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5:00 PM
I don't know what that means.

Can someone help me with this?
https://math.stackexchange.com/questions/3235882/a-simple-complex-function-spivak-27-12a/3235904?noredirect=1#comment6656318_3235904
I think I'm experiencing a language barrier with the person who answered.

Spivak just made a stupid mistake. It's wrong.

Thank you, That's all I needed XD

Most of the errors in earlier chapters got caught by those of us who taught out of the book numerous times (14 times for me), but I never covered that chapter.
The problem itself has the square root in it (at least, in the last edition), but it's a sloppy error in the answer book.

5:05 PM
LOL. Ak's answer is perfectly fine. Give it to him/her. :)

isn't that absolute value given wrong?

I don't think he answered the question.
@TedShifrin If you don't give the answer I'll do it myself

I'm more willing to say that authors made a careless error. (Particularly since I know Spivak and my name appears in his book :P)
I put a comment.

5:10 PM
Happy now?

YOU ARE MISSING NOTHING
THE AUTHORS MADE A MISTAKE. YOU WERE RIGHT

Stupid to make a federal case out of this.

@TedShifrin hand them a lawsuit

Yes :)

LOL, I know whom I'd like to see in prison. That's a different issue.

5:12 PM
rolls inaccessibly many eyes and then ram the lawsuit with an infinite Dedekind finite set so it peels into nonexistence very gradually

@TedShifrin wHoM

My lawyers have been put on notice

@Secret although I don't get this, still : Oof

@Secret: You owe me royalties when you roll eyes ... particularly more than countably many.

@TedShifrin measure zero ?

5:14 PM
@Simone Compris! Très bien :P

Unroll a proper class many eyes
done, royalties overpaid :P

I'm coming for you Michael Spivak!

ok ok.
calm ur ..

Today I learned that Ted has a copyright on the concept of rolling eyes.
3

@TedShifrin In all seriousness, Thank you Ted.

5:16 PM
So, what are your approaches towards learning a certain new topic

I must be more self confident next time

@Simone a balance between overconfidence and self-doubt is essential

You're most welcome, @Simone. There are some excellent problems in that book, so enjoy them.
If the problem is correctly stated, then it becomes obvious that the answer book is just silly. But there have been occasional mis-stated problems over the years.

Go learn math.
same applies to me though

5:19 PM
@Rithaniel He used to roll an integer number of eyes, until I and Leaky started to explode that meme to the reals and then up the cantor's attic

I should write a book and say in the preface "There are errors placed throughout the text as exercises for the reader to correct"

@Daminark I don't like Minesweeper

@Daminark that's beyond science

The new meta

That's one way to cover your sloppy butt, Demonark.

5:26 PM
Make the errors on prime page numbers

I don't think I'm too careless :P

I am totally not careless, but nevertheless I found typesetting errors/printing errors in my first book that drove me crazy. Several books later, I came to accept the notion that even perfect authors are human and make errors.

I wonder if there are books out there with no known errors

Munkres's topology book?
I believe my differential geometry text at this point is error-free.
I actually don't know of an error in baby Rudin, but I have not personally read the whole book.

Hey guys can someone help me with this:
>The following matrix over $\mathbb{Z_5}$ is given:
$$\begin{bmatrix} 2 & 2& 2& 3\\ 1&3&1&3\\ 3&0&2&2\\ 4&1&0&4\\ 1&2&2&0 \end{bmatrix}$$
Specify the dimension of the row space and the column space.
Specify a base BS for the row space and a base BS for the column space.

Row-space:
$$\begin{bmatrix} 2 & 2& 2& 3\\ 1&3&1&3\\ 3&0&2&2\\ 4&1&0&4\\ 1&2&2&0 \end{bmatrix} \iff \begin{bmatrix} 2 & 2& 2& 3\\ 0&2&0&4\\ 0&0&4&1\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix} \implies \dim(Row)=3$$
Now a question: Is

5:30 PM
The more you write the more you error
on average

@Doesbaddel: The vectors in the row space come from linear combinations of the rows. Same for columns. Doesn't that answer your question?

@TEd

@Semiclassical
I finished the course but I still want to know how to do that practice problem, I think we'll enjoy doing it together :)

@TedShifrin so the first row basis is right and the first column basis?

5:34 PM
No.
What is the dimension of the space that the row vectors/column vectors live in?

The dimension is 3

Oh, I'm sorry. I'm not sure, how do I specify this? :(

Count how many entries there are in each row vector and how many entries there are in each column vector.

