@MaryStar Well, you're just telling me the definition of reciprocal. It didn't answer my question. Try examples.

hi @Semiclassic

@AbhishekBhatia: It points in the unique direction in which $f$ increases at the most rapid rate. Of course if the gradient is $0$, we're at a critical point and the function instantaneously doesn't increase at all in any direction. (If your text and instructor don't make this clear, you might watch my YouTube lecture on this.)

Hi everyone, I'm currently reading through the first few chapters on Baby Rudin, and since it is my first 'real' Def-Theorem-Proof book I was wondering if any of you could give me some general advice on how to get the most out of these kinds of books

Read the definitions. Prove the theorems.

I kid. Try to read over stuff as many times as you need to understand it and do lots of examples.

That's not a great book to start with, @Perturbative. It's very terse, there are no pictures, and it's very sophisticated for a first experience with proofs.

Start with a linear algebra text with proofs. Or an abstract algebra text. Or an easier analysis book.

I just looked at your profile. Strang is not a good place to start. He's very sloppy with proofs.

I, of course, prefer my own linear algebra text :P

Its goal is to teach students to write proofs. That's very different from Strang's book.

@Kari, did you ever put any emphasis on memorizing anything? I've found that given the vast number of theorems, it seems nearly impossible to remember everything I've worked through/proven

But analysis is much harder (the quantifiers are much more complicated) for a first experience with proofs.

You have to memorize definitions. Don't memorize proofs. But work them through so you understand them, and do exercises. Lots of exercises. Get someone to criticize your writing.

Heya @Faraad.

Ted is a great source of knowledge on this stuff, @Perturbative! ^

@TedShifrin, I've got a copy of 'Understanding Analysis' by Sean Abbott, that I'm using as a supplement to Rudin. I just used Strang to give myself an introduction to Linear Algebra. I'm going to read through Sheldon Axler's 'Linear Algebra Done Right', for a second rigorous treatment of Linear Algebra

Although I would be interested to take a look at your text :)

Axler is too weird. You're better off with a standard treatment of linear algebra (that covers stuff in Strang) that also emphasizes proofs.

But I'm saying you should do linear algebra proofs before you try reading a hard analysis book.

Hrumph @Faraad

I feel like I should try to read Axler at some point, if only for the different perspective on linear algebra

shrug

@TedShifrin, okay thanks for your advice. Any book recommendations for Abstract Algebra?

There's not much mystery. He wants to avoid determinants (for the characteristic polynomial), so he defines the characteristic polynomial to be the generator of the ideal of $k[t]$ given by the set of polynomials $f$ with $f(T) = 0$. shrug

fair enough.

@Perturbative: If you're pretty sophisticated, I love Artin's book. I also wrote an algebra book, of which I'm reasonably fond, but I don't necessarily push it for high-powered math people. :P

@TedShifrin, I'll give both Artin and your book a look over :). Any comment on Lang's book or Dummit & Foote?

Dummit & Foote is great, but it's too heavy/encyclopedic, I think. Artin shows you how algebra is part of mathematics as a whole. A lot of very unusual, beautiful stuff in his book and nowhere else for undergraduates.

Plus he integrates linear algebra into the whole thing. Doesn't treat it like a cheap stepson.

sounds like D&F is a good reference text but not a great reading text?

@TedShifrin, thanks for those comments. I'll definitely give Artin's book a look over

the ones I have at home are Hungerford, which I used in undergrad, and Gallian, which I have a copy of for some reason

I like it for a grad course, @Semiclassic, but not as a first course. Especially if someone is reading and has no one to guide him on what's important and what isn't. Just toooo much in it.

I think I somehow got a copy of it when I was in high school, probably owing to the fact that Gallian is a UMN prof.

Those are both "easy-level" undergrad texts, @Semiclassic, sorta like mine.

sounds right.

