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12:29 AM
@nerdy A set $S$ in a metric space $X$ is discrete iff for each point x there is $\epsilon > 0$ for which $B(x,\epsilon) \cap S = \{x\}$
 
pedro tamaroof, yep thats the definition
 
Unbounded sets needn't be discrete.
 
but anyways that proposition was wrong
infinite bounded also can be
i was confusing sets that have no limit points with discrete sets
i guess discrete sets are a subset of sets that have no limit points, correct ?
 
That doesn't happen in your usual spaces. Consider the Bolzano Weiertrass property.
For example, in $\Bbb R^n$ every infinite bounded set isn't discrete.
 
it's exactly this property that i'm trying to figure out intuitively
but i guess i'm too newbie, so i need to consider bolzano wiererstrass property in R
which means : "Every infinite and bounded subset of R has a limit point "
I'm tring to get the characterization of those sets that have no limit points.
someone told me, maybe those sets are exactly those sets that have only finite points in every interval in R
do you think its true ?
 
12:44 AM
@PedroTamaroff that sounds weird... "in $\mathbb{R}^n$, no infinite, bounded set is discrete" sounds better.
 
Hi @robjohn how are the bunnies, and have you decided on calling one of them skully?
Just to make them sound more masculine.
Butch Cassidy & the Sundance Kid pops to mind
 
1:40 AM
@IceBoy bunnies are fine. None are called Skully
@IceBoy So does Thing 1 and Thing 2...
 
2:33 AM
anyone still here?
 
 
1 hour later…
3:35 AM
There^ is your answer @DemCodeLines :-)
 
Hey guys, what math do I need to know before I start studying an intro to algorithms class? Thanks :)
 
3:56 AM
@Sawarnik: Yes, I know topology.
@Adam: Not terribly much believe it or not. Algorithms is a very broad field. Obviously if you want to delve into some kinds of algorithms in particular, you need to know much more math. Generally though, for a good basic grasp of algorithms, you mostly need good problem solving abilities and a broad exposure (not necessarily too deep) to various kinds of mathematics.
@Adam: Most of the graphical algorithms requires knowledge of trigonometry and spatial geometry. Algorithms about physics engine are easier to understand if you have some physics basis. If you want your program to help you to take decisions, you might need to study operational research which is a really huge sub-fields of math which includes graph theory, game theory, optimisation (which then includes analysis and linear algebra)
@Adam: Number theory definitely is helpful as is some basic graph theory. If you really want to be able to analyze algorithms, you'll also need a fairly solid grasp of Big O notation.
@Adam: I reccomend quickly going through The Art of Computer Programming which covers the essentials.
 
4:12 AM
Hey @Nick thanks
 
Welcome :D
 
spatial geometry... I'm wondering more and more what the applications are for this type of math
I've recently realized my enjoyment with math and im deciding to go back to school
but I'm taking a year to figure out exactly which route I want to go
thought I know I want it to involve computers
though*
 
@Adam: Hopefully, you mean going 'back to school' in a metaphorical sense. It would be a waste of time to actually fully learn the things I've listed above. It's much better to dive into the class and learn as you go along.
 
I'm doing a year of further math and then a masters in a computer/math related subject
and yeah im not going to FULLY learn them haha
I just want a sense of what they are used for etc.
and to find something that seems interesting etc.
all of the above @Nick
 
@Adam: That's a pretty good idea. Good luck with that.
 
4:17 AM
Thanks, would you mind sharing why you think it's a good idea @Nick
My BSc is in accountancy but I didn
t enjoy it much at all when I worked at Pricewaterhousecoopers
so im 26 now, and I think its as good a time as any to make the switch to doing something I enjoy
 
@Adam: Well, pursuing any kind of math is an endless but very fruitful endeavour. Anyone in this room can attest to that. Plus, computer science needs math to applied (properly).
Well, I have run.
 
@Nick I see. Thanks for the chat. Hope to see you around soon.
 
 
2 hours later…
5:58 AM
hi
could anyone tell me.. is this question unclear? math.stackexchange.com/questions/952446/… ?
I am surprised by the lack of interest in it
 
 
2 hours later…
7:53 AM
All my holy math books have arrived from amazon, yay!
 
@WillHunting: Congratulations.
Now, use them as pillows. There is a good chance osmosis may occur!
<end of wit>
 
@WillHunting Congratulations
 
@rehband: Do you speak German?
@rehband: sprechen Sie Deutsch , Herr?
 
