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00:00
If the brain moves too fast, the result is concussion.
Not advisable.
@DanielFischer tell that to the all-blacks. :)
Rugby is different.
Kernals. Kernals. Kernals. Learning about them Kernals.
@Alizter Kernels???
I have a headache.
00:13
Take a homomorphism $\varphi:R\to S$. The kernal is $\varphi^{-1}$'s pre-image of 0
It is the set of all solutions to $\varphi^{-1}(0)$
@Pedro I should tell you the classification of fgag and have you prove it. :D
00:33
Go ahead,.
@Alizter kernel
@Pedro Thats what i thought....
@Pedro could you check some work for me briefly? I worte it up quite hastily, so it might not be quite correct
@Pedro Every fgag is isomorphic to a finite direct sum of cyclic groups, each of which is infinite or has prime power order.
@KevinDriscoll OK.
I just want to check that its basically correct (if not air tight)
@KevinDriscoll Honestly, that is hard to read.
Can you write it here?
00:40
@Pedro You mean its too small or you cant read my handwriting?
You're trying to prove that a space is regular under certain conditions.
The handwriting.
Namely, the condition is that every nbhd $U$ of $x$ contains the closure of another nbhd $O$ of $x$.
Hi all, sorry to interrupt/bother, but does anyone know if the converse of the following statement is true? If X is an irreducible topological space, then the constant sheaf Z is flasque.
And you want to prove every point $x$ and every closed set $F$ not containing $x$ can be separated by nbhds, @KevinDriscoll.
Now, since $F$ is closed and $x\not\in F$, one can find an open set $U$ that contains $x$ disjoint from $F$.
Then pick by the condition a nbhd $O$ with $x\subseteq O\subseteq \overline O\subseteq U$.
You claim that $N_xO$ works for $x$ and $N_F=X\setminus \overline O$ works for $F$, yes?
@KevinDriscoll
00:45
@Pedro Yup now that looks right
you're much better at summarizing these than I am
@KevinDriscoll Succinctness comes with practice, sure.
arghhhhhhhhh
@Pedro would you say that what you wrote shows that the 2 conditions are equivalent?
at my practice problem... how does that really work :/
oh lol found my problem I forgot to substitute b for -1 XD
@KevinDriscoll Well, you can go the other way around. Pick an open set $U$ containing a point $x$. Then $X\setminus U$ is closed and disjoint from $x$... =)
00:51
complement definition!
there are just elements in X, not U XD
@Pedro @Usukidoll You still need the whole $O$ and its closure thing to find a nbhd of $X \text{ \ } U$ etc etc but ok
@KevinDriscoll Well, but now you know there is an open $O$ and an open $N=X-F$ for $F$ closed such that $x\in O$ and $X-U \subseteq X-F$.
And $O\cap N=\varnothing$.
:O I'm just new @KevinDriscoll I've seen the $X \backslash U$ somewhere in my book and that's the complement definition... with elements in X, but not in U
Can you show $\overline O\subseteq U$?
what is that negation O?
any element not in O is a subset of the U!?1!
hears crickets
00:59
@usukidoll Its the closure. $O$ plaus all of its limit points. AKA the smallest closed set containing $O$
stahp I'm only new at analysis X_X
I mean I've seen whatever you're doing before
but not in an advanced thing like this
oh yay we can be analysis newb buddies ^^
induction is fun... set theory is shiz
like the average scores were like a 5 out of 10
but for induction it's like 9 out of 10
I'm not actually taking an analysis course, ....well kinda I'm taking intro to functional analysis online and learning some basic topology as we go
I'm taking intro to advanced mathematics
I can do easy proofs...hard ones eehehheheh yeah right
01:04
Hello
Are binary math question approved here?
I'm sure it is
if you make an attempt then why not post it
@Pedro $X- U \subset X - F$ can't be right
because $x$ is in $X - F$ but not in $X-U$
I'm reading it as X is not a subset of X
my book has the < without a line as not a subset
@ChrisOkyen Ya its on topic for Math.SE
$x \backslash u \subset X \backslash F$
neverminddd
01:08
Ok
So I have my combinations here: pastebin.com/KDY64nG1
How can I determine with N digits how many combinations there will be?
It looks like a bit shift plus adding the previous total of the digits one lower in some situations but that formula flops overall
Ok I have this function $$f(x)=\begin{cases} 1& x=\frac{1}{n}\\ 0 (other)\end{cases}$$. How is $f:[\frac{\epsilon}{2},1]\rightarrow\mathbb{R}$ a constant function?
