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4:00 PM
Thus the statement becomes:
$$\int_a^u [\mathop{Op}_{i \in S}m_i](x) dx = \int_{g(a)}^{g(u)} [\mathop{Op}_{i \in S}m_i]\circ g](x) g'(x)dx = h \left(\int_a^u [\mathop{Op}_{i \in S}m_i](x) dx\right)+ K(x)\mid_{g(a)}^{g(u)}$$
 
4:12 PM
ok nvm, the final equality can be a nontrivial symmetry, thus even knowing which $g(x)$ remains unchanged under $\mathop{\large{Op}}_{i \in S}m_i$, it is insufficient to determine $K$ in terms of them
Now to try something simpler...
$$\int e^{f(x)} dx = \int \frac{e^{u}}{[f' \circ f^{\leftarrow}(u)]}du = h \left(\int e^{f(x)} dx\right)+\int k(x) dx$$
 
One generic integral form you could play with would be $F(s)=\int_0^\infty e^{-st} f(x)\,dx$ i.e. Laplace transform
 
hmm... it's been 10 hours and still not much progress in trying to find the rules that governs the rules such that integrating by symmetry is obtained. The Lapalce transform might be an example that is tractable enough to handle and figure whether I am missing something obvious...
Anyway, apologies for this LONG workings here, I am going to continue in mathworks so the chat will not be clogged
 
Hi all. How is this blue part obtained?
 
@Secret Thank you so much, secret! I did not see that. Sorry.
 
np
 
4:35 PM
@MatheinBoulomenos mathein ! text me when you see this :D
hey faust
@Faust faust i said hi -.-
 
@KasmirKhaan lol hello
 
haha
 
@LeylaAlkan i dont understand the question did you try expaning out the bottom
 
I didnt understand either. It makes use of binomials but I didnt understand how
 
do you know the binomial expansion?
 
4:42 PM
Yes
 
its defintly not obvious but you try it for like n =3 and just do the first couple terms and you get that the bottom gives you the middle which is easily seen to be the top
the first term is the product of all the x
w8
yes
 
I also didnt understand the uses of $b_n$'s there
 
the line $1b_1 + 2b_2 +... +(k-n) b_{n-k} =k$ is really important too
its a way of making sure you dont over count
do you know where the $ n-b_1 $ come form in the second term?
sorry mayb im not a good person to explain this
i high suggest the example when n=2 and n=3 and just try expaning out the first few terms it makes it easier to see wheere the terms are coming from
 
Oh, okay. Thank you @Faust , I'll try that way too
 
@KasmirKhaan what's up
 
4:51 PM
yeeey :D
mathein finally is back :D
 
rember your looking for the kth term so you need to pick which term your looking for ^^
 
How are you ? long time no see
 
@MatheinBoulomenos ohai o
 
@Faust ohaio gozaimasu
 
it has been dark days since mathein left us -__-
 
4:51 PM
@MatheinBoulomenos guten tag
 
@LeakyNun guten Tag
 
i dont see those crazy things he writes that i dont understand
 
@KasmirKhaan I'm doing fine, thanks. I'm learning French atm
 
@MatheinBoulomenos nihon deska?
 
nice haha
pourquoi voulez vous apprendre le français ?
 
4:53 PM
@Faust doitsu-jin des
 
going to study in france ? ._.
 
@KasmirKhaan Knowing French is useful in the fields I want to work in
 
and vous is a term of respect
 
yeah, I want to do an Erasmus in france
 
what is that ? :D
 
4:53 PM
you can say tu to me :D
 
haha
 
@MatheinBoulomenos du
 
du hasst mich , du hass mich gefragt
 
canada?
 
Erasmus is an European university student exchange program
you can apply for a scholarship and you don't have to pay university fees for a semester if you get accepted
(and you get a bit of money)
 
4:55 PM
Good luck mathein :D
 
thanks
(you could do that if you want to go to Germany for a semester e.g.)
 
did i misunderstand or was your answer to do you speak japanese, where are you?
 