8:09 AM
Ich liebe dich!
Eine kleine Nachtmusik!
 
@WillHunting: Are those songs?
 
@Nick One of them, lol.
 
Are you telling me no one has released a song named 'Ich liebe dich!'
That gives me an idea, I'll translate Bow-Chicka-Wow-Wow, retitle it and become famous in Germany!
... mhh, too bad I don't know enough German.
 
Greetings
 
Greetings
 
8:21 AM
@robjohn I need to show you something crazy awesome (but in private).
 
@Chris'ssis: If you're not too busy could you help me with a very simple type of differential equation. Just give me a hint as what to do: $$x^2\mathrm dy-y^2\mathrm dx+xy^2(x-y)\mathrm dy=0 $$
Apologies if it is too trivial for you.
@IceBoy: Greetings. Could you help me with the above?
 
post it on main and you will get a detailed answer
 
8:36 AM
I don't want a detailed answer.
Just a method.
 
Ask for a hint at the method.
 
I'll search for it first. :D
@IceBoy: So, how's your day?
 
Fine thanks, how are you my friend?
 
@Ice: Wandering in between topics. It's kind of like travelling planets. One moment you're on Venus, the next moment you're on Mars.
 
Topics within math?
 
8:40 AM
... No, topics within math, I consider countries or regions of a planet.
I have a definitive landing points on each planet.
 
Good plan.
 
@Ice: What's your favorite planet? + What's your favourite topic (pick anything)?
 
Mars, because it is red.
afk
 
@Ice: Really, you know about the $A$lliance for the $F$reedom of $K$osovo?
 
@robjohn @Chris'ssis I can't find that paper I mentioned the other day. Has any progress been made on that sum related to the Dedekind eta function at $i$?
 
8:46 AM
@Chris'ssis hey there.
 
@RandomVariable I didn't work on it anymore, I'm not used to theta function involving series.
 
@RandomVariable I tried, but I have it on hold for now.
 
afk
 
@Nick gesundheit!
 
@robjohn: and a Guten Tag to you too.
 
8:49 AM
@Nick whenever I say afk, I feel it is the sound a cat makes coughing up a fur ball :-)
5
 
i thought it meant
4 mins ago, by Nick
@Ice: Really, you know about the $A$lliance for the $F$reedom of $K$osovo?
until I googled it
 
@Nick in some contexts, perhaps, but not in a chat room :-)
 
@robjohn: Could you help me with the following:
22 mins ago, by Nick
@Chris'ssis: If you're not too busy could you help me with a very simple type of differential equation. Just give me a hint as what to do: $$x^2\mathrm dy-y^2\mathrm dx+xy^2(x-y)\mathrm dy=0 $$
 
@Chris'ssis I had fun with this answer.
 
@Nick Natürlich. Du auch?
 
8:54 AM
@robjohn Nice. Let me find a brilliant way there ...
 
@rehband: nicht. könnten Sie mir helfen mit meiner Frage?
 
@robjohn Done. Let me write it up.
 
@Nick Gerne
 
I got superb bounds on a some certain NT problem.
 
@rehband: The above differential.
 
8:58 AM
divide out by $dx$
well, don't look at me! that's the first step =P
 
Well, the limit gets reduced to computing $$\lim
_{x\to 0} \frac{\Gamma(1+x)-1}{x}$$ that's all
 
@RandomVariable Yes
 
@BalarkaSen Yes?
 
The eta function evaluation has been done.
 
@BalarkaSen By whom?
 
9:01 AM
ccorn pointed us at the identities derived in here
Essentially a tweaked form of something that I used here
 
@Chris'ssis Yes, but there is a bit of work beyond showing that. The other answer which shows some work simply assumes that $\Gamma'(1)=-\gamma$
 
@robjohn No, there is no work at all
 
@RandomVariable And I suggested something in here that goes on the line of BrunoJoyal's solution in the link above.
 
@Chris'ssis Oh? there is a "no pencil and paper" way to compute that limit?
 
@robjohn Sure.
 
9:03 AM
@BalarkaSen: Don't you think I tried that already? Then, I'm stuck at $\frac{dy}{dx} = \frac{y^2}{x^2 + x^2y^2 + xy^3}$
 
Eh, I am not going to try, thank you.
=P
@robjohn Stirling?
 
