« first day (3557 days earlier)      last day (1759 days later) » 
00:00 - 18:0018:00 - 00:00

00:03
@Alessandro: Thanks :P Never read that.
@skull: If you were a hobbit, wouldn't you choose to live in comfort?
yes sir
@TedShifrin it's just a dream... you can tell by the rhinoceros taking dictation in the corner.
LOL, thanks. And, you return to your usual disappearing act, happy almost birthday!
@TedShifrin sorry, I get drawn away frequently. I will soon be walking our dogs and then getting dinner. So disappearing again.
00:07
You're so predictable! Glad you're healthy.
How are the rabbits? @robjohn
One passed about a year ago, the other is doing fine.
@TedShifrin yes, things have been pretty healthy in my close friend pool.
Bonnie or Clyde?
@skullpatrol Misty passed, Snowy is doing fine.
@TedShifrin I will be back online probably around 7:30 or so.
PDT (GMT-7:00)
Probably dinner time for me, but we'll cross paths :P
00:42
@MikeMiller Which problem
I am going to read Bing topology today
maybe not
does a transcendental function always output a transcendental number?
Also is there any non-elementary function that no matter how many times we differentiate, is always non-elementary?
@BalarkaSen " can't think about manifolds all day and then sit down and prove a fact about a separable Lindelof space which is T_3 but not T_4"
what is your solution
don't do the latter
@Mathphile That's another really interesting question
00:56
@MikeMiller Actually what I said is bollocks, I just realized
T_3 Lindelof space is T_4
There is no such space
Ok, back to bing topology
So you're proving that such a space doesn't exist
nah i just said something in the heat of the moment
oh you mean the fact is "they dont exist"
thats fair
I'm drowning in a void
@RithikKapoor thank you
@geocalc33 any ideas for my question?
i just think every course should be a graduate course and i should have infinite time to read whatever
01:07
I got a Curious tag badge, but when I click on the badge to see what it means, it says I have not gotten the badge. I bet there is something out of synch in the database until 3:00 GMT.
@Mathphile for your first question no. There's something called an exceptional set
For a given transcendental function the set of algebraic numbers giving algebraic results is called the exceptional set of that function
it's often quite a small set
@geocalc33 so is there any non-trivial function that always outputs transcendental numbers?
By non-trivial I mean some function that is not like $f(x)=e$
01:22
@Mathphile well there do exist transcendental functions that produce transcendental numbers only when given transcendental numbers
but I don't think that there are functions that always output transcendental numbers
I assume this hasn't been proven yet
the reason being this. say $f(x)=e^{\frac{1}{\log(x)}}$
this is a nontrivial function by your standards
...@Mathphile by the way if you prove what kind of number $f(.5)$ is
please tell me
if you can prove the transcendence of f(x) for rational x in (0,1) then you'd prove a new class of transcendental numbers
@geocalc33 no clue about this one
no worries
it's just a hyperbola under a nonlinear mapping
The most I have been able to prove as a transcendental number is the solution of $x^x=n$ where $n$ is any rational number that is not of the form $m^m$ for integral $m$
01:37
transcendental number theory is very difficult
I am now trying to prove that the solution of $x^{1/x}=n$ is always transcendental for all rational $n$
nice
In dimensional analysis, transcendental functions are notable because they make sense only when their argument is dimensionless (possibly after algebraic reduction). Because of this, transcendental functions can be an easy-to-spot source of dimensional errors. For example, log(5 meters) is a nonsensical expression, unlike log(5 meters / 3 meters) or log(3) meters. One could attempt to apply a logarithmic identity to get log(5) + log(meters),
which highlights the problem: applying a non-algebraic operation to a dimension creates meaningless results.
that's kind of interesting
are you mostly into number theory?
@geocalc33 yes
although I lack formal education in it
I am currently studying computer science at uni but I still like to explore math on the side
01:45
@Mathphile ah I see.
I was able to prove that the solution of $x^{1/x}=n$ is transcendental for all rational $n$ that are not perfect powers
@geocalc33 if you are interested in transcendental number theory then maybe you can help me prove that $x$ is transcendental for all rational numbers $n$
I can make a new room for this if interested
02:10
0
Q: An elementary observation that maybe can prove the Twin Prime Conjecture.