Ah, I didn't understand the question, sorry
 
haha ok =)
 
I thought you were asking if I was Japanese
 
4:56 PM
thats nihon jin deska i think
 
i thought u were german
 
oh ok ok we on the same page now
 
that's what I said "doitsu-jin des"
 
4:58 PM
yea im just alittle slow ^^
 
Hey guys!
 
morning
 
@Mathein huh, that sounds nifty! When would you likely be doing it?
@Faust how's it going?
 
still sick but not as bad today
 
@Daminark planning to do it in 1.5 years which will be after the first semester in my masters
 
5:03 PM
I've got a quick poll: how would you interpret $p$ and $i$ in the following passage?: $\oplus R$ is a free $R$-module. Let $p$ and $i$ be canonical homomorphisms $p:\oplus R\to R$, and $i:R\to \oplus R$ such that $pi$ is the identity on $R$.
 
need to do some hw
@MatheinBoulomenos what part of algebra u doing ur masters on
 
I want to finish the one-year algebraic geometry course first
@Faust dunno
 
@rschwieb so it feels like $p$ is the projection and $i$ is an inclusion
 
algebraic number theory, probably
 
hrd
 
5:04 PM
I'm aware of the family of canonical homomorphisms $p_j:\oplus R\to R$ on the j'th coordinate, and I'm aware of the canonical injections $i_j$ from $R$ to the $j$th coordinate, but I'm not sure what this guy is saying.
 
Yeah that's why I'm a bit concerned
 
@rschwieb oh hello! good to see a ring theorist around here
 
@MatheinBoulomenos good luck ur braver man then i
 
@Dami That was my first impression, but projecting onto a particular coordinate does not seem very canonical to me.
@MatheinBoulomenos Hiya!
 
@rschwieb I guess whoever wrote that didn't think much and fixed a coordinate in his head
 
5:06 PM
Later on, the author draws the conclusion that $ip$ is the identity on $R$, so that they're actually isomoprhisms.
 
Is there a book this comes up in? Maybe we can try to read around that passage and see if context gives a clue? I'm guessing it's laziness but otherwise...
 
sanity check $\varphi(n^{k+1}) = n^k \varphi(n)$
 
Whaaaaaa?
 
@MatheinBoulomenos Well, it's a paper from a 1972 Proceedings of the AMS
 
@Leaky assuming this is a homomorphism with stuff denoted multiplicatively, yeah
 
5:07 PM
@Daminark totient function...
 
@LeakyNun looks sane...
 
Okay
 
I'm fairly certain this proof has more than one mistake. It's kind of surprising tho
 
@LeakyNun you tried proving it?
 
The only canonical homomorphisms for the free modules I'm aware of are the family of $2*j$ projections and injections. And I don't think his proof works with those.
 
5:09 PM
@LeakyNun wlog $n$ is a power of a prime and then there's a handy formula for $\phi(p^m)$
 
@Faust did that in my head
@MatheinBoulomenos used that formula
there's also a formula generally, based on the fact that totient is [weakly] multiplicative
 
yeah, sure
 
Yeah that's right I think
Sanity check: tentatively passed as long as I'm sane
 
well thats a negative
 
Fricc
 
5:11 PM
:p
 
But yeah I mean, so I'm coming across this sorta backwards
You have a formula $\phi(n) = n\prod_{p\mid n} (1-\frac{1}{p})$
 
Hi @AlessandroCodenotti
 
@Daminark yes, did that in my head
 
But $n$ and $n^k$ have the same prime factors
 
5:12 PM
I just thought that my formula was too good to be true
because I never saw it
 
How is it going with logic? @Mathei
 
and I do come from ppcg, if you forgot
 
didn't much logic lately
trying to learn French atm
 
@AlessandroCodenotti come va logico?
 
what field $K$ would be with dimension 2 over $\Bbb F_2$ ?
 