@robjohn $$\lim_{x\to 0}\frac{\ln (\Gamma(x+1))}{x}=\lim_{x\to 0}\frac{\ln (\Gamma(x+1))}{\Gamma(x+1)-1}\times \lim_{x\to 0}\frac{\Gamma(x+1)-1}{x}$$ Done.
 
@BalarkaSen: Yeah, it's messy. Atleast tell me if $\int\left(\frac 1{y^2}-\frac1y\right)e^{-y}dy$ is possible.
 
Yes, it is.
Might or might not take gamma. But it is possible to integrate those.
OK, I am out.
 
@Nick Hmm sry, I'm not sure...I don't know much about solving diff eq.s
 
9:06 AM
@BalarkaSen: Well, it's not possible for me. Wolfram's answer is too strange
I've never seen $Ei(x)$ before... ever.
Well, I'm go drown in research now.
@rehband: That's okay :D So, how is your day going?
 
@Nick It only just started! So far so good. Studying some old questions that Chris posted on MSE at the moment
How about yourself?
 
@rehband: I've been obsessing over little questions since morning. I should stop now and go learn something more interesting.
 
@Chris'ssis and how does that give $-\gamma$?
 
@robjohn l'Hôpital? As I said, no need for pen and paper.
 
@Nick What will you learn? =)
 
9:14 AM
@Chris'ssis wherein you use that $\Gamma'(1)=-\gamma$. That is what the other answer used.
 
@robjohn The value of digamma at $x=1$ is well-known.
 
@rehband: Obviously something boring. :D
 
@Nick Lol
 
@Chris'ssis yes, but that is not the point. I have computed that limit in many ways, but I was trying to find the most basic proof, using the least in background.
 
@robjohn Major part of people have no idea of the inequality in $(2)$. I think my way is pretty basic. :D
 
9:18 AM
@rehband: Plus, I solved the diff eq. considering $x$ to be a function of $y$. I'm glad that's over. Now, I'll move onto Chemistry... bis später!
 
@Nick Good stuff. Have fun, tschüss!
 
@rehband: Tschüß :D
 
@Chris'ssis I provide a proof of that. It is slightly different than Gautschi
 
@robjohn ok
 
The definition of Gamma is $x\Gamma(x)=\Gamma(x+1)$ and $\Gamma$ is log-convex. I used both of those in pretty much the most basic way.
@Chris'ssis Your answer is good, it just requires knowing the derivative of $\Gamma(x)$ which, although well known, requires a bit of work to show.
 
9:36 AM
@robjohn Can one learn about Gamma function and nothing about its derivative? I think it's mandatory to have such knowledge once you enter this stuff. :-)
Anyway, I found something interesting here ...
@robjohn I'm trying to find a nice way of proving that $$\int_{0}^{\infty} \frac{e^{a x}-e^{-a x}}{e^{\pi x}-e^{-\pi x}} \cos(n x) \ dx=\frac{\sin(a) e^{-n}}{1+2 e^{-n} \cos(a)+e^{-2n}}$$
 
9:55 AM
$$\frac{\sin(a x)}{\sin(b x)}=2 \pi \sum_{k=1}^{\infty}(-1)^{k+1}\sin\left(\frac{a k \pi}{b}\right)\left(\frac{k}{(k \pi)^2 - (b x)^2}\right) $$
 
10:11 AM
$$\operatorname{arccoth}(3)-\operatorname{arccoth}(5)+\operatorname{arccoth}(7)-‌​\operatorname{arccoth}(9)+\operatorname{arccoth}(11)-\cdots=\frac{\pi}{2} \log(3+2\sqrt{2})$$
There is one more thing that keeps me away from being like Ramanujan ... to master the area of partial fractions. Then, there will be no more limits ... :-)
 
@Chris'ssis why no more limits?
 
@robjohn I think that one can do anything in the area of integrals, series and limits with a very strong knowledge on partial fractions.
 
@r9m :D How are you doing? :-)
 
@r9m killed the complex zeta integral?
 
