« first day (2601 days earlier)      last day (1592 days later) » 

5:00 PM
giving $\lim_{z \rightarrow \infty} \sec^2(z)$
now what's this value?
 
sigma_g :A-->A
Sigma_g (a) = g.a
this is a group G acting on the set A. for each fixed g in G we get a map sigma_g
i) for each fixed g in G , sigma_g is a permutation of A
ii) the map from G to S_A defined by g--> sigma_g is a homomorphism
@LeakyNun Can you help me understand the notations here?
 
morning
 
@Faust morning
 
Good Morning
 
Hows leaky?
 
5:03 PM
@Faust Sup faust
 
@KasmirKhaan how?
@Faust nm
 
Hey kasmir khaan
 
@LeakyNun like in words , what does that say
 
hows ur hw going?
 
I sent the second assigment
 
5:04 PM
@KasmirKhaan which part?
 
deadline is tomorrow
well @LeakyNun let me tell you what I understood
this is a group action from the group G acting on the Set A .
sigma_g = g.a, is a permutation of A. like hmm
how do we prove this?
does 2 sided inverse imply a permutation?
 
verify the axioms?
 
if we have a map from A --> A
 
@KasmirKhaan do you know how a permutation is (formally) defined?
 
let me give it a try =p
its a bijection from a set to itself
 
5:08 PM
correct
 
but what is sigma_g inverse here?
or does that even make sense
 
@KasmirKhaan figure it out yourself
 
okay :D
we want s' (s_g) = a
s = sigma here
s_g = ga
s'(ga) =a
we need something that give is g' on the left
s' = s_g'
@LeakyNun is that correct?
 
right
 
okay :D
 
5:13 PM
although you don't need to define the inverse function in terms of sigma
 
what do you mean?
 
> s' = s_g'
 
oh okay =p
but it is eays to show that
s_g' o s (a) = a
 
alright.
 
and s o s_g'(a) =a
but I dont get why this is enuf to conclude that s_g is a permutation
or wait
is it because we found a map that is bijective from A to A ?
 
5:16 PM
right
 
hmm this is new idea =p
 
@LeakyNun I am on my phone right now and I don't recall which one 7 is.
 
@TobiasKildetoft generated subgroup
 
all righty thats it for kasmir
need to go sleep ><
 
lol
 
5:19 PM
woke up 3 am today
 
what?
Hi, Demonark
 
@LeakyNun were you talking to me ?
 
Not sure what is tedious about that one.
 
unleash the Demon
 
@LeakyNun Ill continue when i wake up =p I got lots more Q's ,ill text you :D
 
5:20 PM
@TobiasKildetoft 4a. $[\forall g \in G\forall s\in S:gsg^{-1} \in \langle S \rangle] \implies \langle S \rangle \trianglelefteq G$
 
hi
my question : can a matrix A be ever similar to the matrix A+I
hint please
 
Hint: How does the trace behave w/r/t a similarity transformation?
 
its the same
so there does not exist such an A
right?
 
tr(AB)=tr(BA)
tr(ABA') = tr((AB)(A')) = tr((A')(AB)) = tr(B)
simple proof :P
 
@NV-US fill in the logic
 
5:31 PM
i know similar matrices have same trace
ok
thinking
 
lol it's just a line
 
Strange so much discussion happening here on a Sunday.
 
there is a countable model of ZF
some day I'll write a program for that model
using lazy evaluation
 
"I'll do it someday" already sounds like lazy evaluation :P
 
"I'll do it someday" is what I've been telling myself about some things for a few years now :P
 
5:34 PM
Your time is better spent writing papers and books.
 
since if A is similar to A+I, then tr(A) = tr(A+I), then sum $a_{i,i}$ = sum $a_{i,i} + n$, if A is a n x n matrix, which cannot be true, therefore such an A cannot exist. is this ok?
 
Yeah, that's what I had in mind.
 
but our professor has written (depends on the field, justify properly) - after the question. Is there something missing?
 
Yep. this reasoning assumes that tr(A) can't equal tr(A)+n. that works if tr(A) is a real number
But what if your field is integers mod n?
 
oooohhhh
then trace will be equal
nice!
ty
 
5:38 PM
I guess I should insist that n be prime if Z/n is a field, but you get my drift
 
i know that, thanks
 
Right
 
Skolem's paradox:
1. ZF has a countable model M. [Downward Löwenheim–Skolem]
2. The axiom of infinity guarantees that $\Bbb N$ is an object in M.
3. The axiom of powerset guarantees that $\mathscr P(\Bbb N)$ is an object in M.
4. Cantor's theorem proves that there is no surjection $\Bbb N \to \mathscr P(\Bbb N)$.
5. Since $\Bbb N$ is countable and our universe is countable, there must be a surjection $\Bbb N \to \mathscr P(\Bbb N)$.
 
