anyone answer my noob question anywhere i go lmao
@ramose how can it be continuous if its a "list"?
the type of data is continuous though. so a small sample of it , will ignore the fact that any data from out of the list could take any value then
ok, i guess that is kind of obvious
19:02
$\{\pi, 2 \pi , \dots \}$ is a discrete set of points, though each point is irrational.
 
2 hours later…
20:34
o/
21:03
Hello guys!
Is $(\mathcal P(A),\subseteq)$ a set? Or is it a system?
It's a (partially) ordered set
Hi @davidlowryduda!
21:29
rip
@MikeMiller How's life?
It's pretty good. I'm graduating this year. Carrying out the finishing touches on what is essentially my thesis, making sure sentences read well and the notion is consistent.
(It's not.)
21:47
Oh man, finishing up. What's your thesis on?
@AlessandroCodenotti no, it's a partial order; P(A) is the partially ordered set
(maybe you're right, why am I talking about useless stuff)
@LeakyNun thank you!
So, in general, what is an ordered pair?
@davidlowryduda attempt at an elevator pitch: there's a recipe to get invariants of smooth 4-manifolds, by counting solutions to certain PDEs. There is a recipe to take this idea and instead get homology groups on a 3-manifold (your differential encodes solutions to the 4-dimensional PDEs on R x Y). There are some deficiencies with the first pass at a definition, as it ignores an SO(3) symmetry of the equations.
@LeakyNun if we want to be annoying it is a set just as everything else in ZFC
I define equivariant homology theories for 3-manifolds following this recipe but respecting the SO(3) action, satisfying some useful relationships. This means I have to do some analysis (elliptic PDE) and some algebra (equivariant cohomology of chain complexes). Occasionally there is a glance at some topology, but not much.
21:54
@AlessandroCodenotti sure
@AlessandroCodenotti but in Lean it's an element of a sigma type (dependent [cartesian] product)
(the type of $\subseteq$ is $\mathcal P(A) \to \mathcal P(A) \to \operatorname{Prop}$ which depends on $\mathcal P(A)$)
22:11
What is the name of a function that maps from all real numbers TO all real numbers, like x or x^3?
it's more than just bijective because log(x) is bijective but it doesn't map from all R to all R
I think you're looking for a surjective function $f\colon\mathbb{R}\rightarrow\mathbb{R}$.
not quite, because if a function is bijective then it is surjective, and I just pointed out the problem in that.
$\log(x)$ is not a function $\mathbb{R}\rightarrow\mathbb{R}$
log(x) is definitely a function for x>0
without a doubt
Suppose $G$ is abelian and $|G| = pq$ where $gcd(p,q)=1$. If $G$ contains elements $a$ and $b$ of order $p$ and $q$ respectively, how can I show $G$ is cyclic?
22:18
yes, which means that its domain is not $\mathbb{R}$
I never said it was R
I said not being R was the problem
x is bijective over all R but ln(x) is only bijective over some R
I want a term that generalizes that to a function that is bijective over all R
and I said it sounds like you are looking for a surjective function $f\colon\mathbb{R}\rightarrow\mathbb{R}$, to which $\log(x)$ is not a counter-example.
or, in that case, a bijective function $\mathbb{R}\rightarrow\mathbb{R}$
it's contrary to my request of the term that defines functions which map from all R to all R.
I didn't ask about a function that is bijective only over some R, I explicitly stated from all R to all R.
The term is still bijective
nope
I explicitly stated the parameters for which that term is not applicable to the general circumstances
bijective is already part of the definition of whatever this term is.
saying bijective is the definition of bijective is circular
Whatever this term is, it has the definition of "bijective over all real numbers".
22:24
You can't talk about mappings without specifying domain and codomain. You are exactly looking for a bijective mapping $\mathbb{R}\rightarrow\mathbb{R}$.
Right, and I am looking for a specific term that describes that explicit circumstance.
There is none, it would be redundant
It clearly wouldn't be redundant if it describes only specific classes of functions like all odd powers of x.
It would be redundant in the sense that you do not talk about mappings without specifying domain or codomain and as soon as those are specified, you are just requiring plain bijectivity.
@VaneVoe You are both wrong and rude, which makes me and I'm sure others less likely to be interested in chiming in.
22:33
Let $n = n_1 + ... + n_r$ be a partition of $n$. I am trying to show that Lagrange's theorem implies that $n!$ is divisible by $\prod_{i=1}^r n_i !$. My thought was to show that $S_{n_1} \times ... \times S_{n_r}$ can be embedded into $S_n$, but this seems rather difficult. What's the best approach?
@user193319 What difficulty do you encounter?
Trying to definition the embedding. I tried the following: $\varphi (f_1,...,f_n) = \varphi_1(f_1) ... \varphi_r(f_r)$, where $\varphi_i(f_i)$ is defined so that it equals $f_i$ when restricted to $\{1,...,i\}$ and the identity on $\{i+1,...,n\}$. Each $\varphi_i$ is injective, but I don't think that $\varphi$ is injective
Note $f_i \in S_{n_i}$.
I find this hard to understand. Let me try to rephrase. Given $f_1, \cdots, f_r$, with $f_i \in S_{n_i}$, you define $$\varphi(f_1, \cdots, f_{r})(j) = f_k(j)$$ if $$n_1 + \cdots + n_{k-1}+1 \leq j \leq n_1 + \cdots + n_{k-1} + n_k.$$ That is, you just apply $f_i$ to the appropriate subset of elements of $\{1, \cdots, n\}$.
Why do you think this fails to be injective? What is an example of $(f_1, \cdots, f_r)$ that you think this will map to the identity?
22:49
Hmm...Is your $\varphi$ the same as mine? I'm having trouble seeing it.
I had trouble understanding what you meant. What you wrote seemed not to make any sense.
First off, $f_i$ has domain a set with $n_i$ elements, not a set of $i$ elements. Further, what you wrote does not seem to define the value of that function on every element of $\{1, \cdots, n\}$, and it does not seem to give a well-defined value for any one element.
23:01
You're about to waste much more time by not understanding the correct answer you received, but no skin off my back.
If you want to get anywhere doing anything, work on your ego.
would it be incorrect to call a linear algebra basis a matrix? for the purposes of my calculations, I'm letting each vector of the basis be a row of a matrix F. then calculating the product between F and another matrix A. just trying to use correct notation in my article
@Collin That is correct, these are not conceptually the same thing. A basis of a vector space is a sequence of vectors $v_1, \cdots, v_n$ satisfying certain properties.
Given any two bases, there is a matrix called the 'change of basis matrix' which takes the first basis to the second basis.
The matrix you are thinking of - the one whose rows are the vectors $v_i$ - is the 'change of basis matrix' whose first basis is the standard one: $(1,0,\cdots, 0), (0, 1, \cdots, 0)$, etc, and the second basis is the one you gave me.
(Or maybe you use the canonical basis as the second basis, I forget the conventions.)
The moral, though, is that a matrix determines another basis once you're given a basis to begin with, and vice versa.
Depending on your audience, this might be unnecessarily pedantic.
stats/cs/ai people. Thanks though. I'll look into change of basis matrices
@Collin I would ignore most of the discussion above, then. Say something like: "A basis v_1, ..., v_n determines an invertible matrix A, whose rows are [blah]. We call this the matrix associated to the basis v_1, ..., v_n."
that would work! thanks
23:10
The conceptual point above is not important so long as you are clear what, precisely, your matrices are.
No problem.

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