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12:00 AM
I never did the symmetric polynomial approach, Artin's is much easier. I don't know that it would be true that every quantic polynomial's galois group would be S_5
quintic*
 
For $p,q,r,s$ in some field, how would I prove from the field axioms that $\dfrac{p}{q}\cdot\dfrac{r}{s}=\dfrac{pr}{qs}$, i.e., $(pq^{-1})(rs^{-1})=(pr)(qs)^{-1}$?
 
every irreducible one perhaps
 
($q,s\neq0$ here)
 
Is there something I can do to make latex commands show up in these chat rooms. i keep getting the source language instead of the rendered output
 
@DavidReed Look up "ChatJax"
 
12:02 AM
ty
 
no, look at the link to LaTeX in chat in the upper right @DavidReed
 
@TedShifrin The top search result (for me at least) gives the same link, so ¯\_(ツ)_/¯
 
oh, I guess it depends where you search, Legion
 
But yeah, would you have any idea how I'd solve my current problem?
The equation seems pretty basic
 
@LegionMammal978 first show that $b^{-1}a^{-1}$ is an inverse of $ab$ (this works in any group)
the rest follows from commutativity and associativity
I guess you also need to use the uniqueness of inverses
but I think if you use the notation $a^{-1}$ you already asumme that
 
12:07 AM
This seems sort of silly to me. Multiplication is commutative and associative ... so ... done.
 
is this the regex for this language: x starts with 0 and does not contain substring 101
 
I'm just not sure how to get from, as @Mathein said, from $aa^{-1}=1$ to $a^{-1}b^{-1}=(ab)^{-1}$
 
0*(1*000*)*1*0
 
@LegionMammal978 what happens if you multiply $ab$ and $a^{-1}b^{-1}$?
 
Well, what is $(ab)(a^{-1}b^{-1}) = a(a^{-1}b^{-1})b$?
 
12:10 AM
Assuming you've defined a field to be a commutative division ring via the notion of group of units, uniqueness is implied. mathein's approach is the correct way.
 
Ah, I see now
Lots of commutativity and associativity
 
That's what I said :P
Granted, I was a bit short.
 
@Theo The string 01 is contained in the language but does not match your regex
 
@LegionMammal978 why not? since (1*000) does not have to be present, it can show up zero times
there we are left with 0*1*0
 
@Theo It does not contain the final 0
 
12:15 AM
@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$
 
I just proved that the set of all zero divisors, along with $0$, in a commutative ring forms an ideal. Is this right, or did I make a mistake?
 
this is wrong
consider $\Bbb Z/(6)$
$3$ and $2$ are zero-divisors
 
products work, but not sums :)
 
but $3-2$ is not
 
Ah...I see...
 
12:21 AM
@MatheinBoulomenos Oh ok. I don't believe I've ever encountered that terminology. Thank you for letting me know.
 
@LegionMammal978 what about this? 0*(1*000*)*1*0*
 
@Theo That matches the empty string, which is not in the language
 
Okay...If $d$ is a zero divisor, then will the principal ideal $(d)$ consist entirely of zero divisors?
 
@LegionMammal978 0*(1*000*)*1
that should do it i believe
 
12:26 AM
@user193319 Good other rules to know for ideals: $(a) = (b)$ iff $a,b$ are associates and $(d) = R$ iff $d$ is a unit
 
@Theo Doesn't match 0.
 
the implication $(a) = (b)$ implies $a$ and $b$ are associates only works in integral domains iirc
 
iirc?
 
if I recall correctly
 
ah
You may be right, when I went through algebra the notion of an ideal was only defined for elements of integral domains
If that's true then I stand corrected.
 
12:30 AM
@LegionMammal978 this? 0*(1*000*)*0
@LegionMammal978 nvm doesn't cover 01
@LegionMammal978 is this possible? 0(0*)(1*000*)*
 
@Theo Doesn't match 01.
 
why not? 0(no 0 after)(one 1, no 000)
 
That would be the regex 00*(1*(000)*)*
And that doesn't match 010
 
is it possible to do: 0(1*0*)*
that way 101 won't occur, and 01, or 0000, or 010 still occurs
 
@Theo That matches 0101.
 