6:04 PM
how to show that, if two real square matrices are similar over C, then they must be similar over R. my try : let $A = P^{-1}BP$, then $P = C+iD$, $C$ and $D$ are real matrices. Comparing real and imaginary parts i get, $CA=BC$ and $DA=BD$. I now need to show that one of C and D is invertible
 
the proof that C and D are invertible.
 
yes, any hints
?
both of them? wont only one will suffice?
 
right, only one
 
hints please
 
1 0
0 i
is invertible
but none of
1 0
0 0
or
0 0
0 1
is invertible
 
6:08 PM
so is the question wrong?
 
no, this just proves that P cannot be
1 0
0 i
 
oh!
 
and before you waste your time with me, I don't know the solution.
@AlessandroCodenotti you might be interested:
24 mins ago, by Leaky Nun
Skolem's paradox:
1. ZF has a countable model M. [Downward Löwenheim–Skolem]
2. The axiom of infinity guarantees that $\Bbb N$ is an object in M.
3. The axiom of powerset guarantees that $\mathscr P(\Bbb N)$ is an object in M.
4. Cantor's theorem proves that there is no surjection $\Bbb N \to \mathscr P(\Bbb N)$.
5. Since $\Bbb N$ is countable and our universe is countable, there must be a surjection $\Bbb N \to \mathscr P(\Bbb N)$.
Are there any logical people here lol
 
5 is wrong
That surjection won't be in the model
 
@AlessandroCodenotti why?
@AlessandroCodenotti nice!
 
6:12 PM
P(N) is uncountable according to the model
 
but why does Cantor's theorem fail outside the model?
 
In the outside metatheory you can see that the model is countable
 
Hey there people!
 
I'm not sure what this question means
 
Hi, Demonark
@AlessandroCodenotti why can't I use Cantor's theorem outside the model to prove that P(N) is uncountable?
 
6:14 PM
Because what the model sees as P(N) is countable according to the metatheory. What the model thinks is P(N) and what the metatheory thinks is P(N) need not to be the same, the model is missing some subsets
 
@AlessandroCodenotti why do you know about all this?
I mean, logic isn't mandatory right
 
@AlessandroCodenotti @Daminark can u help me with : how to show that, if two real square matrices are similar over C, then they must be similar over R. my try : let $A = P^{-1}BP$, then $P = C+iD$, $C$ and $D$ are real matrices. Comparing real and imaginary parts i get, $CA=BC$ and $DA=BD$. I now need to show that one of C and D is invertibl
 
Because I like that kind of math
 
@AlessandroCodenotti this is my solution, maybe you can help me check:
in Logic, 11 mins ago, by Leaky Nun
1. there are only countably many objects x in M such that M(x) "∈" M(2^N).
2. cardinality inside the model no longer has any meaning once you go out of the model, since it's just another first-order logic formula.
3. let SM={ M(x) "∈" M(N) : M(x) "∉" f(M(x))}
4. SM doesn't need to correspond to anything inside M, since SM is a set defined outside M.
you can ignore step 2 :)
 
any hints to show that detC or detD is non zero
 
6:19 PM
I looked around the web for an explanation of combinations with repetitions. And I keep finding explanations that first model the problem in terms of places and then explain it in terms of the permutation of those places. But it doesn't give me a direct intuitive understanding of it.
If I try to figure it out for myself, I end up with: Say I need to select 3 objects from a set of 8 distinct objects. For the first object, there are 8 ways to select, for the second also 8 since repetition is allowed, and so for the third. That 8^3 ways.
Then I somehow need to remove the repeats... I dont know where to go from here..
 
@AlessandroCodenotti but how can any subset be missing?
 
@NV-US I dunno if making one of the matrices invertible is the right idea
 
Show that $$\dfrac{\sec^2 \theta-\tan\theta }{\sec^2 \theta+\tan\theta}$$ lies between $$\dfrac{1}{3} \quad \mathbb{and} 3$$
Attempt:
I used $1+\tan^2\theta = \sec^2\theta$ to find that the denominator is always greater than one. Then I framed an inequality and eliminated the denominator. FInally, I am left with:
$$\tan^2\theta-\tan\theta+1>0$$ . How do I proceed? Any hints?

Please don't answer if it's unnecessary busy work for you/ wastes your time/ etc etc.
 
@Abcd let $y=\dfrac{x^2-x+1}{x^2+x+1}$
 
You may be better off trying to note that det(C+iD) ≠ 0
 
6:22 PM
@LeakyNun I did that.
 
