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15:22
Hey @MeowMix
Tell me what you're working on
4
Q: A minimization problem in function fitting setup

Rajesh DachirajuLet $\Omega$ be a convex, closed, compact set in $\mathbb{R}^d$ with a smooth boundary. Given a data $(x_i,d_i)$, $x_i \in \Omega$,$d_i \in \mathbb{R}$, $i = 1,2,3...N$, and $\sum\limits_{i=1}^N d_i = 0$ I want to find a function $f:\Omega \to \mathbb{R}$, such that, $\int_{\Omega}f(x)dx = ...

Hello! Anyone here knows group theory and can answer this question:
2
Q: Can 2 Rubik's cube faces generate a corner 3-cycle?

user21820 If you turn only two adjacent faces of the Rubik's cube, is it possible to reach a state where only three corner pieces are out of place (and all other pieces are in the original places)? This was a question I thought of many years ago. At that time I was able to prove it by writing a comput...

@user21820 not I captain
@DHMO good morning
@SimplyBeautifulArt hi
@SimplyBeautifulArt: Haha you're watching this chat-room too?
15:28
@user21820 I just got here
@user21820 I'm pretty sure you can make it so 4 corner pieces are changed and everything else is the same using this method
but that's just because I own a rubiks cube
oh I'm thinking of edges sorry
@Dodsy: It's easy to prove that you can preserve all corner pieces (even orientation) while permuting any 3 edge pieces.
ah I see.
strange that it's the exact method.
This suffices to obtain the alternating group of edges and we cannot do more because to preserve the corners you need an even number of quarter turns.
you turn adjacent sides two times each in alternating and will reach a state where two edges from the bottom are on the top and two edges from the top are on the bottom.
15:34
That's a 2-pair-swap. There's a few well-known 10/11-move algorithms to do the edge 3-cycle in the 2-gen group.. But if you don't want to find them by trial and error, you could use the 2-pair-swap in a commutator to get a 3-cycle.
The 2-pair-swap by itself does not always generate the alternating group in a more general setting, but it does here.
so moving two adjacent sides once each leaves the cube in a state where 5 corners are changed.
but you're looking for exactly 3
:)
anyways you know your cubes man
I sit around all the time and find these crazy algorithms
like do you know the one where you can change just the middles?
Yes the problem is harder than at first glance. I didn't solve it for a decade, whereas all other invariance problems I came up with could be solved rather easily.
Yes that's 4 slice turns.
Which is a commutator on centre poles.
you are quite the rubiks theorist my friend.
I don't know any of this terminology :)
7
Q: Solving Rubik's cube and other permutation puzzles

user21820I've seen two questions on solving the Rubik's cube but none of the answers have given a complete solution using mainly mathematical techniques. Furthermore, I've not seen a good explanation of general techniques for solving permutation puzzles in general, including those of the Rubik's cube fami...

I explain them over here! =)
oh cool!
when I first learned to solve it I did not google it
it took me an entire year haha.
15:39
At first it was just a brief post to show how to mathematically solve the Rubik's cube and other general permutation puzzles, and then I later expanded it to include a simple actual algorithm for the cube that does not require any memorization.
That's nice. I first solved it by myself too; took me a long time as well.
Later I found some webpages describing commutators, but only when I had learnt group theory did I realize the technique could be explained and proven rigorously to always work for general permutation puzzles.
there's one algorithm that I use, not to solve it but just for fun.
trivial stuff
it makes these cool patterns on the cube
for instance some red is on the orange, orange on the red
can anybody help me in a question
@Dodsy 6 double turns?
15:42
there's lots of algorithms like this
No, give it a try. You do one turn on each side in opposite directions
two turns on the bottom
I have asked it but didn't get any response
0
Q: Set of values of $m$

search $C$ and $D$ will be lying in each half part of the circle $$x^2 - y^2 -2x=0$$ And $$y = mx+8$$ but now how to proceed . Can anybody please help me in this.