12:35 AM
oh right
okay im a bit lost, I know 0*(1*000*)*1*0* will make it so 101 does not occur in substring
but how do i avoid an empty string
and make sure it starts with 0
 
gimme a minute
 
hey
anyone like graph theroy?
anyone still here? :-o
 
12:56 AM
I haven't gone through graph theory in nearly a decade
My guess is if nobody here is up for it, comp sci ppl use it a lot, you may have luck in stack overflow chat
 
@Theo Okay, finally got 0(0|11*00)*(ε|11*|11*0)
Note that the last union operation includes the empty string
 
do we need the empty string there?
 
Yes, in case the string doesn't end with 11* or 11*0
You can omit the epsilon if you'd like
 
Thanks @DavidReed :)
 
Or expand out the regex into 0(0|11*00)*|0(0|11*00)*11*|0(0|11*00)*11*0
 
1:08 AM
ahh gotcha, thanks so much!
 
Yeah, ended up having to design an FSM and do manual state reduction
 
@alan2here Have to warn you though...they can be mean. Typical comp-sci nerd I know everything narcissism
 
In computer-regex, it could also be written 0(0|1+00)*(1+0?)?
 
and im assuming this is much more difficult to get a regex for
either all the symbols in odd positions within w are
0's, or all the symbols in even positions within w are 0's, or both.
 
@Theo gimme a few minutes
Is w 0- or 1-indexed?
Although I guess it doesn't really matter
 
1:21 AM
w $\in$ {0, 1}
 
As in, is the first position odd or even
 
i'm not too sure what you mean by 0- or 1-indexed
first position is odd
 
Okay
This is definitely possible
 
@DavidReed Including relevant greek mythology?
 
@LegionMammal978 how?
 
1:33 AM
@Theo I've already finished designing the state machine
 
@LegionMammal978 i tried doing (01)*(10)*(0)* but that leaves for an empty string
 
@Theo An empty string should be allowed
But that particular regex matches stuff like 0110
 
@Narcissusjewel Intellectual narcissism. Greek would be closer to somatic narcissism I believe.
 
@DavidReed I was just trying to make myself relevant, you know how we are ;).
 
hahaha.Well now you're relevant! Tell me what motivated that name.
 
1:43 AM
@LegionMammal978 how did you do it then?
 
@Theo As I said, FSM state reduction
I have a regex now, currently simplifying it as much as I can
 
okay
 
@Theo Okay, got 0*|1|01(01)*(ε|0)|10(10)*(ε|1)
It has 4 cases: 0000..., 1, 0101..., and 1010...
 
is 1 a case
?
I thought 0 had to be in the string
whether in even position, odd, or both
 
Whoops, slightly misinterpreted the language specification
Just remove that case, then
Although I suppose "all even positions are 0" could be considered vacuously true due to there being no even positions
Same with the empty string (to remove that you'd change the first case to 00*)
 
1:58 AM
00*|1|01(01)*(ε|0)|10(10)*
we'd end up with this then
 
As I said, you'd also have to remove the 1 case if there has to be at least one 0, so it would be 00*|01(01)*(ε|0)|10(10)*(ε|1)
 
but do we still need (ε|1) at the end?
 
Yes, so we can match strings like 10101
(while still matching 1010 and 101010)
 
ohh right
thank you !
 
2:18 AM
x does not start with 00 and does not end with 11
stuck on this one now
im going to bed after i solve this
 
well, lookarounds can be translated to regular expression :P
 
yeah but how do i translate this to regex?
 
it's not an easy thing to translate lookarounds generally
 
oh, how would i approach it?
 
well
"x does not start with 00" should be easier
 
2:24 AM
(01)*(10)*(1)*
0*(01)*(10)*(1)*
how about the second one?
@LeakyNun
 
I wouldn't build the regex first :)
I literally think about it in terms of states
 
something like this?
now how would i turn this into regex?
there should be a 1 from q1 to a0
q0*
 
2:39 AM
@Theo why do you need to turn into regex?
 
practice for finals
 
just translate your DFA to regex
it's a finite combinatorical problem :P
path findings
 
how? our prof didn't teach us this
we've never gone from dfa to regex
 
you just need to find all the paths from q0 to the accept states
start from q0
do branches
until it repeats
000(q4)
001(q4)
0100(q5)
0101(q3)
0110(q3)
0111(q0)
etc
 
so
(01)*(010(0)*)*(10)*(100(0))*
would that be it then?
 