10:15 AM
Can anyone help,

If $I$ and $J$ are finite indexing sets. Is it true that:

$\bigcap_{i\in I} (\bigcup_{j\in J}A^i_j)= \bigcap_{j\in J}(\bigcap_{i\in I} A^i_j)$ ?
 
r9m
@Chris'ssis exam time ,, pressured !! ^^ :O
 
the one that appeared in Perron, i mean
 
r9m
@BalarkaSen (y) ;) Thanks !! your suggestion really worked !!! awesome :D
 
@r9m Aha, you mean you put pressure on your students ... :-)
 
excellent.
@Chris'ssis no, he means he is taking exam
 
10:16 AM
@BalarkaSen Believe me, I'm very good at identifying people with a strong math background. He is not just a student as you think. :-)
 
r9m
@Chris'ssis ?! :O omg !!
 
yes, r9m is a brilliant guy
 
@r9m ?! :O omg !!
 
i think he could do with some serious NT rather than integrals
he is good at NT
 
@r9m should be the future Prime Minister of MSE.
 
r9m
10:18 AM
@BalarkaSen nah man .. I'm just starting to learn this stuff ! ^^
 
well everyone starts with learning
personally i know nothing of nt except PNT
well, at least in analytic nt
 
@BalarkaSen: How do I find the values of x in the equation: $\sin x + \cos x = 1/x$
 
not solvable using elementary techs presumably
 
I tried using Taylor series expansion and then approximating
 
r9m
@Chris'ssis I'm a college student -_- (btw if I were setting papers .. I'd do my best to make them as enjoyable as possible ... not something scary and intimidating !! :( .. )
 
10:22 AM
@BalarkaSen: Can I haz Best approximation technique,plz ?
 
newton?
 
@BalarkaSen: Faster than newton ...
 
@r9m stick to NT man.
not sure @Nick
 
r9m
@BalarkaSen if only I have what it takes to understand this subject ^_^
 
you do. have faith.
 
r9m
10:25 AM
amen to that !! ^^
 
@BalarkaSen: I just need to estimate the 3 solutions in $x\in(0,2\pi)$
 
r9m
gtg (bbl)
 
me too. byes
 
@r9m: Good luck on your journey to be a NT Master
 
LOL @nick
 
10:26 AM
The result above is wrong for some reason I just noticed now.
 
and don't get attracted too much to integrals @r9m. don't let @Chris'ssis hypnotize you
 
@BalarkaSen lol, that's crazy. How to stay away from the most beautiful area? :D
 
=P
@Chris'ssis NT is not as bad as you think. Everything is beautiful in mathematics.
=)
 
Yeah, don't be a mathcist.
 
now i seriously need to leave
 
10:30 AM
Later pal
 
OK, no I need to have a talk with @DanielFischer. Are you there?
 
Depends, @Balarka, what do you want to talk about?
 
@BalarkaSen: Hopefully, I am not the reason you need to leave...
:(
Apologies for any rudeness or disrespect.
 
@DanielFischer Number theory. We have generalized a decision problem we (me and some other guy) is working on. it looks like this : given an interval [a, b] with real endpoints, produce a rational inside with smallest denominator.
@Nick eh? what did you do?
 
@BalarkaSen Farey fractions?
 
10:36 AM
@BalarkaSen: ... I have no idea. I sometimes rub people like a glass rod on a silk cloth.
 
Oh...?
googling
OK, this blows my mind. Thanks @DanielF
 
Welcome.
 
News flash: Daniel blows mind with two words :-)
 
Ah OK now that I see it, I believe I have read about them. Just not carefully enough to remember what they are and why I should care.
 
10:46 AM
Ford circles! Hyperbolic geometry! What the!
 
@Ice: Meanwhile, In the City of Townsville, Mojojo plots to destroy the Powerpuff girls.
 
Powerpuff girls were so silly.
Never watched them much.
@Daniel Farey sequences approach seems reasonable at algorithmic level, but can anything be done in theoretical setting?
 
@BalarkaSen: It was a reference to @Ice 's narrational skills.
 
oh ah
why the lol @N3buchadnezzar?
 
10:50 AM
@BalarkaSen: He either understood what I meant or he just laughed remembering the funtimes he had as a kid.
...Assuming he was ever a kid.
My dictionary defines a kid to be a baby goat.... I need to update my dictionary.
 
He is not even a human. He is... Nebuchadnezzar.
 
r9m
@BalarkaSen hehe .. I like NT ('cos there are loads of integrals and series) @Chris'ssis _/\ _ !! ;)
 
Just an e misplaced due to the watching of too much NUMB3RS.
 
@BalarkaSen: ... That used to be a TV show on AXN
NUMB3RS
 
@BalarkaSen NANBEERS
 
r9m
10:53 AM
oho !! pun pun (got it !!) :D
 
@r9m Don't be ridiculous. The integrals are just estimated, not detailed out with humongous closedforms with supreme technologies.
 
