Mathematics - 2017-07-30 (page 2 of 4)

Typhon

06:02

@AkivaWeinberger is there any way via the various operations of the collatz conjecture to map some number n to n+3? I mean without going through 1.

pbeentje

I've tried substituting for t both ways round, and both ways round I get a mess. It's been almost 20 years since I learned trigonometric identities so that hasn't helped either

Ted Shifrin

Write $t$ in terms of $x$ and plug it into the $y$ equation, @pbeentje.

@Lucas: Are you using $N(a+bi) = a^2+b^2$?

If so, it looks very false to me.

Daminark

06:17

Note that $N(a+bi) = (a+bi)(a-bi)$

Ted Shifrin

That's good for the converse :)

Lucas Henrique

@TedShifrin do you have a counter-example?

Ted Shifrin

Lots.

Lucas Henrique

:(

Ted Shifrin

Why do you think it's right?

Are you sure you don't mean the converse?

Lucas Henrique

06:21

@TedShifrin yeah.

Ted Shifrin

Try it with $\alpha = 3$.

Daminark

@TedShifrin Wait what?

Ted Shifrin

Twink, are we going around this again?

Lucas Henrique

5 hours ago, by Lucas Henrique

Conjecture: Let $z \in \Bbb Z[i]$. If $\mathrm{N}(z)$ is composite, $\exists z_1, z_2 \in \Bbb Z[i] : z = z_1 z_2$

Ted Shifrin

Wait yourself, Demonark.

Twink

06:23

what? it's just a question

Lucas Henrique

@TedShifrin ?

Ted Shifrin

@Lucas: $9$ is composite, but $3$ is a Gaussian prime.

Lucas Henrique

:(

ooooh. yeah, restriction.

Daminark

Oh... you're not square rooting it

Ted Shifrin

You don't get an integer if you square root it, Demonark.

Daminark

06:24

Lol I'm braindead today

Lucas Henrique

Let $z \in \Bbb Z[i] - \Bbb Z$, blah blah blah.

Ted Shifrin

That won't do, @Lucas.

You should do lots more experimentation before you make conjectures. :)

Twink

@TedShifrin I need to talk to you give me your whatsapp

Ted Shifrin

I don't have whatsapp, Twink.

Daminark

I think you might be able to say something like, you have $a+bi$

Twink

06:25

then your Grindr or something

skype

Ted Shifrin

Start with an email, Twink.

Twink

but that's very slow

I want to talk in real time

Ted Shifrin

I certainly don't take orders ...

Twink

please

Daminark

You separate out if either $a = 0$ or $b=0$ by asking that it's congruent to -1 mod 4, and if they're both non-zero, you want $a^2 + b^2$ to be prime, congruent to -1 mod 4

I could see that, or perhaps a more elaborate scheme, to be true

Ted Shifrin

06:29

The book is not my favorite, but if you know differential forms, it's the only book at that level that uses differential forms.

Daminark

Doesn't yours do?

Or are the books at different levels?

Lucas Henrique

@Ted we were trying to prove that if $p$ is prime then $p \equiv 1 \, \mathrm{mod} \, 4 \iff \exists a, b \in \Bbb Z : p = a^2 + b^2$.

Ted Shifrin

I only use forms in one little section, Demonark. I'm writing for a general undergraduate audience.

@Lucas: You'll find that proved in my book, among lots of other places.

Lucas Henrique

oh.

ha... ha... blushes

Balarka Sen

@TedShifrin I think my temporary goal is to do surface theory (Teichmuller geometry, singular foliations, MCG)

Twink

06:31

but is that the best book to understand this other book? amazon.com/…

Balarka Sen

so I should learn 1) real analysis and dynamics 2) Riemann surfaces

Twink

cause it's the same author

Balarka Sen

I have been told to read Arnol'd ODE book for the first and Forster for the second

Lucas Henrique

a bit off topic, but it surprises me that such an important mathematician as Teichmuller could be a Nazi.

Ted Shifrin

You need some serious background for the advanced book ... experience with multivariable analysis, manifolds, and preferably some Riemannian geometry. Very different levels.

Balarka Sen

06:32

and after i'm done go headlong into Farb-Margalit or soemthing

Ted Shifrin

@Lucas: Not the only one. And there are similar 'surprising' names among composers, etc.