2:44 AM
do you need any unions?
 
that is all the branches that go to accepting states
 
somehow you did it without any union?
 
I don't think i need union
but not 100%
 
01001 does not pass
god, when have I fallen to be a human regex-checking machine :P
 
01001 does pass, q0,q2,q3,q5,q5,q3
q3 and q5 are accepting states
 
2:48 AM
but not your regex
 
ohhh right
(01)*(010(0)*)*(10)*(100(0))*(010(0)*1)*
that works i believe
 
000(q4)
001(q4)
0100(0->2->3->5->5)
0101(0->2->3->5->3)
0110(0->2->3->1->3)
0111(0->2->3->1->0)
1000(0->1->3->5->5)
1001(0->1->3->5->3)
101(0->1->3->1)
11(0->1->0)
paths to 3:
---
2->3
1->3
5->3
---
0->2->3
3->1->3
0->1->3
3->5->3
---
1->0->2->3
1->0->1->3
---
0->1->0->2->3
3->1->0->2->3
 
Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2*sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine"
 
it doesn't have a period
but you might want to say that it oscillates more frequently as time progresses
 
yes indeed. Does that have a formal name?
 
2:55 AM
no idea
paths to 3:
---
1: 2->3
0: 1->3
1: 5->3
---
01: 0->2->3
10: 3->1->3
10: 0->1->3
01: 3->5->3
---
101: 1->0->2->3
110: 1->0->1->3
---
1101: 0->1->0->2->3
1101: 3->1->0->2->3
===
01: 0->2->3
10: 0->1->3
1101: 0->1->0->2->3
---
11: 0->1->0
---
11: 1->0->1
---
---
10: 3->1->3
01: 3->5->3
1101: 3->1->0->2->3
 
hey guys.
 
base form: 01|10|1101
with states indicated: [0](0[2]1|1[1]0|1[1]1[0]0[2]1)[3]
 
so I have a very fundamental question on functions.
 
loops: [0](1[1]1)[0], [1](1[0]1)[1], [3](1[1]0|0[5]1|1[1]1[0]0[2]1)[3]
 
let $f: \Bbb R \ton\Bbb R$ be an arbitrary line. how can I formally prove that its image is the reals?
 
3:04 AM
@LucasHenrique by "line" do you mean $y=mx+c$?
 
@LeakyNun yeah.
 
loops: [0](1(11)*1)[0], [1](1(11)*1)[1], [3](1(11)*0|01|1(11)*1(11)*01)[3]
@LucasHenrique well, what is the definition of "its image is the reals"?
combined regex: (1(11)*1)*(01|1(1(11)*1)*0|1(1(11)*1)*1(1(11)*1)*01)(1(11)*0|01|1(11)*1(11)*01)‌​*
 
@LeakyNun $\forall y \in \Bbb R \exists x \in \Bbb R: f(x) = y$.
 
slightly simplified: (11)*(01|1(11)*0|11(11)*01)(1(11)*0|01|11(11)*01)*
 
injectivity is trivial.
 
3:08 AM
@Theo and this is just half of the regex :P
 
i couldnt advance much
 
@LucasHenrique and how do you prove a statement that starts with $\forall$?
 
@LeakyNun you pick a random?
 
@LeakyNun wow.. thank you
 
holy shit
sorry
 
3:09 AM
Now $y$ is a random number
then you need to prove $\exists x \in \Bbb R: f(x)=y$
how do you prove a statement that starts with $\exists$?
 
@LeakyNun in this case you just solve the equation backwards
 
can you solve it backwards?
 
yeah.
 
I haven't told you anything
you've solved all of the question yourself
literally all that I asked you is to unfold definitions
remember this skill, it's very useful
 
@LeakyNun yeah, it was ridiculously easy
@LeakyNun actually that was my thought from the beggining
but for some reason I thought I was wrong
 
3:12 AM
now, the "proper" term for "random" is "arbitrary"
 
dummie me.
@LeakyNun yeah ikr
 
the proof goes like, "let $y$ be arbitrary. Then, when $x=\dfrac{y-c}m$, we observe that $f(x)=y$. QED"
 
since its surjective and injective then its a map
 
right
 
i’m a high schooler. as a graduate, are epsilon-lambda proofs that beautiful?
(totally unrelated)
cuz I love them
constructions, inequalities, etc
 
3:14 AM
good for you
@LucasHenrique well, beauty is in the eye of the beholder
my answer isn't that useful anyway
@AkivaWeinberger hola
 
@LeakyNun why tho?
 