We have 4 entries for the row vectors and 5 for the column vectors

5:40 PM
Huh?
Do you know what a row is and what a column is?

(2,2,2,3) is one row isn't it?
xD

Oh, apologies.
I stopped paying enough attention. You are correct. I apologize again.

No problem ;D

rolls 13 eyes at self

@TedShifrin I hear Knuth's Art of Computer Programming is error free or nearly

5:42 PM
At least we got an integer this time

I believe that. The TeXBook I bought decades ago that Knuth wrote seemed error-free.

@TedShifrin no worries

Because he sends checks for some small amt to people who found errors (which are never cashed)
Bc people would rather have his autograph than 25 cents or whatever it is

@TedShifrin i cant write a line of tex without errors damn

@Eric: Write a few books. You'll improve.
Like everything else, it's a matter of practice.

5:44 PM
Mike linked it to me yesterday
damn it's purrrrrrrty

I probably shouldn't do this until (unless?) I get tenure but at some point later on I'd like to write books on random topics I have nothing to do with to learn about them. Sounds like it could be fun

Yeah, I've seen it.

@TedShifrin Vectors in the row space come from $\lambda_1(2,2,2,3)+\lambda_2(1,3,1,3)+\lambda_3(3,0,2,2)+\dots+\lambda_5(1,2,2,0)$ right

i now think this is unironically how all math books should look

Demonark: That's a laudable goal, but teach the subject a few times to make sure you've learned it.

5:47 PM
@Eric this guy has similar formatting by the looks of it to one professor I'm interested in at Madison named Laurentiu Maxim

steal his ways dude

The basis for the row space is $B=\{(2,2,2,3)+(1,3,1,3)+(3,0,2,2)\}$ right?

these dudes are sharing a hivebrain

No, no, @Doesbaddel. What is a basis?

5:48 PM
I mean without the plus signs

@Eric, Demonark: I think they're both stealing from John Hubbard.