5:13 PM
I want to skip the first course so I need to do well on the placement test for the language courses
 
@Liad $\Bbb F_2[X^2+X+1]$
 
@Liad $\Bbb F_4$
 
yea what i thought
 
@MatheinBoulomenos that's very helpful lol
 
but how do i build a field with 4 elements :P
 
5:14 PM
@Liad you build it on $\Bbb V_4$
there aren't many options
@Daminark thanks for repeating the above argument :P
 
adjoin a root of an irreducible polynomial of degree $2$ over $\Bbb F_2$ (there's only one)
 
@Daminark aww i was just joking
 
Rip
 
or since there are only four elements you can even try to fill in the addition and multiplication tables by trial and error, that's not too bad
 
@MatheinBoulomenos hmm
 
5:16 PM
I did that once in my first semester
it's doable
 
by "hmm" I mean
1 min ago, by Leaky Nun
@Daminark thanks for repeating the above argument :P
 
Wut
 
Zee
Does a contravarient tensor act on the basis representation of a covectors?
 
@MatheinBoulomenos French pronunciation is the opposite of logical :P
 
@AlessandroCodenotti I agree
 
5:17 PM
@AlessandroCodenotti well it's quite consistent
 
(though English is quite bad, too)
 
why the dimension will be $2$ ?
 
@Liad you can construct a finite field of order $p^n$ by quotienting $\Bbb F_p[X]$ by a monic irreducible polynomial of degree $n$
 
i.e. i can read a french word that I don't know and be able to pronounce it correctly
@Liad because basis is 1 and X
 
@ÍgjøgnumMeg nice thanks.
@LeakyNun ok.
 
5:19 PM
$\int_{0}^{1} s^n(1-s)^n ds =...?$
 
@MatheinBoulomenos True. I like German because there are 2000 rules in the language, but at least they're followed (usually)
 
@Liad In your case you have $\Bbb F_2[X]/(X^2 + X + 1)$ (you know that $X^2 + X + 1$ is irreducible because it's quadratic and it has no roots in $\Bbb F_2$)
 
In mathematics, the beta function, also called the Euler integral of the first kind, is a special function defined by B ( x , y ) = ∫ 0 1 t x − 1 ( 1 − t ) y − 1 d t {...
 
@ÍgjøgnumMeg yea i know it is irreducible i have to write out the details of why this implies that $\Bbb F_2 /(x \ ^ 2 +x +1) $ will be a field of order $4$
 
@Lozansky $\frac{(n-1)!^2}{(2n-1)!}$
 
5:21 PM
@Liad Remember that it's $\Bbb F_2[X]/(X^2 + X + 1)$; $(X^2 + X + 1)$ is not an ideal of $\Bbb F_2$ (whose ideal structure is very simple)
 
yea right it is $\Bbb F_2[X]$ , typo.
 
Zee
Does d/dx1 tensor d/dx2 = ( a dx1 ( d/dx1 ) + b dx2(d/dx1) ) ((c dx1 (d/dx2 ) + d dx2 (d/dx2)) ?
 
@LeakyNun Wouldnt it be $n!^2 /(2n+1)!$?
 
@ÍgjøgnumMeg i cant see why the order of the quotient will be 4 , can you help?
 
@Lozansky maybe
@Liad because $X^2 = -X-1$
so you have exactly 4 distinct elements: $0X+0$, $0X+1$, $1X+0$, $1X+1$
 
5:26 PM
@Liad You have a relation $X^2 + X + 1 = 0$ in your new field, which will help you compute a table
 
or more commonly, $0$, $1$, $X$, $X+1$
 
got it. thank you both :)
 
Np np
@Liad Keith Conrad's notes on this stuff are excellent: math.uconn.edu/~kconrad/blurbs/galoistheory/finitefields.pdf
 
Zee
I can’t tell if you can’t help me or you don’t want to ...
 