Someone like you doesn't need to know everything to pass an algebra qualifying exam :P

in high school I liked Gallian because it has a lot of mini-biographies on mathematicians

lolyes

googling pulls up a pdf copy of the seventh edition on a Boise State page, lol: math.boisestate.edu/~liljanab/MATH508/…

Yeah, I tried to put history in two of my books, too. I think it's a good idea.

Algebra is too hard.

one thing I may have to start reading on for my own curiosity is classical mechanics, namely the contributions of people like Jacobi and Hamilton

Have you looked at Arnold or at Abraham-Marsden?

I've got A-M back home right now, actually.

(Of course there are the standard physics books like Goldstein ...)

@MikeMiller All of math is too hard.

I've tried to read Arnold before. And I have a copy of Goldstein again---one of the graduating PHD students said I could keep his copy

I've spent a day now trying to calculate an orthogonal complement.

I had a copy myself, but I dunno where it is :/

That's not algebra :P

It is when you're doing linear algebra, heh

Though I more mean the intellectual history rather than the actual results

$mic$ $drop$

Linear algebra ≠ algebra

I got thinking about it when 0celot asked why Jacobi fields have that name

@Ted You're right, it's a proper subset. ;D

Just like elementary school arithmetic ≠ mathematics :P

If I can't take an orthogonal complement, how can I do any algebra?

Arithmetic $\subset$ mathematics. xD

Elementary arithmetic != elementary number theory

i just bought some chocolate to console myself with as i struggle through this abstract homework

@Semiclassic: I've never delved into it deeply, but I imagine it's related to energy minimization, yes.

More broadly, I'm a bit curious about some of the pre-history of differential geometry

i feel like these questions are easy. I just ... cannot see how to do them.

@SAW: I'm sure you get excellent chocolate in Hungary!

I mean, there are some obvious signposts---Gauss's theorem egregium (no hope on having spelled that right) or Riemann's work

Milka bars are heavenly. That and the combinatorics are the only reason I'm here. xD

@Semiclassic: Spivak includes quite a bit in his 5-volume opus (including original Gauss translated).

Hmm, nice.

something sort of amusing: One book I ran across while browsing on this stuff came out in 2012, and the author is Goldstine (yes ,that spelling)

so if you want to learn results from classical mechanics, you read Goldstein. if you want to learn some of the history of classical mechanics, you read Goldstine.

@TedShifrin Thanks so much for the reply! Why does it point in such a direction?

(more precisely, it's a history of the calculus of variations in the 17th-19th century. link)

@AbhishekBhatia As I said, your text should derive that (and my lecture certainly derives and explains it carefully). The reason is the directional derivative formula in terms of the gradient. The directional derivative of $f$ at $P$ in direction $v$ (unit vector) is the dot product of $\nabla f(P)$ with $v$. Then use the fundamental formula for dot product (product of the lengths times cosine of the angle).

@TedShifrin Bonsoir

@JeSuis!

@TedShifrin what's up ?

Not too much. What've you been up to?

@TedShifrin I understand the direction derivative is in the same direction as v. But why is that direction increasing?

It should have max. magnitude but why increasing?

$\nabla f(P)\cdot v$ is greatest when the two vectors are pointing in the same direction. You have maximum possible directional derivative, which is $\|\nabla f(P)\|$. That's a positive number, hence increasing.

ignores @Faraad this time

I moved away for a time from Internet, I get my "licence" and now I get into "master". :)

@Semiclassical I just saw this. How funny.

Félicitations, @JeSuis.

yep. it confused me at first, since i saw references to Goldstine and thought "huh, i didn't remember it having all the history in it"

then I realized it and it made sense.

what i'm more broadly curious about is the pre-history, so to speak, of differential geometry.

A lot of that is in a general history book, like Kline's 3-volume book, @Semiclassic.

but when it comes to stuff that's multi-dimensional, i figured that had to wait until Riemann

hmm

I'll have to check that out when I get back

Oh, you're still being wedded in NY?