@GFauxPas He stopped sending real checks because of some fraud issues, but he still sends checks for "one hexadecimal dollar" (2.56$)$B=\lambda_1(2,2,2,3),\lambda_2(1,3,1,3),\lambda_3(3,0,2,2)+\dots+\lambda_5(1,2,2,0)$Oh, @Doesbaddel. I don't know. You don't go back to the original rows unless you're sure they're linearly independent. 5:49 PM Oh stealing is cool and good and i endorse it Ah i made a typo again Sry. @Doesbaddel: You established it's a 3-dimensional space. So you need 3 linearly independent vectors. I wanted to write$B=\{(2,2,2,3),(1,3,1,3),(3,0,2,2)\}$I took every vector from the original matrix that are not all zero Can you tell from your row operations that those rows are actually linearly independent? What if you switched some rows to get echelon form? You take original columns, but almost never original rows. 5:53 PM Argh, I'm so dumb, I wanted to take the columns sry, I think im tired that's why I can't concentrate that much @ÉricoMeloSilva oh btw there's a Venkatesh lecture today at 4 Yeah, we can't take the rows, because it changes if we exchange two rows while doing gaussian elimination Yes, for example. im aware of it and it's a shame i cant go cause i have an appointment then What is an appointment worth? 5:55 PM Rip a lot @Ted, if u recall i have a broken neurology Well, I didn't know what kind of appointment it was. :) I have had a lot of stooopid appointments in my life. Now my final try hopefully:$B=\{(2,1,3,4,1),(2,3,0,1,2),(2,1,2,0,2)\}$I took the first, second and third vector, because these are the non zero rows in the final row echelon form. I hope, this is right xd i also wouldnt understand the venkatesh lecture literally at all not even a modicum of understanding nada You're doing columns now. You have to see in which columns pivots appear in echelon form. I know nothing, @Eric. 5:58 PM me neither Pivots appear in the first, second and third column respectively I need to take the column vectors of these exact rows then? Yeah I guess that's fair. Tomorrow's is gonna assume people know about automorphic forms and it conflicts with a class so I won't be going. I have epsilon hope for today. At least it'll be fun even if I don't get it did u go to that weird lecture they had on monday No, @Doesbaddel, as I said already, you need to check to see in which columns (nothing to do with rows) the pivots appear. the topic looked tangentially interesting but i didnt end up going 5:59 PM Cohen Prize Lecture? yeah that one Demonark and Eric: When you're grad students, I would encourage you to go to every colloquium, even if you don't think you'll understand it. Yeah I went, it was nifty Some will be terrible, but you'll end up being exposed to a breadth of mathematics and "meeting" some great mathematicians. @TedShifrin ill endeavor to do this as best i can 6:01 PM @TedShifrin I had done this before, I guess. Maybe you didn't noticed this. I take the vectors of column 1,2 and 3, right? Good. I think I missed about 5 in my time at Berkeley (not counting time I was out of town). @Daminark What was the topic? Yes, @Doesbaddel. But you wrote "take the column vectors of these exact rows," which makes no sense. That makes sense for sure. Would you say this also applies to traveling to conferences (assuming funding can be obtained)? @AlessandroCodenotti it was about districting and Markov chains Demonark: In general, going to conferences that concentrate on stuff completely out of your area is not necessarily a good use of money/time. But a colloquium that comes right to you for one hour ... 6:03 PM there's a way bigger cost to traveling for something With the way my brain works, I find it easier to take the column space of the transpose than the row space of the original. That's because I tend to think of vectors as columns rather than rows. Ah, I had high expectation after readin "Cohen" :P @TedShifrin sorry, I made a typo. I actually wanted to write columns xD @AlessandroCodenotti it's just named after him cuz he went here Oh, Demonark, was this to do with mathematicians telling politicians how gerrymandering has been abused? 6:03 PM Doesn't make a difference at the end of the day probably OK, @Doesbaddel. Go to sleep :P @ÉricoMeloSilva Ah, makes sense @TedShifrin thank you. Good bye @GFauxPas: No need to do the work more than once. Write the vectors as columns as you want in your final answer. That's what I mean 6:04 PM Basically the speaker was talking about how shapes aren't actually a good criterion for fair districting, and that a possibly better idea is to use Markov chains I actually wanted to write "columns" instead of "rows". Good bye I think you misunderstood this sentence :D My lin alg professor didn't care, though he said that in some textbooks the authors can be somewhat draconian on insisting vectors have to be columns. Demonark, Tufts Univ. ran a course for mathematicians last summer on fairness in making districts. I actually don't know the content, but I know at least one person who went to be trained. @ÉricoMeloSilva turns out it's not the well known Cohen, it's a student who was here and died, so his family gave some money to endow this prize oh really? 