Mathematics - 2018-04-19 (page 1 of 3)

Kasmir Khaan

1:15 AM

Hi anyone here ?

Nicholas Roberts

2:27 AM

If you have a connected manifold of dimension greater than 1, is it still connected if you remove a point?

PVAL-inactive

@famesyasd Take the standard example $f(x)=x$ if $x \in Q$ , $f(x)=-x$ otherwise and modify it in the following way. Send the sequence $\{1/n| n \in \Bbb N \}$ to $\{1/2n \}$ sending $n$ to $n$ in the obvious way. Now pick any countable set A of rationals not close to the origin (for instance the positive integers \geq 2) and on A let f be any bijection from $A$ to $A \cup \{1/{2n+1}}$.

Nicholas Roberts

I know its not true in dimenion 1

Kasmir Khaan

@LeakyNun :d

Here?

:D

PVAL-inactive

Now f sends points in $A$ to points arbitarily close to zero, so $f$'s inverse is not continous at zero.

@NicholasRoberts Yes as it is true for $\Bbb R^n$ for $n>1$.

Nicholas Roberts

Ah ok, so that point that was removed lies in some coordinate path diffeomorphic to R^n, so removing a point there wont change connectivity as long as n > 1?

patch*

PVAL-inactive

2:32 AM

Connectedness isn't exactly a local property, but it still follows pretty quickly from that.

If you use paths, just note theres path between any two patches and that doesn't change after removing a point.

and that the patches themselves remain pass-connected.

Nicholas Roberts

Ok. thats exploting the fact that the patches cover the manifold and are path connected

PVAL-inactive

yeah

Nicholas Roberts

Thatsbetter argument. But why isnt what i said good enough?

PVAL-inactive

There's other arguments but you need to use some serious definition of connectedness.

user76284

Do these bounds look right, if $f$ is monotonically decreasing and non-negative?

I haven't found a source to check this.

Leaky Nun

2:44 AM

@KasmirKhaan ?

Kasmir Khaan

@LeakyNun wonna do some logic ? :D

kinda need help ._.

Leaky Nun

ok

Kasmir Khaan

with natural deduction in proposi logic

we go to logic room ?

or do it here?

Leaky Nun

u decide

Kasmir Khaan

i think logic room is better because i can see messages later :D

i text you there

Semiclassical

3:28 AM

I'm not sure what it signifies that I'm watching a Let's Play and because of that I now have a probability question.

Suppose I flip a coin repeatedly until I get five consecutive heads. How many flips do I expect to make?

...loool

45

Q: Expected Number of Coin Tosses to Get Five Consecutive Heads

A fair coin is tossed repeatedly until 5 consecutive heads occurs. What is the expected number of coin tosses?

Nicholas Roberts

@Semiclassical do you need help understanding that question?

Semiclassical

Nah.

Nicholas Roberts

Hah, just kidding :p

Semiclassical

lol

Though that question isn't actually quite the right one for the original problem

in the game being LP'd, they have a rock-paper-scissors game where the goal is to play until you get 5 wins in a row (with ties not being counted as interrupting the streak).

Nicholas Roberts

3:45 AM

Semi, are you a maths student? Teacher?

You seem very knowledgable

Semiclassical

Physics grad student

Nicholas Roberts

Noice!

Semiclassical

Since ties don't interrupt the streak, we can momentarily ignore them. Then it's just a 50/50 success rate, so this gives the question linked

And from there, you get an expected value of 62 non-tied games

...buuuuut, you still have to play the non-tied games

so what I don't know right now is how many games you'd have to play total

It'll be more, but I don't think it should be dramatically so because the ties don't interrupt the streak

PVAL-inactive

Idiocy: cnn.com/2018/04/18/politics/donald-trump-immigrants-california/…

Semiclassical

That seems entirely too mild a description.

PVAL-inactive

3:51 AM

A serious news source can't understand that the word "breeding" in the phrase "crime infested & breeding concept" is refering to crime.

