Mathematics

12:01 AM

I don't know any set theory without choice, but again that is irrelevant (and adding in extra assumptions and finding counterexamples does not affect the logic of the disproof).

Anthony

I don't know enough set theory to understand the wiki page on inaccessible cardinals

user 1618033

@Semiclassical this is nice to try $$\int_1^{\infty}\frac{\textrm{d}x}{x^2(\log(y)-\log(x))}=\frac{\textrm{li}(y)}{‌y}$$

Semiclassical

mmkay

i forget, what's the definition of the logarithmic integral?

user 1618033

mathworld.wolfram.com/LogarithmicIntegral.html

Semiclassical

$\text{li}(x)=\int_1^x \frac{dt}{\ln t}$?

ah, so based at $0$ not $1$

Karl Kronenfeld

12:06 AM

@Anthony Do you know about the cofinality of a cardinal?

That's all we need to prove that inaccessible cardinals are counterexamples (and to define them)

Anthony

I don't know about that.

user 1618033

@Semiclassical not really

Semiclassical

starting with the substitution $x=y t$ seems obvious

i meant in the sense of being integrated from $0$ to $x$ rather than $1$ to $x$

Anthony

@KarlKronenfeld I think we might just have different definition of something somewhere.

user 1618033

@Semiclassical Depending on the value of $x$, one might like to consider the integral in the sense of the Cauchy principal value.

@Semiclassical Yeap.

Karl Kronenfeld

12:09 AM

@Anthony ok

Semiclassical

eh. start with $0<x<1$, and then do analytic continuation

Anthony

@KarlKronenfeld Sorry. I'm just really naive about this stuff. Thanks for your help.

user 1618033

@Semiclassical By the way, I have a very cool solution for $$\int_0^1 \operatorname{li}(x) \ dx.$$ It's not hard. Yeah, it is definitely neat.

Semiclassical

oh? neat

user 1618033

@Semiclassical Give it a try if you want to.

Semiclassical

12:11 AM

i'll try to see through the other one first

though what i really need to do tonight is finish up with grading, ugggghhhhh

user 1618033

hehe, but just to know that I didn't use the previous result here.

Semiclassical

been procrastinating with that

user 1618033

@Semiclassical Just to notice something easy and you're done! It's a very short integral.

As you wish.

Semiclassical

have to get these quiz grades done tonight

user 1618033

No need to change the integration order in a double integral, it works only by the most elementary means.

OK (try it later then ...)

Well, in a double or in a triple integral (depeding on approach, one can use more integrals --- and again, no need for any special operation)

Semiclassical

12:16 AM

though i do see how the first one works: it's just the substitution $u=y/x$

user 1618033

Tools in high school are enough to get you done.

It looks so cool (no need for reversing the order of integration, no need for using DUIS) $$\int_0^1 \operatorname{li}(x) \ dx.$$

I love it, good to include in any good book (sure, also asking these requirements - maybe it's good to remember once in a while that mathematics is an art).

It's late here.

Mike Miller

@AlexC Presumably "injection of Bialgebras" means an injective Bialgebra map :p

Balarka Sen

12:47 AM

Huh. Integrating forms over compact manifolds is quite a natural thing, then.

Mike Miller

You say this as if it's a corollary to something?

Balarka Sen

Hmm, probably unintentionally. It never actually struck me that one could use the partition of unity to define integration over manifold.

Jessy Cat

2:00 AM

1

Q: Finding residues at a point $a$ where $a$ is a pole.

I am faced with the following problem: (a) Find $\displaystyle res_{a} \frac{\varphi(z)}{(z-a)^{n}}$ where $\varphi$ is a given function analytic at $a$, $\varphi(a) \neq 0$, and $n$ is a positive integer. (b) Suppose $a$ is a simple pole of $f$, and let $\displaystyle res_{a} f = A$. F...

@TedShifrin ?

Mike Miller

2:36 AM

@Balarka Do you understand why this is the same as integration of (compactly supported) funxtions on R^n?

Jessy Cat

@BalarkaSen, I thought that was a pretty standard thing. I remember doing that in a class I took a couple of years back.

Jessy Cat

3:23 AM

So apparently, I don't know how to apply the residue theorem properly:

Please somebody help - it's kind of a necessarily skill.

1

Q: Evaluating Improper Integrals with Residues - don't think I'm calculating the residues properly

I have to evaluate the integrals $\displaystyle \int_{-\infty}^{\infty}\frac{dx}{x^{2}+p^{2}}$, for $p > 0$, and $\displaystyle \int_{-\infty}^{\infty} \frac{dx}{(x^{2}+p^{2})^{2}}$, for $p > 0$ using residues. For the first integral, we have two simple poles, one at $x = -ip$ and one at $x=ip$....

complex-analysis complex-numbers improper-integrals residue-calculus complex-integration

user174558

Hi @JessyCat, lol.

Jessy Cat

@WillHunting, what's so funny?

user174558

Nothing. It's a bad habit of mine to always lol.

Jessy Cat

Mmmhmm

user174558

3:47 AM

Hi @AlexClark, lol.

Jessy Cat

4:06 AM

Jon Snow

Oh, that one should totally get a star.

Mike Miller

@AkivaWeinberger Isn't it past your bedtime?

Adeek

4:27 AM

@MikeMiller here

why is the $(CX \cup · · · \cup CX, X)$ is a good pair with quotient space SX ∨ · · · ∨ SX.

Mike Miller

What's a good pair?

thinker

Hi

is chat a place to discuss maths?

