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21:00
@robjohn Indeed. I think I've kicked the habit.
We also require $s(\tau(g, f)) = s(f)$ and $t(\tau(g, f)) = t(g)$.
@HenningMakholm You mean visiting MSE or just chat?
@robjohn Either. I did pop by to vote for you in the mod election, but it didn't seem to work.
118 messages since I left
this is getting active!
@Jordan The last equation doesn't make sense to me.
21:01
The picture to draw here is a triangle: we have three vertices $x$, $y$, and $z$, and three edges $f$, $g$, and $\tau(g, f)$, where $f$ goes from $x$ to $y$, $g$ goes from $y$ to $z$, and $\tau(g, f)$ goes from $x$ to $z$.
I don't even know anymore
I am so sick of this problem
@robjohn The one about $${s_n(a+1) \over {ns_n(a)}}$$
Then the final axioms relate $\rho$ and $\tau$: if $s(f) = x$, then $\tau(f, \rho(x)) = f$, and if $t(f) = y$, then $\tau(\rho(y), f) = f$.
I have never done one thing for so long, I am just angry I need to do something else
@Jordan Hey, just keep doing. That was quite right and good. Only the last part wasn't correct. Take a look at it now and tell me if you can find where was the mistake.
21:03
I have failed to do this one thing for about 2 hours now
This basically says that $\rho(x)$ is a degenerate edge: joining it to another edge doesn't change anything.
And these are all the axioms for a category.
I fixed the last aprt and the answer is still wrong
Of course, traditionally $C_0$ is called the set of "objects", $C_1$ is called the set of "morphisms", $s$ is normally called $\textrm{dom}$, $t$ is normally called $\textrm{codom}$, $\rho(x)$ is normally written $\textrm{id}_x$, and $\tau(g, f)$ is normally written $g \circ f$... but I find this tends to put certain, uh, preconceptions in people's minds.
@Jordan Hang on, I'll read the rest of it.
I get x/4 - x/16 -sin4x/8 or something but sin is 0 so it doesn't matter what sin is
I need to make the first two terms equal pi/16
which seems impossible, I have tried to solve that equation but it doesn't work
21:07
Very good. Thank you. Ignore my previous comment.
So a category is a graph with some additional structure.
Indeed.
Conversely, every directed graph $\Gamma$ generates a "free" category, whose "morphisms" are just finite head-to-tail strings of directed edges of $\Gamma$.
The "identity morphism" is the empty string (for each vertex separately).
But if you have a strongly connected directed graph (with a unique loop on each vertex), then that's already a category.
@HenningMakholm I was third, but not that close.
@robjohn I couldn't find results in a human-readable format.
No, wait, sorry. Don't ignore my previous comment: $s$ and $t$ give us the start and end point (respectively) of an edge in $C_1$. So why are they maps $C_0 \to C_1$ instead of $C_1 \to C_0$?
@Jordan First: $\int sin^2x cos^2 x$ = $\int (1-cos2x)/2 * (1+cos2x)/2$ = $\frac{1}{4} \int 1 - cos^2 2x$ = $x/4 - \frac{1}{4} \int cos^2 2x$
21:15
@HenningMakholm Bill posted them here
What is wrong with my way?
Ah, now that was a typo...
@Jordan I'm talking about your way! There are much easier ways to do it but we're trying to find out what you're doing wrong.
I changed the last equation there.
@robjohn How is that supposed to be read?
I got 11 votes?
I hope it's correct since it's 0200 a.m here!
21:17
I don't think there are easier ways to do it really, the easier ways involve memorization of very complex idetnities
@Jordan Okay, did you understand what I changed?
@Jordan You want to evaluate $\int \sin^2 x\cos^2 x dx$?
@Jordan Then about your substitution, it was mostly right except you needed to do another substitution.
21:19
@robjohn Thanks. It seems the chatroom constituency isn't as strong as we'd like to think. :-)
$\frac{-1}{8} \int cos^2 u du$ = $\frac{-1}{8} \int \frac{1+cos2u}{2}$ = $ \frac{-1}{16} \int cos 2u$
@Jordan What integral are you trying to evaluate?
@PeterTamaroff Wait, I'll finish talking to him and then you can ask him questions!