Semiclassical

Ah.

Yeah.

PVAL-inactive

I completely disagree with Trump's sentiment, but this kind of straw man by a serious news source is ridiculous.

Semiclassical

Yeah.

It's an odious remark (and a poorly written one, to boot) but it's clearly referring to sanctuary cities breeding crime (not breeding people).

There's still a racial subtext, I think---that undocumented immigrants tend to be sources of crime---but that's different.

It's fine to criticize that implication, but misreading it just makes it easy to call 'media bias'.

user76284

Does anyone know how to show the Franklin graph is nonplanar without using Kuratowski's theorem? I've been racking my brain with this one.

PVAL-inactive

Taking a single statement out of context is dangerous but at least expected. Taking two words out of context in a four word phrase is absolutely insane.

Semiclassical

4:01 AM

well, I think the writing is also just kinda crap

"Soooo many Sanctuary areas want OUT of this ridiculous, crime infested & breeding concept."

if that was in a writing assignment, I'd think it'd get docked simply for being ambiguous as to what 'breeding' referenced.

PVAL-inactive

The extra oo's and the all caps aren't to my taste, but I think he is perfectly understandable.

Semiclassical

I don't object to that.

it's the construction of the end of the sentence

PVAL-inactive

Well if this was in a professional paper for instance, and you contacted the author for a misunderstanding like that you'd be laughed at.

Semiclassical

eh

if that entered into a professional paper, I think the editor would be who gets laughed at

PVAL-inactive

I mean like the slightly ambiguous word usage of breeding

not the entire tweet.

or a similar word

Semiclassical

4:05 AM

eh. I guess 'laughed at' is an overstatement. But I still think that the word usage there would give an editor pause.

If I were to clarify it, I'd probably do "crime-infested and crime-breeding"

because otherwise it is ambiguous as to what 'breeding' refers to

PVAL-inactive

That's probably objectively better, but I think I've seen way worse writing than this example in serious papers.

Semiclassical

So I do count that as poor writing. But criticizing Trump for poor writing is too easy

@PVAL-inactive maybe, but uh. he's the President?

the fact that his tweets have become so normalized at this point is just bizarre to me

PVAL-inactive

I have higher respect for the authors of these papers than I do for any president though.

It's viral marketing

judging by his election its been at least somewhat effective.

Semiclassical

judging by the 2016 election, yes

I'm not sure the judgment will be the same after the 2018 midterms

PVAL-inactive

Maybe, but I don't really think that America is any less vulnerable to the subversive techniques that Trump used to seize the presidency than before the election.

Semiclassical

4:20 AM

Depends on what the audience is, I think

Among his base, I think you're right.

But it's not clear to me how long such tactics can keep being effective among the wider electorate

Sooner

4:43 AM

Hey can anyone help me solve my problem which I posted yesterday? I will appreciate that!

Secret

8 hours ago, by Semiclassical

"I like to think of life as an infinite-branched tree that is slowly being dismembered!”

...with occassioanal mergings

[Random]

A | A

A B | A & B

A | A or B

If A B : A => B

ugh, I really need a good way to pack infinity so that it fits this chat bar

localplutonium

7:37 AM

is this a chat room where I can get hw solution help ?

DarkVampiric AbstractArtist

What question do you have? I won't be able to answer it all. I'm not really that capable.

@localplutonium what are you after?

localplutonium

An (m x n) matrix A that is row equivalent to an echelon matrix U that has (k) non-zero columns with a solution set dimension = m-k for Ax=0.

prove that this is false.

or explain.
Intuitively I would think this is true so its hard for me to figure explain.

DarkVampiric AbstractArtist

If U has k nonzero rows, then rank A = k and dimNul A = n - k by the Rank Theorem. So it's false.

localplutonium

can u link the theorem .

