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00:00
Whenever I get a functional analysis upvote I think "Cool, soon I'll get a badge". But I already got one.
@KarlKronenfeld Do you remember YACP? I think that's him.
@DanielFischer If $p(0,0) \neq 0$,it cannot be $(p^k)(0,0) = p(0,0)^k=0$. But how does this help?
Yes, @Studentmath, we have the right answer.
Hello professor @TedShifrin
00:01
@evinda Which radical are you talking about? I assumed nilradical, $p \in \operatorname{Rad}(I) \iff \bigl(\exists n\bigr)\bigl(p^n \in I\bigr)$.
@Hippa: Use the spectral theorem, of course.
:) I'm still stuck on the same thing I have been trying to prove for over a month.
@TedShifrin uh ? How ?
Hi @skull
@DanielFischer I haven't seen him do any algebraic geometry though. Turns out he has 16 answers on the subject; I am still unsure.
00:02
Well I managed to excel in the proof of the first part of it, got it really neat. I don't think I will ever manage the second part, though..
It's even more annoying that the first thing he called wrong was actually correct and that it took him ten minutes to be right.
@RudytheReindeer You're on your way to silver.
@TedShifrin The Spec Theorem just tells me that $A=PD^tP$
Diagonalize, take the square root, undiagonalize.
GRR
ME STOOPID
@TedShifrin thanks :-)
00:03
Je ne dis rien :)
@KarlKronenfeld The username is YACP's old user id, and he referred to answers (and comments) of YACP a couple of times on meta.
Anyway I gtg to sleep a bit :) see you in your dreams
Later pal
@DanielFischer I am using this definition:

$$Rad(I)=\{ a \in \mathbb{C}[x,y] | \exists n \in \mathbb{N} \text{ such that } a^n \in I \}$$
@DanielFischer Ah, that makes it more convincing.
Hi @Ted
00:05
@DanielF: I know not the drama of which you speak. Hi @Karl
@DanielFischer So, we are talking about the same definition... :)
@evinda Okay, that's the nilradical all right. So we see that if $p\notin \langle x,y\rangle$, then $p^n \notin \langle x,y\rangle$ for all $n\in \mathbb{N}$, and hence a fortiori $p^n \notin \langle x^5,y^3\rangle$.
@TedShifrin Am I speaking of drama? Is this a dagger which I see before me?
@DanielFischer How do we see that if $p\notin \langle x,y\rangle$, then $p^n \notin \langle x,y\rangle$ for all $n\in \mathbb{N}$? :/
Anyhow, I am off, think I will let it be and move to regular works. G'night @Ted @Skull
Bye my friend :)
00:09
@evinda Because $q\in \langle x,y\rangle \iff q(0,0) = 0$.
Given the 9 field axioms, if it's not mentioned that for the identities $1$ and $0$ that $1 \neq 0$, is there a way to prove that $1 \neq 0$?
night @Studentmath
@user130018 No, $\{0\}$ satisfies everything except $1\neq 0$.
So that example doesn't work, but I still don't see why the minimal primes should cover all of the zero divisors @Karl. You're claiming a reduced Noetherian ring has no embedded primes?
@MikeMiller yes, the set of zero divisors is exactly the union of the minimal prime ideals for any Noetherian ring.
00:15
I must have proved this before, then..
So, you're having difficulty showing which inclusion?
Everything's contained in a minimal one. Zorn's easily gives that every element is contained in some prime minimal with respect to containing it.
@DanielFischer So, we use the fact that: $q \notin <x,y> \text{ iff } q(0,0) \neq 0$ and we have the following:

We suppose that $p\notin \langle x,y\rangle$. Then, $p(0,0) \neq 0 \Rightarrow p^n(0,0) \neq 0 \Rightarrow p^n \notin <x,y> \Rightarrow p^n \notin <x^5,y^3> \Rightarrow p^n \notin I \Rightarrow <x,y> \notin Rad(I)$

So, we have proven that $p \notin \langle x,y\rangle \implies p \notin \operatorname{Rad}(I)$.