and then put the sides in their respective places
if you do it twice
or more times
in different places
you can end up with really cool patterns
@search nobody wants to get eye cancer
15:43
I see. You're in the opposite turns group right?
the typesetting in the picture is awful
@SimplyBeautifulArt man, you win
it reminds me of high school
I am not sure @user21820
Meaning that every quarter turn on one face is accompanied by simultaneous quarter turn on the opposite face in the opposite direction.
15:44
@SimplyBeautifulArt maybe you should use $n$ instead of $n+1$
but theres a cool pattern in that where everything is in the right place but 4 edges
well I guess they're in the right place
Yup that's the one.
there
are other patterns in that, like you can make l's
and diagonals
@DHMO your answer is good too though
I wish I could be more descriptive but I've never explained my cube habits
15:45
There's an interesting invariant for that one; you can never get just 4 edges flipped.
You can make 8 edges flipped; can you figure out how?
Also, can you find 7 moves that makes every face have a bit of every colour?
Heheh.
hm.
check out this cool pattern from that thing I showed you
That's 8 double turns. =)
Wait only 6 needed.
8 double turns?
huh maybe I'm going about it the hard way
takes me 10 turns
Find the number of different paths in Kn of length 2 from v1 to v2? anyone can help?
and using that algorithm you can change exactly edges
3
15:52
morning
I feel like I sound really stupid today
morning faust
@Dodsy On the solved cube do L2U2R2L2U2R2.
@user21820 would you discuss Collatz conjecture with me? I know you love it when noob's work on hard problems ;)
0
Q: Proving that a Group isn't Cyclic

JackI'm trying to prove that the group below isn't cyclic: $G = \{ (1, 2, 3, ... n)^a\cdot(n+1, n+2, n+3, ... 2n)^b \mid 0 \leq a,b \leq n-1 \}$ To do this, I'm trying to show that none of the elements of $G$ have order $n^2$ - however I've made little progress. Any help would be much appreciated,...

What does $G = \{ (1, 2, 3, ... n)^a\cdot(n+1, n+2, n+3, ... 2n)^b \mid 0 \leq a,b \leq n-1 \}$ mean?
what's that notation @user21820 ?
what's U?
15:54
@Dodsy I don't see what you mean there.
@user21820 Makes chequer pattern on 2 faces if I remember
Left Up Right.
and the 2 denotes the number of turns?
@BalarkaSen could you answer me? What does $G = \{ (1, 2, 3, ... n)^a\cdot(n+1, n+2, n+3, ... 2n)^b \mid 0 \leq a,b \leq n-1 \}$ mean?
Yeap by convention it is clockwise quarter turn, 2 would be a double turn.
15:54
Also, I sort of proved that the subgroup $\langle ab, bab \rangle$ is also the whole group, so good luck finding other subgroups @BalarkaSen
@DHMO what molecule is that
@user21820 oh cool!
No I have no idea what that question is trying to say.
@BalarkaSen heh
So my algorithms have more algorithms inside of them....
lmfao
15:55
@DHMO I wasn't paying much attention to explicit examples. It is true that there is a subgroup which is isomorphic to $F_2$.
@Dodsy In general, any pattern involving those 8 triplets can be solved by at most 8 double turns.
lol
I'm very impressed
@BalarkaSen alright
15:56
I've never met someone who knew these different patterns.
@BalarkaSen: I see you know group theory! Do you mind giving a thought to my Rubik's cube question? =)
what is $F_2$?
the red part is mispleading
This follows from my nuking the orbit proof. There are probably easier proofs too but I am not going to try to do it.
Free group on two generators.
@user21820 I think no state is more than about 23 double turns from completion.
15:57
@user21820 I've tried explaining how cool this stuff was to someone and I think I made them pretty bored. Never talked to anyone about this stuff.
@user21820 Nah.
@RobertFrost That doesn't make sense. You cannot solve the cube with only double turns.
0
Q: Denoting covers from a set $X$

ZophikelHow would one denote generated sets from a given cover within a set $X$ utilizing the definition in $(1.1)$ $$(1.1)$$ A cover of a set $X$ is a collection of sets whose union contains $X$ as a subset. Rigorously speaking we have if: $$C = \Big\{U_{a} : a \in A \Big\}$$ is an indexed family of ...