DarkVampiric AbstractArtist

Rank–nullity theorem

In mathematics, the rank–nullity theorem of linear algebra, in its simplest form, states that the rank and the nullity of a matrix add up to the number of columns of the matrix. Specifically, if A is an m-by-n matrix (with m rows and n columns) over some field, then rk ⁡ ( A ) + nul ⁡ ( A ) = n . {\displaystyle \operatorname {rk} (A)+\operatorname {nul} (A)=n.} This applies to linear maps as well. Let V be a finite dimensional vector space, and W be...

localplutonium

7:49 AM

and row equivalence implies that they have the same number of rows right?

DarkVampiric AbstractArtist

No, not exactly.

localplutonium

oh wait nvm now I get where u got rankA , since U is in echelon form with k rows ...

and thank you.

DarkVampiric AbstractArtist

No probs :).

Waiting

8:02 AM

Hi @robjohn. How is it going lately? I have a nice integral for you to try with complex integration.

$$\int_0^{\infty} \frac{\sin(a x)}{\sin(bx)} \frac{1}{1+x^2}\textrm{d}x$$

Actually, the contour integration is the way to go.

Anush

8:32 AM

hi.. I was wondering why math.stackexchange.com/questions/2743169/… is not getting upvotes

can anyone suggest improvements?

(thanks to the person who upvoted!)

DarkVampiric AbstractArtist

If it's something profound, clever, or groundbreaking, then yes you'd get a lot of upvotes.

Anyways, I was the person who has upvoted :).

Anush

@DarkVampiricAbstractArtist thanks!

lots of questions get upvotes which are just well.. average

Silent

8:47 AM

Hi

DarkVampiric AbstractArtist

Hi

Silent

What is suitable matrix for linear operator that represents right multiplication by A to row vector: this says it is $A^T$, while this says it is $A$.

Anush

@DarkVampiricAbstractArtist maybe the question is too hard?

user161117

9:28 AM

why do you want upvotes

philmcole

@Silent I am not sure at all but I think it would make more sense to be $A^T$. If you think about it like a linear transformation should give you $T(x)=T(x_1v_1+\ldots+x_nv_n)=x_1T(v_1)+\ldots+x_nT(v_n)$ then by matrix multiplication you can construct either a matrix $A$ with the $T(v_i)$ as columns and have $T(x)=Ax$ or a matrix $B$ with $T(v_i)$ as rows and have $T(x)=x^T B$.

I've assumed $B=\{v_i\}_i$ is your basis

Silent

@philmcole Alright! thank for your reply!

philmcole

9:48 AM

0

Q: Symmetric linear transformation has symmetric matrix

I want show that Let $T: \Bbb R^n \to \Bbb R^n$ be a symmetric linear transformation, i.e. $T(\mathbf{x}) \cdot \mathbf{y} = \mathbf{x} \cdot T(\mathbf{y})$ for all $\mathbf{x},\mathbf{y} \in \Bbb R^n$. If $B$ is any orthonormal basis of $\Bbb R^n$, then the matrix $[T]_B$ is symmetric. Why...

linear-algebra linear-transformations symmetric-matrices

anybody got an idea?

user112495

10:08 AM

I'm trying to model an SIS model on a complete graph with infection rate $\beta$, and recovery rate $\gamma$. However, the endemic steady state I'm getting is different to that the from the mean-field ode system suggests I should be getting

I suspect it is because one is discrete time (equal time steps of one day), while the other is continuous, and I am essentially using the rate as a probability in the simulation. But I'm not sure how to go about doing this properly.

Silent

11:44 AM

I think that this insightful answer has not got enough upvotes.

user8663905

So, I'm new to groups and homeomorphisms. I was hoping someone could help me out with a simple problem. Given $\phi : A \to B$ is a group homeomorphism. We define $ker(\phi) = \{ x\in G|\phi(x)=e^H \}$. Show that $ker(\phi)$ is a subgroup of $G$.

