Mathematics - 2015-03-28 (page 1 of 3)

Committing to a challenge

12:43 AM

Hello

I can answer questions about the Enigma @ThomasA, where was the question?

Alec Teal

@Committingtoachallenge

Committing to a challenge

@AlecTeal

What was this about btw?

Mar 19 at 6:34, by I'mGettingThere

@Committingtoachallenge GTFU?

David Wheeler

Get the fires underway?

Committing to a challenge

It's an acronym for grow the f*** up, but he never clarified what he was referring to

Ropstah

"Get the f*ck uot"

:)

Alec Teal

1:03 AM

@Committingtoachallenge I wanted your help

It's up!

evinda

Does it hold that it is known that all problems in NP, apart from the NP-complete ones, can be solved in polynomial time by a deterministic Turing machine?

Committing to a challenge

@AlecTeal What's up?

Alec Teal

@Committingtoachallenge how are you getting on with books

Committing to a challenge

@AlecTeal Hadn't had much time for most of them recently, I have been doing assignments, but I have recorded in my book my pages skimmed and read for various texts

@AlecTeal How are you doing?

Working hard?

Alec Teal

Ill actually

Committing to a challenge

1:15 AM

That's unfortunate. My household is sick, and I have avoided getting too close to the sick ones since I am so busy

David Wheeler

I have been ill for days FML

Howcan

1:41 AM

I hope you feel better! The weather here has been changing so much recently that my sinuses seem to be constantly getting destroyed.

Alec Teal

@DavidWheeler you're good at stuff

What's the LaTeX for underline and overline

Like you know $x$ with a bar on top (or underneath)

Ropstah

"Good at stuff", except for geometry (sorry for the lols) :)

David Wheeler

\underline{x} for $\underline{x}$ and the same with overline

@Ropstah I haven't seriously done any geometry for like, 40 years

Ropstah

1:57 AM

Fair enough

I've been seriously affected by it for more than 30 years

Alec Teal

2:18 AM

@Committingtoachallenge will this go up or down, math.stackexchange.com/questions/1209635/… bets please

Committing to a challenge

It'll go down

Alec Teal

I bet that one

So you may not

But seriously @Committingtoachallenge I can't find why the argument is wrong! (Not the pi=4 one) but purely from measure theory

Committing to a challenge

Were you expecting me to say, it'll go up and then upvote it to win :P

You should put the question in a > field

I can't even find the question

Alec Teal

?

Committing to a challenge

Put the question at the top aswell, then people know what they are reading for

Alec Teal

2:22 AM

@Committingtoachallenge it's an argument for the perimeter being 4 using measure theory

See I can guess and be like "it's a different metric space" but thatd be a HUGE guess

@Committingtoachallenge math.stackexchange.com/questions/1209408/… I hate these STUPID SILLY REPGRAB questions

Jasper Loy

@Committingtoachallenge Hi Alex.

Committing to a challenge

Hey Jasper

Alec Teal

It makes me want to ask "must 0 and 1 be the same to have a functional number system" - which is just the proof that the additive ad multiplicative identity must be distinct on any field with more than one element.

Committing to a challenge

@AlecTeal What is wrong with the question?

(I mean in terms of him asking it)

Alec Teal

@Committingtoachallenge Find me a book that defines vector space which does not use right after the example of polynomials of basis 1, x and x^2.

find me one that doesn't say "Oh BTW bro, the things in this are called vectors"

Committing to a challenge

2:34 AM

Well now he too knows ;)

How are you doing @Jas

Karim Mansour

hey guys

Jasper Loy

@Committingtoachallenge Not too good. Hanging in there.

user129943

Committing to a challenge

Hi @Kari

Karim Mansour

I am trying to understand third isomorphism theorem without just from it set representation to get better intuition of it.

Committing to a challenge

2:35 AM

@KarimMansour Ahhh I can't help you there haha

Karim Mansour

:D

I will just go dabble a little bit more and if there is anything I am not quite convinced I will ask it as a question :D

its good to dabble xD

Committing to a challenge

Yep, sounds good

@JasperLoy Are you answering questions still?