Wedding was yesterday

but we'll be driving back, and that starts tomorrow morning

Ohhh, getting to miss extra days, I see :D

lol

yeah :/

@TedShifrin merci!

i should be back in time to teach lab on tuesday, though.

@Abhishek: You have it now?

we're planning to keep driving through this time rather than stop for the night at a motel, since it'll be three of us this time instead of two.

Minnesota is not a short drive. Even from Detroit, you're probably another 10 hours?

Hmm, maybe a little less than that.

Congrats if I'm correct in assuming that you've been hitched, @Semiclassical :-)

nope

my sister

unhitching as we speak

Ah, then congrats to her!

You're unhitching, @TedShifrin?

yep, it was a nice wedding

No, @Kari. Don't try to make every comment about someone in here! :P

You seem pretty firmly hitched to this room, @Ted.

Let me see. We left Minneapolis about 6am and got to Toledo at about 7pm

@MikeM: I was pretty much absent for a month. It can happen again.

Haha, I'll try not to!

Fair enough, fair enough.

And then the next day we got delayed by fog until about 9:30 and arrived in NY at about 7:30pm

Just finishing cooking something to take to the debate-fest potluck tomorrow night, @MikeM. I'm going to enjoy the martinis and food even if I don't enjoy some of the rest.

Wow, @Semiclassic. That's a longgggg trek.

yeah.

I remember doing Berkeley to Seattle by myself nonstop in grad school. It was 16 hrs nonstop, I think.

oof.

the unpredictable bit in this is weather

Now I do San Diego to San Francisco and (because of LA traffic adding a few hours) that's plenty for me.

Here I am complaining about going to the store to pick up some biscuits and you guys are travelling for days.

getting delayed on the second day by fog was a real pain.

Yup, imagine January or February, @Semiclassic.

no thx

@TedShifrin I think the point that confuses me what all values the vector $v$ can take.

@TedShifrin I've got too much work to do tomorrow to watch the debate.

All possible unit vectors, @Abhishek. Seriously, you should watch the video I linked. I have 100+ lectures on multivariable and linear algebra up there. There's lots of proofs, but you might find stuff that helps.

oh, f***, the debate

@MikeM: I'll have a few extra martinis for you. Luckily, I only have to walk two blocks.

huh, that statement also works without the commas

A bunch of students haven't had their fees paid, despite the university being contractually obligated to do so (a while ago). Some of them are having their health insurance lapse and their classes dropped. So half my time is going towards that, and the other half calculating this damn orthogonal complement.

I have a cube ABCDEFGH. O is the center of the diagonals of the base EFGH. I needed to calculate the angle OAH and then calculate the angle between AO and the plane ADEH. I calculated the first one, but I don't see the difference between both angles, I imagine them as the same. Can someone help me on how to approach to this exercise?

Ugh @MikeM. I won't even ask about the orthogonal complement. It would take two days of explanation.

@PichiWuana use vectors

Yup, use vectors. Do you know vectors and dot products?

There's probably a more clever approach, but vectors are simple and just plain work

I learned it in physics but not in math (we will learn it in the future). Thus, I can't use it.

unfortunate.

So what exactly are you allowed to use, @Pichi?

you can probably still proceed if you make proper use of the Law of Cosines

So we're allowed trigonometry, not just Euclidean geometry? Just asking what's fair and what isn't.

If we look at a $\sigma$-algebra $\mathcal{A}$ as a $\mathcal{F}_2$ vector space, it follows that if $\mathcal{A}$ is finite then it's a power of $2$, but in the infinite case, can I use this 'view' to prove that $\mathcal{A}$ must be uncountable?

I can use trigonometry, geometry theorems.

@JeSuis no because some infinite vector spaces are countable

So, yes, law of cosines is fine.

@Pichi: How do you define the angle between a line segment and a plane?