Александр Пушкин related to @Александр @AlexanderGruber?
@alex where'd you learn Rep'n theory of finite groups from?
01:13
Russian?!
That was
@ChrisOkyen That doesnt look like binary, but maybe im misunderstanding something
@Mike my favourite book is Huppert's character theory of finite groups
but it is a bit difficult to find. isaacs is also very good and he is easier to find.
you can actually find Huppert on scribd I think, but the site is a bit clunky, I can't figure out how to use it.
@KevinDriscoll What I am saying and asking (simultaneously), is could the problem of finding an equation to figure out the number of combinations of N digits with the rules you see using Binary Bitshifts and Addition?
scribd is garbage, I hate it. but I'll remember those names. thanks.
01:18
@Mike isaacs also wrote my favorite finite group theory book
Hi all, I have the following problem. I can easily show that the constant sheaf Z on an irreducible topological space X is flasque, but I am wondering if the converse is true. Namely, does the constant sheaf Z being flasque imply that X is irreducible?
(huppert wrote one too but it's in German)
Anybody know if binary operations are the way to solve this problem
it looks like I don't need to know a whole lot about f.g. rep'ns, so I don't need a whole course in it or anything
@Mike what do you need?
01:19
@Chris so you want to know working in base $N$ (ie $N$ digits) how many combinations you can make of length $N$ or less (without repitition)?
@dmdmdmdmdmd you'd be better server asking on main
oh sorry
I think that is right
let me read that once more though
@alex here's the list of all the topics I should know
Basic definitions, matrix coefficients, Schur orthogonality, invariant inner products and complete reducibility of representations, characters of finite groups and parametrization of complex representations by characters, character tables, Peter-Weyl theorem.
I seem to vaguely remember parts of all of that except for PW from my $S_n$ representations course
I believe we are on the same page @KevinDriscoll. So it really is where there is N digits it is N base
just like binary or hexadecimal base or octal based
@ChrisOkyen Yes, but I want to be clear that you've excluded repitition. Like 11 appears nowhere on your list.
01:23
yeah repitions like 11 are excluded
reptitions like say I have 123 then I write 321 is also excluded
@Chris Ya when you say combinations the 2nd example you gave is excluded by definition, but not the first
I don't see the obvious answer, so I'd post on main @ChrisOkyen but try and be very clear about the problem you're trying to solve. Its not obvious initially from your pastebin
@Mike it sounds like you need about the first half or so of a character theory book
and then peter-weyl which would be a separate thing
so huppert's first half, say? also yeah, from googling peter-weyl it seems like a completely different flavor than everything else they ask for.
@Mike that's more or less where i did it. i'm sure isaacs would work just as well.
So what your are saying @KevinDriscoll is the first examples 11 may not be known or understood as being defined as repetitious as 123 -> 321 and I should be careful to specify this on main
01:27
orthogonality is the core of it if you learn that well you should have not much trouble with the rest
alright, cool. thanks again, @alex
@KevinDriscoll when you mean main I assume you mean SE or SO
IE stackoverflow.com forums
@Mike oh and btw, you could probably use dumit and foote for a bunch of that, the very last section, if you have it.
I don't, though I do have a copy of lang I'm wary to touch
01:31
has anyone used analysis by bartle?
I have dummit and foote ;)
High five
@Mike Susie is one crazy bitch.
@user4140 i like it
@Mike lang is the worst
@Pedro who the hell is Susie?
The writer.
01:37
it's like he sat down with the explicit intention of writing the least accessible algebra text ever :(
@alex his exercises are decent but getting anything out of his exposition is hopeless.
actually what i heard is that he narrated the book to his secretary
hahaha
which explains why he has so many paragraphs that are 1 or 2 sentences long
poor secretary.
01:38
i hope whoever it was got payed a load of cash
I'd have just shot myself by the fifth chapter or so
Yeah, Lang is pineapple crazy-
The term is "cray cray" @Pedro
He defines what an exact sequence is on page 15.
ain't nothin wrong wit that
@Pedro I've been getting mileage out of Hungerford by just proving his theorems myself
01:42
yay @user4140 is back!
since he has so many lemmas this is usually pleasant and not so bad
hi @usukidoll
hi hi'
No one answered my protein question in biology beta
give it time
01:44
@KevinDriscoll do you mean post a thread?