$\operatorname{Rad}(I) \subset \langle x,y\rangle$

Right? Or have I understood it wrong?
@evinda In your third line, you have a nonsensical $\langle x,y\rangle \notin \operatorname{Rad}(I)$. If that is, as I suppose, just a typing slip, no problem. Everything else is right.
yo @DanielFischer. I am reading about $\wp$ and you seem to have some familiarity.
00:33
Quick question about free product of groups. I am reading through some notes and came across this: $\langle a, b, c \, \vert \, a^3, b^2, c^2, aca^{-1}c^{-1} \rangle$. Is this isomorphic to this: $(\mathbb Z_3 \oplus \mathbb Z_2 ) * \mathbb Z_2$?
Nevermind, @Karl.
in a thread it was said $\wp$ has "order 2" meaning that the map $\wp:\Bbb C/\Lambda\to\widehat{\Bbb C}$ is $2$-to-$1$ if we count images with "multiplicity" (e.g. the value $f(a)$ has multiplicity $2$ if $f'(a)=0$, or equivalently $f(z)=(z-a)^2\times$ a function holomorphic at $a$). I have never heard "order" used this way (I've only seen it used for order of poles or roots). do you have a reference for this more general usage? @DanielFischer
also, I'd like to know why $\wp$ is order $2$, and $\wp$ is order $3$, and how to calculate this sort of thing in general (if there are any standard techniques or theorems). presumably this takes me into Riemann surface theory, so I may have to wait to learn about this.
@DanielFischer Thank you very much for your help!!!!!!!!!!!!!!!!! :)
@anon Count the poles, that's usually the simplest way to determine the order.
@anon Are you there?
00:36
@DanielFischer so is there a theorem that says something along the lines of "if the number of poles is m, then it's m-to-1 everywhere counted with multiplicity"?
@PedroTamaroff no
@RobertCardona yes
I've read this recently: if $f(z)=(z-a)^n g(z)$ with $g(a)\neq 0$, then locally (near $a$) $f(z)$ behaves like $z\to z^n$ near $0$. So each point that is not $a$ has fiber of cardinality $n$.
@PedroTamaroff hmm, that's interesting. so if there is only one pole in the torus C/Lambda of order 2, then it would make sense it's 2-to-1 in a nbhd of that pole. not sure how to patch it up to the whole of the torus though.
@anon $\mathbb{C}/\Lambda$ is a compact Riemann surface, so every non-constant meromorphic $f\colon \mathbb{C}/\Lambda \to \widehat{\mathbb{C}}$ is a (branched) covering, hence all fibres of regular values have the same cardinality, and if you count multiplicities, that also holds for the fibres of non-regular values.
good, now I know what I need to go read
00:40
it's a very nice riemann surface at that
my second favorite, even
2
@MikeMiller DAWG
@anon For rational functions, Ahlfors defines the order as the number of times each value is attained on page 31, for elliptic functions on page 271.
@anon Although, for elliptic functions, we have a more direct way, the argument principle. Just compute $$\frac{1}{2\pi i}\int_{\partial P}\frac{f'(z)}{f(z)-w}\,dz,$$ where $P$ is a period-parallelogram such that neither $w$ nor $\infty$ is attained on $\partial P$. Since $\frac{f'(z)}{f(z)-w}$ is elliptic, the integral is $0$.
And of course, by general theory, it's the number of times $w$ is attained in $P$ minus the number of times $\infty$ is attained in $P$.
ah, the sides cancel out, yes
what's w?
@anon Any complex value. Shows that all $w\in \mathbb{C}$ are attained the same number of times as $\infty$.
ah
so number of zeros equals number of poles
00:49
Yes. And number of zeros of $f-w$.
that shows the cardinality of fibers are uniform, but how would that help us compute the # of poles (with multiplicity) if we don't already know the # of 0s?
or were we just computing the integral to do the first thing
@anon Typically, you know the poles (if anything).
ah, duh, I was mixing up zeros and poles
The integral shows that the $n$ in "$f$ is $n$-to-one" is well-defined.
@anon Or you were standing upside down.
BADUM TSS
00:54
Aka on your head
@anon a problem!
i need you to check a solution
actually
separability again
@PedroTamaroff ^^
field theory!
@SwapnilTripathi Yes.
I need to show that a separable extension of a separable extension is separable
OK, where are you stuck?
When the characteristic of base field is 0, i did that part.
when char = p. This is the part
where i get stuck.
00:59
let's see, I remember thinking about separability over the summer
sorry to hear that, @anon
@SwapnilTripathi do you get to assume both are finite? I remember the infinite extension theory being daunting.
@MikeMiller heh
that joke has two meanings
@anon: I don't know. That's the statement my teacher gave me. If it's that bad, i don't think we would be needing that