@Daminark
 
4 hours ago, by Leaky Nun
y=(x^2-x+1)/(x^2+x+1)
(x^2+x+1)y = x^2-x+1
(y-1)x^2 + (2y)x + (y-1) = 0
find the range of y such that the quadratic equation has solution.
 
and then what can i write?
 
So the polynomial isn't the 0 polynomial
Meaning it has only finitely many zeroes, so you can find some real number k such that C+kD is not 0
 
@LeakyNun lol, same question :p on same day
 
6:25 PM
C+kD how? i can say that Pk = (C+iD)k $\ne 0$, which implies Ck $\ne 0$
 
Is it necessary for y to be real @LeakyNun?
 
@Abcd you tell me
 
no one can point me in the right direction?
 
@LeakyNun yes because $\tan \theta$ always $\in \mathbb{R}$
 
right
@ˆ-ˆ stars and bars?
 
6:27 PM
@daminark
 
@NV-US that's not what I'm saying. Det(C+XD) for X some complex number is not the 0 polynomial, right?
 
@LeakyNun That's exactly what I don't understand.. Instead of modeling it like that.. is there a way to understand it direct terms..
 
X is iota, yes?
 
@LeakyNun $2yx$? Are you sure?
 
@Abcd right, it should be (y+1)x
 
6:29 PM
Yeah.
 
@NV-US iota?
@ˆ-ˆ well, you just need to use stars and bars to model each situations
let's say you label the objects from 1 to 8
1,1,2 would be xx|x||||||
 
$sqrt{-1}$ @LeakyNun
 
2,3,3 would be |x|xx|||||
@NV-US no, X is a indeterminate variable
 
how can u say such an X exist?
 
x is the variable in f(x)=x^2
@ˆ-ˆ then convince yourself that it is a one-to-one correspondence, so you come up with 3+(8-1) choose 3
 
6:35 PM
@lea
@LeakyNun thanks..
actually I found this:
5
Q: Why does this intuition for combinations with repetition fail?

Arnold DovemanWhen you are learning the difference between combinations and permutations without repetition. The logic for each of them is: 1) Permutation without repetition: Selecting 4 objects from 10, gives $10*9*8*7$ choices. 2) Combinations without repetition: Selecting 4 objects from 10, gives $10*9*8*...

My question is exactly that!!
 
I see
 
Going through the answers.. think i have some understanding as to why you need to model it differently
 
well
the proof of n choose k goes like this:
there are nPk permutations
every k! permutations are the same
so n choose k = nPk / k!
the combination with repetition fails at "every k! permutations are the same"
112 only has 121 and 211
 
@LeakyNun because you can't tell inside the model
 
@AlessandroCodenotti is this related to "there are only countably many definable objects"?
there are only countably many subsets that I can ever describe
 
6:42 PM
Hm, I don't know
 
lol, finally something you don't know :c
but step 2 was very crucial to the understanding
in Logic, 40 mins ago, by Leaky Nun
1. there are only countably many objects x in M such that M(x) "∈" M(2^N).
2. cardinality inside the model no longer has any meaning once you go out of the model, since it's just another first-order logic formula.
3. let SM={ M(x) "∈" M(N) : M(x) "∉" f(M(x))}
4. SM doesn't need to correspond to anything inside M, since SM is a set defined outside M.
 
@LeakyNun i see! so there is no clear way to get at the repetitions... so use the stars and bars or equivalent model to change the view..
 
@ˆ-ˆ right
 
@LeakyNun So the number of repeats would be k^n - Combination(k + n - 1, n - 1)?
 
@ˆ-ˆ sure
 
6:51 PM
It seems like there would be some intrinsic pattern that would make them countable outside of this way no? but thats mere hope i think..
 
@LeakyNun So, did you get through the tediousness?
 
@TobiasKildetoft no, I skipped that part :P
it's one of those things you clearly can prove but would take up a whole page
 
@TobiasKildetoft can u help me with : how to show that, if two real square matrices are similar over C, then they must be similar over R. my try : let $A = P^{-1}BP$, then $P = C+iD$, $C$ and $D$ are real matrices. Comparing real and imaginary parts i get, $CA=BC$ and $DA=BD$. I now need to show that one of C and D is invertibl
 
like the fact that V* is a vector space
 
@LeakyNun It should only take like a line if you use exercise 2 (assuming this was 7(4) you mean)
 
6:56 PM
@TobiasKildetoft a line :O
 
if you mirror a graph across the $x$-axis - is that a horizontal or vertical one?
 