I don't know Arnold's ODE book, Balarka.

Twink

:@

Lucas Henrique

yeah. IIRC there's even a question about Nazi mathematicians here.

not here, but on history of science and math:

7

A: Famous scientists in the Nazi party

One can get a good general idea by looking at the scientists brought over by Operation Paperclip (an operation to bring over top German scientists to America, many of whom were members of the Nazi party or even had leadership roles in the Nazi party). A short list of the more famous scientists ...

Balarka Sen

@TedShifrin Ah, OK

Ted Shifrin

It's also interesting to learn about some of the mathematics that was done in concentration camps.

Twink

06:36

my favorite mathematician is Alan Turing

Ted Shifrin

favorite because he was gay? ... or favorite for the mathematics he did?

2

I know some modern-day gay mathematicians whose mathematics is far more interesting to me, personally.

Twink

for both

for example who?

Ted Shifrin

Robert Bryant is a super-star, just finished being president of the American Math Society.

Amazing mathematician and great person.

Robert Macpherson is another.

Twink

and how do you know they're gay? it's not on their wikipedia's biography

Ted Shifrin

They're both openly gay. They didn't write Wikipedia. Bryant talked about his partner in an interview in the Notices of the AMS that appeared recently.

They've both been friends of mine for decades and decades.

Twink

06:43

wow :D

Lucas Henrique

@TedShifrin "Abstract Algebra: (...)"?

Ted Shifrin

Yeah, Lucas. Section 3 of Chapter 4.

Daminark

@Ted are the sections in your book smooth?

Ted Shifrin

Remember that one-month hiatus, Demonark? :D

Daminark

I thought it was to be reduced to 45 seconds!

Ted Shifrin

06:46

No, I never agreed to any such nonsense.

Daminark

I dunno if I can do a whole month, like at that point I have no choice but to try and make my existence substantial or something

Twink

nice but Alan Turing is the father of computer science and has his own film

Ted Shifrin

It isn't a competition, Twink. Turing was very impressive and endured horrible things (including death). I merely said his mathematics is of less interest to me personally.

Lucas Henrique

How do you think of those, professor?

Ted Shifrin

Huh?

Lucas Henrique

06:52

I just read the section and the lemmas/theorems.

Balarka Sen

I don't think I have a favorite mathematician.

Lucas Henrique

It would take me a lifetime to realize those things.

Ted Shifrin

Lucas, I learned this proof, basically, when I studied algebra as an undergraduate from Mike Artin. Most of it stuck with me. But there are lots of books with lots of knowledge in them :)

Balarka, I have a few, I guess, but no single one.

Lucas Henrique

Also, I'm loving your linear algebra book, @Ted.

Ted Shifrin

Oh, thanks :)

There's some unusual stuff in there on projective geometry and computer graphics in the last chapter.

Lucas Henrique

06:55

I've seen books that are too geometrical or too algebraic

Balarka Sen

@TedShifrin I dunno, I think I used to have a list and it got bigger and bigger as time progressed

Lucas Henrique

Your book shows clarity in both approaches.

Ted Shifrin

To me, @Lucas, interplay among different aspects of mathematics has always been the most exciting thing.

Daminark

It's especially exciting when homotopy theory, posets, and the category of small categories all come together... Lol jk :P

Lucas Henrique

It's happy to hear that. I think the same as you, @Ted. The (small) difference is that you're a doctor and I'm a highschooler. :P

Ted Shifrin

06:57

Blech, Demonark.

Daminark

You'll at least be pleased to note that Peter May isn't into $\infty$-categories

At one point he introduced them and was like "Yeah I dunno, you can't really calculate anything with these" and the other guy was like "Uh... except like not really"

Balarka Sen

he dissed Akhil Matthew on MO for precisely that reason

Ted Shifrin

@Lucas: I have lots more pictures in my algebra book, too, for the same reason that some students learn more visually than others.

Daminark

Oh Akhil is the new guy

Lol this'll get fun

Balarka Sen

oh he came in the bootcamp?

Daminark

07:01

No, the one in the REU is Dylan

Who also thinks you can calculate using infinity-categories

Balarka Sen

ah

Daminark

Akhil is the other new person

Who I guess is coming in September

Balarka Sen

Ah

Ted Shifrin

And he's on MO?

