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12:13 AM
@user193319 I believe the natural extension to multigraphs is just ensuring that $\#(u,v) = \#(\sigma(u),\sigma(v))$ where $\# : V \times V \rightarrow \mathbb{N}$ counts the number of edges between $u$ and $v$ (which would be zero).
 
12:53 AM
@user76284 If you're interested, I think mihaild help me clear this up here: math.stackexchange.com/questions/3197468/…
See the comments on his answer.
 
Ah, that's for en.wikipedia.org/wiki/…. My comment above applies for en.wikipedia.org/wiki/….
There are two kinds of multigraph.
 
1:37 AM
hi @ted
 
 
1 hour later…
2:43 AM
I have this exercise:
Consider the ring $R$ of polynomials in $n$ variables with integer coefficients. Prove that the polynomial $f(x _1 ,x _2 ,...,x _n ) = x _1 x _2 ···x _n$ has $2^ {n+1} −2$ non-constant polynomials in
R dividing it.
But, for $n=2$, I ca'nt find any other non-constant divisor of $f(x,y)=xy$ other than $x$, $y$, $xy$
oh! -x, -y, -xy
sorry
 
 
2 hours later…
4:38 AM
@TedShifrin Hi all and hi Ted! Classic question. But I decided to learn on that. I got a lot more more pickier recently to what I agree to review.
 
 
3 hours later…
7:15 AM
Hey y'all
Heile @Rudi
 
7:44 AM
Let $G$ and $H$ be groups. Is there a name for elements $\langle a, b \rangle$ of the wreath product $G \wr H = G^{\oplus H} \rtimes H$ such that $a(b) = e_G$?
 
 
4 hours later…
11:48 AM
So, just to make sure I'm not missing something, $-\sqrt{-30}$ and $\sqrt{-30}$ are automatically irreducible in $\mathbb{Z}[\sqrt{-30}]$, right?
 
12:03 PM
Why do we say that complex holomorphic functions are analytic locally? LeakyNun gave me an example, but I am sorry to say that i lost it.
 
12:29 PM
hello folks. i have to solve an ODE but i dont know how to go about it
$\lambda^2 f + f'' = A\cos\big((\lambda - 2)\theta\big)$
im given some sine products manipulations as hints but i dont know how to use them. if somebody could point me to some ressource or give me a hint i would be grateful
 
TP: $N_n = \frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 14 + \dots$ the function $N$ has no Integers as range except for $1,3$
2. Where does the series converges? $$\sum \limits_{n=1}^{\infty}\frac 1 {3^n + 1}$$?
 
@Rithaniel what is $\Bbb Z[\sqrt{-30}]/(\sqrt{-30})$?
 
12:45 PM
Wouldn't that be the trivial object?
(Had to work it out a little bit in my head)
 
No, what does an element of $\Bbb Z[\sqrt{-30}]$ look like?
 
$a+b\sqrt{-30}$ of course
 
with $a, b \in \Bbb Z$
when is an element equal to $0$ in the quotient?
 
When it is in the ideal generated by $\sqrt{-30}$ (which is what I was thinking about. I'm not sure what that ideal looks like)
 
$(\sqrt{-30}) = \lbrace r\sqrt{-30} : r \in \Bbb Z[\sqrt{-30}]\rbrace$
 
12:50 PM
The thing I'm not sure about is that $(a+b\sqrt{-30})\sqrt{-30}=-30b+a\sqrt{-30}$, so we have anything with first part equal to a multiple of 30, right?
So would that make this quotient isomorphic to $\mathbb{Z}/30\mathbb{Z}$?
 
err 'ang on
 
@Silent $\sum z^n$
$\Bbb Z[\sqrt{-30}]/(\sqrt{-30}) = \Bbb Z[X]/(X^2+30,X) = \Bbb Z[X]/(30,X) = (\Bbb Z/30\Bbb Z)[X]/(X) = \Bbb Z/30\Bbb Z$ @Rithaniel @ÍgjøgnumMeg
 
Right that's what I was expecting
so ignore me
lol
I'm so rusty
 
Well, I got the trivial object my first work through.
Though, why are we talking about the quotient by that particular ideal? I'm not familiar enough with this stuff to know what that was supposed to indicate.
 
it's supposed to be irreducible iff the quotient is an integral domain
 
1:00 PM
Well for some reason I was expecting $\Bbb Z$ to come out which would make the quotient an integral domain
but I was mistaken lol
For reference @Rithaniel, an ideal $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is an integral domainj
 
Ooooh, okay. So, this actually tells us that $\sqrt{-30}$ is reducible?
 
well $\Bbb Z/30\Bbb Z$ isn't an integral domain hahaha
oops misread
jesus christ
 
Lol, you're good. It's early morning (at least where I am)
 
it's 14:03 here and I'm at work lol
 
Sometimes the morning lasts well into the afternoon.
 