EnjoysMathLet $P$ be the set of positive prime numbers. Let $P^k = P\cdot P \cdots P$ ($k$ times, elementwise). In other words $pqr \in P^3$ whenever $p,q,r \in P$. Let $P^0 = \{1\}$. Now $\Bbb{N} = P^0 \uplus P^1 \uplus P^2 \uplus \cdots$. Notice the union is disjoint and fully covers $\Bbb{N}$. Thus...

@geocalc33 please help save my post
from Arturo
@geocalc33 alright
@geocalc33 got downvoted :(
@EnjoysMath here's what I would do
I know break it into smaller questions, right?
02:17
But I still want a record up of who completed the proof, so I may not close this one
@enjoys you know how a rope sits in 3d? can you "fatten" an analytic function like that?
Nope, no idea
I just do crazy ideas :P
okay i think you can
it's a fibration
like a torus fibration. you break a sphere into tori
take the reciprocal of each tori's radius
and then reasemble to get a complex geometric shape
note: the sphere is assumed to be a symplectic shape
so if I wanted to fibrate the plane with hyperbolae minus the origin I would just break the plane up into infnitely many "fattened" hyperbolae?
02:36
When operating in the domain of integers, are changes from one integer to the previous or next integer considered continuous? From Wikipedia, a continuous function is a function that does not have any abrupt changes in value. Historically for example, when integers were the only numbers recognized, a transition from $1$ to $2$ would not have been an abrupt change, right, because the numbers are "right next to each other"?
 
2 hours later…
04:18
user image
3
@ÉricoMeloSilva
 
1 hour later…
05:56
Hi I have a question in real analysis: we want to find the limit of infinite series $\frac{n^3-1}{4n^5-3n^2+3}$ using comparison test, show that this series is converge. Now, divide all terms by n^5, we get 1/4. So it is finite non-zero limit. Now, it is enough to show that either denumerator or numerator is diverge or converge to show that the other is also the same thing.
I took n^3-1 <=n^3. Now, the partial sum of s_n=n^3 is 1^1 + 2^3 + 3 ^3 + ..... + n^3 which is equal $\frac{n^2(n+1)^2}{4}$ and if you take the limit of s_n, then we get infinite, so it is diverge. It should be converge. What is wrong here.
06:20
@robjohn that remined me of the definition of recursion :-)
@geocalc33 basically, because the prevalence of using computer algebra systems in mathematics is only going to rise, I am pushing for the formation of Stack Exchange communities for each of them separately, because there is a lot of ambiguity that would be resolved this way regard the interpretation of various exotic relational notations that are specific to the software used, etc
There is already one for Mathematica but not the others I believe
Like seriously the fact that this is not already the case is pretty embarrassing their sales and marketing departments must be comatose
07:22
Hi
Is k≡1 or 5 mod 6 is the same as saying k is not divisible by 3?
@skullpatrol recursion: see recursion
@northerner no... $k\in\{1,2,4,5\}\pmod6$
07:52
Hello Robbie
@robjohn who is john and how much money did you take from him
4
If we have to find the residue of $(z^2+1)e^{(1/(1+z^2)}$ at $z= i$ whose series representation is $\sum_{n=0}^\infty \dfrac{1}{(z-i)^{(n-5)} (z+i)^{(n-5)}\times n!}$, can we conclude that the residue is $\dfrac{1}{2i \times 6!}$ even though the series isn't in powers of $(z-i)$ only?
Or do we need to make it in terms of $(z-i)$ only before concluding anything?
08:30
Anyone?
@robjohn seems like Taxi Driver was one of your favourite movies
@Knight not really, there are just some lines
@robjohn Robbie please tell me how did you go through Analysis when it was it was first introduced to you
Your analysis is really awesome
@Knight not sure how to answer that.
@Knight Thanks
@robjohn Are you an undergrad student?
08:50
@Knight Ha! no... I taught math at UCLA in the late 80's.
@robjohn Oh my God! Now I know why you are so humble and kind
I bow thee sir
Can anyone explain in simple terms why the typical partial derivative notation can be ambiguous? I had a multivariable calc class where the instructor introduced his own notation to remove ambiguity, but I don't really get why $\frac{\partial z}{\partial y}$ is ambiguous in the first place? I understand it has something to do with function z depending on x, y but then in turn y depends on x?
I think the notation specified which "path" to take down the function tree to the variable not being kept constant.
 
2 hours later…
10:50
@robjohn Do you mind seeing my residue question above?