Balarka Sen

AH

Daminark

07:03

The two stages of epiphany: reaction, and then caps lock

Ted Shifrin

I don't think Demonark will make it a month.

Unless we all ignore him.

OK, night, all.

Balarka Sen

G'night, @Ted

Ted Shifrin

Welcome home, Balarka.

Daminark

I mean didn't I say earlier that my existence more or less reduces to this style of humor, like I'd need to go into some magical fountain or whatever to have anything else of substance in my life?

8 hours at most, while I'm asleep

And lol see you @Ted

user462339

Let $A,B$ be $\mathfrak{g}$-modules, and $C$ be a $\mathfrak{b}$-module. (The Borel subalgebra.)

What does the subscript mean here:
$$Hom_{\mathfrak{g}}(A,B)=Hom_{\mathfrak{b}}(C,B)$$?

These maps factor in some way?

Lucas Henrique

07:15

4am here. g2g, goodnight chat.

user462339

Goodnight.

Twink

who reported me? ¬¬

user400188

negation eyes

Martin Sleziak

07:32

@aminliverpool In fact at least one other user (namely @Xam IIRC) registered for the event.

The Raiders of Las Vegas

Hi @MartinSleziak

Akiva Weinberger

08:00

The curve equidistant between two circles is an ellipse whose foci are the centers of the circles

3

I don't know what happens if one circle isn't inside the other… my guess is you get a branch of a hyperboloid

pbeentje

@TedShifrin thanks, I tried again and got further than the last time

Akiva Weinberger

(or a straight line if they're the same size)

BAYMAX

There is no Integral equations tag in the main site while asking question?

Martin Sleziak

@BAYMAX There is and 531 questions are tagged with this tag: integral-equations.

BAYMAX

oh sorry

Got it!

Akiva Weinberger

08:33

@TedShifrin I thought this was true

For $z\in\Bbb Z[i]\setminus\Bbb Z$

Akiva Weinberger

08:49

Oh no

I also need to exclude pure imaginary things

$\Bbb Z[i]\setminus\Bbb Z\setminus\Bbb Zi$

Alessandro Codenotti

09:05

Aren't the irreducibles in $\Bbb Z[i]$ the elements whose norm is a prime or the square of a prime?

Akiva Weinberger

Yeah

@LucasHenrique There's an alternate way to end it that doesn't need that fact

We just need to show that $p$ isn't a Gaussian prime

To do that, note that $p$ is a factor of $(x+i)(x-i)$, but it's not a factor of $x+i$ or $x-i$

($x$ being the solution to $x^2\equiv-1\pmod p$)

However, primes have the property that if $P$ is prime and $P$ is a factor of $ab$, it's either a factor of $a$ or a factor of $b$

This fact is true for the Gaussian integers as well

So $p$ can't be a Gaussian prime.

To finish from there, note that since it's prime, you can factor it into Gaussian integers... Organize them into conjugate pairs and take one from each pair

You'd end up with a bunch of things that when multiplied satisfy $z\bar z=p$, so if $z=a+bi$ then $a^2+b^2=p$ and we're done

user462339

Do weights of a Borel subgroup factor through weights of a maximal torus contained in the Borel?

Akiva Weinberger

09:23

I'm still relying on previous lemmas… Assuming Bezout's identity works for the Gaussian integers, suppose $P$ divides $ab$. If it doesn't divide $a$, by Bezout, $Px+ay=1$ for some $P$ and $a$ (a prime and something not a multiple of that prime will always be coprime).

Then $P$ will be a factor of $(P)bx+(ab)y$, which equals $(Px+ay)b=b$.

So if $P$ divides $ab$ then $P$ divided $a$ or $P$ divided $b$.

