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9:00 PM
actually might take a few days break from MSE again
 
Slow and steady wins the race, but the hare really shows no grace...
3
 
:D
 
:DDDDDDDDDDDD
 
user19161
@JayeshBadwaik Where did that come from?
 
@WillHunting His brain
 
user19161
9:02 PM
@Charlie Oh!
 
@WillHunting you jealous of my skill in limericks? ;-) :-P
anyway, its 02:30 now and I really have to sleep. gn.
 
user19161
@JayeshBadwaik Well, the people in the Eng room got sick of my jimericks. I think I will write one now.
 
@JayeshBadwaik Geez, it late for you!
 
@JayeshBadwaik Good night, Jayesh! sleep tight!
 
@Argon Yeah....
 
9:03 PM
@JayeshBadwaik later
 
@Charlie Good night, Marilia... sleep tight!!
 
@JayeshBadwaik :D
 
@aDangerousIdea later.....
 
'Night!
 
Everybody stops!
 
9:05 PM
The hare rarely shows grace upon winning the race.
Instead he jumps up in your face.
 
Classic poetry
 
And leaves without a trace.
 
user19161
@amwhy loves to use ellipses
As well as too many commas
Reputation always soaring
As answers are never boring
She likes pretty blue pictures!
 
Hey gang I have a physics questions
 
and who doesn't?
 
user19161
9:08 PM
@Jordan Welcome back to chat!
 
I have the formula for something to do with tempurature and energy. $m_1 * c_1 * \Delta T_1 + m_2 * c_2 * \Delta T_2 = 0$
well that isn't delta
 
\Delta ($\Delta$)?
$\delta$ is lower case
 
I am not sure how this formula works but it seems to say that $ T_f = T_1 + T_2$
Is that correct?
$m_1 * c_1 * (T_f - T_{i1`}) + m_2 * c_2 * (T_f - T_{i2`}) = 0$
 
@OldJohn You'll leave me as OldRob...
 
@robjohn Ah - you would then be the elder :)
 
user19161
9:12 PM
@robjohn Wow, you actually read the transcript, spooky!
 
@WillHunting transcript, it's just up a little bit
 
$T_f = \frac { $m_1 * c_1 *T_{i1} + m_2 * c_2 * T_{i2} } {m_1*c_1 + m_2 * c_2}$
So then the mc cancels out
 
Pity Lubin doesn't come to chat - he would make us all feel young :)
 
user19161
@robjohn Congrats on reaching 40k!
2
 
user19161
@OldJohn Who's Lubin?
 
9:14 PM
 
@WillHunting It just happened! Since you noticed it so fast, I will say thank you :-)
 
Good evening, ladies.
 
@robjohn wow! precisely 40K - well managed!
 
hi @MattN.
 
Um... Good evening!
 
9:15 PM
do you know about the Hom functor?
 
A bit.
 
@robjohn Congratulations!
 
@OldJohn I've been standing off from answering so that it might not go too fast :-)
 
4
A: For a finitely generated abelian group are the homomorphisms into the integers finitely generated

Zhen LinYes. 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...

 
9:16 PM
@robjohn Neat
 
@OldJohn Now I can answer some more :-)
 
this is what I wanted to link
@MattN., he uses Hom(-,Z) being left exact, what do oyou think?
 
@robjohn and get to 50K ... 100K ...
 
Zhen Lin, I assume.
 
yes
 
9:17 PM
I discovered new physics
 
@OldJohn I will try. We've had a number of people break the 100K barrier recently
 
do you know why it is left exact?
 
$T_f = T_1 + T_2$
 
@robjohn I have the more modest aim of achieving the "generalist" badge
 
but my new physics doesnt seem to work
 
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
 
@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?
 
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
 
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.
 
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 $N$
 
I wonder if there is any money in discovering new physics
 
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.
 
Let $N = \mathbb Z$.
 
@OldJohn I'm working on the Epic badge. I am 72% of the way there.
 
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
 
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.
 
9:24 PM
well maybe it shouldn't the 0 at the end
since it's a contrafunctor
 
Oops, typo. Sorry. Corrected.
 
user19161
@Jordan What stuff is not covered in the textbook?
 
(at least you know I'm paying attention now :)
 
I'm glad : )
 
but still
 
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
 
shouldn't it be 0 --> Hom --> Hom --> Hom
 
@WillHunting Energy transfer and work
 
@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.
 
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
 
9:26 PM
@cassandra0 Yes, of course. Excellent! : )
 
@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
 
@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.
 
@cassandra0 Meh. I was impressed there for a minute....
 
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.
 
@WillHunting Take that Arturo! >B(
 
@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.
 
I barely graduated high school
 
Now we need to show the following:

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

(ii) $\mathrm{im} \bar{v} = \mathrm{ker} \bar{u}$
 
9:28 PM
@robjohn >8( has been replaced?
 
First (i):
 
@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
 
@robjohn I spent years doing nothing but analysis (of a sort) - now I am relaxing mostly with a bit of number theory :)
 
and matches the eyes more closely
 
\bar v = Hom(v,Z) = - o v
 
9:30 PM
@OldJohn I've been doing some number theory, I've played around with the basics
 
@aDangerousIdea perhaps not :-)
 
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 ...
 
@robjohn it is a very alluring branch of maths, isn't it?
 
