Mathematics - 2012-11-29 (page 4 of 5)

Yes. In fact we can say even more: $\textrm{Hom}(L, \mathbb{Z})$ will be a finitely-generated free abelian group. This uses the fact that $\mathbb{Z}$ is a principal ideal domain. Indeed, take a finite presentation of $L$, say $$\mathbb{Z}^{\oplus a} \longrightarrow \mathbb{Z}^{\oplus b} \longrig...

Old John

9:16 PM

@robjohn Neat

robjohn

@OldJohn Now I can answer some more :-)

cassandra0

this is what I wanted to link

@MattN., he uses Hom(-,Z) being left exact, what do oyou think?

Old John

@robjohn and get to 50K ... 100K ...

Matt N.

Zhen Lin, I assume.

cassandra0

yes

Jordan

9:17 PM

I discovered new physics

robjohn

@OldJohn I will try. We've had a number of people break the 100K barrier recently

cassandra0

do you know why it is left exact?

Jordan

$T_f = T_1 + T_2$

Old John

@robjohn I have the more modest aim of achieving the "generalist" badge

Jordan

but my new physics doesnt seem to work

cassandra0

9:19 PM

by the way it is quite a nice proof I like it

but I wouldn't have understood it if I didn't do my own proof first

Matt N.

@cassandra0 Well, the "why" is a rather philosophical question. But the proof is rather mechanical. (diagram chase, short)

Are you asking whether I can prove it?

cassandra0

and I'm weak at visualizing exact sequences

@MattN., yes, well, if you like to

if not I don't wante to bore you with it

Matt N.

So we're talking about $\mathrm{Hom}(\cdot, N)$.

@cassandra0 Not bored at all, just a bit tired. But I think I can still manage this one.

cassandra0

yeah that takes us into Set which isn't an additive category, so it might be awkward. Better may be to consider this as a functor into a subcategory of Set

$\mathbb Z$ rather than $N$

Jordan

I wonder if there is any money in discovering new physics

Matt N.

9:22 PM

Let $$A \xrightarrow{u} B \xrightarrow{v} C \to 0$$ be exact.

user19161

@cassandra I saw your answer but decided to post another slightly different one.

Matt N.

Let $N = \mathbb Z$.

robjohn

@OldJohn I'm working on the Epic badge. I am 72% of the way there.

Jordan

fuck I hate physics, shit never makes sense and they always assume you know way more than you are suppose to for an intro level class

Matt N.

Now we claim $$0 \to \mathrm{Hom}(A,N) \xrightarrow{\bar{v}} \mathrm{Hom}(B,N) \xrightarrow{\bar{u}} \mathrm{Hom}(C,N) $$ is exact.

cassandra0

9:24 PM

well maybe it shouldn't the 0 at the end

since it's a contrafunctor

Matt N.

Oops, typo. Sorry. Corrected.

user19161

@Jordan What stuff is not covered in the textbook?

cassandra0

(at least you know I'm paying attention now :)

Matt N.

I'm glad : )

cassandra0

but still

Old John

9:25 PM

@robjohn excellent! - it is very tedious finding out what I need to do to get the generalist badge - don't think I am too far away now

cassandra0

shouldn't it be 0 --> Hom --> Hom --> Hom

Jordan

@WillHunting Energy transfer and work

robjohn

@cassandra0 better than me. I don't understand it at all. It has arrows, I know that :-)

user19161

@Jordan Are you sure it is not in it? It should be.

cassandra0

I mean really I'm just going by what Zhen Lin wrote, I have to check which direction these things go from scratch each time

Matt N.

9:26 PM

@cassandra0 Yes, of course. Excellent! : )

Jordan

@WillHunting It is in the book but we never covered these parts in class but an in depth understand of them is required, and the associated formulas, for working with temperature

Old John

@robjohn I now understand a tiny bit of category theory - I took some notes on "abstract nonsense" on my last holiday :)

user19161

@rob I just realised that my sentence ending with 40k ends with a !, which makes your rep surpass everyone on SE.

Matt N.

@cassandra0 Meh. I was impressed there for a minute....

Jordan

so now I have to go back, stop doing my homework, and study for 14 hours or so before I can finish my homework

also I hate that math isn't a thing that works in physics, there is always some obscure physics concept that trumps math

user19161

9:27 PM

@Jordan Well, then you just need to read up on your own I guess.

robjohn

@WillHunting Take that Arturo! >B(

Jordan

@WillHunting Yeah that is why I hate school really, it isn't for normal people, it is for people who had private tutors their whole life and learned all these things in high school

user19161

@Jordan No, you need to understand the physics first, and the rest is just modelling it with math and manipulating equations correctly.

Jordan

I barely graduated high school

Matt N.

Now we need to show the following:

(i) $\bar{v}$ is injective

(ii) $\mathrm{im} \bar{v} = \mathrm{ker} \bar{u}$

a Dangerous Idea

9:28 PM

@robjohn >8( has been replaced?

Matt N.