It seems impossible to separate (z-i) and (z+i)
 
1 hour later…
12:00
@robjohn hey you familiar with Sidi's method for root-finding btw?
13:02
This place looks like it is full of math nerd.
Can I be more off topic in this chat.
I am self teaching myself math lol.
I don't even have a professor which hit hard.
And get downvote when asking something stupid here which hit hard.
Am I the only one who has Average IQ? Well I am 17 yr old now. Anyway who the fk cares.
Wait this place doesn't even looks like chat. It's even worse than 9gag chat system lol.
Any mathematicians play ark? And observe human behavior in games? Or everybody here are wierd bookworms? I am bookworm but most of the time I stare at theorem than actually prove it. And I mostly proove them in my mind which is sign of laziness.
For God's sake can somebody chat please? I am having anxiety here from wierd guys like you. And when I go out I think people are weirder. I am like the global minima on my wierdoness function where people having higher and lower intelligence have higher wierdoness than me. Well I think I am the most normal human ever.
@RealDumbfoggybrain there's no reason to expect others to be chatting or to reply immediately
I just study real analysis which too me long lazy time to complete. I self study 2 books like for 4 months and get complete understanding of it. Look I am such a loser.
@SimplyBeautifulArt Thanks you got my anxiety out of my mind.
So I can keep chatting.
@RealDumbfoggybrain IQ is a terrible measure of intelligence, and honestly comparing yourself to others like that isn't helpful to anyone
I haven't speak for week because I am introvert.
@SimplyBeautifulArt Don't you think I am so slow at learning real analysis lol.
@RealDumbfoggybrain no? A lot of people find real analysis difficult.
13:17
Math is hard.
@SimplyBeautifulArt It is tricky it is not difficult but consume lots of time when self teaching.
I got 110 IQ in online test.
Surprisingly I am learning analysis and get complete intution of it but the only thing is When talking with hardcore mathematician I am kinda lazy and sloppy.
Which gets me fkin downvote.
And I cry like a baby.
Then watch life sucks memes for inspiration I mean get more depression lol.
...are you complaining over one downvote?
@SimplyBeautifulArt Yes because I am sloppy writer. I imagine number and watch behaviour.
@SimplyBeautifulArt You know mathematician are fkin serious about rigor. They want no imperfections.
And you get downvote.
You guys are seriously example of smart people.
You guys don't talk a lot everywhere and just observe.
Anyone good at measure theory? I wanted to understand something ...
I don't see what would suggest that but that's definitely not true at all
13:22
I just shut my mouth up in front of real people and I speak like idiot here.
if nothing else, it's definitely an over-generalization
Well I will learn measure theory then topology after freakin real analysis. Hopefully it will get less painful because I proved all the scratch now.
@MoreAnonymous Just ask, if somebody knows the answer they'll answer you
You know guys My life is full of pain. I save money without eating lunch to buy books. And my parents think I am spoiled brat lol.
(I do know some measure theory, but it depends on what your question is)
13:25
@AlessandroCodenotti I wanted to know if this question (of mine) made sense: math.stackexchange.com/questions/3419921/…
5
Q: What is the measure of $\int_{A}^B a_{\frac{x-A}{dx} } f(x) dx$?

More Anonymous Mathematicians like to ask covering various sets with open intervals and the answers to these riddles have strange tendencies to become strange lemma's or theorems Heine-Borel Theorem, lebesgue's Number Lemma, Vitali Covering Lemma, Besicovitch Covering Theorem, (etc) -3blue1brown By...

Is it okay to talk here about oneself and share pain?
Do you feel okay with it pal?
@MoreAnonymous I want to try it but I need to sleep.
@RealDumbfoggybrain I think it's okay. Personally, I think its quite common for people to experience things not working out in life. Just remember you should preserve!
@RealDumbfoggybrain you can ping me another time :)
@MoreAnonymous Thanks. I thought math lovers don't have emotion but I think this gives contradiction lol.
I feel good when chatting.
13:28
:)
@RealDumbfoggybrain what makes you say that? We don't.
The good feeling of being ahead of my classmates is I can teach them pretty much anything except undergraduate and higher mathematics.
@SimplyBeautifulArt May be I stared book too long lol. So everything was like emotionless for me. My teacher once told me May be I have asperger's syndrome.
Tell your teacher to teach, not practice psychology
I feel bored of everything except math. It is both addictive and painful.