9:31 PM
$$\bar v = \text{Hom}(v,\mathbb Z) = - \circ v$$
 
$$0 \to \mathrm{Hom}(C,N) \xrightarrow{\bar{v}} \mathrm{Hom}(B,N) \xrightarrow{\bar{u}} \mathrm{Hom}(A,N)$$
 
@OldJohn Often the problems seem that they should have easier solutions than they end up having
 
okay
 
@robjohn >|B(
 
user19161
I gave up on number theory after high school.
 
9:32 PM
@aDangerousIdea it makes me think of someone with a hat
 
$\bar{v}$ 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).
 
oh that is easy!
 
@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
 
Yes!
 
9:33 PM
That's what I meant by mechanical.
A bit tedious perhaps.
 
but that's such a coincidence that we just happened to have the exact condition we needed..
 
No!
 
user19161
@OldJohn Are those also called diophantine?
 
That was one of the assumptions we started with: it is contained in the diagram we start with.
 
@WillHunting not really - more like "quadratic forms"
 
9:34 PM
@WillHunting Hard diophantine
 
Next we want to show $\mathrm{im}\bar{v} \subset \mathrm{ker} \bar{u}$ and $\mathrm{im}\bar{v} \supset \mathrm{ker} \bar{u}$.
 
@OldJohn I usually think of diophantine as those questions involving only integers, but that may be too broad
 
Remember: the following diagram is exact
$$A \xrightarrow{u} B \xrightarrow{v} C \to 0$$
 
>>>B ((
 
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 ...
 
9:35 PM
@MattN., yeah
 
Hence we have that $v$ is surjective and $\mathrm{im}u = \mathrm{ker}v$.
 
I guess that surjectivity comes from the --> 0
 
Yes.
 
@robjohn I tend to think of diophantine as mainly being solving equations by diverse and seemingly unconnected methods :)
 
Exactness means image equals kernel at every object in the diagram.
 
9:36 PM
@OldJohn Pell's equation is given as an example of a Diophantine equation on Wikipedia, not that that means anything
 
If it is the zero map, its kernel is everything.
 
@robjohn yep - a very good example too
 
@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.
 
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!
 
@anon could I bother you a little?
 
9:39 PM
@robjohn I have gained a few from that too :)
 
Slightly. A tad.
 
@OldJohn It was fun when the answer dawned on me math.stackexchange.com/questions/54547/…
 
@robjohn very nice!
 
The book by Cox
 
@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
 
9:40 PM
?
 
why the fuck didnt anyone tell me my formula was wrong?
 
@cassandra0 that's the one
 
@Jordan inner peace
 
depending on $d$ it can be very deep
 
@Jordan Easy on the f
 
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.
 
@aDangerousIdea yeah, formula is a 7 letter word!
 
@Charlie You're going to have to specify the function for anyone to say anything.
 
@anon $f(x,y,z)=xyz$ subject to the region $$x,y,z\geq 0, xy+yz+xz=2$$
 
@robjohn How can I have inner peace if I still suck at everything I do?
 
Now we have $\mathrm{im}\bar{v} \subset \mathrm{ker}\bar{u}$.
 
9:42 PM
oh that was neat
 
Let me repost the diagram, I forgot the bars on the maps.
 
@MattN., I understand it
 
Also, it's scrolled off.
 
@robjohn Yeah that is one and three quarter 4-letter words
 
@anon we should be able to give upvotes to comments here, too :-)
 
9:43 PM
@cassandra0 Okay, so you don't want to see the other inclusion?
 
@MattN., I think I can prove it similarly!
 
@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
 
Yes. Iirc it's slightly less obvious but still very easy.
So we're good?
 
thank you !!
 
I need to learn fourteen years of math in six weeks, wut do?
 
9:44 PM
@cassandra0 You're very welcome.
 
@Jordan Study hard that's all you can do, right?
 
@aDangerousIdea I guess, I think I need to take like four years off of school and just start over
 
@Charlie What method are you using? Lagrange multipliers?
 
@anon Lagrange multipliers
 
math 70, 80, precalc 1, 2, trig, calc 1 and then I can start math again
3 years
 
9:46 PM
Sure, why not?
 
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
 
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!
 
this is more important
 
@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
 
I'll show one of my favorite even though it's very very easy
 
@Jordan Maybe then you will really learn something. From the beginning; your way.
 
3
Q: how to find integer solutions for $axy +bx + cy =d$?

Loers AntarioHow 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.

 
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
 
@Jordan Get out of the system.
 
@OldJohn That is pretty involved. Nice!
 
@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
 
@robjohn I am sure there must be a neat way to do it - but I couldn't find a slick way
 
@OldJohn, that is the neat way :)
it's a very high degree equation
 
9:54 PM
@OldJohn Erdos may have worked on something similar. It has that feel
 
@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
 
If I was smart enough to be able to make it on my own I wouldnt struggle through school
 
@Jordan You can build up your smarts and learn to work hard on your own, if you really want it.
 
@OldJohn and generalizations of that, too
 
9:56 PM
I am not that motivated unfortunately, I have no idea what I want in life
 
no idea is not good
 
@Jordan take your time, no pressure.
 
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
 
@robjohn I have a slight regret that I never wrote a paper with my supervisor - it would have given me Erdős number 3 :)
 
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.
 
@OldJohn Ooh, that would be neat!
 
@robjohn undeserved, but neat :))))
 

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