6:05 PM @GFauxPas: In my multivariable math book I used only columns (for reasons I can justify), but in my linear algebra book I was sloppier. Ted: the speaker we had for this was from Tufts actually Oh, might be the same. isnt it called like the paul j cohen prize Do you feel like saying those reasons now? came up with this problem yesterday and I'm wondering what kind of math to classify it under 6:07 PM I think Paul R Cohen Semi! Hello friend! @TedShifrin i always thought this was weird cuz it should be well known that people actually in practice have drawn districts to expressly make them less fair oh wild In principle, Eric, there are federal laws about this. Not that politicians follow laws. Or the Constitution. For some positive integer n, generate a 3-row array such that 1) each row contains each number from -n to n exactly once, and 2) each column sums to 0. Yeah there's a lot that's deliberate, but some states now are apparently trying to get it right 6:08 PM I have a construction for that, which I'm happy enough with i think one of the carolinas ruled their gerrymandering unconstiutional but fixing it is gridlocked But I'm wondering if there's a better way of looking at it than that Pennsylvania, I think California? North Carolina is the famous one, I think. I don't know about anything going on in CA. Maryland is another highlight in that regard 6:09 PM Matrices akin to that have names, Semiclassic, but I don't remember details. I'm thinking of ones made out of$1$s and$-1$s, I think. yeah, that sounds like something hadamard That's it. I've heard claims that California ungerrymandered some time back but I'm not sure. Pennsylvania is more doing it now I think either that or (bi)stochastic Speaking of appointments, I have one with my eye doctor, so I'm disappearing. 6:10 PM later goodbye Have a good day, all. Safe travels, Semiclassic. See you Ted! Bye, Demonark. et al @Daminark we should gerrymander so hard that technically no one lives in any of the districts 6:12 PM Galaxy brain strat unironically an improvement we need more mathematicians and scientists in office nah we're all just as stupid as them for the most part 6:37 PM Just got details of my scholarship, soooo so excited to get paid to do mathematics all day lol yeah it would be nice hi chat Hey @Balarka welcome back 6:43 PM hey here is a great charity to donate to www.drugpolicy.org isn't it nice to see logic and common sense at work as opposed to some open ended statement that has no real impact and a bunch of guys in blue shiny costumes that only the elderly look up to bragging about stealing a small amount of cash and drugs, being hail by the media as heroes literally every week on the news www.drugpolicy.org Hey everyone Could anyone tell me why cos(x) is of bounded variation Is it like some little trick, just like monotonically increasing functions are of bounded variation or is there a bigger proof behind it Hi @Balarka to be honest its a risk asking any question on any social media network but this is more expert that facebook messenger in that sense you are in the right place good sir you should probably hire a Stack Exchange team for that no problem getting hedge funds to pile in for the return in investment look they are literally paying people for mathematics in this room * actual currency of payment may or may not be worth actual money distributed under the brand name reputation. Reputation may or may not necessarily imply actual credibility. individual results may vary 7:20 PM @S.Crim try to compute the variation of a C^1 function 7:31 PM Is there geometric intuition for what the pullback of a vector bundle looks like? Hi chat Hi @Astyx Hi @Astyx @Alessandro @Daminark :) @AlessandroCodenotti If$f : X \to Y$is a map,$E$is a vector bundle over$Y$, then essentially$f^* E$is the vector bundle over$X$with fiber over$x$being the fiber of$E$over$f(x)$, so to speak. You can make this idea somewhat precise. Do you know how to build a vector bundle from the data of transition functions? I know, but that doesn't help me understand how "twisty"$f^\ast E$I know that$(f^\ast E)_x=E_{f(x)}$Like what if I take$E=S^1\times\Bbb R$and$f:S^1\to S^1$,$z\mapsto z^n$? 7:41 PM Right, so the twist comes from the twists of$E$composed with$f$, more or less. Is$f^\ast E$twisted? I think it should be No, it's still the trivial bundle. hmm wait let me think about this So$f^\ast E=\{(v,x)\in E\times X\mid \pi(v)=f(x)\}$by the definition we're using Pullback of trivial bundle is always trivial - this isn't too hard to prove Ah, I think I'm seeing it, if$f$is$z\mapsto z^n$pulling back the trivial$1$-dimensional$S^1$bundle along$f$looks like rolling a cylinder until it overlaps with itself$n$times Which is still a cylinder 7:46 PM Yes that's it By uniqueness of fibered product you can argue it has to be$S^1 \times \Bbb R$, since that fits in the commutative square Wait which fibered product are you thinking about? There's another way to do this. If I have a global section$s \in \Gamma(Y, E)$, can I get a section in$\Gamma(X, f^* E)$in some natural way? @AlessandroCodenotti This thing$f^* E$is just limit of the diagram$X \to Y \leftarrow E$in TOP @BalarkaSen Aha, that's what I was missing, but it's pretty obvious now that you wrote it out explicitely Fibered products in Top are so much nicer than in Sch Ah yeah I reckon they must be I don't know what it is in Sch Weird mostly 7:50 PM For affine schemes it's dualizing tensor product, which is pushout in Aff_k, right? People care about them because Grothendieck's relative point of view says you should consider Sch/S, schemes over a fixed scheme S and then products in Sch/S are fibered products in Sch @BalarkaSen Yes, for non affine stuff you need