@Kirill vertical
 
vertical mirroring (reflction) proceeds along the horizontal axis and vice versa, @LeakyNun?
 
yes
@TobiasKildetoft don't you need to expand the product?
 
@LeakyNun good to know, thank you
 
6:59 PM
@LeakyNun You mean given some element in the subgroup, you write as a product like in the previous part? Yes
 
@TobiasKildetoft how is that one line
which line?
 
@LeakyNun Let $s\in \langle S\rangle $ so $s = \prod_{i=1}^ks_i^{j}$. Then $gsg^{-1} = \prod_{i=1}^k(gs_ig^{-1})^j\in \langle S\rangle$.
Just add a few words about where everything lives and you are done
 
Why $\prod_{i=1}^k(gsg^{-1})^j\in \langle S\rangle$?
 
@LeakyNun typo, I had forgotten the subscript there
 
@TobiasKildetoft still why?
hmm, closure
 
7:04 PM
@LeakyNun product of powers of elements in that subgroup
 
alright
 
Got a nice email from Jim Humphreys with some comments on my latest paper yesterday.
 
@TobiasKildetoft what's your paper about?
 
@LeakyNun An algorithm for obtaining the characters of simple modules for algebraic groups in positive characteristic, assuming one knows those of the tilting modules and assuming Donkin's tilting conjecture
 
I see.
pretends to understand
 
7:18 PM
Heh, yeah, the topic of the paper goes somewhat beyond what one would encounter in pretty much any course at all (well, almost)
 
7:30 PM
@TobiasKildetoft 8.2 is really cool. I enjoyed it.
 
@LeakyNun Yeah, that is a neat one
 
I used three squares.
Hi, @Danu
 
if a given transformation is triangulable, how can i determine the triangular matrix?
 
I think that was what I needed too, but I don't recall my solution any longer
 
ask wolframalpha :P
 
7:32 PM
@PVAL-inactive So I decided to treat myself (I'm graduating from my master's degree soon) and got the Ardbeg Corryvreckan & Laphroaig 18... I'm psyched!
 
@TobiasKildetoft it's really easy
 
answer please
 
@NV-US there should be a way with flags
 
what flags?
 
Salut @GabrielRomon
 
@GabrielRomon read it, just now. we have done this in class. i took any non zero vector in v1 in V, then calculated T(v1)= v2, and subsequently found other vectors.
am i on the right track? @GabrielRomon
 
@NV-US You should start with an eigenvector.. That gives you the top left entry. Then consider the action on the quotient and get a new one. Repeat.
 
@Danu I do love Corryvreckan
 
top left entry? action on quotient, what quotient? help
@to
@TobiasKildetoft
 
@NV-US You are looking for a basis such that the corresponding matrix is upper triangular. An eigenvector gives the its eigenvalue as the top left entry in such a matrix
Then quotient out by the subspace generated by that eigenvector
 
7:42 PM
ok, but i did not understand "Then quotient out by the subspace generated by that eigenvector"
 
Ahh, write what the matrix does to a vector as $av + w$ where $v$ is that eigenvector.
 
i dont understand, sorry
 
@MikeMiller It's the best
(to be confirmed; I'm waiting for my defense in 2 weeks to try it LOL)
 
Hahaha
If you like Ardbeg you'll like this bottle
It's my favorite of their varieties
 
@NV-US So find an eigenvector $v$ and pick a basis with that as the first vector. Write the matrix in this basis and consider the part of the matrix you get by removing the first row and column. Continue doing this with the new matrix
 
7:49 PM
I've previously had the 10 (several bottles) and the Uigeadail
I don't remember much about the Uigeadail though... It was several years ago.
 
I don't really remember my detailed thoughts on - yeah exactly haha
 
I'm drinking the 10 right now
I was also very tempted by the Talisker 57 North
I guess that'll be my next "fancy bottle".
 
It's been at least a year for me since either of those and I have a really bad memory for most things
I've only had their one that has like storm in the name
 
I used to take notes back when I lived in Amsterdam and had a close friend I used to drink 'em with
Right, dark storm I think.
 
Update: the name is just storm
 
7:52 PM
Oh, just storm
There's also a Dark Storm :P
 
Haven't had that. Though now that I try to remember how to describe a single bottle of scotch, it wouldn't have made a conversational difference if I had
 
:)
I just know that the Talisker 10 is amazing value
 
a liquor store near me has it for around or under 40 USD
 
It's only 6 euros more expensive than a Jameson on the website I get it from (26 vs 19)
25*
 
I don't know how the euro is flagging against the dollar
That would have been like $35-40 a year ago I think
 

« first day (2601 days earlier)      last day (1592 days later) »