Mathematics

Nobody recognizeable

12:44 AM

Someone please help me.

Semiclassical

1:34 AM

apparently i'm blind

oh, not blind, but this is still weird

this is a fairly short paper at 21 pages: siba.unipv.it/fisica/ScientificaActa/volume_2_1/Regazzini.pdf

but for some reason section 2 has its own list of references, separate from section 3

so when you get to the end of those references, it looks like the paper has ended already without ever getting to section 3 :/

skullpatrol

That is weird, I guess they want each section to be self contained?

Since the first section is just an introduction.

CroCo

Is this equilibrium point stable from this phase portrait

Semiclassical

1:50 AM

@skullpatrol yeah, i dunno. possibly it's just typical of that journal

Nobody recognizeable

Can someone help with the aforementioned question?

Semiclassical

2:10 AM

@Nobodyrecognizeable not going to say a lot, but one issue with that problem: the solution to that ode + boundary conditions does not have bounded first derivative at t=0

which creates some issues with solving it correctly. my advice would be to ignore the boundary conditions until you've got the general solution

otherwise you're very liable to get an answer which makes no sense

Nobody recognizeable

3:10 AM

@Semiclassical thanks by trial I got $ x=e^{t/2} +c$. Now the boundary conditions do demand different values of c.

Semiclassical

the way you'd get that answer is precisely by enforcing boundary conditions along the way.

which, as I said, you should not do

if you don't, you'll end up with two integration constants in your general solution for x(t)

and that's enough to satisfy both boundary conditions.

Nobody recognizeable

@Semiclassical two integration constants how?

Semiclassical

by not enforcing boundary conditions until the end.

Nobody recognizeable

@Semiclassical actually I put $x=e^{mt} $ and got m=1/2. Am I in the right way?

Semiclassical

No. Guessing a solution is not a good approach here.

Nobody recognizeable

3:16 AM

@Semiclassical then please give me a reference.

Semiclassical

My suggestion: Start by multiplying the entire ODE by $x$, and see if you can recognize the left-hand side as a total derivative

i.e. is there some F(x(t),t) such that the left-hand side is dF/dt

If you can do that, you can make progress.

Nobody recognizeable

@Semiclassical OK trying.

skullpetrol

your username begs the question is it possible to recognise Nobodyrecognizeable without experiencing the sensation of recognition?

6

Nobody recognizeable

@Semiclassical $x^2/2=e^{t+c} +c_2$?

@skullpetrol I've met you earlier. You didn't recognise me. So my name makes sense.

@Semiclassical I got c.

@Semiclassical thanks for the help. You solved my problem. Have a nice day. Good bye.

Balarka Sen

6:59 AM

Hi @Mathein, @Ryan

Alessandro Codenotti

7:24 AM

Hi @Balarka

Balarka Sen

Yo @Alessandro

skullpatrol

\o @y'all

Alessandro Codenotti

I have to learn how connections, the Clifford algebra and the spinc group work

Any good references?

Balarka Sen

I have heard Michelson-Lawson is good

I was reading a little bit from some GTM I can't remember the name of

Alessandro Codenotti

I see, let me order it on my favourite online Russian bookstore real quick

GTM? Geometric ??

Balarka Sen

7:33 AM

Ah yeah Friedrich, "Dirac Operators in Riemannian Geometry"

Graduate Text in Mathematics :3

Alessandro Codenotti

(are you familiar with those topics if I have questions?)

Balarka Sen

Nope!

Mike is though

Alessandro Codenotti

@BalarkaSen ahhh makes sense

Balarka Sen

And I know how to conjure him up

Alessandro Codenotti

Draw a circle of salt on the ground and start computing its fundamental group

Balarka Sen

7:39 AM

You need to repeat "Heegaard-Floer homology, awaken instanton-eously" thrice while doing so

Mornin'

@BalarkaSen LOL

Hi @ÍgjøgnumMeg

Morning @ÍgjøgnumMeg

I'm messing up some computation but also feel lazy to sit down and do it

Nick

8:05 AM

I just spent two hours explaining to little brother why $a+b=a-(-b)$ is true for all integers even when one of $a$ and $b$ is positive and the other is negative. His proof contains recursion apparently.

It's enough to say lhs=rhs when when you have brought both side to the same form, but he's encouraged to do the evaluation entirely which means at a point (-)(-) = (+) which he can't do without the initial equation.

Mathematic is a belief apparently.

Skies burn

9:49 AM

How young is he @Nick?

Nick

@Skiesburn 12

Skies burn

Jun 19 at 6:24, by skullpatrol

What's so baffling about negative numbers? by a Field medalist.

Show him that ^

See page 117 i.e. pdf page 5

Nick

yeah, I did try to explain it as number line hopping over multiples of +1 increments to the right and -1 decrements to the left and each number being a leap from 0 which condenses the hops. So, that substantiates $-a = (-1)\cdot a$ for every value of a.