1:06 PM
no it doesn't since $\Bbb Z[\sqrt{-30}]$ is not a principle ideal domain
 
right
 
Ah, so this quotient theorem only applies in PIDs?
 
no it applies in general but ideals don't behave the same way that elements do in non-PIDs, so maximal ideal $\iff$ generated by irreducible element doesn't hold, I think
 
Ah wait, an irreducible doesn't have to be prime and the theorem talks about primes, right?
 
can we factorize $(\sqrt{-30})$ then
 
1:10 PM
Don't think so, I think it's maximal among principal ideals
@Rithaniel I've hijacked your question a bit because it's highlighted some stuff I need to refresh lol
 
Ah, you're fine, that's how a conversation will usually go. I think I can show $\sqrt{-30}$ is irreducible by using the norm function, though.
 
1:28 PM
2
Q: Open Sets in the Wedge Sum and a Homeomorphism

user193319I am presently working through example 1.21 in Hatcher's book on wedge sums of topological spaces. He makes a few claims which I am having trouble verifying. First, let me set-up some notation. Let $\{X_i\}_{i \in I}$ be a collection of topological spaces. Then $\amalg_{i \in I} X_i := \cup_{i ...

 
Please help in the proof: I can't see how tail going to zero says series convergent.
I know that converse is true.
 
@Silent it's the basic theorem of analysis.
$\sum a_n$ converges $\implies a_n\to 0$.
Or is that not your problem?
 
Can anyone help ?
 
No, no! In the slide, professor says: $\sum_{N+1}^{\infty}|c_n|\to 0$ hence $\sum_0^{\infty} |c_n|$ converges. I can't follow that reasoning.
@anakhro yes, this is not my problem :)
 
1:44 PM
Each of the six faces of a die is marked with an integer, not necessarily positive. The
die is rolled 1000 times. Show that there is a time interval such that the product of all rolls
in this interval is a cube of an integer. (For example, it could happen that the product of
all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include
at least one throw!)
 
Isn't that clear, @Silent?
Since $\sum_{N}^\infty |c_n|$ converges to 0, then it is bounded. So you just take the finite sum at the start of the sequence and add it to the bound.
 
oh!!
thank you very much
 
Once again, Travelocity customer service can go f itself
 
ok so I just want to know why this post is going to be closed math.stackexchange.com/questions/3200274/…
 
How do you know it is going to be closed?
 
1:57 PM
Told me on Friday that they couldn’t get in touch with Icelandair until Monday since it was a holiday for Icelandair. Okay, fine
 
Well it only needs two more votes
 
On Monday, I ask for an update and get told they’re working on it. On Tuesday I get an updated itinerary!... which is exactly the same as the old one. I tell them as much and am told they’ll review my case
 
I mean I will just post it again there is nothing wrong with it
 
@Adam Can you not see what the votes are saying?
 
No response on Wednesday at all. Contacted them again today and they said they’ll review my messages...
 
2:01 PM
it says it's unclear what I am asking and I clearly am asking for someone to provide a counter example
 
so, fingers crossed that they actually book the corrected flight this time
 
2:15 PM
@Adam I don't actually see you ever ask for a counter example in the question. That would be a good edit to make.
You only make a vague comment about "counterexample sought" in the title.
So it's not clear at all that that is what you are asking.
So I highly recommend you include a bold comment in the question body that says explicitly what you need help for.
 
@anakhro tail of harmonic series not go to zero?
 
@Secret if $\lim_{N\to\infty}\sum_{n = N}^\infty a_n = 0$, then you can find an $N_0$ such that $\sum_{n=N_0}^\infty a_n < M <\infty$.
Sorry, made a lot of typographical errors. :P
 
2:39 PM
@anakhro did you read the question title?
 
2:54 PM
@Adam no. Quite frankly, I never read the title. The title should not contain additional information to the question.
Moreover, the title is vague and doesn't clearly ask a question.
And even more so, your insistence that your question is blameless with regards to the reports indicates more than ever that your question probably should be closed.
If all it takes is adding a simple "My question is that I want to find a counterexample to _______" to your question body and you refuse to do this, even after someone takes the time to give you that advice, then ya, I'd vote to close myself.
 