Are you following @Jordan?
$\tau$ gives us $\circ$ for morphisms -- how do we get $+$?
@PeterTamaroff The way I read it, you were first choice for 11 voters and second choice for 6 of the voters who put Bill as first choice.
21:21
I did 2 subsitutions
@MattN: One step at a time! Abelian categories is still some way off.
@ZhenLin I was aiming for pre-additive categories only : )
@Jordan What did you get? $s = 2u $\
Here's a fun example of a "abstract" category which isn't really all that abstract.
21:22
something like that
$C_0 = \mathbb{N}$.
Let $C_1$ be the set of all matrices over some field $k$.
@PeterTamaroff Then afterwards, you were not second choice for the single voter who put checkmath first (who didn'r provide any alternative vote, by the way), nor third choice for any of the two voters who voted 1-Bill, 2-checkmath (their third votes went to Eric and Benjamin).
Define $s : C_1 \to C_0$ to be the number of columns of a matrix, and $t : C_1 \to C_0$ to be the number of rows.
Define $\rho(n)$ to be the $n \times n$ identity matrix, and $\tau(B, A)$ to be the product matrix $B A$.
@gigili I dont think that changes the answer though since sin goes to 0
@HenningMakholm I can't understand it well.
Nevermind.
21:24
I am evaluating from 0 to 2pi
I see. Wait: $\rho(n)$ could be any $n \times n$ matrix, no?
This is a category! And it's equivalent to a well-known one: the category of finite-dimensional vector spaces over $k$ and $k$-linear maps.
@Jordan You missed the "1"
And it just so happens to be a preadditive and even abelian category.
@MattN: In principle. But then the axioms wouldn't be satisfied here.
21:26
@Jordan If you're evaluating the integral in $(0,2 \pi)$ then you should be explicit about it.
Put the limits of integration in the integral.
it doesnt matter until the end
I really do not have time like this, I can't believe i just wasted 2 hours on a single problem, I still have 6 sections to study
@Jordan $\frac{-1}{8} \int cos^2 u du$ = $\frac{-1}{8} \int \frac{1+cos2u}{2}$ = \frac{-1}{16} \int 1+\cos u du=\frac{-1}{16} \int 1 du+ \frac{1}{32} \int \cos s ds $
It's $1+\cos 2u$
@ZhenLin Which (category) axiom wouldn't be satisfied? I assume you are talking about vector space axioms now?
@Jordan Then you should plug in that into the previous integral!
@MattN: The one about the composition of identities, of course.
21:30
You missed some constant which will help to have the $\pi/16$ or whatever is the final answer supposed to be.
Not whining won't help me get my homework done
@ZhenLin By composition of identities you mean this?
@Jordan I mean this: positive thinking.
Yes.
21:32
@gigili your answer doesn't format correctly
@Jordan If you give me your question I'll try and help you out.
I posted it I just want to know what i am doing wrong
Umm.
@Jordan Link
$\frac{-1}{8} \int cos^2 u du = \frac{-1}{8} \int \frac{1+cos2u}{2} = \frac{-1}{16} \int 1+\cos u du=\frac{-1}{16} \int 1 du+ \frac{1}{32} \int \cos s ds $
It had too many dollar signs.
21:34
Oh, I see.
What's the final answer supposed to be @Jordan?
0
Q: Integral of $\int \sin^2 x \cos^2 x dx$

JordanThis seems pretty simple to me but I can't get it. $$\int \sin^2 x \cos^2 x dx$$ $$\int (1-\cos^2 x) \cos^2 x dx$$ I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an id...

@ZhenLin Cool, thank you!
But the objects in this category happen to have an additive structure already.
Nope
Matrix addition?
21:35
@Gigili $\frac{ \pi}{16}$
The set of objects is $C_0$.
Ah. Right, that's why I thought that $\rho (n)$ could be any matrix at first.
But yes, matrix addition makes this into a preadditive category.
@Jordan 0 to 2\pi?
This is probably the fundamental example of a "ringoid". You can't add arbitrary matrices, and you can't multiply arbitrary matrices either.
The $s$ and $t$ tell you which ones you can add or multiply.
21:37
@ZhenLin Yes, excellent, thank you for this!