Face turn != double turn.
ok maybe 23 singmaster moves
@user21820 the point is, many moves are never necessary
15:59
@Dodsy Now you found one. If you solve my 7-move puzzle, let me know! =P
@user21820 how about this one? imgur.com/a/AKXP4
I like this pattern a lot.
What is Kn ? @Smit
@RobertFrost You should look up Wikipedia first:
There are two common ways to measure the length of a solution to Rubik's Cube. The first is to count the number of quarter turns. The second is to count the number of outer-layer twists, called "face turns". A move to turn an outer layer two quarter (90°) turns in the same direction would be counted as two moves in the quarter turn metric (QTM), but as one turn in the face metric (FTM, or HTM "Half Turn Metric", or OBTM "Outer Block Turn Metric"). The maximum number of face turns needed to solve any instance of the Rubik's Cube is 20, and the maximum number of quarter turns is 26. These numbers...
16:00
@user21820 imgur.com/a/AKXP4
did you see it?
it's a beaut.
@user21820 There we go, 26 moves. I was jsut saying it's fewer than you would think
@Dodsy You linked the earlier one.
I wonder if you have a different algorithm to get to this state.
no that's a new one
see the lines don't extend to the bottom!
@RobertFrost 20 double/face turns.
@Dodsy I don't have one to hand but try moving the centre up 2 turns then the bottom 2 turns. Then rotate the whole cube to the left a quarter turn and repeat 4 times - at a guess.
16:02
@Dodsy Oh I didn't reload the page; it was the same link.
oh I see.
If I have the series $f(u) = \displaystyle \sum_{j=0}^\infty c_ju^j$ where $\dfrac{c_{j+2}}{c_j} = \dfrac{j(j+1) - c}{(j+1)(j+1)}$, it is convergent or it diverges at $x=\pm 1$?
0
Q: Under What Conditions is an ideal $I \subset R$ in the given ring $R$ prime?

ALannisterI am working on the following problem: Let $R$ be an additive abelian group. Turn $R$ into a ring by defining $xy = 0$ for all $x$, $y$ $\in$ $R$. Under which conditions is an ideal $I$ of $R$ prime? I have already shown that if $I$ is prime, then $R - I$ is at most a singleton by showing ...