I mean, the real thing you want to prove is that $\Bbb Z[i]$ is a Euclidean domain

All the nice things about it come from there

Secret

10:36

[To be checked] Proof of existence of solutions to integral equations as a formulae of the form of integral operator involved

More generally, given a functional equation, proof of existence of solution of the functional equation depending on the functions involved

Ideally, proof of the existence or nonexistence of solutions to a functional equation, as a function of the functions and arrangement involved in the functional equation

A possible layout for the proof might be as follows:

Secret

11:01

Let $F$ be the class of formulae written in some given language $\mathcal{L}$ and said language contains relation operators such as $=$, $<$, $>$, $\geq$, $\leq$, $\equiv$, $\sim$, $\approx$. Let $E$ be the class of equations or inequalities. Then $v \in E$ is an equation/inequality or set of equations/inequalities (for systems).
Let $f : E \mapsto Op \subset F$ be a map which picks out the operators (formulae with free parameters in any orders of logic) and arrange them into an ordered set in Op. Let the solution class of $E$ be $X \subset F$. Now, the existence of solutions is defined by …

For $v$ where solutions $x$ exists, computation of $f(x)$ will give the ordered set of operators and hence the dependence of the solution to the structure of the given equation/inequality $v$

If $f(x)=\emptyset$, then it means the solution has no dependence to the structure of the equation/inequality

Actually, the above framework will lead to an obvious problem: It is easy to show that there are many equations will have the same solution $2$ which is just a number and thus there are no free variables in its set/class builder notation

and said equations are not necessary "isomorphic" wrt each other since e.g. some contains nonlinear operators while some contains linear ones (the action of linear operators on a set can never be isomorphic to the action of nonlinear operators)

What is needed is some notion of "variation" that works on the language level, such that when one letter of the formulae is replaced systematically by other letters, how the solution class changes

Another thing to note is that operators are themselves part of a solution class to some equations. E.g. a linear operator $L$ always obeys $L(x+y)=Lx+Ly$ for any set x,y (possibly under some restrictions)

Therefore, the solution set map $S$ need to have two arguments i.e. redefine $S : E \times Op \mapsto X$

Then for the above example, say a certain $v \cong Lx+Ly=L(x+y)$ from some ring $R$. Then $f(v)=\{L\}$, $S(v,L) = L$, $S(v,x) = R$

user84215

11:29

@Secret I think you will become an outstanding mathematician with creative and novel ideas.

Fawad

@AkivaWeinberger radical axis?

Secret

That's still a long way, I might have novel ideas, but my maths speak is not good enough to describe them yet, but with sufficient hardwork and collaboration with others, it will come I believe

Secret

11:46

hmmm...

9

Q: Are there any existing problems that wouldn't be solvable with a halting oracle?

I understand that most problems are trivial if a halting oracle is available (or, I think equivalently, hyper-computation). However, applying the argument that shows the Halting Problem is impossible for a Turing machine also shows that it is impossible for a Turing+oracle to decide the Halting P...

I wonder if my above framework is still within turing degree $0$... (except for any equations/inequalities that are "isomorphic" to the halting problem)

BAYMAX

any1 on Integral equations can help me out!

Here is the question

lattice

12:03

Hey chat

The Raiders of Las Vegas

Hey pal.

lattice

A short question: Let $R$ be a ring (commutative and with 1). In the category of $R$-Algebras, what are the morphisms?

user462339

If I say $f$ is a $k$-morphism between schemes $X$ and $Y$. Then $X$ and $Y$ are schemes over $k$ right? It seems directly reading off the defintion from Hartshorne, I should have then that $f:X\to Y$ compatible with $\psi_1:X\to k$ and $\psi_2:Y\to k$.

But should that be $\text{Spec}(k)$?

@lattice This is an $R$-linear, ring homomorphism.

lattice

Okay thanks

user462339

(I.e. multiplicative, additive, and satisfying $R$-homogeneity or however you say it $\psi(r\cdot x) = r\cdot \psi(x)$ for $r\in R$.)

Okay nvm my question above, I am happy with this:

17

Q: Scheme of finite type over a field $K$ v.s. $K$-scheme

I'm lost in some definitions about schemes. I have some trouble about two definitions of a scheme of finite type over $K$, for an alg.closed field $K$. Version 1 (Hartshorn) : a scheme of finite type over $K$ is a scheme $X$ together with a morphism $X \to K$, where $X$ is a scheme (a locally ri...

algebraic-geometry schemes

Secret

12:29

[Random] The set with freewill elements

Let $S$ be a nonempty set. This set is defined as $S = \{x| x \not\in S\text{ is undecidable}\}$

The result is that $S$ will simutaneously exists and not exist since it is undecidable whether a contradiction is formed

Things get even more interesting when some notion of fuzzy logic is employed

$S_{y} = \{x|x \not\in S_y\text{ is undecidable y% of the time}\}$

Now you have a set which has some probability to not exist

Secret

12:53

5

Q: An image of the hierarchy of algebraic structures

Hello! Does anybody know an image of a graph featuring the hierarchy of algebraic structures? Something rather complete. So far I've found similar images describing the hierarchies of classes/categories in various programming languages. For example Haskell's basic algebra library Coq's math cl...

mathematics-education abstract-algebra

Double fields and triple fields -> a case where someone put too many axioms in an algebraic structure

and this other one, I wonder if we can create an algebraic structure out of all of these...