I can't figure out the second meaning
I guess I do. ;)
01:04
@MikeMiller depends if you're talking about separability in math, or separating in romantic relationships
the latter being how a nonmathematician would read it
ehhh
I think they'd raise an eyebrow at the word separability
but fair
@anon I'll write what I've done.
wait a minute
over 9,000 hours later...
??
Was that for me?
wait a "minute"?
01:11
yes. I was writing the proof on a paper
*am
@Swapnil try to prove a tower of simple separable extensions is separable
then see if you can use that as a lemma
@anon
*Suppose, if possible, k(x) is not separable.
@Sarah I see you have not been coming to chat, no emails from you either, lol. Good luck.
@AlexanderGruber That pic looks stupid, lol.
@skullpatrol 9,000 hours too soon? :P
01:21
@Alizter Not sleeping yet? I just woke up after sleeping 8 hours.
You just don't understand my sense of humor, @Jasper. ;)
hahaha. @JasperLoy
I think the picture is very funny.
@AlexanderGruber I hate all things Chinese, like I said before. It's so sad to be born one this life.
@anon Where art thou? I'm desperately waiting for your opinion on the proof.
01:22
He's gone
@JasperLoy Good thing I'm French :p
thanks @Mike.
OK, at least there are many Chinese math journals, lol.
@skullpatrol: Does that to me all the time. :P
@SwapnilTripathi just eyeballing it, what is ln f(x) in char p?
I think Chinese will become a major math language soon. It already is one of the options in some grad schools.
01:24
I hope so
I'd never run out of letters to name things
I don't know!! That's all that came to my mind when I saw f'h+fh'=0 @anon
I am not going to touch Chinese or Russian, hopefully. I only want the Latin script, lol.
Haha. So it seems it is again a dead-end!
Anyone used Assimil here before to learn a language?
@JasperLoy is that like duolingo for borgs?
01:26
@AlexanderGruber Duolingo is a computer program. Assimil is a real book with CDs, lol. I have been reading reviews of over 10 different lang programs the past few weeks, still deciding on the best for myself.
@anon I think it will be all if we show the finite case. :D
Hey, is there a good way to show that a ring is a PID in general?
well, if it's noeth, show (a,b)=(c)
show that it's a dimension 1 UFD?
Not Noetherian.
01:30
Could someone please explain to me why equation (3) is true? :S
I don't understand :(
@Fargle What is your ring?
D is domain and R is range
@Fargle you mean you don't know a priori that it's noeth :-)
anon got cutesy before me
$\{\frac{a}{p^n}|a \in \Bbb Z\}$
p prime
01:31
@anon is actually a famous actor, lol.
@anon And yeah, anon, you're right.
localizations of PIDs are PIDs
@Fargle That's a localization.
NIC EJOB PEDRO
YA GOT IT
01:32
I prefer globalisation to localisation, makes the world better.
if you say so...
@PedroTamaroff Right, but I'd like to prove it directly. Do I just have to show that every ideal is principal manually, I guess?
@Fargle hands-on approach: split into two cases: the ideal has elements with denominator exponent bounded, or not bounded
@anon I think we could use f(x)h(x)=g(x^p) to get some contradiction!! I have that gut feeling.
@Fargle yes
01:34
@Fargle If you want to prove it directly, i.e. you're not willing to use any theorems... then yes, what else can you do?
:'(
Hi pal
@PedroTamaroff Please check my proof above and let me know if you can find a contradiction considering the statement f(x)h(x)=g(x^p) , please?
Hi @twink how is life?
hi pal
01:37
:-)
@JasperLoy life sucks
@MikeMiller Yeah. Ask a stupid question, I guess, haha.
@Twink If you wanna die, talk to me first.
Can someone please explain to me (3)?
lol I don't wanna die
I just want to know why (3) holds
01:38
Have you been there @JasperLoy?
@Twink that should be a song.
@skullpatrol I have wanted to die many times, lol.
It's sad what they did to Alan Turing.
@AlexanderGruber lol
@JasperLoy yes it's sad
@Twink Some naughty person, lol.
@JasperLoy lol
01:43
I guess nobody wants to help me with my doubt :(
@SwapnilTripathi Are you from India?
Yes. @JasperLoy
2 messages moved to Temp room
now they're (removed)
@anon can you please help me? :(
dunno what D and R stand for
01:45
D is domain and R range
$\lambda \in \Bbb C$ is such that $\Im \lambda >0$
02:10
@alizter Are you sleeping?
02:40
@JasperLoy now I wanna die
:(
02:58
@robjohn Ohhh most helpful , I see :) I changed the accepted answer to the most helpful :)
 
3 hours later…
06:00
When is a Frobenius endomorphism an automorphism? I see when it's injective, but not when it is surjective.
 