@Zee formatting what you've written in Mathjax would be a good start (I cannot help you btw, this is a general tip)
absolves self of responsibility
 
5:33 PM
what a name, lol
 
Anyone knows how is the operator's norm on the space of homomorphism between $\Bbb R^d$ and $\Bbb R^d$ defined?
 
Zee
Ya I know , I dont have the motivation to learn math Jax frankly , computers are beneath me
 
i.e. on the space $Hom(/Bbb R^d, /Bbb R^d)$
 
lol
I feel like you may be misusing the phrase "beneath me"
 
@Zee ...as you write that on an internet chat, presumably using a computer.
 
Zee
5:36 PM
Ya, but atleast it’s mindless , learning math Jax feels like learning math without any of the honor
 
riiiiiiight
 
The problem is that you're in a chatroom where most people use mathjax to communicate mathematics in a readable way
 
Zee
I wouldn’t be here if I had people to chat with about math IRL
 
So basically you can't be arsed to format your math in a way that's readable but still expect people to do the labor of figuring out what you're asking in order to get help.
 
Zee
ya? Am willing to do it myself
 
5:38 PM
wtf hahaha
 
Zee
But that also has to do with the fact that I can’t compile math Jax either :p
 
uh
link in the room desc?
 
math jax rocks
 
So learn? It takes like 5 minutes to learn what you need to know
 
Zee
Ya I been there like 20 times
 
5:39 PM
you dont even need to learn
you just have second tab with all the commands
 
...that doesn't seem like the most productive comparison.
 
Hi @Semiclassical
 
how are you ?
 
Zee
Mathematics is about the Honor of the mind , and computers take that away
We should ban them
 
5:41 PM
jesus you're talking crap
 
Zee
You think Grothendieck would learn math Jax ?
 
non sequitor
@Tanuj alright. about to head out for a bit
 
@Semiclassical I've got a question to ask from Magnetism , now be a good time ?
 
If it's quick.
 
5:43 PM
@Semiclassical Yea sure.
 
Zee
Mathematics should be done as if there is a possibility a mathematician would need to recreate all of it on a deserted Island , computers won’t help with that at all
 
you're literally asking for help on a computer
 
if we didnt have computers
 
plus there's the implied comparison between you and Grothendieck, which uh
 
aah nvm
 
5:44 PM
@Semiclassical Let $B$ and $\epsilon$ denote induction to magnetic field and energy density at mid point of a long solenoid carrying current $i$ . Then I have to tell which of the given graphs correctly represent relation between $B$ and $\epsilon$.
 
yeeeeah
 
dont get baited :p
 
Zee
Me and Grothendieck are made from the same clothes
We are blood brothers
 
Insofar as you both presumably wore clothes composed of matter and had blood pumping through yours veins, sure.
 
5:46 PM
@Semiclassical Okay , so $B=\mu n i$ , for a solenoid , $n$ being the number of turns per unit length
 
Zee
It’s just a matter of time , you’ll see
 
@Zee lol wat this is actually crazy
 
perhaps u mean you are cut form the same cloth?
 
Zee
Ya
 
I think he means that he and grothendieck were actually golems made from clothing
 
Zee
5:49 PM
Am not as smart as he is , but we have the same soul
 
brother from another mother
 
reminds me diablo II could make a golem outa anything
leather shirt = golem
 
Zee
Yeaaa boy
 
well
at least this is interesting
 
Grothendeckard Cain
 
5:51 PM
Guys, let $\mathbb R$ have the finite-complement topology. I want to show that open sets of $\mathbb R$ can be compact. So let $A\subset\mathbb R$ such that $\mathbb R\setminus A$ is finite. Then $\mathbb R\setminus A$ is compact. Let $\{U_i\}$ be an open covering of $X\setminus A$.
I should somehow use the fact that all $U_i$ have infinite elements, because then I could somehow find a reduction of the open covering, leaving possibly only finitely many points to not be covered, which we can resolve easily. Any idea?
 
@Tanuj Okay. What do you have for the energy density at the center of the solenoid?
 