Well my teacher has majored in neuroscience.
Had
@RealDumbfoggybrain @RealDumbfoggybrain why is it painful?
@AlessandroCodenotti any first impressions?
(on the question)
13:35
Neuroscience is not psychology @RealDumbfoggybrain
@MoreAnonymous It is painful when you need to post question. I feel very unoriginal when asking other.
@RealDumbfoggybrain I've asked loads of dumb questions (which I think is worse). And there's no need to compare. Who cares if the question has been asked before? The important thing is you understand the answer.
@skullpatrol Physiology is applied neuroscience lol.
True
Not Psychology
@MoreAnonymous I am not comparing. I think of famous mathematician like Gauss and Euler. They sound like they never struggled and were always original.
13:38
Hello
Guys don't you think the pandemic is making things worse like you can't buy books from bookdepository anymore.
@RealDumbfoggybrain Well theoretically it is possible to only have effort but not struggle. But I haven't been that lucky.
@NoName yo!
@MoreAnonymous But a great professors is someone who have struggled. I have heard of people who is successful in this field due to their persistent.
@MoreAnonymous Which you best of luck in this field lol.
'sup @MoreAnonymous. Your handle is even 'More Anonymous' than NoName. xD
@RealDumbfoggybrain I actually did 2 masters in physics :P
13:44
@MoreAnonymous Lol.
@MoreAnonymous How old are you lol.
Maximillian Janish graduate at 16 yr old.
From math department.
You know wut I didn't even understand plus and minus operation at age of 14 lmao
@NoName haha ... And done!
I was thinking about philosophy before 14.
@RealDumbfoggybrain What kind of philosophy?
After 14 I learned high school Mathematics in weeks.
@MoreAnonymous Natural Philosophy
@RealDumbfoggybrain Ah ... I engage in that from time to time
13:48
I accelerated quickly to undergraduate course yr 2 in years but wasted years playing fps game.
@MoreAnonymous I discovered lots of thing when I was elementary school kids.
I discovered things that were already discovered.
@RealDumbfoggybrain Yea ... Computer games can be addictive. Though they stopped being so for after a while
@RealDumbfoggybrain Tell me about that. I think thats everyones story
Though atleast I have one original formula :)))
@MoreAnonymous Like einstien I also thought of time is not absolute
To be fair I was quite lucky I asked myself an interesting question
Also in math trigonometry and calculus that I were thinking abou also include probability
@MoreAnonymous tell about yours too
@RealDumbfoggybrain I like Einstiens equivalence principle. I get confused when I try to extend it to QM though (for operator rather than vectors)
13:52
Well I have several thing going in my mind most of which I vaguely thought. Not with rigor
I had sorry
Not have
11
Q: The Definite Integral Problem (with a twist)?

More AnonymousThe Definite Integral Problem (with a twist) In the Riemann integral one essentially calculates the area by splitting the area into $N$ rectangular strips and then taking $N \to \infty$. Here's something I asked myself related to the Riemann integral. Let's say I split the area into say $3$ st...

@MoreAnonymous I am relearning physics lol. Staring which classical Mechanics again.
Classical Mechanics by R.Douglas.
It is pretty good book. But most of the time they just skip lots of things so I need to rethink how it was derived.
Classical mechanics is actually cool. I had it beaten into me from my high school days. Since, I studied for a highly competetive exam
@MoreAnonymous sounds like you study in prestigious university XD.
Until now everything is pretty intutive for me. Am sure those unsolved problem will Knock me out soon. I mean in math field.
@RealDumbfoggybrain Well Im not in uni anymore. And I think Durham university is kind of prestigious. But honestly I wish I was at Cambridge or Oxford in my uni days.
@RealDumbfoggybrain Why don't you try your hand at some open problems
?
14:03
look this isn't a subject to choose if your only motivation is to be considered intelligent and or fish for sympathy. I can assure you I have been just as young and emotional at some point. This life is not entirely merit based, and you need to learn to survive on your own satisfaction of what you study
@MoreAnonymous I will try it and I am sure I can solve any problem given good amount of time. But it will be waste of time since I need to learn. When I finish yr 4 course I will try it.
@AdamL "survive on your own satisfaction of what you study" - liked this part alot. I'm studying Ted Shiffrin's differential geometry these days via youtube. It kind of keeps me chirpy ...