4 hours later…
09:37
is there anyone here?
@HamedGasemi yes
@DanielFischer can answer most of your doubts ;P
10:01
hi
10:32
can i just ask why this rafflearnold guy has 10000 downvotes? does anyone know?
also does he change name? i have seen other guys with 8000 downvotes and stuff that seem to disappear
Why has noone talked today, is the chat inactive?
per day 1.2k today 212
wut
 
1 hour later…
11:56
@DanielFischer That won't be for another while...
Unless people start asking more FA questions.
It may take a bit, but it will happen.
It seems too far in the future, I think you cannot be so sure.
I was going to comment something on the question you just answered about huge font but then I couldn't think of something appropriate so I decided not to.
sdf
sdf
Anyone who can tell me what this terminology means: G is an abelian group of type (4,2,2)? What does the "of type (4,2,2)" mean?
If one knows how, one can scale the images a bit down so that they have more fitting size. But I don't remember how, @RudytheReindeer.
I got a chemistry problem that I'm stuck but I don't think it's chemistry.SE worth. Can someone help me here?
12:38
Okay. Wolfram makes no sense.
I am pretty sure wolframalpha.com/input/… goes to 0, if not to positive infinity.
The latter term is obviously smaller than the first term.
@Studentmath Since in the latter term you have a smaller denominator, it is not obvious. Apparently, it is larger.
12:57
hi
hello
good morning
hi @robjohn
is now a good time to discuss by the probabilities we were talking about are not always monotonic?
@dorothy since the conditions on which they are based are monotonic, the probabilities should be monotonic. My guess is that the variance for large numbers makes it easy for the results to become skewed.
13:15
I am not sure I follow your reasoning... sorry. We are asking P(A_i| A_j = 0 for all j < i). So it is quite possible that if two previous A_i's are 0 then the next one is unlikely to be 0 but if three previous A_i's are zero then the next one is very likely to be zero isn't it?
and vice versa
what is the condition that you are saying is monotonic?
and when you say variance are you talking about random sampling? My results are from exhaustive enumeration
n =4 is the simplest case that shows the non-monotonicity
hi. I have a question regarding "almost everywhere." If it is not the case that property $P$ holds a.e., can I conclude that $P$ fails on a set of positive measure (hence a measurable set) or can I simply conclude that $P$ fails on a set with interior measure greater than $0$ (which need not be measurable)?
Hello everyone! Not supposed to spam you, but I am looking for an efficient way of computing a certain product of $3$ matrices as stated here. Thank you very much for your consideration!
@dorothy You're asking that given the $n$ previous dot products are $0$, what is the probability that the next dot product is $0$, right?
@robjohn yes.. but not n as that is the length of the vector
@robjohn so given the $i$ previous dot products are $0$, what is the prob that the next dot product is $0$?
13:19
if $i=n$ then the answer is 1
@dorothy if $i=n-1$, then the answer is $1$
I am not sure that is right
the first dot product is with the unrotated vector
oh..oops
my mistake
you are right.. n-1
@Daniel ach so..
@robjohn actually.. take n = 1011. n =4 . Now consider the inner products with 1011, 1101, 1110. Why does this imply the inner product with 0111 is zero?
@robjohn this isn't the important question in any case :)
One more T.M. Sow clone here: DV & CV, if you don't mind.
2
13:30
@dorothy Assuming 0=-1 in the sum, the transitions are 1001, 1100, 0110. This means that the elements of the original vector must be 0101 or 1010.
@robjohn You are up early, lol.
@dorothy so it seems that there is no vector whose inner product with all three of those is 0
@robjohn ah yes.. my example was bad. (And 0 = -1, sorry about that)
@dorothy no problem... I've been using binary, but I use 0=1 and 1=-1 so that xor works