Jasper Loy

@Committingtoachallenge Not at the moment.

Committing to a challenge

@JasperLoy You should answer some!!!11111

Jasper Loy

@Committingtoachallenge What is 11111?

Committing to a challenge

2:42 AM

Oh, it's sort of a pseudo-meme to post exclamation marks mixed with ones.

Jasper Loy

@Committingtoachallenge I am going to take a nap, see you.

David Wheeler

back in the day, 1337-speakers would type !!!!!11111, in homage to the incompetent typing of early gaming as a way of showing hyperbole in enthusiasm.

Jasper Loy

@Committingtoachallenge I hope my miracle comes soon, keep me in your prayers.

Committing to a challenge

|34c|< !N 7h3 |)4Y U5 1337357 94|\|\3Rz \/\/0u|d |)357R0Y

@JasperLoy Enjoy!

Karim Mansour

3:00 AM

0

Q: Visualizing Third isomorphism theorem

That is the statement of the third isomorphism theorem. Let K and B be normal subgroups of a group G with N $\subset$ K $\subset$ G. Then K / N is normal subgroup of G / N, and the quotient group $(G / N) / (K / N)$ is isomorphic to G / K. I have read and understood the proof but I am trying to...

what is wrong with the font in MSE

it sucks now

Kaj Hansen

3:11 AM

It's my 365th consecutive day on here :D

ᴇʏᴇs

3:22 AM

Congrats @KajHansen

Committing to a challenge

@KajHansen GRATS!

@KarimMansour I know. It hurts my eyes.

Karim Mansour

yeah !

congrats @KajHansen I will reach that soon in 334 days

Ropstah

Anyone know if these sort of translations are even possible?

Committing to a challenge

Possible in terms of what?

@Ropstah Obviously this isn't graph theory, so I don't know what it is

Ropstah

In terms of Coxeter polytopes

Consider this change

Committing to a challenge

3:35 AM

These are symmetric groups or something?

Ropstah

Yes these are symmetries

David Wheeler

@KarimMansour what exactly troubles you about that theorem?

Karim Mansour

yeh looks like it

@DavidWheeler I am just trying to get the intuition I don't know I feel I am missing something.

Committing to a challenge

What is the presentation?

Ropstah

What do you mean?

Karim Mansour

3:37 AM

@DavidWheeler know that if we consturct a surjective homomorphism from G/N to G/K whose kernel is K/N then use first iso theorem then it will give us the result

David Wheeler

Well, here is how I see it. $G/N$ is $G$ "assembled" into $N$-sized pieces, instead of "single-element" pieces.

Karim Mansour

but I want to see the logic

yeh

David Wheeler

A certain amount of those $N$-sized pieces will form $K$, and that subset of $N$-sized pieces is $K/N$.

Karim Mansour

but then what do we have when we do $G/N$ / $K/N$ I want to visualize that specific one.

Committing to a challenge

in Complex & Functional Analysis, 29 secs ago, by Committing to a challenge

Let $T$ be a mapping from $\Bbb C \to \Bbb C$ with a fixed point of $T$ being $T(z)=z$

Why can a mobius transform have at most two fixed points in $\Bbb C$

Karim Mansour

3:41 AM

yes

David Wheeler

Well, $K/N$ (the $N$-sized "clumps" that make up $K$) is a subgroup of $G/N$ (the $N$-sized clumps that make up all of $G$).

Karim Mansour

oh

I see

I see that is perfect @DavidWheeler yeh thats very nice way to think about it.

thank you

David Wheeler

The thing is, our normal intuition makes us see $G/N$ as "division" of some sort-in reality it's the OPPOSITE, it's "consolidation"

Karim Mansour

yeah

David Wheeler

All we're really saying is that $gK$ corresponds to $[g][K]$.

Karim Mansour

3:45 AM

yeah

David Wheeler

what makes this possible is that group multiplication imposes a certain "uniformity" on a group's underlying set-everything lines up nicely.