 
Analysis is used alright [ =( ] but not at the level of finding closed forms.
The analysis part is the reason I left analytic NT. I am in love with algebra.
 
r9m
closed forms are nice !! (you know how important closure is in people's lives =P)
 
Clopen forms <3
 
10:56 AM
@r9m eh, well, I am not interested in analytic NT atm anymore.
algebra algebra algebra
 
Al Jibr!
 
r9m
@BalarkaSen algebraic NT (I plan to read it this winter if possible :-) .. )
 
yes
@r9m Are you familiar with commutative algebra?
That's what I am studying right now. Hard stuff. Mandatory while reading algebraic NT.
 
r9m
@BalarkaSen nope .. thats why its postponed to winter (I don't have the prerequisites)
 
Have fun with that!
 
r9m
11:00 AM
_/\ _
 
afk
 
r9m
$\left| \right|$
 
I'm Bach.
 
@DanielFischer
 
Dang, I need to go again.
afk
 
11:12 AM
@BalarkaSen What do you mean? "In theoretical setting", it is clear that every nondegenerate interval $[a,b]$ contains at least one rational with smallest denominator. If you want to find it (or one of them, if there are several), that's an algorithmic problem, isn't it?
 
How can i show that
$$ \int_0^\pi \frac{(\cos x)^2\,\mathrm{d}x}{1 + \cos x \sin x}=\int_0^\pi \frac{(\cos x)^2\,\mathrm{d}x}{1 - \cos x \sin x} $$ ?
 
@DanielFischer Algorithmically it turns out that the run-time in Farey approach is polynomial, something like $O(n^{1+\varepsilon})$. We have just found out $O(\log(n)^k)$ algorithm using CFs. So that settles it.
 
@BalarkaSen Yes, if the interval is short, the naive way is too slow. But if you start with a reasonable guess (a convergent of a number in the interval), and consider the Farey neighbours with smaller denominators until you leave the interval, it's pretty snappy.
 
@N3buchadnezzar $x\mapsto \pi-x$?
 
@Chris'ssis Sorrey
@Chris'ssis It should be sine in the nominator!
$$ \int_0^\pi \frac{(\cos x)^2\,\mathrm{d}x}{1 + \cos x \sin x}=\int_0^\pi \frac{(\sin x)^2\,\mathrm{d}x}{1 - \cos x \sin x} $$
 
11:24 AM
@N3buchadnezzar How is that? $\cos(\pi-x)=-\cos(x)$
 
@Chris'ssis Look it is sine not cosine
 
@N3buchadnezzar I think you mix up some trig formulae
 
@Chris'ssis But they are the same, eg
$$
\int_0^\pi \frac{(\cos x)^2\,\mathrm{d}x}{1 + \cos x \sin x}
= \int_0^\pi \frac{(\sin x)^2\,\mathrm{d}x}{1 - \cos x \sin x}
= \int_0^\pi \frac{(\cos x)^2\,\mathrm{d}x}{1 + \cos x \sin x}
= \int_0^\pi \frac{(\sin x)^2\,\mathrm{d}x}{1 - \cos x \sin x}
$$
 
@N3buchadnezzar $$\cos (\pi-x) \sin(\pi-x)=-\cos(x)\sin(x)$$ I just showed you why they are equal, or I missed something?
 
@Chris'ssis Which works for the first to the third.
 
11:29 AM
@N3buchadnezzar Some signs failed there.
Anyway, back in 20 min.
 
$$\int_0^\pi \frac{(\cos x)^2\,\mathrm{d}x}{1 \pm \cos x \sin x} = \int_0^\pi \frac{(\sin x)^2\,\mathrm{d}x}{1 \pm \cos x \sin x} $$
 
11:57 AM
Attention all Wikipedians, Please fix this page. en.wikipedia.org/wiki/Range_(mathematics)
2
 
12:11 PM
what don't you like about it?
 