I have to go afk for a bit bbl
13:36
@robjohn bye
back
13:54
Any one interested in a problem on perfect fields?
Nothing is perfect, except imperfection.
@JasperLoy: woh! That was a bit too much philosophical.
:P
@SwapnilTripathi Well, you just need to get used to me, lol. I am in this world, but I am not of this world, lol.
@JasperLoy Coming here daily would do the trick!
14:19
Could someone clarify what they mean by here? 'The derivative of a 3-by-3 determinant is the sum of three 3-by-3 determinants obtained by differentiating the first,second,and third rows,respectively.'
A={a,b,c}{d,e,f}{g,h,i}. Then A'=det({a',b',c'){d,e,f}{g,h,i})+det({a,b,c){d',e',f'}{g,h,i})+det({a,b,c){d,e,f‌​}{g',h',i'})
@link
Ah! @SwapnilTripathi Thank you.
@Link you're welcome! :)
14:37
Hi @Swap !
@Sawarnik Hi,
How have you been?
Fine :)
@SwapnilTripathi What are you drinking in your avatar?
It sure gave me wings. ;)
@Sawarnik
14:39
@SwapnilTripathi Oh :D :D
@SwapnilTripathi Try to press Shift + $\uparrow$
So it went to my latest comment, i guess?
@SwapnilTripathi So??
@N3buchadnezzar Is there any way to delete that post!
14:42
@Integrator Ask a mod.
@Sawarnik I tried to ask robjohn, He said you should ask OP to UN-accept it and wait!
Probably you meant unaccept?
@Sawarnik oops, Edited!
Yeah, so then you must write a comment to the OP.
And sadly, wait.
@Sawarnik I did!
14:45
Bye all, off to dinner!
@N3buchadnezzar What should I do?
@SwapnilTripathi Bye, Save me some!
@Integrator Dunnp
And @Integrator, I don't know what happened. I edited my latest post?
@N3buchadnezzar Dunnp??
@SwapnilTripathi Bye! :)
Too early though :/
14:47
@Integrator Dunno
Don't know.
Hostel timings. Can't complaint! :D @Sawarnik
Hey I am kinda stuck with the local truncation error, it is very trivial but I am just looking for a name for the method used
anyone familiar?
@SwapnilTripathi Oh sure :P
@Integrator About that problem, I wrote up the log sums, but had problems evaluating them
14:49
I have (Yk-Yk-2)/dt expanded as 2yk' etc etc
is this a taylor expansion?
@μακακας Is there any way to ping you without the hard way?
I see the chat all the time no worries
I probably should change my name in the future
@Sawarnik You are familiar with the local truncation error?
@Sawarnik I have the above term which is then expanded as a taylor polynomial but one of the coefficients makes no sense whatsoever
@μακακας Not at all, sorry :(
@μακακας :D
@Integrator I got the following
$$ \frac{f(n)}{f(r)f(n-r)} = \exp\left( \sum_{k=r+1}^n k \log k - \sum_{k=1}^{n-r} k \log k \right) $$
15:13
@PedroTamaroff Nope. Prof's teaching basics of algebraic topology. But I don't like it.
It's full of computations.
@N3buchadnezzar What about modulus ?
Hamis.
@Integrator you just take the modulus of the expression abouve?
15:30
@Pedro If you are willing to restate the point-set topology problem you posed like a month ago, I am willing to try it.
@μακακας your name can be considered offensive
@UserX not at all
I know it's a kind of chimpanzees but we both know it's a childish way to swear in Greek.
@UserX lol nice. Mind if you help me with Numerical Analysis :)
@μακακας if I could I would.
Anyway bbl got class
15:36
later pal
@UserX i read it as MAKAKAC
that's not at all offensive, no.
@N3buchadnezzar oops, I meant Modulo
lol I really don't care about my name I need help with numerical analysis please :/
$$ \int \frac{5 \cos x \ \mathrm{d}x}{\sqrt{\sin \,x} - 3 \sqrt{ \sqrt{\sin x \, } + 4\,} \,} $$
local truncation error
15:40
@Integrator Yeah so ?
@N3buchadnezzar Why did you say 'Be ashamed'?
@N3buchadnezzar I thought you said so, because I've posted almost almost un-implementable method to a programming related question.
@Integrator Yeah. I saw that when I tried to take the modulo
@N3buchadnezzar But did you noticed OP's behaviour?
15:56
@Mike testing if I am on ignore
def fun( n ):
P = 1
for i in range(1,n+1):
P = P*i^i
return P

print(fun(3))
Where is my mistake ?

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