Ropstah

@Committingtoachallenge: also consider this, the edges of the dodecahedron (green arrows) 'comply' with rules, but the vertices (red arrows) don't

Karim Mansour

yeh since its closed under operation unlike for example if we just take arbitrarily set

and deal with that

Mike Miller

@Committingtoachallenge: Do you already know all the mobius transformations when you ask this? Because if so, you could just check them.

Committing to a challenge

I meant what is the group presentation

@MikeMiller What do you mean Mike?

Mike Miller

3:49 AM

I mean what I said. Do you know what the set of Mobius transformations $\Bbb C \to \Bbb C$ are?

David Wheeler

@KarimMansour It's really easy to see this with vector spaces in low dimensions, the cosets correspond to "natural clumps" we understand well-lines, planes.

Committing to a challenge

@MikeMiller Would the answer be different if we were looking at the set from $\Bbb C_\infty \to \Bbb C_\infty$?

Mike Miller

Yes.

Karim Mansour

oh yes right and for R^0 for example its just points

Ropstah

I wish i could speak as fluently as you haha

Karim Mansour

3:50 AM

yeah

Ropstah

Mathematically haha

Mike Miller

The latter are the Mobius transforms that also fix $\infty$. Actually, considering just Mobius transforms $\Bbb C \to \Bbb C$ have only one fixed point, if you don't county $\infty$.

Ropstah

i'm combining two groups @Committingtoachallenge

I think...

David Wheeler

So $\Bbb R^3$ can be thought of as a "deck of planes", or a "plane of thin straws"

Ropstah

One two-dimensional spiral based on Fibonacci and golden ratio

Committing to a challenge

3:52 AM

@MikeMiller Wait so we can only have one fixed point $\Bbb C \to \Bbb C$

So I should assume my lecturer just changed notation

Ropstah

The second is the group which corresponds with regular polyhedra, a dodecahedron in this case

@Committingtoachallenge: does that make sense, or does it need to be more mathematical in it's phrasing?

Karim Mansour

@DavidWheeler going back to vector spaces then there must some natural definition for cosets for vector spaces right?

Committing to a challenge

@Ropstah Well normally symmetric groups have a presentation in terms of cycles

Ropstah

I'm calculating points based on a single point of origin

Ah i see

There is a recursive pattern in the code, but i'm cheating with the polyhedra

David Wheeler

@KarimMansour Yep, they are TRANSLATES of subspaces (cosets that include (pass through) the origin)

Mike Miller

3:55 AM

@Committingtoachallenge Anyway, do this. Start with a Mobius transformation $\widehat{\Bbb C} \to \widehat{\Bbb C}$. Suppose it sends $\infty$ to $w$. Compose with $z \mapsto \frac{1}{z-w}$ to get a Mobius transformation that sends $\infty \to \infty$. This must be of the form $az+b$, so has precisely two fixed points unless $a=1$ (otherwise it has fewer, or if $b=0$ every point is fixed). Use this to conclude that your original Mobius transformation has at most 2 fixed points.

Karim Mansour

I see yeah that makes sense

Ropstah

@Committingtoachallenge: I'm starting with a radius as a parameter. The rest is generated based on that.

David Wheeler

The simplest example is our old friend the line $y = ax + b$-the $y = ax$ is the subspace part, and the $b$ is the translation part

Ropstah

So operations in terms of translation/rotation

Karim Mansour

its easier to see it that way once we express it in terms of low dimensional vector spaces

Ropstah

3:57 AM

Are mobius transformations related to what I'm doing?

Committing to a challenge

@Ropstah No, unfortunately noone uses those delicious subrooms and we get 3 conversations at once :)

Karim Mansour

@david it becomes more concrete

Clarinetist

Do any of you have suggestions on blog websites that use LaTeX/Mathjax/what have you?

Committing to a challenge

I have used wordpress and it renders latex with the addition of type $latex as the first dollar sign, rather than just the dollar sign

Clarinetist

Does it look nice and readable though, that is the question.