@Semiclassical idk what you mean .. I know energy density is $B^2/2 \mu $
 
all I meant was that you hadn't actually said what the energy density was.
But now you have :)
So, what are you confused about?
 
@Semiclassical Yea , just use that , right ? I'm stupid !
 
Zee
Dostoyevsky is part of my team too
 
5:54 PM
eh, you were just being a bit silly
 
@Semiclassical I know , I somehow managed to complicate it and left it during my test today
 
ah, dang
 
nvm about my Q btw
 
@Semi so apparently I can't take quantum bc of conflicts and am stuck in a lab class I wasnt planning on taking
Rip me in fizix
 
.......
yeah, that sucks
 
Zee
5:57 PM
You shouldn’t take physics
That’s also against the honor of the mind
 
I won't get the opportunity to do our quantum sequence before I graduate now cuz it's spring-fall
I'm p bummed on it
 
Oh, what a nice bird I see outside, flipping around in the air
 
go play with the bird
 
@Zee tfw learning is against the "honor of the mind"
 
@Eric what are you likely gonna end up doing on the whole? What with the quantum conflict and no harmonic?
 
5:58 PM
tbf I feel like learning accounting might be like that :P
 
Zee
Ya accounting and to a lesser level physics , I mean who cares about this universe ? Maybe I’ll wake up tomorrow in another one and all my physics just changed
2
 
@Daminark take honors waves and complex and then take a history class to fill the hole in my heart where schlag would be
I'm doing enough math between my research assistant schtick and complex so I'm not even mad
 
That's fair yeah
 
Plus my physics Prof is a big adorable nerd so I'm ok with takes waves w him
Shit quantum conflict is a good band name
 
Lmao
 
6:13 PM
Physics gives mathematics a real meaning.
 
Is "orbit type" just terminology for the orbits of a group action under the natural equivalence relation?
 
Zee
Mathematics does not need a real meaning , nor does it have one , it is meaningless , we do it as a game or a form of art, nothing more
 
@ÍgjøgnumMeg thanks ! im reading it now.
 
@EricSilva "Quantum supremacy"
 
@Liad No problem, he has a lot of those papers and they're all very useful math.uconn.edu/~kconrad/blurbs
 
6:17 PM
which is an actual phrase
 
WAT
@Semi ur rocking my world rn
 
Quantum supremacy or "quantum advantage" is the potential ability of quantum computing devices to solve problems that classical computers practically cannot. In computational complexity-theoretic terms, this generally means providing a superpolynomial speedup over the best known or possible classical algorithm. The term was originally popularized by John Preskill but the concept of a quantum computational advantage, specifically for simulating quantum systems, dates back to Yuri Manin's (1980) and Richard Feynman's (1981) proposals of quantum computing. Shor's algorithm for factoring integers,...
 
@ÍgjøgnumMeg seems great! i will use those with my course's book.
 
Sticking quantum in front of anything just makes it sound cooler
Quantum drugs
 
quantum suicide
 
6:20 PM
Oh dude that's good
 
In quantum mechanics, quantum suicide is a thought experiment, originally published independently by Hans Moravec in 1987 and Bruno Marchal in 1988, and independently developed further by Max Tegmark in 1998. It attempts to distinguish between the Copenhagen interpretation of quantum mechanics and the Everett many-worlds interpretation by means of a variation of the Schrödinger's cat thought experiment, from the cat's point of view. Quantum immortality refers to the subjective experience of surviving quantum suicide regardless of the odds. Keith Lynch recalls that Hugh Everett took great delight...
 
Zee
Quantum quantum theory?
 