Silent

@user8663905 note that identity is in kernel, and show that if $x$ and $y$ are in kernel, so is $xy^{-1}$.

user8663905

Gotcha. That's exactly what I needed. Should I also show associativity?

ÍgjøgnumMeg

You need only show that $\ker \phi$ is non-empty and that if $x,y \in \ker \phi$ then $xy^{-1} \in \ker \phi$

the so-called "one-step subgroup test"

Tobias Kildetoft

@user8663905 Just a note: They are called homomorphisms, not homeomorphisms.

@ÍgjøgnumMeg That is a terrible test that practically never saves time

ÍgjøgnumMeg

11:50 AM

@TobiasKildetoft it works here though!

and how come it's terrible=

?*

Tobias Kildetoft

@ÍgjøgnumMeg It always works

It just doesn't make things any easier

ÍgjøgnumMeg

fair enough

it's just

something simple from an intro group theory course

idk

not an expert

@Tobias what's something better/preferable?

Tobias Kildetoft

@ÍgjøgnumMeg Just applying the definition.

ÍgjøgnumMeg

Fair

Szabolcs

I noticed, by playing around, that any random polygon triangulation I can make up seems to be 3-colourable. Is this generally true? I was able to find discussion about "Eulerian triangulations", which these are not, yet they are 3-colourable. Can someone given an example of a non-3-colourable triangulation?

Tobias Kildetoft

11:54 AM

The one-step test is not really that bad, but I usually advice against using it because students tend to get so enamored with it that they fail to see that it doesn't save them any time, and they forget what it really means to be a subgroup

Szabolcs

ÍgjøgnumMeg

@Tobias I see, I don't think I've ever done anything difficult enough that it hasn't been trivial to use that test though

Silent

@user8663905 product is associative in $G$, so if you restrict that to some subset of $G$, then it is associative there, too. So, no.

@TobiasKildetoft i will keep that in mind. Sorry that i recommended it.

Szabolcs

12:10 PM

I guess I should have asked on main instead. Here it is: math.stackexchange.com/q/2744487/12384

Akiva Weinberger

@Szabolcs So I guess the main thing distinguishing these from arbitrary maps is that each of these regions only have three neighbors (at most)

Szabolcs

@AkivaWeinberger Yes.

Sha Vuklia

Let $f_k(z)=(1+k)^{-1}z^{k+1}$. My book says that $f_k$ is holomorphic on $\mathbb C\setminus\{0\}$/. But it’s also holomorphic on 0 right?

Szabolcs

@AkivaWeinberger Not being familiar with this topic, I can't immediately see the implications of that, though.

Tobias Kildetoft

@AkivaWeinberger That is not enough, as that also holds true for the complete $4$-graph.

Szabolcs

12:21 PM

@AkivaWeinberger I found a lead: en.wikipedia.org/wiki/Brooks%27_theorem

Akiva Weinberger

@TobiasKildetoft Ah, I see. So I guess the condition is that every region has at most three neighbors, and every region on the boundary has at most two neighbors.

@Szabolcs I don't either

@Szabolcs A-ha.

So that solves it, then.

Szabolcs

@AkivaWeinberger It says that if the maximal degree is D, then it's D-colourable, except in two cases: 1. cycle graphs of odd length. Cycle graphs has max degree of 2, so they're still 3-colourable. 2. Complete graphs. I think the only planar complete graph is the one with 4-vertices, but that can't correspond to a triangulation, I think.

Akiva Weinberger

Right, it can't, because the triangles at the edge of the triangulation would have at most two neighbors, but there is no vertex in the 4-complete graph with at most two neighbors.

Tobias Kildetoft

@Szabolcs Right, to see that it does not correspond to a triangulation, note what happens when you glue three triangles to a triangle

once you make two of them have an edge in common, they cannot share one with the last one

Alek Murt

Hey guys, could someone help me interpreting this answer math.stackexchange.com/questions/2742350/…, i really don't see how to get the inequality from the last part of the answer?