Karim Mansour

3:59 AM

@DavidWheeler actually now that I have intuition the motivation of the proof or the proof itself is much more intuitive now I can see how the person who made it thought about this theorem.

very nice indeed

Ropstah

I see

:)

David Wheeler

The nice thing about the isomorphism theorems is they cut "deep", they apply to a WIDE range of structures.

Committing to a challenge

@Clarinetist It looks bad

Karim Mansour

yeah

Committing to a challenge

@Clarinetist It looks really light and small, I hate it

Clarinetist

4:02 AM

Dang :/

Any other suggestions here?

Committing to a challenge

@Clarinetist Use pictures always

Karim Mansour

yeah I attended once lecture on category theory there is once branch of like generalized version of isomorphism called morphism that apply to many objects

Clarinetist

@Committingtoachallenge I can see why now...

Hmm

Karim Mansour

but didn't understood much though that lecture was way above my head

Committing to a challenge

Look how damn good that looks as a picture

Clarinetist

4:04 AM

It looks nice... but I gotta say, MathJax looks quite nice on Tumblr...

I might consider getting a Tumblr account, lol

Ropstah

@Committingtoachallenge: "That's not the way to communicate math through the internet"

David Wheeler

category theory can get very abstract, very fast.

Committing to a challenge

Well that page I linked you has mathjax on tumblr as well yes, looks really good @Clar

Karim Mansour

yeah

Clarinetist

Lol my sister is going to be baffled once she finds out I have a Tumblr account

Ropstah

4:05 AM

Maybe serverside processing was my thought -> but images are bad, sound reasonable... tex.stackexchange.com/questions/19036/…

Committing to a challenge

@Clarinetist Are you a trans-female, pan-sexual, who identifies as a wolf?

David Wheeler

the basic objects are "arrows", and naturally, we think of arrows as going "from" something "to" something.

Clarinetist

Lol am I missing a reference here? :P

Committing to a challenge

@Clarinetist Apparently you don't know tumblr, so yes hahaha

Clarinetist

Lol

Karim Mansour

4:06 AM

yeah I understood that and for example for groups we take isomorphisms

David Wheeler

we can take isomorphisms, but homomorphisms are more "useful" as arrows

Clarinetist

No, Tumblr, I do not want to be MysticMagazineBouquet.

Karim Mansour

I see

because we relate more objects that way I guess

it would be nice to study category theory but over the summer I want to finish dummit for algebra

with problems before moving on to more abstract things in math.

David Wheeler

4:21 AM

that's fine, it sometimes helps to have a sense of what is to be gained, first

Karim Mansour

yeah

Committing to a challenge

@MikeMiller Oh I am just using $\Bbb C\to \Bbb C$, but no, I don't know all such transforms

Mike Miller

They're just the Mobius transformations that fix $\infty$. Show that they're all of the form $az+b$.

Committing to a challenge

What does 'fix $\infty$' mean sorry

Mike Miller

It doesn't move $\infty$.

Committing to a challenge

4:29 AM

Ahh alright

David Wheeler

As we discussed before, if $f(z) = \dfrac{az+b}{cz+d}$, its clear $f(\infty) = \dfrac{a}{c} \neq \infty$.

Mike Miller

sure that can be $\infty$.

David Wheeler

Only if $c = 0$, which is what you said above.

Mike Miller

i see your point now

Committing to a challenge

5:00 AM

I will get back to that stuff when I work through enough of the text I have grabbed, so I can actually 'get' it, since clearly I don't

Kaj Hansen

@Committingtoachallenge, what are you studying?

Committing to a challenge

Doing complex variables by James Brown and Ruel Churchill, seems really good so far, and I am 40 pages in

@KajHansen Trying to catch up on Complex analysis

3401 is complex analysis so I am really behind

Kaj Hansen

Ah, I'm doing CA from Stewart & Tall

Really not super interesting though tbh

Committing to a challenge

I am finding this CA really fun though, but not algebra level of fun

But my CA I am learning is trivial to everyone else here

Kaj Hansen

I liked algebra, and I'm also really enjoying my topology course

Committing to a challenge

5:02 AM

I can't wait to get into topology

Kaj Hansen

Oh, my CA is mostly computational due to the course being watered down here.