Sorry if I'm bothering you @user21820
@Topologicalife what is the equation this time?
16:04
@Dodsy I could do it in 13 moves.
I don't know if it is optimal.
I counted each slice as 2 moves.
@user21820 would you discuss Collatz Conjecture with me?
I think I did it in 10.
yeah 10 moves.
@DHMO it is an associated differential equation to Legendre's ODE.
@RobertFrost Yes I would discuss anything with you if and only if you first learn logic as I've recommended to you. =)
But I don't remember how to check if it is convergent or if it diverges at $x=\pm 1$. I think that diverges but I don't know how to prove it.
16:05
haha! come on I have a problem I need some help with
Otherwise sorry, find someone else to ask!
what's a slice, @user21820
@user21820 You any good with metric spaces?
@Dodsy Turning the middle layer is a slice turn. I counted it as 2 moves each. Did you do any slice turns?
middle layer
16:06
@RobertFrost I know about them; until I see the question I can't tell whether I can answer it.
R1L-1B2, turn cube quarter turn; R1L-1B2
the negative denotes that we don't turn the left in the same direction as the right
@Dodsy: Oh that trick.
I found that before but forgot.
Lol.
It's a commutator by the way.
@user21820 Forget this question math.stackexchange.com/questions/2232774/… but look at my own answer to it, which says this is a work in progress.. I need to prove the highlighted bit.
a commutator
ah!
cool!
I didn't even know people knew about these things honestly
but that was pretty infantile to believe.
I thought people just solved it and went on their way.
@RobertFrost I'm not going to look at Collatz conjecture in any shape and size.
@Dodsy: If you ever solve my 7-move puzzle, let me know!
16:10
@RobertFrost take any number $n$. If $n$ is even divide by $2$. If $n$ is odd, multiply it by $3$ and add $1$.
@user21820 This is definitely a novel approach. Go on; just look at the highlighted bit in the answer I've linked. It's a metric spaces problem. If you just give it 30s and hen say no I'd really appreciate it.
My rule is: logic first.
I learned logic
Nope. Don't argue.
@Astyx Complete Graph of size n denoted as $Kn$
16:12
@Dodsy: Going off soon anyway. See you next time!
@user21820 I will, thanks!
And I guess $v1$ and $v2$ are just two random vertices ? @Smit
@RobertFrost Are you a college student?
No. Just a hobbyist
I left school years ago
ah I see.
I guess I'm sort of an amateur hobbyist at the moment, until I go to school next year.
16:16
@Rüdiger eye cancer ??
yeah it is on high school level
@search I'd hardly say that that is high school level....
@Astyx In the question they didnt mention. So i guess we can assume that,
but maybe I went to a shitty high school.
@Smit Well any path $(v_1, x, v_2)$ for $x\in K_n$ satisfies that, so you get $n-2$ paths
Or $n-1$ if $v_1=v_2$
Meh I don't remember almost nothing about series.
16:20
@Dodsy actually it is high school competetive exam
I think it should be easy prove that diverges.
@Dodsy anyway can you help me in that
If you don't remember almost nothing, do you remember almost everything ?
I don't have the knowledge to help you, I can't even visualize the problem.
why
16:21
What series are you talking about @search ?
what visualisation @Dodsy
That one. I posted it here a few minutes ago.
39 mins ago, by search
0
Q: Set of values of $m$

search $C$ and $D$ will be lying in each half part of the circle $$x^2 - y^2 -2x=0$$ And $$y = mx+8$$ but now how to proceed . Can anybody please help me in this.