First (i):

robjohn

@OldJohn I need to look into more algebra. I have concentrated heavily on analysis

@aDangerousIdea I am trying out >B( I think it looks meaner

Old John

@robjohn I spent years doing nothing but analysis (of a sort) - now I am relaxing mostly with a bit of number theory :)

robjohn

and matches the eyes more closely

cassandra0

\bar v = Hom(v,Z) = - o v

robjohn

9:30 PM

@OldJohn I've been doing some number theory, I've played around with the basics

a Dangerous Idea

>E(

robjohn

@aDangerousIdea perhaps not :-)

Matt N.

Sorry just noticed another typo. Let me repost the second diagram.

user19161

@robjohn Don't forget geometry and topology, and set theory and logic, and combinatorics and graph theory, and ...

Old John

@robjohn it is a very alluring branch of maths, isn't it?

cassandra0

9:31 PM

$$\bar v = \text{Hom}(v,\mathbb Z) = - \circ v$$

Matt N.

$$0 \to \mathrm{Hom}(C,N) \xrightarrow{\bar{v}} \mathrm{Hom}(B,N) \xrightarrow{\bar{u}} \mathrm{Hom}(A,N)$$

robjohn

@OldJohn Often the problems seem that they should have easier solutions than they end up having

cassandra0

okay

a Dangerous Idea

@robjohn >|B(

user19161

I gave up on number theory after high school.

robjohn

9:32 PM

@aDangerousIdea it makes me think of someone with a hat

Matt N.

$\bar{v}$ is the map $(f:C\to N) \mapsto (f \circ v): B \to N)$. Assume $(f \circ v) = (g \circ v)$. Then since $v$ is surjective it follows that $f = g$.

So we have (i).

cassandra0

oh that is easy!

Old John

@robjohn exactly - I am amazed by how much hard maths gets used when trying to answer the question of which primes can be represented by $x^2+dy^2$ for instance

Matt N.

Yes!

a Dangerous Idea

<|B-(

Matt N.

9:33 PM

That's what I meant by mechanical.

A bit tedious perhaps.

cassandra0

but that's such a coincidence that we just happened to have the exact condition we needed..

Matt N.

No!

user19161

@OldJohn Are those also called diophantine?

Matt N.

That was one of the assumptions we started with: it is contained in the diagram we start with.

Old John

@WillHunting not really - more like "quadratic forms"

robjohn

9:34 PM

@WillHunting Hard diophantine

Matt N.

Next we want to show $\mathrm{im}\bar{v} \subset \mathrm{ker} \bar{u}$ and $\mathrm{im}\bar{v} \supset \mathrm{ker} \bar{u}$.

robjohn

@OldJohn I usually think of diophantine as those questions involving only integers, but that may be too broad

Matt N.

Remember: the following diagram is exact
$$A \xrightarrow{u} B \xrightarrow{v} C \to 0$$

a Dangerous Idea

>>>B ((

Old John

I have a whole book on the topic, and it includes stuff like Fermat's first ideas, then Gauss, then elliptic curves and modular functions ...

cassandra0

9:35 PM

@MattN., yeah

Matt N.

Hence we have that $v$ is surjective and $\mathrm{im}u = \mathrm{ker}v$.

cassandra0

I guess that surjectivity comes from the --> 0

Matt N.

Yes.

Old John

@robjohn I tend to think of diophantine as mainly being solving equations by diverse and seemingly unconnected methods :)

Matt N.

Exactness means image equals kernel at every object in the diagram.

robjohn

9:36 PM

@OldJohn Pell's equation is given as an example of a Diophantine equation on Wikipedia, not that that means anything

Matt N.

If it is the zero map, its kernel is everything.

Old John

@robjohn yep - a very good example too

robjohn

@OldJohn I got a good number of points for answering a question by knowing when a number can be written as a sum of two squares.

Old John

the book on $x^2+dy^2$ is mainly about the problem of trying to caracterise the primes representable by various forms, rather than solving the diophantine equations - it gets too hard for me really, but I am enjoying the tussle!

Charlie

@anon could I bother you a little?

Old John

9:39 PM

@robjohn I have gained a few from that too :)

anon

Slightly. A tad.

robjohn

@OldJohn It was fun when the answer dawned on me math.stackexchange.com/questions/54547/…

Old John

@robjohn very nice!

cassandra0

The book by Cox

Charlie

@anon I have to discuss about the existance of max and min ofa certain function. That it have min i found, but i cannot show if it has max

cassandra0

9:40 PM

?

Jordan

why the fuck didnt anyone tell me my formula was wrong?

Old John

@cassandra0 that's the one

robjohn

@Jordan inner peace

cassandra0

depending on $d$ it can be very deep

a Dangerous Idea

@Jordan Easy on the f

Matt N.

9:41 PM

Ok. Let $f \in \mathrm{im}\bar{v}$. Then there is $g:C\to N$ such that $f=g\circ v$. Then $\bar{u}(f) = g \circ v \circ u$ and since $\mathrm{im}u = \mathrm{ker}v$ this maps to zero.

robjohn

@aDangerousIdea yeah, formula is a 7 letter word!

anon

@Charlie You're going to have to specify the function for anyone to say anything.