Committing to a challenge

I have only done rudin level stuff and this complex analysis does some

Kaj Hansen

So you aren't the only one feeling that way.

Committing to a challenge

Yeah we have non-math students taking it

But this semester for the first time they have real-analysis as a prereq

So it isn't anywhere near as bad as it was in the past in terms of being watered down, which is good.

Mike Miller

@KajHansen: Try reading from one of the masters later; Ahlfors, say. The Riemann mapping theorem is one of my favorite theorems.

Kaj Hansen

5:13 AM

Thanks for the advice @MikeMiller. I do feel like analysis in general will be one of my weak points going into grad school, so I could really benefit from going through this stuff with a finer comb.

Mike Miller

For sure. Everyone needs to know a good chunk of analysis - even people whose interests are pretty algebraic.

But who knows, you might end up a topologist that uses PDE in his work. :)

(Anyway, even algebraists should know some complex analysis - the geometry of Riemann surfaces is a beautiful theory, that leads to algebraic geometry!)

Kaj Hansen

:D Historically, I've been very intrigued by what algebra has to offer, but I'm also really enjoying my topology course, so we'll see. I'll probably wait until I've actually added some depth to what I already know before I really choose a particular path.

Mike Miller

Just know that what a topologist does today is, essentially, completely and totally different than what's in your course.

Kaj Hansen

Yeah, so I've been told. I'll have to try out, say, algebraic topology in the near future.

Mike Miller

Reading Guillemin-Pollack's book on differential topology and taking an algebraic topology course will give you a good flavor.

Committing to a challenge

5:36 AM

@DavidW I can't find it now, but I was under the impression that 'matrix multiplication' in $\mathfrak{sp}_4$ is the commutor bracket, is this correct?

Disciple of Barney

It commutator @comm

Mike Miller

@Committingtoachallenge: When you're talking about a Lie algebra, you do not want to think about the bracket as a multiplication operator. It's just a Lie bracket. (The prototypical example of what a bracket should be - and what it is for the 'standard' matrix algebras - is the commutator.)

Committing to a challenge

Oh woops, I have read it many times and every time I read it as commutor rather than commutator

If I am looking to show that $X,Y\in \mathfrak{sp}_4$ and thus show that $XY-YX\in\mathfrak{sp}_4$ can I find an arbitrary matrix form for $X,Y$ and just use matrix multiplication?

Disciple of Barney

They have to be arbitrary, you can't pick a specific one. The multiplication $XY$ and $YX$ is normal matrix multiplication

Mike Miller

@DiscipleofBarney What was your previous username?

Disciple of Barney

5:42 AM

@MikeMiller Why do you ask? (Its not a secret just curious before I tell you)

Stan Shunpike

Hey y'all

Mike Miller

Well, I assume you've been here a while, since people don't usually become super active outta nowhere, so I assume I know you. I'd just like to know who I know :)

Committing to a challenge

I was told that what I want to show is that because $\mathfrak{sp}_4$ is the set of all matrices $X$ such that $X^t M + M X=0$ that I want to get $(XY-YX)^T M + M (XY-YX) =0$

Stan Shunpike

Anybody know anything about the curvature tensor? I got a conceptual question I'd like to clarify with someone

Committing to a challenge

But I can just find my arbitrary $X$ that is $4\times 4$ and fill it with arbitrary values(which I have done) and just do matrix multiplication to show that it is of the same form

Disciple of Barney

5:44 AM

@MikeMiller I actually havnt been super active till recently. My most recent name was Paul Plummer (my real name). I was sort of active on chat a couple year or two ago but I think I used a different name than my real one.