Second quantization, also referred to as occupation number representation, is a formalism used to describe and analyze quantum many-body systems. In quantum field theory, it is known as canonical quantization, in which the fields (typically as the wave functions of matter) are thought of as field operators, in a manner similar to how the physical quantities (position, momentum, etc.) are thought of as operators in first quantization. The key ideas of this method were introduced in 1927 by Paul Dirac, and were developed, most notably, by Vladimir Fock and Pascual Jordan later. In this approach,...
(that's not quite 'quantum quantum theory' but close enough)
 
This sounds like one of those weird things philosophy majors do where they throw in quantum into things to make stoner arguments
 
hahaha
 
Zee
6:24 PM
I don’t like quantum theory , it makes no sense
 
TIL if something doesn't make sense to you it's not your fault, it's the theory's fault for not making sense
7
 
Good news. I now know what I will be doing 7 months loner than I did previously. I have been offered an extension of my postdoc so I can teach algebra again next term.
 
Congrats!
Gratulerer
lol
 
6:41 PM
thanks
 
6:52 PM
Nice!
 
@TobiasKildetoft Congratulations!
 
@BalarkaSen thanks
 
@TobiasKildetoft I offer you my earnest greeting and best wishes for the following good event that has happened to you
@TobiasKildetoft I, son of my motherly mother, offer you my heartfelt and heartpiercing greet and wish you all the best for an event, extremely happy and good, which you speak of as "good news" has taken place and therefore I must express and utterly utter my utter joy at hearing the well-being of you
 
@TobiasKildetoft congrats!
 
7:02 PM
Sorry I probably shouldn't try to be more verbose than three panels
 
@Balarka where's the accompanying picture?
 
@Daminark There
(For reference, I was making a verbose meme: examples include eg this or this)
 
Lmao
 
reminds me of the old sinfest calligraphy comics
 
Yeah that's close enough
 
7:11 PM
Wow these anti-plagiarism things are scary
 
Zee
@0celo7
Does a contravarient tensor act on the basis representation of a covactor by inserting itself as an argument?
 
7:29 PM
How is the coeff of $\frac {x^n}{n!}$ in $\sum_{n\geq 0} \frac {x^{n+1}}{n!}$ is n?
 
it would be x wouldn't it?
 
The coefficient of the term $x^n$ is $1/(n-1)!$, so the coefficient of $x^n/n!$ is $n!/(n-1)!=n$.
you have to mentally shift the index by 1 so that you're looking at the term $x^{n}$ rather than $x^{n+1}$
@JoeShmo that'd be like saying the coefficient of $x^4$ is $x$ since it's $x^4=x(x^3)$.
 
nope.
the coefficient of x^3 in x^4 is x
silly, yes.
 
No. The coefficient of x^3 in x^4 is zero.
Otherwise you'd have, for instance, that the coefficient of x^2 in ax^2+x^4 is a+x^2
 
true
but it's funnily phrased in that paragraph to begin with
 
7:37 PM
yeah, but I'm not sure how else to say it
 
"the coefficient is x" is my attempt at a joke
 
heh, okay
it's annoying to use n twice for both the summation index and for the outer index
I guess I'd have done the summation over k not over n.
 
k to the rescue
 
if I have a vector field on a surface, restricted to a curve and parallel transported: $X(t) = x(t)r_a + y(t)r_b$ how do I find the angle between $X(t)$ and $X(t+\delta)$. Do I just take the dot product of the two? In some way it's as if the basis is actually x(t), y(t) and the tangent vectors themselves are the components of the vector.
 
I wrote it this way : $\sum_{n\geq 0}\frac {x^{n+1}}{n!}=x\sum_{n\geq 1}\frac {x^{n}}{n!}$ So from here how do you say "the coeff of $x^n$ is $1/(n-1)!$" @Semiclassical
 
7:40 PM
eh, that doesn't really make your life easier.
 
what's the difference between a plane and a subspace?
 
also, the index should still start at n=0 in what you wrote
If you wanted it to start at n=1, you'd need to shift $n\mapsto n-1$ as well
in which case you have $\sum_{n\geq 0}\frac{x^{n+1}}{n!}=\sum_{n\geq 1}\frac{x^n}{(n-1)!}$
 
Ah, okay I got now :)
Thanks @Semiclassical
 

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