Akiva Weinberger

12:31 PM

How did you make those images? @Szabolcs

user8663905

Could someone give me a quick rundown of exactly what this is saying? $ker(\phi) = \{ x\in G|\phi(x)=e^H \}$

Akiva Weinberger

The kernel of a homomorphism $\phi$ from $G$ to $H$ is the set of elements in $G$ whose image is the identity element of $H$

user8663905

Oh, the sup-H denotes that it's a specific element in H? I was interpreting it as "raised to the power of H"

Wouldn't sub-H be a better syntactic choice?

Akiva Weinberger

Yeah

Szabolcs

@AkivaWeinberger I am working on a Mathematica package for graph theory and network analysis. (I need the network analysis part, the graph theory is mostly for learning and entertainment.) It's mostly an interface to igraph, but it has other stuff too. It already had functions for converting geometric meshes to graphs in various ways, and I just added supports for finding minimum vertex colourings. I'll try to make a release in 1-2 weeks with this extra functionality.

@AkivaWeinberger Here's the package if you're interested, and if you have Mathematica: IGraph/M

DarkVampiric AbstractArtist

12:41 PM

Let A = P({1, 2, 3, 4}). Let f, g and h be the following functions.
f : A → A defined by f(X) = X ∩ {1, 4}.
g : A → Z defined by g(X) = {-1 if x={} }, {the smallest amount of X if x=/={}}
the smallest element of X if X 6= {}.
h : {1, 2, 3, 4} → A defined by h(x) = {1, 2, 3, 4} − {x}.2. (i) Is f one-to-one?
(ii) Is g one-to-one?
(iii) Is h one-to-one?
My attempt: (i) f is one-to-one. If f(x1)=f(x2) for x{1}, x{2} in R. then x1 ∩ {1, 4}=x2 ∩ {1, 4}, which makes x1=x2.
(ii) g is not one to one, since g({1,2})=1, g({3,4})=3

Can someone check that?

Silent

@AkivaWeinberger, will you please have a look at this?

Silent

12:59 PM

@TobiasKildetoft, I am confused here, don't we get contradictiction in first paragraph? so we can stop there, right?

user8663905

We define the order of $x (|x|)$ to be the smallest positive integer $n$ such that $x^n=e$ and if no such positive integer exists, we say that $|x|= \infty$.
So, if I am reading this correctly, this means that $n$ is the $\infty$ for all elements $x$ that are not the indentity?

Silent

@user8663905 What? I can't understand this sentence. We say that x has infinite order if there is no n such that $x^n=e$. For example, $1$, in additive group of integers, has infinite order, since for no $n$, $n\cdot1=0$.

user8663905

Looking up the wikipedia page, I forgot about cyclic groups. It seems that integer under addition does have infinite order for all elements which are not 0.

Silent

yes.

Tanuj

1:17 PM

@Rick Hi !

Rick

@Tanuj hi

Tanuj

@Rick how much are you getting in MAIN ?

Secret

[Random]

Evaluate $$\int^{\int^{\int}} e^{B_0(x)+J_{-1}(x)}\frac{H^{(5)}(x)}{\ln x} dx$$

Secret

1:41 PM

Evaluate

$${}^{\int}\int^{\int^{\int}} e^{x} dx$$

Evaluate

user193319

2:27 PM

1

Q: Upper Derivative and Increasing Function on a Compact Interval

Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $\overline{D}f(x)$ is defined as follows: $$\overline{D}f(x)=\lim_{h\rightarrow0}\left[ \sup \left \{\frac{f(x+t)-f(x)}{t}: 0<|t|\leq h \right \} \right]$$ I am ...

real-analysis measure-theory proof-explanation

2

Q: Vitali Covering Lemma Proof

Why may we assume that each interval in $\mathcal{F}$ is contained in $\mathcal{O}$? What warrants this reduction? Why is statement (4) true? If $x \in E - \bigcup_{k=1}^n I_k$, then $x \in E$ and $x \notin I_k$ for every $k=1,...,n$. Given some $\epsilon > 0$, there exists $I \in \mathcal{...