@LucasHenrique is "beauty is in the eye of the beholder" a useful answer to the question "are epsilon-lambda proofs that beautiful"?
 
exactly
 
3:40 AM
My professor spent at least 20 minutes today proving that a trefoil isnt an unknot, it was fun
 
@LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$
I enjoyed developing the lebesgue integral though. I thought that was cool
 
@GFauxPas how did your professor prove it?
 
3:55 AM
put the trefoil on a torus, put the torus in $\mathbb S^3 \cup \{\infty\}$, took its complement, broke it down into factors of spaces with the free product of amalgamation,
used van kampen's theorem to find $\pi_1$ of the trefoil
and then found a sub-group of its $\pi_1$ that wasn't abelian
which it would be, if it were the unknot, we showed
took the knot's complement, not the torus's complement
 
4:15 AM
hmm
 
 
2 hours later…
6:30 AM
Exposing open problems to high schoolars to solve
 
 
4 hours later…
10:51 AM
I suspect $\{0\}$ is actually closed, since it is the complement of the union of all nonzero singletons, all being open sets
In addition, the sets $\Bbb{R} - \{a\}$ is also closed
The topology is almost discrete with a (suspected) limit point of $\{0\}$
As Leaky suspect, any set containing $\{0\}$ will be closed since the complement is a union of open sets
 
Hello
someone knows the Lagrange differential equation ?
 
In particular, this topology does not contain neither (open nor closed) sets
The topology also seemed to be second countable, with a countable base of the form all open intervals not containing zero
 
This space doesn't look second countable, almost every singleton is open
 
But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons?
uh wait a sec...
... I need arbitrary intersection to produce singletons from open intervals...
ok nvm then
 
11:07 AM
According to your argument the discrete topology on $\Bbb R$ should also be second countable while it's not
Anyhow this space is metric, so second countable iff separable and it shouldn't be hard to show that this space isn't separable
 
About first countability, is a constant sequence e.g. a,a,a,a,a,a,... a valid neighbourhood sequence for the point $\{a\}$ $a\neq 0$?
because it seems the neighbourhood of any nonzero singleton can only be itself in this topology
 
You can just take $\{a\}$, if a point has a minimal nbhd you don't even need a sequence
 
I see
 
@Secret $\{a\}\cup\{b\}$ is an open nbhd of both $a$ and $b$ if they are distinct and not the origin
 
also my mistake earlier: 0 is actually not a limit point since the neightbourhood $\{0\}$ contains only 0 and no other points
 
11:22 AM
This is not a nbhd of $0$, it's not open
 
right, cause there are no open balls of 0 radius
hmm... 0 does not even have a nbhd, since any set containing 0 is closed
I have no idea how to deal with points having empty nbhd
o wait a sec...
the open set of any topology must contain the whole set itself
so I guess the nbhd of 0 is $\Bbb{R}$
Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…)
Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time
 
11:48 AM
@Secret That's not a problem, the open balls centered in $0$ are open as well
 
hmm... something is not really adding up...:
$d(a,b)=|a|+|b|=|a|+|0|+|b|+|0|=d(a,0)+d(b,0)$
So going from a to b directly is no more efficient than go from a to 0 and then 0 to b
hmm...
$d(A \to B \to C) = d(A,B)+d(B,C) = |a|+|b|+|b|+|c|$
$d(A \to 0 \to C) = d(A,0)+d(0,C)=|a|+|c|$
so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination
But if the starting point is nonzero, then the distance increases quadratically
Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius)
only when travelling via the origin, will such qudratic penalty in travelling distance be not apply
More interesting things can be said about slight generalisations of this metric:
 
12:13 PM
Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ...
To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ...
2
Q: Analog of Vandermonde determinant for fitting a quadratic form?

Jarek Duda1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d...

However, there remains the question when fitting ellipsoid is well defined (det != 0) - it seems basic for e.g. algebraic geometry ... ?
 