Committing to a challenge

@DiscipleofBarney Oh, hey Paul

Disciple of Barney

@Committingtoachallenge Hey

Mike Miller

Ah, gotcha. Thanks, appreciate it.

Disciple of Barney

@Committingtoachallenge You can fill them with arbitrary values, in a way that the matrix would still belong to $\mathfrak{sp}_4$ Although I feel like that might not be the easiest way (maybe it is).

Mike Miller

not convinced you can just do it like that. might have to be a little clever.

Committing to a challenge

5:51 AM

Sorry, my $M=\begin{bmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{bmatrix}$ if that makes it seem possible

Wait a sec, I have some pictures on my camera of my matrix of sadness lol

Mike Miller

shouldn't matter what $M$ is. pretty sure this will work for any $M$.

Committing to a challenge

I rubbed out my matrices without a photo, I cri everytym

What is the notation for the vector space of all $n\times n$ matrices

Over $\Bbb C$

is it $M_{n\times n}(\Bbb C)$?

Disciple of Barney

I am not sure if there is a universal notation, but that is pretty common

Mike Miller

6:08 AM

I'd use $M_n(\Bbb C)$

unless you're in a context where you want to also think about $m \times n$ matrices

Disciple of Barney

$Hom(\mathbb{C}^n, \mathbb{C}^n)$ If blocks of numbers make you sleepy, so you decide to say screw bases

$End(\mathbb{C}^n)$

Mike Miller

well, since it's $\Bbb C^n$, you've still got a basis :)

Disciple of Barney

Only if decide to use the coordinates as a basis, but yah

@Committingtoachallenge Thought about it a bit and there is definitely an easier way than filling than actually calculating the matrix

Kaj Hansen

@DiscipleofBarney cdn.slowrobot.com/2232015031586.jpg

2

Disciple of Barney

@KajHansen Hahaha.

Kaj Hansen

6:21 AM

:)

Disciple of Barney

@MikeMiller How much complex analysis is needed to understand the Riemann mapping theorem (I really need to learn some complex anlaysis)

Mike Miller

@DiscipleofBarney: Its proof is sort of disjoint from techniques you learn earlier in complex analysis. Normal families is the keyword; flavor is closer to real analysis than complex analysis in my opinion, but the theorem itself is very much complex-only. It can't hurt to find an exposition (my favorites as usual are Ahlfors and Conway) and go back to learn about the things it uses that you don't know.

Committing to a challenge

@DiscipleofBarney Can you give me a hint for this easier way?

Disciple of Barney

@MikeMiller Thanks, by Ahlfors you mean the complex analysis textbook or an expository paper

@Committingtoachallenge $X^t M + M X=0$ gives $X^tM=-MX$,

Mike Miller

His complex analysis book. Conway has two; you would want the first one.

Ahlfors might be too hard, as he's a bit terse. It's my favorite but that's because I've seen what he does before. His proofs are short and elegant. It might be a bad choice for a first go, if this is your first go. Might ask the regulars (especially Daniel Fischer?) for recs tomorrow.

Disciple of Barney

6:31 AM

@Committingtoachallenge Due you think old unaswered proof verifications / work checking questions should be deleted. They seem really specific specific and of no use to anyone but the OP, and if they are old they are probably of no use to the OP anymore.

@MikeMiller I will definitely take a look at both of them. I will probably also take a look around for an expository (with proof) paper for the Riemann Mapping theorem, a lot of big results tend to get pretty good expository papers written about them.

Committing to a challenge

@DiscipleofBarney Yes I do

@DiscipleofBarney Noone is going to bother answering them and the site would be better off with someone reasking the question if it is relevant to more people regardless

Disciple of Barney

@Committingtoachallenge That what I was thinking, just wanted a second opinion.

Disciple of Barney

8:17 AM

@Committingtoachallenge Are you MK?

Committing to a challenge

@DiscipleofBarney What is MK?

Disciple of Barney

Makato

Committing to a challenge

@DiscipleofBarney The big time troller?