For example $\Bbb F_2[X]$ is a countable vector space

(also I got an anonymous downvote and will now proceed to lose sleep over it and whine on meta)

@TedShifrin Well, if it'll take you two days to explain what the orthogonal complement is, I can wait :)

That's not what I said, @MikeM. At least, it's not what I meant to say. scrolls up

@TedShifrin We learned that I need to choose a point on the line segment and put a height to the plane through that point (sorry for my English)

;)

@mercio Then the answer (I was able to find a countable family but perhaps it contains necessarily an uncountable one) here (math.stackexchange.com/questions/918195/…) is not correct

@Pichi yes you need to orthogonally project the line on the plane in order to compute the angle between them

OK @Pichi, thanks. So you're working in the plane of the line segment and the normal vector to the plane. You're thinking that AH lies in the plane of OA and the normal to the ADHE plane. I'm not sure.

you can't just pick any point on the plane

but I haven't looked at the picture so maybe it does the same as question 1

times like these, I wish Geogebra worked in 3D :/

@mercio: I don't think so, but I'm trying to think carefully about the geometry.

@JeSuis well it could be lacking an explanation of "an infinite sigma-algebra is always uncountable" which I don't know right now wether it's true or not

@TedShifrin I thought that ADHE would be the plane (not AHO) and the line segment would be AO.

but since it IS the origina lquestion

yes that answer could be bad

@Pichi: You misread my sentence. I agree with what you said.

I am thinking about it differently, but it's the same thing.

@mercio the result it's true but it's not a proof here

If you lay the plane ADHE flat, @Pichi, is O directly above the midpoint of AH or not?

yeah

@TedShifrin No. O is the intersection of diagonals of EFGH.

Yes, I know that.

But if you project it onto the plane ADHE, where does it land?

@TedShifrin Si on a une application continue $f[0,1]\to\mathcal{L}(\Bbb{R}^n,\Bbb{R}^p)$ telle que $f(t)=L_t$ si on suppose que $L_0$ est injective, comment avoir une idée géométrique de pourquoi $L_t$ sera aussi injective pour $t$ petit ?

@JeSuis: Maintenant je suis en train de faire qqch d'autre.

@TedShifrin I think things just don't work like I thought they did. I'm scared.

@MikeM: Is that politics or orthogonal complements?

@TedShifrin Quand vous aurez le temps si vous avez le temps, comme toujours ;).

The latter.

But both.

@TedShifrin Directly above midpoint of EH...?

No, that's not right, @Pichi.

Or maybe I'm misunderstanding what you mean

@mercio $L_t$ is a linear map, and $f:[0,1]\to\mathcal{L}(\Bbb{R}^n,\Bbb{R}^p)$ is continuous

with any natural topology on the matrices representing the linear maps I assume ?

yes sure

if it's injective then some minors are nonzero

by continuity they stay nonzero in a neighbourhood of zero

@PichiWuana You're correct. Sorry.

@TedShifrin When you say if you lay the plane ADHE flat, what do you mean by that?

Oh

or something like that

minors ?

So when you drop the perpendicular from O to the plane, it does not land on AH, so you are getting a different angle, @Pichi.

some $n \times n$ determinant with carefully selected columns

@mercio ok thanks

Ohhhh I see

@mercio Yes, the all-important semicontinuity of rank!! :)

I guess so yes

Good night to all :)

Bonne nuit, @JeSuis.

@TedShifrin Thank you. Now I understand

Je suis ravi que mercio t'ait aidé.

owrrrrrr

@Pichi: I didn't do much, but I'm glad you figured it out.

@TedShifrin as do I

So glad to have more helpful people around :)

hey does anyone have a rough idea of what like the "easiest" differential topology book is?

in this question: math.stackexchange.com/questions/439370/…

Guillemin and Pollack's book is a standard first book.

Oh lol... that's the book i'm using and i'm pretty lost.

but just to continue what I was about to say, matt says: "To answer this question, you will need to have a much stronger understanding of basic ideas like derivatives"

and like, I am more lost than the questioner.

So I guess if Guillemin and Pollack's book is the (roughly speaking) the easiest, I'll just have to go back and review more questions. Thanks.

