/mg/ = /math/ general: Completions Edition

This is shit

Another good textbook. So long user.

Howard anton's kill this in one shot

How do I into algebraic geometry?
t-brainlet with only an undergrad courses.

Found the engineer.

...

Just started this one recently, I rather like it so far. Short, and too the point.

I'm studying introduction to differential equations.

I got an 85 on my first midterm. really bummed about it because I made a few stupid mistakes.

Why do people like this book? I don't really get it. I've given it a try many times but it's just hard to read, even now after knowing plenty of abstract algebra. His treatment is painfully concrete, this is most apparent in how he treats the determinant in the first chapter by just giving a formula, no motivation at all. Hoffman and Kunze has served me much better, although his treatment of determinant still isn't enlightening. Evan Chen's napkin has a section on determinant and trace which is extremely nice, though you have to know a bit of algebra.

That is a good book. Though later on it goes a little crazy with the characteristic polynomial for everything. I'm not convinced getting trace and determinant from it is really all that sensible. You could take a look at Hoffman and Kunze as well, it's a bit harder than Axler but not that much.

Milne has the notes you want
jmilne.org/math/CourseNotes/
Here's some more books with AG background material and basic AG
dmat.cfm.cl/library/ac.pdf
homepages.warwick.ac.uk/staff/Miles.Reid/MA4A5/UAG.pdf
Did you say you want more possible books to read you fuck? Well let me tell you, when I wanted to learn AG I went to mathstacks and found a pretty good set of answers that worked out for me.
math.stackexchange.com/questions/285201/path-to-basics-in-algebraic-geometry-from-hs-algebra-and-calculus/285355#285355
mathoverflow.net/questions/2446/best-algebraic-geometry-text-book-other-than-hartshorne/57019#57019

>muh motivation
If your textbook doesn't use this definition it is shit tier:
math.stackexchange.com/questions/21614/is-there-a-definition-of-determinants-that-does-not-rely-on-how-they-are-calcula/21617#21617
(I'm being facetious.)

that's in hoffman&kunze...
multilinear forms and even the grassman ring

Just a formula is how determinants were historically defined in the first place. Abstract stuff like this is a recent development. True of most things. Are you also dissatisfied with textbooks that define the derivative by the limit of a formula?

That's also the definition in Chen's napkin I mentioned
usamo.files.wordpress.com/2017/02/napkin-2017-02-15.pdf
I actually really like definition. All you need to do is define a vector product which is bilinear and preserves area signature (v x w = - w x v), construct the space of r products over a vector space of dimension n, then consider a generic linear map f:V->V and define a map on the product space via mapping v x w to fv x fw. Now the dimension of the space is the binomial coefficient n!/(r!(n-r)!) so when n=r it just has dimension 1. Now the product space derived map is a linear map on a 1 dimensional space so it's just multiplication by a constant, which is what we call the determinant of f.
The historical development of a subject is not always the best way to go about teaching it. There's plenty of simpler motivation for the determinant besides the admittedly abstract one above. Just consider each column as a geometric vector and consider a a signed area of the shape created.
Also I am dissatisfied by defining the derivative with a limit, at least initially. A limit is a fairly complex notion compared to the simple notion of change in y over change in x that the first users of calculus used. Thompson's calculus made easy treats it using differentials which is very easy for students to grasp and merely requires algebraic manipulation and geometry. After this you can easily state limits and it will be a well motivated rigorous definition.

How should I learn homological algebra? Also, how much category theory do I need for homological?

The motivation is how determinants are used. If you can't develop your own intuition behind concepts then maybe math isn't for you.

intuition is passed along
concepts should be motivated
expecting newcomers to a topic to develop all the intuition themselves is ridiculous. that's why talks are given

But isn't using determinants at all. It's just some assumptions, a wedge product, which leads up to both an interpretation of the determinant in general vector spaces and it also gives a method of calculating them. Nowhere does it attempt to use the determinant to solve a problem.

I'm sorry, but you simply cannot teach intuition.