$d(x,y) = \begin{cases} |x|+|y| & x \ne y \\ 0 & \{(x,y) \in ([a,b] \cup [0,c] \cup [d,\infty),[a,b] \cup [0,c] \cup [d,\infty)) : x = y\} \end{cases}$
Here, the shortest route become dependent on whether the desintation in question and everything in between is isolated (have x,y coordinates lie on the null line)
 
Here describing two points as intersection of such ellipses adds only symmetric (-x) points - the question is when it is true (also in higher dimensions)?
 
Any alg geom guys on? I know zilch about alg geom to even start analysing this question
Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit
To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — balu Apr 13 at 18:24
grr, thought I can get some more intuition in SR by using open balls
 
1:02 PM
@Secret, if you want to understand SR, study sine-Gordon model: extremely simple (lattice of coupled pendula) but already has Lorentz contraction, SR scaling of mass/momentum, time dilation for breathers: en.wikipedia.org/wiki/Sine-Gordon_equation en.wikipedia.org/wiki/Breather dropbox.com/s/aj6tu93n04rcgra/soliton.pdf
 
1:17 PM
“Theorem X: (1) and (2) are equivalent. Proof: Clearly, (1) iff (2).” Haaate youuu book author
 
At least omit the proof then. Don't get anyone's hopes up
 
tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two.
The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$.
And all it has in the proof is the assertion that the above is clearly true.
Haaaaate
 
1:33 PM
@Semiclassical but it’s clearly true :p
 
(The argument i see is that their claim amounts to the matrix having a Cholesky factorization, and that’s true whevever the matrix is PSD. But they say nadda )
Haaaate @LeakyNun
 
Hey everyone!
 
@Semi this reminds me, in my analysis book the differentiation chapter started with a review of differentiation in $\mathbb{R}$
And the first theorem it gave... The proof said "Look in a rigorous calculus textbook"
 
1:40 PM
In fairness it was basic stuff and by that point everyone knew Spivak but like, you can't write that and then proceed to still put the pirate face qed
 
Though in that book, the real annoying thing was that half the proofs were exercises, half were "this is trivial", and half had typos
 
To fill in my parenthetical: if such vectors exist, the above matrix has elements of the form $v_j^Tv_k$
So we can write the above matrix as $M=(v_1,v_2,v_3)^T(v_1,v_2,v_3)=V^T V$
And this is psd since $x^T M x=x^T V^T V x = \|Vx\|^2\geq 0$
 
0
Q: Numerical computation of divergence and curl on a discrete surface.

user8469759I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k...

 
1:57 PM
Let $r_0,r_1,r_2,..$ be irrationals
Consider $a,b$ both irrational and the interval $[a,b]$
Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$
It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint
This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will
2
Q: How did Irrationals Stay in [0,1] with small open cover of Rationals removed?

MathStudentIf I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals. One way for an irrational number $\alpha$ to be in this new set is b...

5
Q: Complement of a countable open cover of the rationals

ArtDSuppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals. Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat...

 
2:12 PM
Suppose that $(X,d)$ is a metric space equipped with addition. Does it follow that this addition operator is continuous with respect to the metric topology?
 
2:32 PM
I felt like I need to think this through again. I don't think I understood the open cover of the rationals and irrationals
 
3:19 PM
hmm...
(For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs)
Let $\lambda^*$ be the Lebesgue outer measure
We previously proved that $\lambda^*(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero.
We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^*([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting,
or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi
W also knew that one way to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^*(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^*(\Bbb{I}\cap [-1,1])$ is to use $\lambda^*(\Bbb{Q}\cap [-1,1])+\lambda^*(\Bbb{I}\cap [-1,1]) = \lambda^*([-1,1])$ and thus deducing $\lambda^*(\Bbb{I}\cap [-1,1]) = 2$
However, what I am interested here is to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below:
We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages:
First denote an enumeration of the rationals as follows:
 
3:41 PM
The name was actually chosen via this chain of events:

1) Milne released a beautiful textbook: Algebraic groups (Theory of group schemes of finite type over a field)

2) He included an image of a butterfly and a quote from Grothendieck about viewing a reductive algebraic group

3) I took the picture and needed a name.

4) I googled a bunch of butterflies

5) Found some that looked good, and eliminated ones whose names were boring.
 