Categlorical logic might allow me to hop between algebraic sturctures, by adding,deleting, or operating with axiomatic systems, interpreted as logic propositions

Mahmoud

13:50

Hi

Fawad

14:04

@Mahmoud السَّلَامُ عَلَيْكُمْ

Mahmoud

Hey $:)$

Fawad

How are you?

Mahmoud

Fine I guess, you?

Fawad

Good

Fawad

14:16

@Fawad $$\int\frac{x^2}{\sqrt{1-x^2}}dx=\int x\frac{x}{\sqrt{1-x^2}}dx=-\int x\frac{d\sqrt{1-x^2}}{dx}dx$$ — BAI 18 mins ago

Can someone explain me how is this possible?

I am newbie in integration

Semiclassical

@Fawad First step is just algebra, second is recognizing that $\displaystyle \frac{d}{dx}\sqrt{1-x^2}=\frac{(1/2)(-2x)}{\sqrt{1-x^2}}=-\frac{x}{\sqrt{1-x^2}}‌$

Mahmoud

I was able to figure out that: $xy-x_0y_0=y(x-x_0)+x_0(y-y_0)$

However, I don't know what he meant by the first and second arguments, it doesn't add up for me

Felix.C

14:35

Hey a book of mine states, I don't quite trust the book, that for the implicit function theorem the point we use as starting point has to be regular. Neither my script nor wikipedia lists this as requirement and I think there could be some examples where this is not the case but, but all other requirements are set. Did you ever hear of this?

Felix.C

14:49

Actually, it is true I just remarked...if $f_y$ is invertible then for sure $df$ is surjective.

Secret

isn't invertibility requires a trivial kernal?

Joe Stavitsky

15:10

Where do they get the multiplication by 2 inside the brackets? ctrlv.in/983510

Astyx

What's $r$ ?

Oh right

What $r$ are you talking about ? Which line ?

Joe Stavitsky

o n/m stupid

I didn't notice it in the original function

Mahmoud

15:26

I really don't seem to have the ability to solve the problem (it's starred), I'd appreciate any help.

Felix.C

15:37

@FuzzyPixelz Multiply with the denominator of both inequations and perhaps you have to use triangle inequation but this shouldn t be hard

FuzzyPixelz

I don't understand the need for the two equation of $|x-x_0|$, and yes I've used the triangle inequality but it just doesn't work out correctly, there is something I'm missing

@Felix.C

gian

16:10

Hey chat.

Mats Granvik

hi @gian

user2457324

Hi -- can someone please help me with this Q: https://math.stackexchange.com/questions/2376054/satisfying-the-following-determinant-inequality
thnaks

thanks**

Ted Shifrin

16:33

@Felix.C: Yes, of course, you need the derivative to have maximum rank. But it's usually stated with an explicit choice of coordinates so that the $\partial f_i/\partial y_j$ portion of the matrix is invertible, and then you get that the level set $f(x,y)=0$ can locally be written as $y=\phi(x)$.

FuzzyPixelz

Hey @TedShifrin ! $:)$

Ted Shifrin

um, hi, Fuzzy ... did you change names?

FuzzyPixelz

Yes! I was formally known as Mahmoud, just a few hours ago :P

Ted Shifrin

Ohhh ...

formerly

FuzzyPixelz

Oh thanks for that, I've been spelling it wrong for ever ..

Ted Shifrin

16:35

They are two different words :)

FuzzyPixelz

That's a shame

Ted Shifrin

So, are you learning more with a new name, Fuzzy?

FuzzyPixelz

I'm not sure I understand what you mean.. but no, my display name here is irrelevant, I believe.

Ted Shifrin

Oh.