12:32 PM
@robjohn here is a cute result
$$\operatorname{arctanh} \left(\frac{1}{2} \right)- \operatorname{arctanh}\left(\frac{1}{6}\right)+ \operatorname{arctanh}\left(\frac{1}{10}\right)-\operatorname{arctanh} \left(\frac{1}{14}\right)+\cdots=\frac{1}{4} \log(3+2\sqrt{2})$$
 
12:47 PM
@Chris'ssis I wonder if that can be related to the sum of the arccot of squares that I computed a while ago.
 
$$\operatorname{arctanh}\left( \frac{x}{2}\right)- \operatorname{arctanh}\left(\frac{x}{6}\right)+ \operatorname{arctanh} \left(\frac{x}{10} \right)-\operatorname{arctanh}\left(\frac{x}{14}\right)+\cdots$$
$$=\frac{1}{2}\log\left( \frac{\displaystyle \cos \left(\frac{ \pi}{8}x \right)+\sin\left( \frac{\pi}{8}x\right)}{\displaystyle\cos \left(\frac{ \pi}{8}x\right) -\sin\left(\frac{\pi}{8}x\right)}\right)$$
 
@Chris'ssis That looks very close to the answer I got...
 
hi @robjohn
 
1:04 PM
People say that it is called Gnaussian elimination. Actually that is not true. Swedish speaking people know that it is all about eliminating the gneiss from the equations. Without removing the gneiss you are in big trouble.
 
@user2179021 hello
 
Apparently it is gneiss in english too.
 
let me come up with something marvellous ...
$$\int_0^1 \operatorname{arctanh}\left( \frac{x}{1}\right)- \operatorname{arctanh}\left(\frac{x}{3}\right)+ \operatorname{arctanh} \left(\frac{x}{5} \right)-\operatorname{arctanh}\left(\frac{x}{7}\right)+\cdots \ dx=2\frac{G}{\pi} $$
 
1:19 PM
@DanielF
 
@DanielF Apparently a separate problem but been wondering about this one : Consider the Riemannsurface of $w^3 = z^2 - 1$. I get branches at $[-1, 1]$ and $[-i\infty, i\infty]$. But this is incorrect, right?
The branch points aren't connected at all!
 
What do you mean with "the branch points aren't connected"?
 
Well, the branch cuts must connect the branch points over P^1.
Not?
 
You have three branch points (in the sphere), $\pm 1$ and $\infty$. If you make a branch cut connecting the three, you get three slit sheets, and you glue them together to get the surface (of which you already have a concrete realisation in $\mathbb{C}^2$).
 
1:25 PM
But I am getting weird branches. $[-1, 1]$ and $[-i\infty, i\infty]$. You can check those numerically.
 
@BalarkaSen You're letting a stupid computer compute, are you?
 
@DanielFischer I am using them to verify, yes.
 
The computer doesn't know where the value it computes the third root of comes from.
So it can't distinguish between the $1$ arising from $z = \sqrt{2}$ and the $1$ arising from $z = -\sqrt{2}$.
 
@DanielFischer Heh? Well, I am using the fact that at the branch cuts the (single-valued, the first conjugate) function $(z^2-1)^{1/3}$ jumps.
You can even verify it by hand. Use binomial.
 
So it produces an artificial branch on the imaginary axis. You need to determine/choose the branch-cut yourself, and adjust the result accordingly.
@BalarkaSen For the computer, it is "compute $\sqrt[3]{w}$". It doesn't know that you want a holomorphic branch of $\sqrt[3]{z^2-1}$ and that that's where the $w$ comes from.
 
1:31 PM
I am not sure what's the fuss about computer. Why not verify where function is discontinuous by hand? That's what I am doing.
@DanielF What's your visualization for having $[-1,\infty]$ as a branch cut? (Yes, I am reading Arnold's lectures)
 
@BalarkaSen By computer or by hand, if you compute w := z*z-1; r = cube_root(w);, you get an even function, which has an artificial branch-cut somewhere. If you use the principal branch of the cube root, that artificial branch-cut is the imaginary axis.
@BalarkaSen I don't visualize. (I don't even know what that means.)
 
Well, by having $[-\infty, 0]$ as a branch cut of $\sqrt{z}$ I understand that the action of the positive branch of the square root acts over points close together in $\Bbb C$ separated out the by the negative real axis as this :
And by hand calculations by use of binomial theorem, I get $[-i\infty, i\infty]$ and $[-1, 1]$ as branch cuts for $w(z)$ mentioned above. I can't visualize $[-1, \infty]$
 
@BalarkaSen You don't "get" branch-cuts. You choose the branch-cuts.
 
That's what is not making sense to me.
I understand branch cut for $w(z)$ over $\Bbb C$ as being an arbitrary complex line segment such that if two points are close and separated by that segment, then under the action of whatever $w(z)$, the points are moved away, i.e., jump discontinuity occurs.
 