@user21820 the key bit is the bit highlighted yellow in my "work in progress" answer.
I wanted to say "at $u = \pm 1$". Not $x$.
16:24
@Adeek u any good with metric spaces?
@search For instance, if you know something enough you can see where everything goes and how it comes together. You can visualize how it works. Can you visualize two points in the complex argand plane? Can you see what the satisfied equation means for the points and what it is coming from?
Mm would be divergent because it is a geometric series?
Also, I feel like I can't understand most notation on this website.
@Astyx Why n-2 ? is it because of length 2?
16:25
Once you can visualize these points, you'll see how it all comes together, which way things are going and what it means
Because there are $n-2$ possible values for $x$
math isn't about solving for x, it's about understanding how things work.
@Astyx by doing this will my question surely get answered
It will have better chances yes
I can't understand your question for the moment, let alone answer it
@Astyx oh so it's not just me? great!
16:29
Hi @Sha
How are you ?
Hi @Astyx
pretty good! how are you?
I guess I'm good
I should be stressed by my exams (in two days) but strangely I'm not
hmmm... well, you'll get stressed eventually:P don't worry
XD
I'm not sure wether that's a good thing :p
haha XD (stress is good in moderation :P)
16:32
$G = \Bbb Z/2 * \Bbb Z/3 = \langle a, b | a^2 = b^3 = 1\rangle$
Can anyone find subgroups of $G$?
Trivially $\langle a \rangle$ and $\langle b \rangle$ are subgroups
but how many subgroups can we find?
which 4?
trivial ones and the one you said
apart from those 4?
I'm sure there are more
None other
16:34
what's your favourite functor
The cardinal of a subgroup must divide the one of the group
here 6
So it's either 2 or 3
@Astyx but $G$ is infinite
Is it ?
Why ?
So, any idea to check the convergence of that series?
compute the first few terms
like euler would
16:36
@Topologicalife What series ?
@Astyx $abababababab\cdots$ is an element of the group
@LeGrandDODOM ah right, thanks! using your hint, I get: $\mathbb P(X+Y\geq x+y)\leq\mathbb P((X\geq x) \cup (Y\geq Y))=\mathbb P(X\geq x)+\mathbb P(Y\geq y)-\mathbb P((X\geq x)\cap(Y\geq y))\leq P(X\leq x)+P(Y\leq y)$
$*$ is free product
My bad
@Astyx $f(u) = \displaystyle \sum_{j=0}^\infty c_ju^j$ with $ \dfrac{c_{j+2}}{c_j} = \dfrac{j(j+1) - c}{(j+1)(j+1)}$ at $u = \pm 1$.
16:38
Well then the one generated by $ab$
And $ab^2$
And $b^2a$, $ba$, $bab$
with $c \in \mathbb{R}$ and for $c \neq n(n+1)$ where $n = 0,1,2,3,\dots$
(otherwise it would be convergent since it is finite)
Any of those to the power of something (except $a$ and $b$)
@Astyx $(b^2a)(ab) = e$ so they are the same group
$(ab^2)(ba)=e$ so they are the same also
and, hey, @DHMO, you told me yesterday the recurrence relation it doesn't depend of where we center the solution of the differential equation. Actually it does.
@Topologicalife I see
16:41
In fact take any sequence of $a$ and $b$'s
That is because the coefficients of the Taylor series are not the same.
Simplify it till you can't anymore
@Topologicalife For Euler's sake just do it
It'll give you a list of all the elements of $G$ in their simplest form
@Astyx could you show me an example?
16:41
And seek relations between those
@Rüdiger which thing?
the one with the convergence
euler didn't do this either
So any element of $G$ can be written $a^{n_1}b^{m_1} \dots a^{n_k}b^{m_k}$
I don't know how to to prove if it converges or if it diverges, that is what I'm asking.
@Astyx $n_i$ must be $1$ lol
16:42
Where the $m_1, n_1$ are any positive integer
Figure out $n_i\in \{0,1\}$, and that $m_i \in \{0,1,2\}$
Suppose none are 0 except the extremities (which can be 0)
and then?
I need paper otherwise I'll end up saying nonsense
@Astyx I am your paper
So you get a list of "elementary products", ie $a,b,b^2, ab,ba,b^2a,ab^2$
agreed
16:47
You can figure out $b^{-1} = b^2$, $(ab)^{-1} = b^2a$, etc
agreed
So you finally get 4 elementary products $a,ab, ba, b$
@DHMO no spamming the same message repeatedly please
@Astyx and then?
6 mins ago, by DHMO
and then?
16:50
@SimplyBeautifulArt how does that bother you?
The you get cyclic subgroups of $G$
By taking the span of these and their powers
@DHMO oh no, I'm just being silly with you
I have a question: if $Tx = /lambda x$ for a bounded linear operator $T: X \to X$ where X is a complex Hilbert space, how would I go about proving $\text{Im}(\lambda I - T) + \text{Ker}(\lambda I - T)$ is dense in $X$?
eff Hilbert space
would be very grateful for any hints, thanks very much
16:51
ie $a, b, (ab)^k, (ba)^k$
@Astyx are there no other?
I agree, eff Hilbert space, but there is revision to be done
There are others
Giving a full list might be ambitious
@Astyx for example?
For each simplified sequence we stated above, you get a different span
And then there are subgroups generated more than one element, which are (much) more complicated
16:58
I'm curious in subgroups generated more than one element
because up to now I can't construct any example that isn't equivalent to the whole group
@DHMO for instance $abab$ and $baba$

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