to glue together a bunch of those That makes sense Just take \C-valued points (or any points) and your fibred product is just the same as the set theoretic / topological one So the forgetful functor Sch -> Top takes fibered products to fibered products nono not the forgetful functor I meant taking Hom(Spec \C, -) 7:54 PM Oh that thing So what's an example where the forgetful functor doesn't do that Forgetful functor wouldn't work, if you consider the fibred product of Spec (\C) x_{Spec \R) Spec \C I think for affine schemes Spec(R otimes_A R') is still homeomorphic to Spec R x_Spec A Spec R' = Spec( \C \times \C) = 2 points Ah this is roughly because basechange is happening. I see the point now @BalarkaSen I'm actually not seeing that 8:01 PM @Alessandro Given a section$s : Y \to E$, take$sf : X \to E$, and upgrade that to$s' : X \to f^* E$by defining$s'(x) = (sf(x), x)$. Does that work? Hmmm sanity check: since$f^\ast E\subseteq E\times X$the map$f^\ast E\to X$is the restriction of the projection$E\times X\to X$? Yup @BalarkaSen Alright, looks like this works then. If$\pi:E\to Y$then$\pi(sf(x))=f(x)$since$s$is a section, so$s'(x)$lands in the right place, and it is also a section of$f^\ast E$, just because it does nothing to the second coordinate Great Thanks! 8:09 PM So now you can argue as follows:$E$is trivial over$Y$iff there are sections$s_1, \cdots, s_n$such that for every$y \in Y$,$s_1(y), \cdots, s_n(y)$spans$E_y$. Pull these sections back to$X$to obtain a basis of sections for$f^* E$as well. That proves pullback of trivial bundle is trivial no problem Ah, that's a cool argument Very clean Given any vector bundle$E$over$Y$, choose a trivializing atlas$(U_i, \varphi_i)$for this. Then for every pair$i, j$of indices, denote$U_{ij} = U_i \cap U_j$. You have the various maps$\varphi_j \varphi_i^{-1} : U_{ij} \times \Bbb R^k \to U_{ij} \times \Bbb R^k$which is of the form$x \mapsto (x, g_{ij}(x))$where$g_{ij}(x) \in \text{GL}_k(\Bbb R)g_{ij} : U_{ij} \to \text{GL}_k(\Bbb R)$are called the transition functions of this atlas Oh, right, we talked about this in AG The data of the vector bundle$E$over$Y$and the data$(U_i, \varphi_i, g_{ij})$of the atlas and the transition functions are equivalent in the following sense;$E = \bigsqcup U_i \times \Bbb R^k/\sim$where$(x, v) \sim (x, g_{ij}(v))$whenever$x \in U_i$and$j$is such that$U_i \cap U_j \neq \emptyset$, and the projection$E \to Y$is given by forgetting about the$\Bbb R^k$factor$g_{ij}$encodes the twists for$E$, so to speak Right We went through this to see the analogy between vector bundles and locally free sheaves 8:15 PM Aha In both cases locally they look trivial, but there can be twists hidden in how they patch together Right So I claim that for any map$f : X \to Y$, a vector bundle$E$over$Y$with a trivializing atlas$(U_i, \varphi_i)$and transition functions$g_{ij}$,$f^* E$has a natural trivializing atlas and transition functions Do you want to try to figure it out? Not right now to be honest, but I want to understand$f^\ast E$better so I'll try to work it out and come back to you! For sure! 1 hour later… 9:34 PM if$f_n:\mathbb{R}\to\mathbb{R}$are Lebesgue integrable functions, the integrals are uniformly bounded, i.e. there is such$M$that$|\int f_n|\leq M$for all$n$, and the pointwise limit$f$of the sequence$f_n$exists and is finite almost everywhere, then is$f$necessarily integrable? 9:46 PM @happyEddie Yes, because$f = \limsup f_n$why is$\limsup f_n$integrable? Depends on how you construct Lebesgue integrals for nonnegative functions, integrable functions are ones that can be approximated almost everywhere by nondecreasing sequence of simple functions well i guess i need to see if it follows from the definition that$\limsup$of integrable functions is integrable at least the assumption on uniform bound is needed because the limit of$\chi_{[-n,n]}$is not integrable 10:10 PM Wait I mistook integrable for measurable But it's still true I think yeah the pointwise limit$f$will be measurable, that i know fatou's lemma says that lim inf (which is lim in this case) is integrable if$f_n$are nonnegative, let's see if i can decompose$f_n$into positive and negative part and use fatou's lemma twice or maybe look at the sequence$|f_n|$instead... anyway i'm off for now, thanks 10:34 PM actually it doesn't hold with those assumptions, counterexample:$\chi_{[-n,0)}+\chi_{[0,n)}$, i think it needs$\int |f_n|\leq M$instead... ok now i'm off for real :) 11:12 PM @happyEddie,$\int f = \lim \sup_{n \in \mathbb{N}} \int f_n$assuming$f_n \nearrow f$if i remember correctly 11:43 PM Does this differential equation make sense?$f(x)=f'(x)+g'(x)$11:57 PM Depends what you’re asking. As an equation which involves derivatives, it’s perfectly sound. However, the typical context in which you’d see that DE is where g(x) is known and f(x) is unknown (You can solve that ODE to get g(x) in terms of f(x) easily. It’s harder if you want to get f(x) in terms of g(x), which is the more interesting case) hey guys if its okay with you, can someone give me feedback on the following proof 1 An infinite set A is$\textbf{countable}$if$\mathbb{N} \sim A$. (There exists a bijective function from the naturals to A). WTS: Let A be an infinite set. A is countable$\iffA=\{a_1,a_2......\}$, where$a_i \neq a_j$for$i \neq j$. proof Assume A is countable. So$\mathbb{N} \si...

@Semiclassical ah okay good to know

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