real-analysis measure-theory proof-explanation

2

Q: Subsequence of Measurable Functions

Given a sequence $\{f_n\}$ of measurable functions, why does there exist a subsequence $\{f_{n_k}\}$ such that $\lim_{k \to \infty} \int_E f_{n_k} = \liminf \int_E f_n$? I need to use this in a theorem I am proving, but I don't see how to justify it. Just for your information, I am trying to prov...

real-analysis sequences-and-series measure-theory pointwise-convergence

user8663905

Question: Do you guys generally place 0 in the natural numbers, or not?

Semiclassical

That's come up quite a few times in discussion on this chat.

The tricky thing is that, in some contexts, including 0 is necessary to make things simple; in others, it produces unnecessary complications.

Eric Silva

you do whatever makes you have to write less

Semiclassical

^

My own attitude is that, if I only care about addition, then I include 0

For instance, when writing out Taylor series I definitely think of n=0 as being a natural number

Eric Silva

anyone who takes a strong philosophical stance on it is just choosing the wrong hill to die on i think

user8663905

2:35 PM

Fair enough, it's a fairly pragmatic approach.

Semiclassical

On the other hand, if I care at all about multiplication / number theory then I really don't want 0 in there.

(I don't count stuff like modular arithmetic in there since you include negative numbers there)

Eric Silva

@Semi i guess i think, "if i exclude 0 am i gonna need to write down 0s" or "if i include 0 am i gonna need to exclude it"

Semiclassical

Right.

Eric Silva

if both become problems i just stick with whatever one i did first and just be consistent

Semiclassical

A good point of comparison, building off my Taylor series example, is that when you do a Dirichlet series you start at n=1. And that's precisely because the coefficients in Dirichlet series reflect multiplicative/number-theoretic properties of the positive integers.

Eric Silva

2:38 PM

does it really matter tho unless u choose to index like $\sum_{n \in \mathbb{N}}$

Semiclassical

Not really, no. It's more a philosophical point

Eric Silva

the only time i do something like this is with fourier series

i usually index with $\mathbb{Z}$ tho so then it doesnt matter anyway

Semiclassical

namely, that the 'natural' numbers to use in the Taylor series context are nonnegative integers, whereas with Dirichlet series the 'natural' numbers are positive integers

Anush

is there any way I can improve math.stackexchange.com/questions/2743169/… ?

it isn't getting much love

Semiclassical

The simplest way to make it unambiguous, though, is to just stick with positive integers vs. nonnegative integers...or, at least, that's the simplest way to do it in English

I think the French have different conventions as far as that goes?

Eric Silva

2:41 PM

probably

they hate being similar to english speaking math people so

Semiclassical

lol

I know there's a big difference between how people typically talk about continuity in the French tradition vs. the American tradition

Eric Silva

really? that i didnt know

Semiclassical

yeah. though tbh I'm having trouble finding evidence for that now :P

see the back and forth starting here:

Nov 9 '16 at 19:54, by Astyx

I've heard that when anglosaxons define limits for functions they don't consider the point where the limit is calculated. For instance $\lim_{x\rightarrow 0} \delta_{x,0} = 0$ (where $\delta$ is the Kronecker delta)

Eric Silva

ANGLOSAXONS

Semiclassical

lol

Eric Silva

2:48 PM

i am no saxon

philmcole

Hello

abenthy

Hi

Semiclassical

some apparent context here, though it seems a bit biased in its framing: reddit.com/r/math/comments/7rx8q3/…

abenthy

If $G$ is a semisimple Lie group, $K$ a maximal compact subgroup of $G$. Is the intersection of $K$ with a reductive subgroup $H$ of $G$ a maximal compact subgroup in $H$?

philmcole

I have difficulties understanding the difference between a quadric and a quadratic form. A quadratic form is a polynomial with every term having exactly degree 2. Is a quadric then just the solution set of points satisfying a certain quadratic form?