Charlie

@anon $f(x,y,z)=xyz$ subject to the region $$x,y,z\geq 0, xy+yz+xz=2$$

Jordan

@robjohn How can I have inner peace if I still suck at everything I do?

Matt N.

Now we have $\mathrm{im}\bar{v} \subset \mathrm{ker}\bar{u}$.

cassandra0

9:42 PM

oh that was neat

Matt N.

Let me repost the diagram, I forgot the bars on the maps.

cassandra0

@MattN., I understand it

Matt N.

Also, it's scrolled off.

a Dangerous Idea

@robjohn Yeah that is one and three quarter 4-letter words

robjohn

@anon we should be able to give upvotes to comments here, too :-)

Matt N.

9:43 PM

@cassandra0 Okay, so you don't want to see the other inclusion?

cassandra0

@MattN., I think I can prove it similarly!

Old John

@robjohn math.stackexchange.com/a/169120/32441 is my favourite of my answers, but I am still not convinced it is absolutely watertight - and reckon there has to be a shorter way

Matt N.

Yes. Iirc it's slightly less obvious but still very easy.

So we're good?

cassandra0

thank you !!

Jordan

I need to learn fourteen years of math in six weeks, wut do?

Matt N.

9:44 PM

@cassandra0 You're very welcome.

a Dangerous Idea

@Jordan Study hard that's all you can do, right?

Jordan

@aDangerousIdea I guess, I think I need to take like four years off of school and just start over

anon

@Charlie What method are you using? Lagrange multipliers?

Charlie

@anon Lagrange multipliers

Jordan

math 70, 80, precalc 1, 2, trig, calc 1 and then I can start math again

3 years

a Dangerous Idea

9:46 PM

Sure, why not?

Jordan

every semester of school I take is the same

I study really hard, stay ahead, take a test and then do really poorly because i can't take tests and then I dont feel like studying anymore because it never really pays off

I never retain any knowledge

100 credits in and I haven't really learned anything

awesome @OldJohn

@cassandra0 what is?

your proof

@Jordan Just start over then.

3 mins ago, by Jordan

@aDangerousIdea I guess, I think I need to take like four years off of school and just start over

Old John

9:50 PM

@cassandra0 took me 3-4 days to work out all the details - would probably have been more productive to have answered a bunch of low-hanging fruit - but thanks!

cassandra0

this is more important

Jordan

@aDangerousIdea I am too old to start over and I certainly dont have the money, I just want to learn what i came to college for, I am almost done with the bullshit and I can finally start to learn what I want

cassandra0

I'll show one of my favorite even though it's very very easy

a Dangerous Idea

@Jordan Maybe then you will really learn something. From the beginning; your way.

cassandra0

3

Q: how to find integer solutions for $axy +bx + cy =d$?

How can I find the integer solutions for the diophantine equations $axy +bx + cy =d$ ? the smallest particular solution ($x_0$,$y_0$) and a way to generate the rest.

Jordan

9:52 PM

I just hate school so much, the whole system is set up to stress you out and make you feel like shit

you mess up slightly and forget some arbitrary formula slightly and you lose 50 percent of the points on the test, that is so irrelevant to any real life situation

a Dangerous Idea

@Jordan Get out of the system.

robjohn

@OldJohn That is pretty involved. Nice!

Jordan

@aDangerousIdea I am not smart enough or hard working enough to make it on my own, I am the type of person that needs a degree to have a chance

Old John

@robjohn I am sure there must be a neat way to do it - but I couldn't find a slick way

cassandra0

@OldJohn, that is the neat way :)

it's a very high degree equation

robjohn

9:54 PM

@OldJohn Erdos may have worked on something similar. It has that feel

Old John

@robjohn I bet he would have found some really clever way to tackle it, too!

@robjohn he did something nice with products of consecutive integers not being squares, I think

Jordan

If I was smart enough to be able to make it on my own I wouldnt struggle through school

a Dangerous Idea

@Jordan You can build up your smarts and learn to work hard on your own, if you really want it.

robjohn

@OldJohn and generalizations of that, too

cassandra0

en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture

Jordan

9:56 PM

I am not that motivated unfortunately, I have no idea what I want in life

a Dangerous Idea

no idea is not good

Charlie

@Jordan take your time, no pressure.

Jordan

I am just sick of doing nothing, just wasting my time in school while everyone I went to high school with is out making tons of money and doing something with their lives. people a lot younger than me are graduating and I am just starting

cassandra0

en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Graham_problem

Old John

@robjohn I have a slight regret that I never wrote a paper with my supervisor - it would have given me Erdős number 3 :)

Charlie

9:58 PM

@Jordan what "people" do doesn't matter. what matter is what you do to yourself. only Jordan matters to Jordan. Forget what others achieve.

robjohn

@OldJohn Ooh, that would be neat!

Old John

@robjohn undeserved, but neat :))))

Transcript for

$Mathematics$ Mathematics