Your motivation should be the drive to understand, as fully as possible, the topic you're trying to learn. If that isn't your motivation, then you probably should be in some other field.

people usually learn both at the same time. grab any algebraic topology book and it will teach both

Either a high schooler or engineer. I read Anton is high school bro

not motivation as drive to keep going, you asshole
motivation as in why was this definition the right one? etc etc

you cannot teach general ability to grasp intuition, but you sure as hell can communicate the main idea of a topic and the gist behind the main results that lead to the right definitions quite concisely

is there any difference between [math]\all x,y[/math] and [math]\all x \all y [/math]

I see the first as quantifying all pairs x,y while the other quantifies for each x all values of y

but they to me seem to result in the same pairings of x and y

Shilov's book is pretty comprehensive for its size and introduces notions as they are used. It also has a neat little go-between with category theory at the end. Hoffman's and Kunze's textbook is very elegant but is not how most mathematicians actually use the tools of linear algebra. I wouldn't recommend it as a first book on the subject.

Also, Shilov's book has an [math]aesthetic[/math] cover that fit that image.
Give this a read later on, after you have some notions of how algebraic geometry is served.

>I see the first as quantifying all pairs x,y
No, it's just shortened notation for [math] \forall x\forall y [/math].

I see, thanks.

So my professor in Real Analysis keeps writing the real line as [math]\phi \neq s \subseteq R[/math]

Is this notation correct? Because its annoying me and I'm not sure you can combine them like that.

Obviously not, the correct symbol for the empty set is [math] \emptyset [/math]. Also, it is standard to write R as [math] \mathbb{R} [/math] to make it clear that it is the real numbers we are talking about not some random ring [math] R [/math].

>the real line
>Is this correct?
No.

I'm not sure how to get the Real symbol in latex which is why I didn't put it.

Thanks for noting the difference between the empty set symbol and phi. I've been saying its phi for years and only now realized that its different.

In what way? Is it wrong to say that the nonempty subset of reals is the real number line? I was just thinking the notation is funky and needs to be separated into two statements.

Atiyah-MacDonald and then Hartshorne.

Maybe skip chapter 1 of Hartshorne and read something better for classical algebraic geometry.

>In what way?
"""Real""" numbers don't really exist.

tex.stackexchange.com/questions/58098/what-are-all-the-font-styles-i-can-use-in-math-mode
Also, there is nothing wrong with that notation. It's a contracted form of [math] (\emptyset \neq s) \wedge (s \subseteq \mathbb{R}) [/math].
(Since we're on the topic of autism, stop reddit spacing or fuck off.)

read Vakil's notes

>can understand all the random shit I've talked about here
>still think Wilson-Kadanoff renormalization group is black magic
When will we have a mathematical theory of renormalization group(oid)s? I feel like all I'm doing is stumbling around a dark room when all I have to go on are examples.

You might be interested in what Hairer is doing then
youtube.com/watch?v=5ZTRqVeSUKI
arxiv.org/pdf/1303.5113.pdf

I got it:

first, we rewrite the curve equation as a polynomial [math] f(x,y) = x^3 - 4x - ky^3 + ky [/math]. It is clear that this polynomial is irreducible over [math] \mathbb{C} [/math] so by the genus-degree formula its genus is 1.
next consider the homogenized polynomial
[math] F(X,Y,Z) = X^3 - 4XZ^2 - kY^3 + kYZ^2 = X(X-2Z)(X+2Z) - kY(Y-Z)(Y+Z) [/math]
it is obvious that [math] F [/math] has a rational point, for example [math] (1 : \frac{1}{2} : \frac{1}{2}) [/math].
all is left to show that it has no singular points so we need to find if there are values of [math] k [/math] for which the partial derivatives
[math] \dfrac{\partial F}{\partial X} = 3X^2 - 4Z^2 [/math], [math] \dfrac{\partial F}{\partial Y} = -3kY^2 + kZ^2 [/math] and [math] \dfrac{\partial F}{\partial Z} = -8XZ + 2kYZ [/math] don't vanish.
let [math] X,Y,Z \neq 0 [/math]
from the first derivative we get [math] X = \pm \cfrac{2Z}{\sqrt{3}} [/math] from the second [math] Z = \pm Y\sqrt{3} [/math] (if [math] k \neq 0 [/math]) and from the third [math] X = \cfrac{kY}{4} [/math] which gives [math] k = \pm 8 [/math]

so the answer is [math] k \in \mathbb{R} \setminus \{ -8, 0, 8 \} [/math]

has posted the answer?

what i proposed

I'm kinda surprised it took almost a whole week for someone to post a solution to the easiest problem. I was under the impression that there are a lot of curious high-schooler and college freshmen here, given how often people asks for textbook recommendations.