Eric Silva

2:51 PM

@Semiclassical idg why intuition should necessarily be the guide, the whole point of the definition is that intuition wasnt good enough historically

Semiclassical

shrug

I'm used to the American definition of limit

Eric Silva

im still hung up on anglosaxons

Semiclassical

lol

our anglosaxon mathematical heritage

Eric Silva

ic bespeak thaet westsaxona theode

but only a little

Semiclassical

I think the only old english I know is from choir experience

"Sumer is icumen in, Lhude sing cuccu"

I can actually remember how to sing the version we did

Eric Silva

3:01 PM

the thing that's really annoying about trying to like actually study the language is that the spelling wasnt consistent

and the words changed a lot over the course of the existence of the written language because of christianity

so then you're used to seeing heofon for heaven and then out of nowhere u get a slightly older text and suddenly see neorxenawange everywhere and youre like wtf is that

Semiclassical

lol

I suspect trying to read through the old English text of Chaucer would be interesting

Eric Silva

chaucer is middle english

Semiclassical

hmm

what am I thinking of then

Eric Silva

modern english speakers can just read chaucer

it's hard and you need a middle english dictionary but you can do it

poetryfoundation.org/poems/43521/beowulf-old-english-version

here's old english

Meow

Does anyone know what Borel subgroups of $\text{SO}(n),\text{O}(n),\text{SU}(n),\text{U}(n)$,... are? I'm having trouble finding anything other than $\text{GL}_n(\mathbb{C})$ and $\text{SL}_n(\text{C})$ (and I guess the answer over $\mathb{R}$ is probably similar).

Semiclassical

3:06 PM

@EricSilva nooope

Eric Silva

i took a two month class that only read beowulf

it was my favorite class in undergrad

Silent

Please someone tell how we got $A'=Q^{-1}AP$

I don't get this because $X$ is not (necessarily) invertible

Eric Silva

@semi dvusd.org/cms/lib/AZ01901092/Centricity/Domain/2891/… as you can see reading canterbury is almost no diff

Semiclassical

hmm

I think I must be thinking of an older work, then, whose text I saw in a book and was just bowled over by how almost-English-but-not-quite it was

Mike Miller

There are parts that are hard to parse, most of it will be ok but there will be crappy parts now and then

You're probably thinking of a harder to read segment you'd seen

Eric Silva

3:17 PM

i think i had an easier time reading canterbury than i have reading joyce lmao...

Mike Miller

lol

Meow

@Silent You've only inverted $Q$.

Semiclassical

no, I think it was an older work

0celo7

@EricSilva the calibrations guy?

Eric Silva

no u cretin

Semiclassical

3:18 PM

Joyce is too confusing for me

Eliot is just confusing enough :P

0celo7

algebraic geometry is pretty rough, semic

Hmm, I must be missing something

oh, that Joyce

Eric Silva

this is a literature chatroom clearly

0celo7

amazon.com/Compact-Manifolds-Holonomy-Mathematical-Monographs/…

:P

Eric Silva

3:35 PM

tfw my professor meant to say "what is impedence" but said "what is penis" instead

user8663905

3:47 PM

Could a finite group be constructed $x^n|n \in Z$ such that, any $x^n=n^{-2}$ where $n \leq -2$ and $x^n=n^2$ where $n \geq 2$ ?

Semiclassical

@EricSilva and now I note that "impedance matching" is a thing in electrical engineering...

Mike Miller

i don't at all understand the question. what precisely is the set the group operation is on?