3:17 PM
meh it doesn't really matter I've found them anyway
 
Well now you know the reason people were voting to close.
 
3:35 PM
but if a title inherently states what the op is looking for I hardly see the fact that it has been explicitly restated as a reason for it to be closed, no it was because I orginally had a lot of errors in the expressions when I typed them out in latex, but I fixed them almost straight away
lol I registered for a forum on Australian politics and it just hasn't sent me a confirmation email at all how bizarre
 
4:39 PM
@Adam I don't think so; I don't know anyone by that name.
 
5:31 PM
Nevermind
 
6:05 PM
Hi, I have the following problem for a math contest:
For which value of b is there only one intersection between the line y = x + b and the
parabola y = x^2 - 5x + 3?
The answer key says the answer is -6
How did it get that answer?
 
Combine the two equations and rearrange them to get a quadratic equation, then calculate the discriminant.
 
I have a nother problem: If Train A leaves at noon from San Francisco and heads for Chicago going 40 mph.
Two hours later Train B leaves the same station, also for Chicago, traveling 60mph.
How long until Train B overtakes Train A?

I got 6 P.M. but the answer key says 4 P.M.
 
illinois is 2 hours behind California ;)
 
Isn't Illinois 2 hours ahead?
but I get what your saying
 
yeah you're right lol, was just making a joke anyway
I'm not from the US so idk
 
6:19 PM
oh ok
 
7:01 PM
@swagbutton8 as a check, suppose the answer were 6pm. Then train A will have travelled at 40 mph for 6 hours, giving 240 miles. Similarly, train B will have travelled at 60 mph for only four hours, giving 240 miles. So that checks out
By contrast, the answer key result of 4pm would mean that train A has gone for four hours (so 160 miles) and train B for 2 hours (so 120 miles). Hence A is still ahead of B at that point
So yeah, at first glance I’d say the answer key is wrong. The only way I could see it being correct is if they’re including the change of time zones, which I’d find pretty annoying
But 240 miles seems waaay to short to cross two time zones
So my inclination is to say the answer key is nonsense
 
ive got a confession to make folks
i dont actually understand math i just think the starred messages are funny
6
 
 
2 hours later…
8:59 PM
Hi swagwagon
Hi chat
 
9:15 PM
r there closed forms for the sums of the prime zeta function
 
10:05 PM
3
A: Will all solutions of this ODE look like this?

ChappersYou can actually show this using only that the derivative of a function is zero if and only if it is constant, the exponential function differentiates to almost itself, and some ingenuity. Suppose that the equation starts in the equivalent form $$ y'' - (r_1+r_2)y' + r_1r_2 y =0. \tag{1} $$ (Obvi...

Hi there,
I'm currently going through a proof of why all general solutions to second ODE look the way they look. I have a question mark regarding the linked answer.
Where does the term e^{(r_1-r_2)x} come from?
It seems like it is taken out of the blue, but it yields the desired result.
 
you forgot to pop that in $ signs schn
 
Sorry.
 
11:02 PM
@schn the term for that is that that exponential is an integrating factor
 
Okay. From what I've learned up to this point, integrating factors have only appeared when solving first ODE. Why does it pop up when trying to solve $\u'' + (r_1-r_2)u' = 0$, which is a second ODE?
 
Well, it’s a second-order ode in u(x). But if you define $v(x)=u’(x)$ then the ODE is $v’+(r_1-r_2)v=0$, which is first order
 
Suppose $U$ is open in $X$, $V$ in $Y$, does it follow that $U \vee V$ is open in $X \vee Y$.
 
@Semiclassica True, so substitution is totally legitimate in this case?
 
Sure. You solve the ODE for v(x). Then u’(x)=v(x), so you can antidifferentiate to get u(x)
 
11:12 PM
hey joe shmo
 
But the point is really that the same integrating factor idea works here
 
Okay. Thanks a lot!
 
is analytic continuation of a sum like zeta function generally pretty hard to do?
 
Suppose you’ve got a second-order ODE like $y’’+p(x)y’+q(x)y=0$
 
I know there is only 1 analytric continuation, if it even exists
 
11:15 PM
You could still look for a function g(x) such that $g(x)y’’+g(x)p(x)y’+g(x)q(x)$ is the derivative of some first-order ODE
And that’s a useful idea, in fact
But with second-order odes you typically get something leftover that you can’t eliminate. (Look up “canonical form of Sturm-Liouville equations” if you want more on that.)
 

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