I was just at a university summer school to study mathematics, because I have no formal education in the subject and they wanted some proof of my abilities before they would accept me. I am some 95% sure I failed, but in that month I learned more about mathematics than I had in the rest of my entire life. I'm going to study basic mathematics over the summer, and if I really did fail, I suppose I'll move on to linear algebra and single variable calculus to fill up my free time over the next year.

What is there beyond topology
Is it the end

>there are true statements which are not axioms which cannot be proven
Does this bother anyone else?

what level of algebra do i need to start doing proper alg topology

God is hiding behind them.

No, because the only unprovable statements are self-referential.

Be careful with that word "only".

Anybody here had any experience with fragrant sets?

>Anybody here had any experience with fragrant sets?

No. Things like "fragrant" are made up definitions for competitions. No one studies fragrant sets. Just look at other competition math problems in number theory and you'll see many define some concept.

> the only unprovable statements are self-referential
Which I might add, are largely vacuous. Mathematicians have actively tried to avoid self-referential statements since the dawn of time anyway. It's pretty interesting how it has always been non-mathematicians who made a fuss over Gödel's result.
Are you trying to say that there are statements that are true but unprovable that are not self-referential?

>No. Things like "fragrant" are made up definitions for competitions.

That would explain why I can't find anything on it~

>That would explain why I can't find anything on it~

Yeah, it's pretty common. I like competition number theory problems so I'm working on it. i already proved that b has to be bigger than 1. I'll post if I reach something.

The problem is that some statements could be "stealth-ly" self-referential in that they could depend on self-referential statements in convoluted ways.

Anyone can post that math iceberg infographic

I'm not sure if I'm on the right track at all as I don't have much experience with this kind of math but, I'm guessing that a 'fragrant' set would not include prime numbers, which would mean that the elements for the above set we need to find would be located somewhere in between two prime number solutions for x^2+x+1.

I tried setting a = 0 and then kind of brute forcing through to see what I could find but I didn't get anything.
I found two sets where it would have been fragrant if not for one out of place element each time.

I really have no idea how else to go about this problem, was thinking of picking it back up after I sleep.

How do I begin to learn algebraic topology?
t-undergrad

>Are you trying to say that there are statements that are true but unprovable that are not self-referential?
Not that guy, but there definitely are. The whole point of Godel's theorem is that it produces unprovable statements that are just innocent number-theoretic properties, along the lines of "there is an x such that for all y there is a prime p such that F(p, y) = x". There properties also have a self-referential *interpretation*, which is why they are unprovable; but at the same time, there's a boring and innocent number-theoretic property about prime numbers that is unprovable yet true.

I forgot to mention but my thinking was if I could find this set of solutions that satisfy the fragrant property with a = 0 then a would equal whatever b-1 is and b would equal the cardinality of the set.

Hopefully this isnt all obvious~

The self-referential aspect is there whether you want it to be or not. The scare-quotes are unnecessary and misleading.
It's Gödel btw. or Goedel if you can't type an umlaut.

>I'm guessing that a 'fragrant' set would not include prime numbers
Is there a reason you don't think it could include exactly one prime?

let f(n) = n^2 + n + 1 for brevity
use polynomial gcd and modulo arithmetic to find out the following:
f(n) and f(n+1) are always coprime
f(n) and f(n+2) can only have common factor 7, and thats only if n=2 mod 7
f(n) and f(n+3) can only have common factor 3, that happens when n=1 mod 3
f(n) and f(n+4) can only have common factor 19, that's when n=7 mod 19
now go through b from 1 to 5 and see that none can work:
b=1 obvious
b=2 cant work since f(n) and f(n+1) are always coprime
b=3 same reason when you look at the middle term
b=4 and b=5 similar reasoning

for b=6 you can match the numbers like this:
i want f(a+1) and f(a+4) have common factor 3
f(a+2) and f(a+6) common factor 19
f(a+3) and f(a+5) common factor 7
so all i need is
a+1 = 1 mod 3
a+2 = 7 mod 19
a+3 = 2 mod 7

which has some solutions (chinese remainder theorem)
so the answer is 6

Nice.

I am not the guy who asked the question but I was already up to proving that for b=4 it was impossible. I was already considering the problem in terms of congruences so I guess if I had kept going at it for 20 more minutes I would have got it.