@MoreAnonymous The only thing I feel absent is graduate course in my list. What you guys do at graduate.
?
@RealDumbfoggybrain You mean what did I do in physics? I did particle physics, general relativity, qft, ... pretty rusty at that stuff now
@MoreAnonymous yeah I have Ted's lectures on my you tube
14:09
@AdamL You uploaded them
?
@MoreAnonymous How much of that stuff you remember lol.
@RealDumbfoggybrain GR a bit... QFT I'm abysmal at. I must say if you understand something fundamentally it doesnt slip as easily with time
@MoreAnonymous I want to hear QFT experience more. I heard lots of people's arse has got kicked by it lol.
@MoreAnonymous Like analysis.😂😂😂
@RealDumbfoggybrain Yes. I would be one them (who got their arse kicked). Umm ... It's more about the language used. It's really confusing
@MoreAnonymous is QM intutive? Or it is just shut up and calculate?
14:16
I blame Feynman for this :P
@RealDumbfoggybrain Well I think I can grasp QM as long as we don't talk too much about the measurement
@MoreAnonymous Feyman didn't understand it himself lol.
@RealDumbfoggybrain well he had some crazy intuition which worked out
Its like Dyson said its actually a pertuabative series
@MoreAnonymous Sounds like I have lot of learn lol. My ego has been monotonic decreasing since I started learning undergraduate math and physics.
@MoreAnonymous Bye. Had nice conversation with you. Wish see ya.
@RealDumbfoggybrain Thanks you 2
@MoreAnonymous no I just created a playlist of what is already up there
14:24
@AdamL You have done good heed for humanity!
lol I don't know if I would describe my contribution to humanity as good perse. necessary in some regards but certainly not good
14:46
Hi everyone, I have a doubt please see if anyone can helop
I am paying 59 for a contest which would give 100 as a price if I win that contest how would the percentage of the charge they are charging and how should I calculate it?
There are only two participants in the contest so the price should have been 118
15:25
@Archer I worked on it last night. The residue is $$-i\sum_{n=0}^\infty\frac1{2^{2n+1}(n+2)!}\binom{2n}{n}=-i\frac{e^{1/2}}3\left(I_0\!\left(\frac12\right)-2I_1\!\left(\frac12\right)\right)$$ where $I_\alpha$ is the modified Bessel function of the first kind. I can post the whole derivation, but it is kinda long to spam the chat with unless you really want it.
@RealDumbfoggybrain college and grad school have that effect...
15:40
Hey Robbie! Sir what you had today in your lunch?
15:55
@Knight Just barely had breakfast recently. No idea about lunch yet.
Very good to see you back here, @robjohn! Maybe change the color orange to floral, for May Day tomorrow! ;-)
@amWhy thanks. Poppies are orange, and there are many of them around here.
@robjohn Dat true!
@robjohn Which computer do you have? What’s the model and company ?
@Knight The laptop I am on now is an Apple Macbook Pro
16:02
@robjohn “barley” why? Are you sick sir?
@robjohn Wow! You’re a rich man :-)
@Knight "barely", not "barley"
3
@robjohn Why you had barley “barely”?
@SimplyBeautifulArt no I hadn't seen that method before. Interesting. It's sort of like polynomial fitting meets the Secant Method.
Anonymous
Hi. Could someone point me to a proof of: "a subgroup having prime index is maximal"?
@Triskelion $[F:H]=[F:G][G:H]$
16:09
I still don’t understand the purpose of the word “barely” in this sentence “I barely had a breakfast recently ...”
Really LOL
I didn’t get that reply
Anonymous
@robjohn Oh, I see. If $[F:H]$ is prime then it cannot be factored that way. That is, there cannot possibly be a legitimate (strict) subgroup $G$ of $F$, of which $H$ is a subgroup. Thanks!
@Knight I just recently had breakfast = I barely had breakfast
@robjohn Okay!
You have taught a new use of the word “barely” (not to be confused with barley :-) )
Thank you! English is my second langauge so sometimes it becomes hard for me to understand colloquial terms
Where barley is in cereal and beer.