Oh, so you're working on limit proofs.

FuzzyPixelz

So.. I've challenged myself to do every single problem in Spivak's Calculus after I was just running over the chapters.. and it has been totally frustrating, I've spent the last three weeks and haven't even finished the first chapter problems.

I'm now stuck at problem 21

Fawad

16:39

Why $\int4x^2e^{2x^2}\,dx$ cannot be evaluated?

Ted Shifrin

Because $e^{x^2}$ has no elementary antiderivative, @Fawad. That's a theorem.

@Fuzzy: So you wrote two terms. You want to show that the absolute value of each of them comes out $<\varepsilon/2$.

The second one is immediate, right?

Fawad

@TedShifrin can i know which theorem it is? Or theorem name

Balarka Sen

Windows 10 tells me US bombers are flying to N Korea

m8 why

Ted Shifrin

It doesn't have a name. But it follows from results in a field called differential Galois theory. Probably goes back to Abel.

Because N. Korea keeps shooting off ICBMs and we have an idiot for a president, Balarka.

But why Windows should be giving you news is beyond me.

FuzzyPixelz

Haven't thought about that at all, there is humility at every stop when you're studying sciences, but mathematics makes me feel that I'm just too stupid to deal with logic..

Secret

16:43

reuters.com/article/us-northkorea-missiles-idUSKBN1AF02K

Balarka Sen

I know the latter as a fact, and that N Korea was being a bit of a (add appropriate adjective) but didn't know they were shooting ICBMs

Ted Shifrin

@Fuzzy: The whole point of the estimate game is to break complicated things into bite-size pieces which you can control.

Balarka Sen

no good will come off this. bleh

Ted Shifrin

Balarka, it's been going on for over a month.

Secret

My attitude to maths: Teach me how to be more weird and I will gladly follow you into the unknown

2

gian

16:45

More weird @Secret

?

Balarka Sen

@TedShifrin it has a News panel on the start up menu that gives the most crummy news from all over the world

Ted Shifrin

Ah, Balarka. There are various things like that in OS and iOS, but one chooses to use them or not.

Secret

@gian Recall the impression when you first touch topology, infinite sets, various nonintuitive theorems in analysis, category theory and much more

Ted Shifrin

@Secret: gian is just starting.

gian

Yeah, I started topology a week ago...

FuzzyPixelz

16:47

@TedShifrin But.. Why are there two inequalities for $|x-x_0|$ ? And just one for $|y-y_0|$

gian

I see what you mean though; it's a very different perspective of looking at mathematics.

Ted Shifrin

@Fuzzy: Notice that in your first term you have $y(x-x_0)$, so you need an estimate on how big $y$ can be. But to answer your question, it's like other arguments you've done (like limits with $x^2$), where you need to stipulate that $x$ can't be too far from $x_0$ in order to get a bound on $|x|$. Saying $|x-x_0|<1$ forces $|x|<|x_0|+1$. You should have done lots of proofs like this.

Zophikel

$$\begin{cases} f(z) & \text{if $|z|=1$} \\ \\ f(\psi)& \text{if $|z| < 1$} \end{cases}$$

$$F(z) =
\begin{cases}
f(z) & \text{if $|z|=1$} \\
\\
f(\psi)& \text{if $|z| < 1$}
\end{cases}$$

Ted Shifrin

@Zophikel: As it stands, that makes no sense. You have to say what $f(\psi)$ means.

Zophikel

@TedShifrin I know i'm checking if it renders

mathb.in dosen't do latex very well sometimes

Ted Shifrin

16:59

Oh, I see why you're confused, @Fuzzy. It doesn't fit the way you broke the sum up. Spivak is thinking of $xy-x_0y_0 = x(y-y_0) + y_0(x-x_0)$.

Secret

Lol, that's the first Do Nothing Technique i.e. of the form $a-a=0$

Ted Shifrin

Yes, I always taught it as "add a clever $0$" or "multiply by a clever $1$".

Secret

For me, the former I usually called the first Do Nothing Technique, and the latter the Second Do Nothing Technique. There's actually a 3rd which appears in contour integration (which is basically the same as your add a clever 0, except that zero is so cleaver it is part of a contour integral that will vanish

In undergrad, they are all simply referred by professors as "by cheating, by a trick"

I vaguely remember there's a 4th Do Nothing Technique, but I currently don't recall the details. It only appears in certain proofs

gian

Wait which one is the contour integration one?