Now that you mention it I do think it could include exactly one prime, thank you.
Thanks for the interest in the question guys i'm loving the replies.

>Thanks for the interest in the question guys i'm loving the replies.
Where did you get it from?

2016 IMO questions

>The self-referential aspect is there whether you want it to be or not.
Yes. But so is the number-theoretic aspect. Self-reference or no, the theorem still shows that there is a clearly-meaningful true property on natural numbers that is unprovable.

>The scare-quotes are unnecessary and misleading.
What scare quotes?

>It's Gödel btw. or Goedel if you can't type an umlaut.
I know. I'm just lazy.

Send help, I made another one of these images.

Nice. Keep making more.

does that have any implications on actual research?
even in the unlikely event that such unprovable statement would be of interest to number theorists, it would be only unprovable in some first-order theory for example PA
but literally no one cares about proving things in PA so the question would be eventually resolved by other means

can you do one of these, but with dumb girl holding some babby-tier book

Well this is awkward

>Well this is awkward
What about it is awkward?

The whole point of Godel's first incompleteness theorem is that you can move to a stronger theory than PA (say ZFC), but then within that stronger theory you can apply a similar construction to get a new set-theoretic statement that ZFC can't prove.

The second incompleteness theorem then gives a constructive example of such a statement, usually written as Con(ZFC), and establishes that it cannot be proved in ZFC unless ZFC is inconsistent (i.e. unless ZFC also proves 0=1). And so on.

And the existence of Con(ZFC), or Con(T) for a general theory T, has pretty much shaped the development of modern set theory (via model theory), so I'd say it has had significant "implications on actual research".

>Self-reference or no
The self-reference is crucial.

>what are you studying
I just learned that you can solve cubic equations in planar geometry with the mathematics of paper folding
my dick is harder than neutronium. so hard that it violates all known laws of physics.

Anyone read Byrne's Elements?

archive.org/details/firstsixbooksofe00eucl

Overall it's a very good piece of design. In most cases, demonstrations are either confined to a single page, or occupy two facing pages. Things get a bit silly in book V however as the graphical devices are made to stand in for algebraic variables, which introduces a little semantic confusion.

Also found this cheeky fucker (pic related).

I took the GRE yesterday. This question on the math was the reason I didn't get a 170: "How many integers from 1 to 2000 (inclusive) are both the squares and cubes of integers?"

The answer choices ranged from 3 to like 44 or something. Please show your work.

The first is 1 = 1^6

Then 2^6 = 8^2 = 4^3

Then 3^6 = 27^2 = 9^3

Then 4^6 > 2000 and thus there are only 3.

Is GRE an american exam? Because this shit is literally first grade arithmetic. I can't believe american grad students can't do it. The US is a fucking joke.

>he literally can't calculate [math]\lfloor 2000^{\frac{1}{6}} \rfloor[/math]
Woe is you.

[math] 2000 = 2^4 \cdot 5^3 [/math]
[math] x^2 = y^3 [/math
[math] x = y \sqrt{y} \in \mathbb{Z} \implies \exists a \in (1, 20 \sqrt{5}) \cap \mathbb{Z},\
y = a^2 [/math]
[math] \lfloor 20 \sqrt{5} \rfloor = 44 [/math]

What the fuck are you doing, retard?

No, what the fuck are YOU doing. Why'd you answer the babby's question?

I'm not doing anything.
Now I'm even more confused.

Good.

is it?

Thompson has proofs with infinitessimals that have
to be completely refactored to work with limits.
E.g. derivative of sine goes something like:
[math]dy=\sin(x+dx)-\sin(x)=\sin(x)\cos(dx)+\cos(x)\sin(dx)-\sin(x)[/math]
And then by small angle approximation:
[math]dy=\sin(x)+\cos(x)\cdot dx -\sin(x)=\cos(x)\cdot dx[/math]
and then divide by [math]dx[/math]. The standard(?)
proof given by Wikipedia is completely different.

Why in the world would I edit pictures of 3DPDs?

It's actually the same idea. It just makes the small angle identities for sinx/x and (cosx-1)/x as being 1 and 0 respectively rigorous by using limits and squeeze theorem.

He might mean some ugly 2d girl like Nozomi.

Fuck off.

>Nozomi
>ugly
Faggot.