Adeek

(X,A) are called good pair

Mike Miller

What would you say if it was $(CX,X)$?

Adeek

we would have $\rightarrow \tilde{H_n(X)} \rightarrow \tilde{H_n(CX)} \rightarrow \tilde{H_n(CX/X)} \rightarrow ...$

the tilde is over the H

Mike Miller

4:35 AM

I mean in the context of your question

WHy is $(CX,X)$ a good pair?

Adeek

I am don't know why it has a long exact sequence ?

I am just trying to understand why $(CX \cup · · · \cup CX, X)$ is a good pair with quotient space $SX \cup · · · \cup SX$.

Mike Miller

Which is why I asked you to think about it in the simplest case.

Adeek

oh I understand why it is a good pair

Mike Miller

Why, then?

Adeek

by excision theorem

Mike Miller

4:41 AM

A good pair is an entirely topological property. It says that some neighborhood of $A$ deformation retracts onto it. That's not accessible via excision, which is an algebraic theorem.

You need to find a neighborhood of $X$ in $CX$ that deformation retracts onto it.

Adeek

I see

we can just take the bottom part of the cone

Mike Miller

So what do you do for $(CX \cup_X CX,X)$?

Adeek

what does $CX \cup_X CX$ represent

shouldn't it be $CX$ again ?

that is one thing that I was confused by

?

also for the case of (CX,X) why is $CX/X ~= SX$?

Mike Miller

1) What is $SX$ (resp. $CX$ defined as?

2) It means $(CX) \times \{0\} \sqcup CX \times \{1\}/(x,0,0) \sim (x,0,1)$. (That is, I take two cones and glue them together along their bases.)

Adeek

yes

but why does translate to CX/X?

Mike Miller

4:54 AM

These are two different questions, @Adeek. The first is responding to you asking what $CX/X$ is. The second is responding to you asking what $CX \cup_X CX$ is.

Adeek

oh

oh ok I see

For a space X, the suspension SX is the quotient of
$X\times I $obtained by collapsing $X\times{0}$ to one point and $X\times{1}$ to another
point.

hi @AkivaWeinberger

I will leave this for tomorrow as I am kinda of sleepy

@MikeMiller I will discuss with you tomorrow. Its bad to discuss something if I don't have the definitions in my head, because that means I didn't see the definition properly.

I am just solving the following problem

More
generally, thinking of SX as the union of two cones CX with their bases identified,
compute the reduced homology groups of the union of n cones CX with their bases
identified.

anyway nights thanks @MikeMiller again

I have an idea where to start when I wake up tomorrow

Mike Miller

5:10 AM

good night!

you probably will not discuss it with me tomorrow since i will be in the air

zoom zoom

Eric Stucky

6:24 AM

such fast

Huy

much eric

user174558

6:50 AM

Hi @Huy, lol.

Pallas

7:03 AM

help me please

cant do this

how to solve??

is this right?

pδ^_M(q_0m, x) = δ^_M((q_01, q_02), x) = (δ^_2(q_01, x) , δ^_2(q_02, x)) ?δ^_M(q_0m, x) = δ^_M((q_01, q_02), x) = (δ^_2(q_01, x) , δ^_2(q_02, x)) ?

* δ^_M(q_0m, x) = δ^_M((q_01, q_02), x) = (δ^_2(q_01, x) , δ^_2(q_02, x)) ?

is this definition of delta m

John Steinbeck

7:35 AM

if $G$ is a group and $V$ a vector space then $G\times V$ is again a vector space right? with addition and scalar multiplicatoin $(g_1,v_1)+(g_2,v_2)=(g_1g_2,v_1+v_2)$ and $\beta(g,v)=(g,\beta v)$

Huy

@JohnSteinbeck: I think a vector space should have commutative addition, so this only works if you assume $G$ to be abelian.

ok shit

you can call me Huy

ok Huy

8:19 AM

That's not a vector space. If it were, $0(g,v)$ would always be zero.

Eric Stucky

8:33 AM

I feel like I am being classically conditioned by my Lie theory course.

On every homework assignment, I repeatedly bang my head on the problems and make no progress

Except for the problems about multilinear algebra, in which I repeatedly bang my head on them, and then I solve them

And now I've come to be very happy every time I see multilinear algebra show up in pretty much any context...

Mike Miller

ehehe

what's a root system?

Eric Stucky

we just don't know

Eh no I might got this

There's an ambient vector space, and it's gotta span, it's gotta have $(a,b)$ be an integer, and then something about reflections.

$s_a(b)$ is a root, where $s_a(b)=b-2\frac{(a,b)}{(a,a)}a$?

Tobias Kildetoft

@EricStucky it is the thing you multiply $a$ by in that map that needs to be an integer

Eric Stucky

oh that makes a lot more sense :/

Tobias Kildetoft

often we define a new inner product that way instead to save on notation

Eric Stucky

8:38 AM

this is something something coroots, right?

Mike Miller

I'm lost

I want to go to bed

Tobias Kildetoft

also you need that whenever $a$ is a root then the multiples of $a$ that are roots are precisely $\{a,-a\}$

Eric Stucky

Nah, nonreduced root systems are okay

Tobias Kildetoft

@EricStucky I suppose. They just never show up in my work

Eric Stucky

Fair enough :)

Tobias Kildetoft

8:39 AM

coroots can be pretty much ignored if you are doing Lie theory. They become quite important for algebraic groups

@MikeMiller root systems seem strange at first, but there is nothing particularly weird going on with them.

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$Mathematics$ Mathematics