Alessandro Codenotti

@BalarkaSen Is it possible that there is a typo in the very first definition? I guess it should be $v\otimes v+q(v)1$ rather than $v\otimes v+q(V)1$?

Nick

I made $(-a-(-b))$ as $-1(a-b)$ so he'd see whatever the numbers, the difference between negative integers is the same as the difference between their equivalent positive integers.

Balarka Sen

10:02 AM

@Alessandro Yeah sounds right

I don't have the book with me at the moment but it is $v \otimes v + q(v)1$

Nick

@Skiesburn anyhow, thank you, I will forward this to him. There's definitely a substantial amount of insight in this that he might enjoy reading.

Alessandro Codenotti

That's the ideal one needs to quotient the tensor algebra by to get the clifford algebra, $q$ is the quadratic form on $V$

Subhasis Biswas

10:53 AM

.does the prime avoidance hold for fractional ideals?

@AlessandroCodenotti

faculty.math.illinois.edu/~r-ash/Ant/AntChapter3.pdf

Balarka Sen

@Alessandro Right, you're formally taking the square root of the quadratic form

Subhasis Biswas

Balarka, what do you think?

Balarka Sen

Which is why I think this construction is relevant in the construction of the Dirac operator, which I have been told should be thought as the square root of the Laplacian

I don't know what prime avoidance is

Subhasis Biswas

Prime avoidance lemma

In algebra, the prime avoidance lemma says that if an ideal I in a commutative ring R is contained in a union of finitely many prime ideals Pi's, then it is contained in Pi for some i. There are many variations of the lemma (cf. Hochster); for example, if the ring R contains an infinite field or a finite field of sufficiently large cardinality, then the statement follows from a fact in linear algebra that a vector space over an infinite field or a finite field of large cardinality is not a finite union of its proper vector subspaces. == Statement and proof == The following statement and argument...

Balarka Sen

I've better work to do that reading a wikipedia page :P

Subhasis Biswas

10:57 AM

@BalarkaSen wait

Let E be a subset of R that is an additive subgroup of R and is multiplicatively closed. Let $I_1, I_2, ..., I_n$ be ideals that are prime for $n \geq 3$. If $E$ is not contained in any of $I_i$'s then $E$ is not contained in the union

Balarka Sen

I really don't have the time or patience to read this. I am sure people who actually know this stuff can address your question better.

Subhasis Biswas

I still can't get a hold of the problem that I asked you to check

what is your advice in this type of situation. Should I leave it?

skullpatrol

Why not ask on the main site?

Balarka Sen

I don't think it's true but I didn't spend much time thinking about it

Subhasis Biswas

@skullpatrol I did. No proper reply so far

skullpatrol

11:02 AM

Patience is a virtue for a good reason :-)

2

Subhasis Biswas

Consider an infinite commutative group $G$. Let $<a_1>, <a_2>, ..., <a_m>$ ($<a_i>=C_i$) be a finite collection of infinite cyclic subgroups of $G$ such that $C_j \not\subset \displaystyle\bigcup_{I \setminus \{j\}}C_i$ , $I=\{1,2,3,...,m\}$. Then $\displaystyle\bigcup_{i=1}^mC_i$ is never a subgroup of $G$.

I have restricted it even further.

skullpatrol

Update your main site question with that^

Subhasis Biswas

okay

skullpatrol

good luck my friend

Balarka Sen

@SubhasisBiswas That should be easy, look at the subgroup of $G$ generated by $a_1, \cdots, a_m$. That's a $\Bbb Z^m$, and $C_i$'s are your coordinate axes. So equivalent to proving that the union of coordinate $\Bbb Z$'s is not a subgroup of $\Bbb Z^m$, which it isn't

Flesh out the details of this sketch and you're done

Subhasis Biswas

11:07 AM

@BalarkaSen the statement is true, then?

Balarka Sen

For a commutative group yes

Subhasis Biswas

@BalarkaSen So, what are you saying is the subgroup is isomorphic to $Z^m$?

i am not quite getting it

Balarka Sen

What's unclear about what I said? $\langle a_1, \cdots, a_m \rangle$ is $\Bbb Z^m$.

Subhasis Biswas

@BalarkaSen How do I prove this? By mapping $\phi((a_1,a_2, a_3, ..., a_m)) = (1,1,1,...1)$ [m times]?

Balarka Sen

By thinking.

2

Subhasis Biswas

11:23 AM

$<a_i>$ is isomorphic to $\mathbb{Z}$. So, $<a_1> \times <a_2> \times ... \times <a_m> \cong \mathbb{Z} \times \mathbb{Z} ...$m times $\times \mathbb{Z}$ ?

@BalarkaSen well, I know you are busy. But a few more minutes to help out a poor soul won't hurt, I guess.

hello @skullpatrol

skullpatrol

re-hi

Subhasis Biswas

:D

what are your insights?

skullpatrol

try sleeping on the problem

your brain will be refreshed

:-)

Subhasis Biswas