$\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short:
 
@Narcissusjewel Nice. I clicked on "sign up via facebook" and my full name and profile were automatically migrated with very little in the way asking my permission
way of*
 
$\{p \in \Bbb{N}, q \in \Bbb{N} - \{0,1\} : \pm \frac{p}{q} \in \Bbb{Q} \land \text{lcm} (p,q)=pq\}$
Next:
$C_0=(-1,1)$
typo: Forgot 0 in the enumeration, which begins before $\frac{1}{2}$
$C_1 = (-1,0) \cup (0,1)$
$C_2 = (-1,-\frac{1}{2}) \cup (-\frac{1}{2},0) \cup (0,\frac{1}{2}) \cup (\frac{1}{2},1)$
After this point, we are going to use an n-tuple as a shorthand of the union of intervals
$C_3 = (-1,-\frac{1}{2},-\frac{1}{3},0,\frac{1}{3},\frac{1}{2},1)$
$C_4 = (-1,-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},0,\frac{1}{3},\frac{1}{2},\frac{2}{3}‌​,1)$
$C_5 = (-1,-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},0,\frac{1}{4},\frac{1}{3‌​},\frac{1}{2},\frac{2}{3}‌​,1)$
$C_6 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},0,\frac{1}{‌​4},\frac{1}{3},\frac{1}{2},\frac{2}{3}‌​,\frac{3}{4},1)$
$C_7 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},
-\frac{1}{5},0,\frac{1}{5},\frac{1}{‌​4},\frac{1}{3},\frac{1}{2},\frac{2}{3}‌​,\frac{3}{4},1)$
$C_8 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{2}{5}, -\frac{1}{4}, -\frac{1}{5},0,\frac{1}{5}, \frac{1}{‌​4},\frac{2}{5},\frac{1}{3},\frac{1}{2}, \frac{2}{3}‌​,\frac{3}{4},1)$
$C_9 = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{3}{5},-\frac{1}{3},
-\frac{2}{5}, - \frac{1}{4‌​},-\frac{1}{5},0,\frac{1}{5},\frac{1}{‌​4},\frac{2}{5},\frac{1}{3}, \frac{3}{5},\frac{1}{2}, \frac{2}{3}‌​,\frac{3}{4},1)$
$C_{10} = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{3}{5},-\frac{1}{3}, -\frac{2}{5}, - \frac{1}{4‌​},-\frac{1}{5},-\frac{1}{6},0,\frac{1}{6},\frac{1}{5},\frac{1}{‌​4},‌​\frac{2}{5},\frac{1}{3}, \frac{3}{5},\frac{1}{2}, \frac{2}{3}‌​,\frac{3}{4},1)$
Therefore: $C = \lim_{n \to \aleph_0} C_n$
Observe $C$ will eventually exclude all rationals. We now want to check the infinum of the intervals in the nth stage
actually... I am not very sure how to check whether all consecutive pair of intervals do have a length tends to zero as $n \to \aleph_0$
 
4:28 PM
@Daminark whenever i hear someone here speak about Sally it's always "Sally was a great man but I have some problems with his pedagogy"
 
Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $|p-q| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero
However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works...
 
hey there
I'm almost sure this is wrong
In fact, Lotta visits client $N$ on day $2^{N-1}$
My approach was a recurrence:
Let's say that for $N$ clients, Lotta will take $d_N$ days to retire.
For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$
Where $d_1$ = 1.
Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.
 
4:56 PM
@DavidReed That's certainly easier :P.
 
My result is that $d_N = 2^N - 1$
 
Hi! How can we solve the equation: sinx+cosx=1,2 ?
 
try squaring both sides
and using the identity $2 \mathrm{sin}x \, \mathrm{cos}x = \mathrm{sin} 2x$
 
5:12 PM
@LucasHenrique Thanks!! :-)
 
@MaryStar np
 
Hi guys. Is this valid? I'm skeptical about the $(5^2)^{30}$ because it equals $5^{60}$, and it was $5^{30}$ before. Why would they double the exponent?

https://i.gyazo.com/a0b49487716b1adcf52ebb13ce96e2cd.png
 
5:34 PM
The intermediate step should be $(5^2)^{15}\cdot 5$, but then the argument proceeds the same and the final answer is correct.
 
Okay, just what I thought. Thanks :)
 
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