I feel chat is a better place over all for me becuase I don't always specify right and in forums we go back and forth clearing stuff up for a long while.In chats its quicker and wont get me donwvoted for not being clear
why don't you just ask questions clearly?
From past posts, it seems its very hard for me
@ChrisOkyen time to learn then, communication is an important skill.
Lang gives a four line proof of JH, @Mike.
@Pedro I believe it
01:51
bunny
and Hungerford's proof I longer, but telegraphs what he's doing from the start :)
Well, he's not that bad, one just needs to digest the material before moving.
@PedroTamaroff ... on to a better book.
i have to show that if $P_n(x)$ is the interpolating polynomial for $e^x$ at the points $\{j\frac{b}{n}|j=0,\ldots, n\}$, $$\lim_{n\to\infty}\operatorname{max}_{x\in [0,b]}e^x-P_n(x)=0$$
what has my life come to
I'm so sorry
01:56
You guys realize that when you say @alex
it tags me, right?
I had like 9 notifications
oops
@AlexYoucis it tags me whenever anybody talks to an @alex too
hi @alex
from now on it's alexander and alexy from me
@AlexYoucis HAHAHA. Poor thing! =D
Now I want a cookie
01:58
I will, from now on, be known as The Shadow.
that's a pretty good icon for it, @TheShadow.
Hello!
@LitheOhm hi there.
Anyone up for a statistics question?
02:00
@alex well one of you has to become TheOtherAlex, I say. now fight over it.
hello both :)
@Mike The Shadow rejects your premise.
easy to choose
Whats the statistics of a statistics question happening in chat./math?
highest rep gets alex
done
02:01
I can answer that
Highest latitude Alex is now Alex
(and The Shadow)
my name is Alexander.
:)
So is mine.
gruber wins the alex title lol
The Shadow claims domain over all variants of Alex.
02:02
I get the concept about mean growth annual rate, but I am curious -- when calculating it, after the rates are multiplied, why does one raise it to the 1/n power? Where n is the number of elements
by 700 points
Why not multiply the values then leave it as is?
@LitheOhm because you're looking for the mean growth rate.
what if it were just two rates?
they'd be multiplied, then the square root of that number would be it
@LitheOhm what would it be if you just multiplied the values and left it as is?
say i grow 200% one year and 300% the next
02:05
stabilizers are also called isotropy grops
fancy
3 times height first year, then 12 times original height at the end of the second year
even when Alex is spelled @L3X @AlexYoucis
sqrt 12 ~ 3.something
AKA Меня зовут Александр
@usukidoll Especially then.
02:06
@Mike Hey, Lang is not that bad, but certainly not a choice for a first read.
@LitheOhm what would the mean rate be, by year?
I'm tired...proofs take so much of my energy
@Pedro I like it more than I let on. It's a great reference.
Now I need to go back to solving this problem I'm definitely being really dumb about.
2 radical 3 over both years
i think it would be radical 6.
02:09
I'm glad this chatroom is getting more radical n
now we just need some bodacious dudes in here.
but from year 0 to year 1 you've grown 2, and from year 1 to year 2 you've grown 3
@Mike What is the problem?
I read it as +200% and +300%, so tripling and then quadrupling. k, I can follow radical six instead. My misunderstanding
so if we were to just multiply and leave it as is it would just be 6
but 6 certainly can't be the average rate of growth because it's larger than both growth rates we observed
ugh I wish there were hunger control shakes... I just ate a decent lunch and I'm still hungry -___-, but I want to practice my proof things geez
02:11
true
@Pedro If $x$ has maximal order in a finite abelian group, prove $\langle x \rangle$ is a direct summons of the group
to reconsile this, we take the square root.
what about incorporating negative growth rates? That doesn't apply to things like +/- interest rates
nibbles on Alex I don't know which one sorry T__T
+5.5%, -1.1%, stuff of that nature. Across multiple years
02:13
@LitheOhm the growth rate associated with 5.5%, for example, is 1+0.055
this should come from first principles, nothing fancy.
for -1.1% it would be 1-0.011
@Mike You mean direct summand, yes?
I have five values in this problem (all percentages): 5.5, 1.1, -3.5, -1.1, and 1.8.
Right. I understand the how through to the solution. I also understand the why, all the way until the part where we use the radical
The worth of something after the interest/depreciation value is applied. Er, rather, growth rates are the interest/depreciation occurring
@LitheOhm what it really comes down to is that you're taking a type of average, a geometric mean. you're used to taking averages by summing and then dividing by n=number of terms. with rates, you're multiplying and then square rooting by # of terms.