Dair i hate you

so much

So I get that $G/G'$ is abelian (G' being the derived group), but is it simple?

If we consider a subgroup $H\leq G/G'$, we automatically know that it is normal because $G/G'$ is abelian. So this basically boils down to showing that $G/G'$ has no proper subgroups. No idea how to show that, though.

If $G$ is commutative then $G/G' = G$

you're going to have a hard time proving that every abelian group has no proper subgroup

Lets make our lives harder by assuming that $G$ isn't commutative.

:(

what's your favorite non commutative group

because I don't know much, probably $D_{2n}$.

if $n$ is even, then $D_{2n}'$ is all even powers of $r$, right?

well then just pick $H = \Bbb Z/4 \Bbb Z \times G$ where $G$ is any non abelian group

and $H/H'$ will not be simple

then.. what ... why are you talking about powers of something that doesn't exist

wait I'm dumb

$rsr^{-1}s^{-1}=r^2$

kk

and $r^n = 1$ ?

lol then yes, it will be the powers of $r^2$

if odd, then all powers

the even powers would just wrap back around

I think?

if $n$ is odd then $G/G'$ will be cyclic of order $2$ which is simple

yes

but if $n$ is even then $G/G'$ will be the product of two cyclics of order $2$

and that isn't a simple group

dang

I was just kinda hoping that $G/G'$ would be simple. It would make this assignment much less difficult.

well no

lol

that's a weird question to ask because abelian groups are much much easier to handle than simple groups

Wait, why do you hate me? I'm confused.

well mr 6 FRAME SPIKE

do you not feel any remorse to what you did

in reality i play jiggs.

my life is a lie

:o

ive been trying to play marth though

i live in europe so marth dair is a meteor and is not that annoying

but then i switched to project m lol

oh i haven't played pm so much.

just american melee

i like zelda though, so i've been considering doing pm since she is better there

she is pretty weird in pm

but so many things are weird in pm that it balances out somehow

well she sucks in melee and can't recover

she has downb

so like she is unusable against a skilled edge-guarder

no sheik, only zelda

I wonder if RyokoYaksa still plays

he bopped me a bit

but i was young and dumb

idk, well I need to go, getting four stocked by diff topo class lol.

aw lol

have you done any calculus before

yes

now excuse me i have to curl up

WAIT

NONONO

That doesn't make sense.

remove the $-1$ on the RHS

You've just got $x(2^{y}-1)$ on the right.

There.

Why'd you go about re-writing that anyway?

note that $2^y-1$ is a common factor in the sum, so you can factor it out and divide it to the other side

$$\frac{2^{xy}-1}{2^y-1}=\sum_{i=0}^{x-1}2^{yi}$$

and that's just a finite geometric sum.

(if you started from that, of course, then what i've said is pretty pointless. oh well.)

it may help to write $z=2^y$, so that the above formula becomes $$\frac{z^x-1}{z-1}=1+z+z^2+\cdots +z^{x-1}$$ where I've written out the right-hand side explicitly

if you multiply both sides by $z-1$ (returning to your original formulation), what happens to the right-hand side?

sure. what would that look like term-by-term if you wrote that out?

actually, $z^{i+1}-z^i$, and the index runs from $i=0$ to $x-1$

Whoops, didn't notice the minus sign in the above sum.

where i was heading is that $$(1+z+z^2+\cdots +z^{x-1})(z-1)=(z+z^2+z^3+\cdots +z^{x})-(1+z+z^2+\cdots +z^{x-1})=z^x-1$$

Right.

which is exactly what's needed for the other side to make sense

Makes a lot of sense. Thanks.

if you want to do this at the level of the summation, without expanding

$$\sum_{i=0}^{x-1}z^i(z-1)=\sum_{i=0}^{x-1}z^{i+1}-\sum_{i=0}^{x-1}z^{i}=\sum_{i‌​=1}^{x}z^{i+1}-\sum_{i=0}^{x-1}z^{i}=z^x-1$$