Secret

The typical example is contour integrating $$\int_0^{2\pi} \frac{\sin x}{e^x + something}dx$$

which is done by picking a semicircular path and adding a $$\int_0^{2\pi}\frac{\cos x}{e^x+ something}dx$$. This is the part that vanishes to zero

the resulting integral, can then be integrated easily because you have both the cos and sin to form an exponential function

Let me dig for the actual example...

Example 2 – Cauchy distribution in wikipedia link:

Contour integration

In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane. Contour integration is closely related to the calculus of residues, a method of complex analysis. One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. Contour integration methods include direct integration of a complex-valued function along a curve in the complex plane (a contour) application of the Cauchy integral formula application of the residue theorem One...

(Showed how much my memory decays for a subject that I don't fully master...)

gian

17:16

I haven't formally taken complex analysis, but isn't evaluating a contour integral analogous to taking a line integral in $\mathbb{R}^2$ ?

But just in a complex plane.

Ted Shifrin

It is a particular sort of line integral, yes.

gian

Interesting. I'll leave that for later though. Gotta commit to topology right now.

Ted Shifrin

Howdy, Eric.

Eric Silva

heyo

Ted Shifrin

Howdy, DogAteMy.

Akiva Weinberger

17:21

Hello

@BalarkaSen You're back

About the "deforming metric spaces" stuff, at one point you asked about its $\pi_1$

Turns out the space of metrics is contractible, though, so it's gonna be 0

Ted Shifrin

Yup, convex.

Akiva Weinberger

Also, the deformation from $(X,d_1)=:X_1$ and $(X,d_2)=:X_2$ can take place entirely in $CX_1\times CX_2$

so if they can be embedded in Euclidean space, so can the deformation

In fact, it can all take place in the smash product of $X_1$ and $X_2$ (that thing that turns lines into a tetrahedron), which embeds into $CX_1\times CX_2$

Emolga

17:52

If $\Bbb EX_n \to 0$ and $\Bbb EX_n^2 \to 0$, then $\Bbb E|X_n| \to 0$?

lattice

18:42

Hey

Let $R$ be a commutative ring with $1$. We have defined the tensor product over $R$-modules via the universal property. I am wondering about the difference between this and defining it over $R$-algebras.

Can I just replace "module" in this definition by "algebra"?

Tobias Kildetoft

@lattice As an $R$-module they are the same

lattice

Maybe my question was not quite exact, the thing is I do not know how the tensor product over $R$-algebras is defined.

Tobias Kildetoft

@lattice It is the unique $R$-module (or more generally $\mathbb{Z}$-module for noncommutative rings) satisfying the universal property

to show it exists, one usually does a construction

lattice

Yes

we have done that construction as well

but only for modules

Tobias Kildetoft

same construction for algebras (note that the construction does not care that there is a product)

lattice

18:49

Okay thanks!

Tobias Kildetoft

One thing that is somewhat different is the categorical interpretation

Since for algebras the tensor product is actually the coproduct in the category of unital algebras, and it is neither the product nor the coproduct in the category of modules

So I suppose it was incorrect of me to claim that the definitions are the same, as one needs to only consider the algebra as a module to be able to apply the precise same definition

lattice

Hmm...

Maybe just to make sure that I understood it right, let me explain why I was asking

So I have to do the following exercise:

Let $I=\{a,b\}$ be ordered by the discrete preorder (that is already confusing me a bit, what is the discrete preorder?). Let $R$ be a commutative ring with $1$, let $\mathcal{C}$ be the category of $R$-algebras and let $(R_i)_{i\in I}$ be a family of $R$-algebras. Show that there exists an isomorphism of algebras $lim{R_i} \cong R_a \bigotimes R_b$

(where $lim R_i$ denotes the inductive/direct limit of the inductive system)

Tobias Kildetoft

@lattice What an odd way to phrase these things. So the discrete preorder is the one where everything is related to everything (otherwise it would not be a directed system)

lattice

so... $a\leq b$ and $b\leq a$?

Tobias Kildetoft

hmm, but that would mean that there should be maps from each algebra to the other which is strange

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