Hello! this might sound stupid. but i did not study integration and derivation theory. Whenn i derive this function u = x + y. So i derive du/dy . do i get 1 or do i get dx/dy +1 ?
when x is a variable and not a constant
16:24
@robjohn Barley is in this song also
@MadSpaces You would get 1 if $x$ doesn’t depend on $y$
@Knight and this song also
well how do i know if ti does? i am solving a differntial equation. on the left side y is derived to x and on the right side i have a term combining x and y .. i subtitue this term with a function u and i want to derive u to y.
and now i am in this situation
This should mean however that y and x depend on each other right?
well i believe the answer is yes because when i do such the equation solves it self elegeantly!
@MadSpaces In that case $x$ doesn’t depend on $y$
but y on x ?
If y depends on x cant one just reform such that x depends on y ?
@MadSpaces Yes
16:32
for example y = 2x we can reform x = y/2 so why you say no?
No, we always have to have an independent variable
@MadSpaces What will you input? Will you give $x$ or will you give $y$ ?
its just a differential equation there is no spesification
Anyway its not a big problem i can just do the same trick but i will derive after x instead of y
@MadSpaces In situations like that $x$ is always the independent variable
Probably yes
16:45
question
Anonymous
Is there any easy way to see that $A_4$ has only one order $4$ subgroup, without having to explicitly write down all subgroups of $A_4$? Also, can we determine the unique order $4$ subgroup without writing all the subgroups explicitly?
yoyo
I have a question
go ahead @MEcho
@Triskelion You don't need to write down all subgroups. A few considerations of what elements can be in one of order 4 will suffice
Anonymous
@TobiasKildetoft Well, an order 4 subgroup of $A_4$ can either have one generating element of order 4 producing a cyclic group (isomorphic to $C_4$), or it could have 3 non-identity elements having order 2 (isomorphic to $C_2 \times C_2$). But not sure what to do after this. Do you have some other hints to get started?
16:53
Can you list all order $4$ elements in $A_4$?
Consider the graph of the quadratic equation 2x^2 -3x + 1. Translate the (x,y) coordinate horizontally and vertically so that the new origin of the coordinate system is located at the point with old coordinates (5,4). The new (X,Y) coordinates are therefore related to the old coordinate system by the equations, X = x-5, and Y = y-4. Substitute these into the formula y = f(x) in order to express Y as a function of X.
(I realise this is pretty easy, but I keep getting the wrong answer, having trouble interpreting word problems atm)
Anonymous
@Thorgott Ummm, would they necessarily be 4-cycles?
yes, are any of those in $A_4$?
Anonymous
@Thorgott Well, I think 4 cycles are odd permutations, so they can't be in $A_4$. However, I'm not sure why an order 4 element has to be a 4-cycle (?).
do you know that any permutation possesses a decomposition into disjoint cycles?
Anonymous
17:00
@Thorgott Ah yes, I think I know that. How does it follow from there though (that every order 4 element has to be a 4-cycle)?
@Triskelion Do you also know how to find the order of an element given that decomposition?
do you know how to compute the order of a product of disjoint cycles?
Anonymous
@TobiasKildetoft Well, not directly. Given that decomposition, I could keep composing it with itself till it hits the identity permutation. But then I could do that with the original permutation too. I'm not sure how (and if) it follows that every order $n$-element has to be an $n$-cycle.
@Triskelion In general, it does not
But take a look through your book and see if it doesn't mention how to do this
Anonymous
@TobiasKildetoft Oh, so what's the required condition?
Anonymous
17:05
@TobiasKildetoft I'm not really using a textbook :P. Lecture notes only, and they're a bit haphazard (and don't really deal with this topic in detail).
You need to get a solid grounding in the basics of symmetric groups first
ok, let's do this step by step, do you know the order of an $n$-cycle?
Anonymous
@Thorgott Shouldn't it be $n$ always?
Anonymous
Right, okay. Let's proceed then :)
17:08
so, say you have two disjoint cycles $\sigma,\tau$, can you rewrite $(\sigma\tau)^k$ into something more tractable?
Anonymous
@Thorgott Hmm, I'm not sure if I can write it as something more tractable without further information. Are you perhaps referring to the fact that the order $\sigma\tau$ is the LCM of the individual orders of $\sigma$ and $\tau$?
well, if you know that, then you should already be aware of what I'm trying to get at
Anonymous
@Thorgott Well, not really. I don't entirely see how an "order 4 element in $A_4$ is necessarily a 4-cycle" follows from "the order of $\sigma\tau$ is the LCM of the individual orders of $\sigma$ and $\tau$"
Anonymous
Say, if an order 4 element of $A_4$ exists that isn't a 4-cycle
you know that every permutation can be written as product of disjoint cycles and you know how to compute the order of a product of disjoint cycles
the rest is trial and error
Anonymous
17:20
@Thorgott If an order 4 element of $A_4$ exists that isn't a 4-cycle, I guess it has to be of the form $\sigma\tau$ where $|\sigma|=1$ and $|\tau|=4$ or $|\sigma|=4$ and $|\tau|=4$
Anonymous
@Thorgott I'm not quite sure how to do the trial and error in the case of $A_4$. There are just too many possible elements and subgroups.