Disciple of Barney

Well some guy using his name, this made me think of you

Committing to a challenge

I did become a big member when he got banned haha math.stackexchange.com/users/28422/makoto-kato

Kaj Hansen

8:20 AM

What'd he do?

Committing to a challenge

@DiscipleofBarney Haha yep that's my exact assignment question

Disciple of Barney

Oh that is not what you asked above

Committing to a challenge

@DiscipleofBarney (1 of many)

Disciple of Barney

Maybe in your class

Committing to a challenge

@DiscipleofBarney Yeah I asked part 3 of the question, he has asked part 1

@KajHansen He trolled constantly for months and raged at people until he was banned

Parts are i) Show sub vectorspace ii) Find dimension iii) what I was working on above

Disciple of Barney

8:23 AM

Okay, just wondering, you mentioned you had multiple accounts and I thought I "caught" you in the act

@Committingtoachallenge

What? Why?

Kaj Hansen

wut

Committing to a challenge

What?

xD

I don't know I just wanted an account for different things, so each could ask questions on one topic

Disciple of Barney

Did you frame all those people who got banned recently for vote fraud? @Committingtoachallenge

Committing to a challenge

This was my 5th account

@DiscipleofBarney Of course not haha, I have done zero fraud voting

Disciple of Barney

Is Makoto K your algebra account? @Committingtoachallenge

Committing to a challenge

8:27 AM

My total reputation in my excel document is a little over 5k

@DiscipleofBarney No haha, I have another account for some of them

@DiscipleofBarney I have already done that question he has put up, and our class has 10 people, so he will probably get caught with too similar a solution

Disciple of Barney

Did you post your solutions, how will they be similar (plus its not like there are a ton of ways to prove it)

Committing to a challenge

@DiscipleofBarney Nah it's due on Wed

Doing complex analysis atm

(because it's due on mon)

Disciple of Barney

Makes sense

Committing to a challenge

28 I just counted

Disciple of Barney

That is crazy, how do you even keep track of them

Committing to a challenge

8:36 AM

Excel spreadsheet lmao

I better get back to work

Committing to a challenge

9:04 AM

What are some important trivial functions I should play around with in the complex plane?

Like $\frac1z$

Kaj Hansen

@Committingtoachallenge $e^z$

Check out how regions, like rectangles, get mapped. (I don't know where you are in CA, but it's kind of cool to do that as an intro exercise).

Committing to a challenge

Really early

Like pretty much just started

Kaj Hansen

Yeah, so that's a good one.

Balarka Sen

finally. i think i have understood homology after all.

Kaj Hansen

But have you done your homework for your schoolteacher yet @BalarkaSen ? ;)

Committing to a challenge

9:08 AM

What defines a mobius transformation @kaj?

Or is that too hard to answer?

Is that rectangle thing a mobius transform?

Balarka Sen

@KajHansen i usually do my homeworks.

Kaj Hansen

I'm not sure @Committingtoachallenge. Never heard of such a thing.

@BalarkaSen, I was the opposite in high school. Did very, very well on my exams, but never did my homework. Was typically a waste of time given that it was computation I'd long since mastered.

Committing to a challenge

Maybe you heard another name for it" homographic transformations, linear fractional transformations, bilinear transformations, or fractional linear transformations."

Balarka Sen

and while the other students enjoy themselves by evaluating trig functions at weird angles from first principles, me and a few other have already proved addition formulas while doing calculus, so it's no big deal

Kaj Hansen

Sorry @Committingtoachallenge. Perhaps you're a lot further along in CA than I presumed you were :P

Committing to a challenge

9:11 AM

I'm really not haha

Balarka Sen

@KajHansen oh i rather prefer not getting detentions :P

Kaj Hansen

We didn't get detention, but my grades suffered. And my parents were contacted on a regular basis...

I take homework pretty seriously now that I'm in university though.

Balarka Sen

right

Disciple of Barney

Yah! $f(x)$ homework

Balarka Sen

my problem is that i almost never study history and language, @Kaj.