02:16
Why is a geometric mean more applicable here? It's the multiplication/compounding, isn't it?
@LitheOhm yeah.
it's because you're working with rates, which are inherantly something you multiply by.
right
that makes sense
>.< feelin' kinda silly now lol. I did read that section.
My first statistics course :)
You know what really grinds my gears: when female athletes are celebrated because of their beauty rather than their skills.
@AlexanderGruber thank you
@LitheOhm stats is actually pretty neat :)
there's a lot of BS they give you in elementary statistics courses which makes it not as fun, but i'm glad you're liking what you see so far
@LitheOhm no problem
02:18
I'm quite excited by it. My bookshelf has a bunch of math books on it, Taleb, Mlodinow, etc.
I actually dropped in here because some students asked for help on the school discussion forum and I could explain all the stuff, save for the geometric mean part lol. Still a bit green I suppose, but meh :)
@LitheOhm one example that should help should be this, if you triple your costs, that should be the opposite of third-ing your costs
but if you were to use the arithmetic mean you'd get (3+1/3)/2 = like 1.5. you'd expect the average of tripling and thirding to be 1.
@Pedro yes
ADR
ADR
hi, is this inequality true $\int f(t)g(t) < g(t)$ if $g(t)$ is monotonic non decreasing and $g(t)$ tends to zero?
hm
It seems odd that suddenly the exponent becomes the object of division instead of the base number.
02:23
@LitheOhm you mean when going from arithmetic to geometric means?
ughhhhhhh! I'm trying to prove $\frac{n(n+1)(2n+1)}{6}$
@PedroTamaroff I'm not so sure there's much difference. Being beautiful and being skilled at sports (or math) aren't so different. It requires a combination of the right environment, the right talent, and consistent practice.
well, if you're looking for the real reason, it's this, but it requires a bit more thought.
my target result should be $\frac{2n^3+3n^3+n}{6}$, not $\frac{n^4+2n^3+n^2+4n+4}{4}$
02:24
@Mike Finished watching ze movie.
I'm there
Well, present lol. Not there yet. Although I hope to be by Summer
@Pedro I could understand if you mean ONLY because they're beautiful
@LitheOhm the basic idea is that you can use exp/log to convert between addition and multiplication
--__--
02:26
you'll catch on to it after some rhumination
like how does the 6 come into play
Thank you for your help. Am off
@user4140 you know induction?
02:31
@Pedro Did you solve my problem?
@Mike The one about maximal order?
Yah. I'm being dumb.
sigh always the Fibonacci whatever it is proof. there are so many of them
$G$ is an abelian group and $x$ has maximal order in $G$, then $\langle x\rangle$ is what?
@Pedro Uh... The cyclic subgroup generated by $x$.
02:33
Don't be silly, you wanted to prove $\langle x\rangle$ is something.
A "summons" you said, but I think you meant summand.
A direct summand, right.
What do you mean by that?
yo dawg, I recognize that one
does anybody understand how the hell to write a $\xi$ by hand.
Gimme a second Pedro
02:35
@Mike Want a hint, or do you want me to torment you?
@AlexanderGruber Know how to write zeta?
@KarlKronenfeld sort of.
@AlexanderGruber I used to suck at it, but now my $\xi$ is quite decent.
First, you have to be good at writing $\zeta$s
Then $\xi$ is $\zeta$ with a twist. BAD-UM TS.
@usukidoll i think I do
zeta is just a C with some extras
02:38
why?
@user4140 I'm trying to prove $\frac{n(n+1)(2n+1)}{6}$
ok
what did you do?
@usukidoll You're trying to prove what about that?
Hold up
That it is an integer?
02:40
its the sum of the first n square I would guess
@Pedro I cn't write greek letters for shit
@KevinDriscoll lol, me neither, only phi
I have to use the proposition as a guideline to prove that argh that sum thing $k^2 = \frac{n(n+1)(2n+1)}{6}$
neither could I
$\vartheta$ is hard for me to get right..
I wanna draw the proposition and what I did
02:41
ok, so you can assume it is true for n. Using that you want to prove it is true for n+1.
@Karl da fuq is that!?
@Karl I'd like a (minor) hint. I'm slow to respond since I'm working
@Mike You want to be tormented?! Great!
:P
02:44
Hey, I'm dumb here, so I'll take the pain.
Bye y'all.
@Karl So can I get the hint along with my torment? I've made no progress.

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