Eric Silva

4:17 PM

@Semiclassical lol

Waiting

4:28 PM

$$\int_0^{\infty} \frac{\sin(a x)}{\sin(bx)} \frac{1}{1+x^2}\textrm{d}x$$

With contour integration works perfectly.

$$\int_0^{\infty} \left(\frac{\sin(a x)}{\sin(bx)}\right)^n \frac{1}{1+x^2}\textrm{d}x$$

Semiclassical

I think you have to be a bit careful about the signs of a,b there to figure out how to close the contour, but yeah

Waiting

Say, $|a|>|b|$

Semiclassical

Hmm, okay

In principle one should be able to do contour integration for the second, but uh

doesn't look fun

actually, one has to be a bit careful about the meaning of that first integral since 1/sin(bx) will have poles on the real line

Waiting

Precisely.

Semiclassical

right way to understand is is probably to consider a contour which runs below the positive real axis, winds around the origin, and then comes back around above it

that still lets you use the machinery of the residue theorem

hmm

Waiting

4:36 PM

Or the contour runs above the positive real axis.

Semiclassical

sure. I'm being careless about the direction

Maybe it's easier to just take the integral as $\int_{-\infty+i\epsilon}^{\infty+i\epsilon}$? I'm not certain tho

Secret

chat.stackexchange.com/…

can he stop that?

Waiting

@Semiclassical Yeap

@Semiclassical Although unanimously accepted as one of the most powerful integration tools (the contour integration), its limitations are painful for some even simple integral problems, and those limitation are often given by the complexity of applying the method.

Secret

Thinking:

Waiting

On the other hand, for some integrals it's a splendid way.

Secret

4:50 PM

$$\int [f(x)]^n \frac{1}{g(x)}dx$$

Given $I=\int \frac{f(x)}{g(x)}dx$ is known, what criteria will ensure the above integrals to be expressed in terms of polynomials of I

(actually no, I need to think of definite integrals, it is too easy for indefinite integrals to break symmetry)

$$\int_0^{\infty} f(x) dx= \int_0^{\infty} g(f(x)) dx$$

Waiting

I studied lately tons of integrals and forced at maximum (based on my knowledge) the power of the contour integration. The conclusion is that eventually one needs to fruitfully combine the contour integration with the real methods for getting good solutions. Often remaining stuck on the contour integration area is pretty painful.

Secret

I have the opposite problem: I am stuck at real integrals

Waiting

For instance, how about integrals with complicated integrand containing say, 2 different polylogarithms and maybe 2 or 3 different logarithms?

Secret

my complex analysis is still not fully on track yet

My most recent play with integral symmetries is deriving something that looks like the lapalce transform, which semi then suggest I should focus on laplace transforms to see what happens

Mar 24 at 16:16, by Secret

$$\int e^{f(x)} dx = \int \frac{e^{u}}{[f' \circ f^{\leftarrow}(u)]}du = h \left(\int e^{f(x)} dx\right)+\int k(x) dx$$

where $\leftarrow$ is preimage

Waiting

Ah, and I remembered some with arctan raised at integer powers that almost put me down when trying contour integrations. No chance. The limitations are depressing.

But: I love contour integration (sometimes it is divine).

Secret

4:59 PM

The above semi failed attempt is a generalisation attempt from the following observation of this integral:

Mar 24 at 14:50, by Secret

$$\int_0^{\pi} \frac{x \sin x}{1+ \cos^2 x}dx = -\int_{\pi}^{0} \frac{(\pi - u) \sin (\pi - u)}{1+ \cos^2 (\pi - u)}du = \int_{0}^{\pi} \frac{(\pi - u) \sin (\pi - u)}{1+ \cos^2 (\pi - u)}du = \int_{0}^{\pi} \frac{- u \sin (-u)}{1+ \cos^2 (- u)}du + \int_{0}^{\pi} \frac{\pi \sin u}{1+ \cos^2 (u)}du = -\int_{0}^{\pi} \frac{u \sin (u)}{1+ \cos^2 (u)}du + \int_{0}^{\pi} \frac{\pi \sin u}{1+ \cos^2 (u)}du$$

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