A branch-cut is a curve that you remove to get a domain on which a holomorphic branch of the desired function exists.
 
1:41 PM
That's a big math jargon to me.
 
As a consequence, it so happens that the phenomenon of a jump occurs when you cross the branch-cut.
 
A branch cut is a cut in the tree branch.
 
@DanielFischer Well, yeah that make sense. If you remove it from $\Bbb C$ then you'll get an everywhere complex and analytic function, yes.
But that's just essentially what I said.
In a jargon-wrapper.
 
@BalarkaSen Choose e.g. $[-\infty,1]$ for the branch-cut. If $\lvert z\rvert > 1$, say, then you get a holomorphic branch of $\sqrt[3]{z^2-1}$ via $$z^{2/3} \sqrt[3]{1-\frac{1}{z^2}},$$ where you take the principal branch of $z^{2/3}$, and the principal branch of $\sqrt[3]{1-z^{-2}}$.
 
Ahhh. That must make sense.
Yes, that makes perfect sense @DanielFischer. This is so weird!
@DanielFischer Is there any way to rigorously prove via definition of branch cuts that any union of segments which connect the branch points of some certain holomorphic function $f(z)$ is a branch cut for $f(z)$?
 
2:00 PM
Hmm, if $f = \log g$, or $f = \sqrt[k]{g}$, then every component of the complement (in the sphere) is simply connected [well, at least if you take sufficiently regular curves for the cut, straight line segments are no problem], so a single-valued holomorphic branch of $\log g$, and hence of $\sqrt[k]{g}$ exists on that component. I don't know if you can prove it for abstract $f$.
 
@DanielFischer You can do this for algebraic functions, non? Take a bunch of algebraic functions $f_i$ such that the branch cuts are segments connecting the ramification points and show that branch cuts of $\sum f_1$, $\prod f_i$ and $\sqrt[k]{f_i}$ satisfy those too.
 
@robjohn I ask myself if we can get the closed form of $$\operatorname{arctan}\left( \frac{1}{1}\right)\log\left(1+\frac{1}{1}\right)+\operatorname{arctan}\left( \frac{1}{2}\right)\log\left(1+\frac{1}{2}\right)+\operatorname{arctan}\left( \frac{1}{3}\right)\log\left(1+\frac{1}{3}\right)+\cdots$$
 
In any case, I'll think about it. Thanks for the excellent help @DanielFischer!
 
@BalarkaSen I expect that you can do it for algebraic functions, but I don't know if there maybe lurk some problems.
 
2:16 PM
I posted a new question
0
Q: Closed form arctanlog series

Chris's sisWhat tools would you recommend me for $$\operatorname{arctan}\left( \frac{1}{1}\right)\log\left(1+\frac{1}{1}\right)+\operatorname{arctan}\left( \frac{1}{2}\right)\log\left(1+\frac{1}{2}\right)+\operatorname{arctan}\left( \frac{1}{3}\right)\log\left(1+\frac{1}{3}\right)+\cdots$$? Adding some par...

Sustain it by upvote ;)
 
=P
Hello @rehband
 
@BalarkaSen Hi Balarka
 
That's the correct way to spell my name, yes.
 
I have finally learned
 
{a, r, l} has nontrivial permutation group acting over it, so beware.
 
2:26 PM
Lolol
 
@BalarkaSen All groups of order less than 8 are trivial.
 
@DanielFischer @rehband @BalarkaSen @Anastasiya-Romanova what would you recommend me for the question I just asked on main?
 
@DanielFischer That's stronger than the Goldbach extremely strong conjecture, also called woodbach conjecture.
 
@BalarkaSen Depends on how one defines "trivial" ;)
 
2:30 PM
@Chris'ssis I'd recommend nothing.
 
@BalarkaSen :-)))))))))
 
Haha mouseovering gives a gem "The tautological twin prime conjecture states that the tautological twin prime conjecture is true"
 
2:52 PM
Theorem. Let $A=k[x]$ and let $V=k[x]/\big((x-\lambda )^{n} \big)$ for some $\lambda \in k$ and $n\in \Bbb{N}$. Then $V$ is indecomposable.
My book says that it suffices to prove that any proper submodule $U\subset V$ is contained in $I:=\overline{(x-\lambda )}$.
But I don't get why it suffices to prove that.
 

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