Chris's sis

9:17 AM

Greetings

@BalarkaSen @BalarkaSen I solved that problem finally and found the explanation with the primes. It was as @Semiclassical suggested.

Balarka Sen

i have lost interest since you didn't even post the integral.

Committing to a challenge

@DiscipleofBarney Is this going to be a thing now lol

@BalarkaSen That's exactly how I feel :\

Ahhh all the primes appear in order when you count $1,2,3,\dots$ of course!

Chris's sis

Yeap, there was nothing special.

Committing to a challenge

...

Balarka Sen

prime generating formulas are fun, they're never anything special.

but good find nonetheless, @Chris'ssis

Chris's sis

9:21 AM

@BalarkaSen They appear because of the factorization inside the log. That log covers all natural numbers when n tends to infinity. Hence it convers all primes.

Balarka Sen

yes, i get it

Disciple of Barney

@Committingtoachallenge Maybe...

Balarka Sen

you can just use $\sum_{k = 1}^n \log(k)$ instead :P

Committing to a challenge

What does $f(z)=\frac1z$ do? as a plot of $\Bbb C$

@BalarkaSen :'(. All that anticipation, next time I will just ignore her until she knows for certain

Balarka Sen

:P

Chris's sis

9:24 AM

Even with some coefficient in front.

Balarka Sen

@Committingtoachallenge What d'you want it to do?

Committing to a challenge

I have read that $f(z)=\frac{1}{z}$ inverses it over the real axis and keeps the magnitude the same, but when I compute it, this doesn't happen

Balarka Sen

I am not sure what you mean.

Disciple of Barney

Here is a very memorable (to me) analysis homework assignment I was given though here

Committing to a challenge

$\frac{1}{z}$, let $z=1+i$, $\frac{1}{1+i}=\frac{1-i}{2}=(\frac12,-\frac i2)$

But apparently it should just do $(x,y)\mapsto(x,-y)$

Balarka Sen

9:26 AM

no, why should it?

Committing to a challenge

Ummm some website said it should

Oh wait nvm

Disciple of Barney

@Committingtoachallenge $f(x)$ what websites say

Committing to a challenge

I have to go to dinner, but I just reopened the website and I see the image is just a little badly drawn

and the 1/z is a point on the plot

clarku.edu/~djoyce/complex/div.html

f(x) sake, that was a waste of time

Chris's sis

@BalarkaSen As regards your suggestion to one of my problems, isn't it more fit to say "linearly independent with the terms present in the integrand"?

Balarka Sen

sure. that'd be better.

Chris's sis

9:28 AM

Thanks. I'd like all be clear.

@BalarkaSen That problem is an incredibly beautiful problem, but I think you don't take such cups (I mean you're not interested in them).

Balarka Sen

Yes, I don't think I am.

Committing to a challenge

10:16 AM

I wonder if there is a point where you study too much and it negatively impacts you

Chris's sis

AWESOME

BBL

Committing to a challenge

Do you know any linear algebra? What about theory of manifolds?

Pretty sure both would help on your quest for integrals

Chris's sis

I did linear algebra in high school. I might send you some pictures from that textbook one day.

Committing to a challenge

I mean real linear algebra

Chris's sis

BBL

Committing to a challenge

10:23 AM

k

Ramanewbie

Is there a forum about Python on stackexchange ?

Committing to a challenge

11:30 AM

Some buffoon asked a trivial question, free rep for anyone for this LHF!!!

evinda

11:44 AM

Hello @cirpis
Could I ask you something about an algorithm?

Mary Star

Hello!! Is someone of you familiar with the Knapsack algorithm??

-1

Q: Application of the Knapsack Algorithm

I want to apply the Knapsack algorithm at the following: n = 4 (# of elements) W = 5 (max weight) Elements (weight, benefit): (2,3), (3,4), (4,5), (5,6) The first version of the algorithm is the following: K(0)=0 for w=1 to W K(w)=max_{w_i \leq w} {K(w-w_i)+v_i} I have found the follo...

algorithms computer-science

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$Mathematics$ Mathematics