Anonymous
I will think about it though. If you have any ideas do let me know. I'll be back in a while!
There really are not that many elements to consider
Given the above, you ought to be able to give a complete description of the possible cycle decompositions that have order $4$ in any symmetric group
I added this picture to mathoverflow:
what you need to consider are the cycle types of a permutation on four elements (the cycle-type being the number of cycles of each length in the cycle decomposition of the permutation)
17:24
@Thorgott are you familiar with the theorem that won the war?
Thm: A permutation is the product of two proper involutions, if and only if its cycle type is matched
How do you solve such a differential equation $\frac{dy}{dx}+\frac{y}{x}-2=0$
@geocalc33 What does it mean to be matched in this context?
Try $y=u+x$
@TobiasKildetoft ah good question, if all cycle lengths occur in an even number, the cycle type is matched
17:28
Show this, show that, the show must go on...
@geocalc33 And what is a proper involution then?
a permutation is an involution, if it has order 2 as a group element
in $S_n$, or alternativly if its cycle decomposition consists of transpositions
(and fixed points) only. An involution is proper, if it has no fixed points.
Of course this is possible only, if n is even. I love this theorem
@MadSpaces $$u= y+x \\ \frac{du}{dx} = \frac{dy}{dx} + 1 \\ \frac{dy}{dx}= \frac{du}{dx} -1$$
Try replacing this value of dy/dx in the original differential equation
Yes ive done such
I have however reached then no further.
$\frac{du}{dx}=3- \frac{u-x}{u-y}$
17:44
I have got the answer when I used $u=y-x$
When we use the relation $u=y-x$ variables are getting easily separated
Anonymous
@Thorgott So there are 3 possible cycles types on 4 letters: 2-cycles, 3-cycles, and 4-cycles (apart from the trivial permutation). Now can an order 4 element in $A_4$ be some composition of 2-cycles and 3-cycles? I guess we don't need to consider 4-cycles as they don't exist in $A_4$.
Anonymous
Anyway, I should confess here that I don't know how to systematically write down all the elements in $A_4$ (and I'm not sure how to write the generating set either. )
that's just the types of cycles
but there are elements in $A_4$ that are not cycles
Anonymous
@Thorgott Umm, for starters I think all possible 3-cycles on 4 letters should exist in $A_4$ because all 3-cycles are even permutations (?)
Anonymous
Now what are the rest of the elements in $A_4$ that are not 3-cycles and how to explicitly write them down?
17:52
Say you have any 2 transversal lorentzian submanifolds consisting of infinitely many transversal dim. 1 manifolds, embedded in a compact subset of the real plane. Call this $M.$ Take all these 1 manifolds, and fix the intersection points. Define $N$ to be a 2-manifold, whose orthographic projection yields $M$. Is the manifold $N$ always unique?
can you think of an order $2$ element in $A_4$?
Anonymous
I guess all disjoint double transpositions must also exist in $A_4$
right
Anonymous
@Thorgott $(12)$?
is that an even permutation?
Anonymous
17:54
@Thorgott Oops sorry. I think only the disjoint double transpositions
Anonymous
$(12)(34)$
right
a double transposition is its own cycle type
Anonymous
Now can anything other than 3-cycles and disjoint double transpositions exist in A4?
a transposition has cycle type $1,1,2$, a double transposition has cycle type $2,2$ (what I'm doing here is listing the orders of the cycles in the cycle decomposition)
@Knight actually, it should be $y=u+x$
Then the equation becomes $\frac{\mathrm{d}u}{\mathrm{d}x}+\frac ux=0$
17:58
you should try putting together a complete list of possible cycle types
@Knight where the solution is $y=x+\frac cx$ for some constant $c$.
00:00 - 18:0018:00 - 00:00

« first day (3557 days earlier)      last day (1759 days later) »