Now, for my follow-up act, I will talk about how categories are really matrices. :p
@Gigili Yes
Then we end up with graphs and matrices being the same things : )
Who the heck says arabesque? If someone used that word infront of me I would punch them in the throat.
@Jordan It probably means "manipulation" "solution" "trick" or any context fitting word.
21:43
But I'm too lazy, I'll just point you all to Chapter VII of my incomplete notes
@MattN: Well, yes, recall that any ordinary graph can be represented using adjacency matrices...
Oh, yes.
@ZhenLin Thank you!
It's not really in any decent state. It's too basic in places, too advanced in others...
If I read it and see a typo, should I just ignore or should I contradict you?
Send me an email or something.
21:46
Hey @MattN did you sort out your problem
@PaulSlevin Well, yes: I now know an example of a pre-additive category in which objects don't already have $+$.
Well I am done with the problem, if I see it on a test I will not even attempt it
@Jordan Well, I get $\pi/8$, there should be a $\frac{1}{2} or something we missed somewhere!
@MattN I am glad
@PaulSlevin Me too.
21:48
@Jordan
The formulas you really need to know (like a poem) are
@Jordan: You're very welcome, by the way! No need to thank me.
Do you see why if I have a set $\{ x, y\}$ thought of as a category then it can not be preadditive
$$\sin(2x) = 2 \sin x \cos x$$
$$1 = \sin^2 x+ \cos^2 x$$
When is your exam @Matt?
$$\cos(2x) = 2\cos^2(x)-1$$
21:50
@Gigili Some time in August, I don't know the dates yet.
@Gigili Thank you for the help, I just wish I didn't have to waste 3 hours on it in the first place. A normal person would have just spent 10-20 minutes on it
Jordan I think a normal person would not be brave enough to even try the problem. at least you tried you need to get more confidence
@Jordan And I'm serious about the "like a poem" thing. Dead serious.
yeah if you know those 3 jordan you will be fine you just need to learn them
I know the half angle, double angle and the basic identities
21:52
hey guy!
It's too hot here, I'm suffering. : ( At least the teddy bear won't have cold feet.
I love math!
math is alright I guess
@Jordan But the guys are giving answers using those! Why do you say require "incredibly deep knowledge of trig"
@MattN I thought it was on Monday or something!
21:53
Honestly Jordan jst relax and stop belittling yourself
I guess so too.
@Gigili No, luckily...
@MattN Okay, you have time. Good luck anyway.
I have an exam on monday and I think my brain has completely shut down from new information
@Gigili Thank you!
21:54
@Peter Tamaroff, I now, because mylove $\ge$ yourlove. Meaning there is room for equality.
no LaTex?
:(
@PeterTamaroff Because I do not recognize the complex transformations they did to the identities to make them math. It isn't something I would see on a test. I would try something more simple first, and hten if that fails another simple method I am comfortable with. I would likely never start to experiment with wacky transformations of poorly understood identities
@PaulSlevin That sucks, I know that : (
@Jordan The tranformations are algebra Jordan.
I am bad at algebra
I have gathered about 37 proofs i hope to have memorised by monday, i know most of them but I am just so very sick of looking over them again and again
21:56
@Jordan I hate to give hope to others, but you were trying to find out what you're doing wrong which is the most right thing to do. Since you might make same mistakes over and over.
what are your favorite fields in math?!
@HowardRoark Finite fields and Galois fields.
I figure that it doesnt matter whether or not i work tonight or tomorrow , If I don't know the course by now I'll be screwed anyway
@HowardRoark I am interested in category theory althougth I am just learning the basics
I quite like $\mathbb R$ and $\mathbb C$, too.
: )
@PaulSlevin No, if I were you I'd do as much work as possible.
@MattN Therein lies the rub my friend. I jst cant. I am literally exhausted . It worries me, because I need to be keen and alert in the actual exam
21:58
@HowardRoark On a more serious note: I don't have a favourite but I know what I don't like: numerical maths, statistics and probability theory.
That is awesome. @PaulSlevin category theory? I need to look that up.
@PaulSlevin Ah, I know that feeling, too. If you're already exhausted then it's probably a good idea to relax.
relax is always a good idea (Gautama Buddha)
i have 9 hours of exams monday - wed

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