« first day (1511 days earlier)   

12:01 AM
Oh wow, 7:20 pm ... I was in a cab driving by the White House almost exactly then, on my way to dinner.
 
You should have stopped him.
 
@PedroTamaroff Imagine a matrix with just two diagonals, say 1 on the main diagonal, and -2 in the diagonal below. Solving a system with that matrix involves successive multiplications with 2 (and additions). Numerical errors thus get exponentially amplified, the computed solution is unusable. Same can happen in an IIR filter due to its feedback loop.
 
@TedShifrin you were the guy
only in a different space-time.
 
@ccorn Oh, OK.
 
You're asleep, @BalarkaSen
 
12:08 AM
he is.
i am his cat.
 
 
1 hour later…
1:18 AM
@Chris'ssis I was at least able to get an asymptotic series that verifies my computational claim.
 
1:45 AM
Hi I'd like to check something. I think I found a mistake in one exam. This says: Check that all the elements in $Q/Z(Q)$ has order 4. But this doesn't make sense (Q is the quaternion grouo) because the quotient is isomorphic to the klein 4 group doesn't? and all the elements different to the identity has order 2
AM i right?
 
Maybe they meant order $2$?
 
I suppose is a typo.
YEs i suppose that the wanted to say that all the elements are of order 2.
I need to take a shower. Thanks as usual.
 
2:10 AM
@MikeMiller Hello there.
 
@JoseAntonio what is the exact wording? it might also mean the number of elements in Q/Z(Q), or the order of Q/Z(Q), is 4.
pre-emptive hi to pedro too
@BalarkaSen trivial.
 
@anon HARRO!
 
how's things and stuff?
 
No, sorry the problem says that the order of each of the elements. I'm sure is a typo. Of course the order of the quotient is 4. I think someone makes a mistake.
Here is other I really like show that any subgroup of the S_5 of order 10 is in the alternating group. I don't sure but to avoid the terrible computation I think that the easiest way is proving that all the groups 2p where p is a prime are either cyclic or dihedral. In this case is trivial to show that the group is the dihedral and is generated by any 5 cycle which of course is in the alternating group and any other group of order 2 since the group is dihedral
the element of order two should satisfy the relatation t p t = p^-1
WLOG suppose the 5 cycle is (12345) it invers is (54321) so the element t must be of the form (15)(24) which indeed is in the alternating grouo
I think is the easiest way which I know so far.
@anon I'm completely sure is a mistake and the wanted to say the order of the quotient and not the order of the elements. Anyway the exercise is very simple in any case.
 
2:27 AM
@anon Glad you asked! Academy-wise, I'm taking one course -- complex analysis -- and looking forward to sit for at least two more finals. Also slowly chugging on Pete's algebra notes.
 
@JoseAntonio once you classify the groups of order 2p, you simply need to disprove there is an elt of order 10 in S5. (also you mean dihedral not alternating in first sentence.)
 
The exercise is to show that any group of order 10 in S_5 are in A_5.
 
A_5 has order 60
 
I'm off to sleep. Cheers @anon
 
night
 
2:31 AM
Adiós, Jose.
 
Good night
Yes, I know
 
so, you're trying to prove groups of order 10 is a particular group of order 60?
 
No no
 
oh, are in A_5
 
that the groups of order 10 in S_5 are in A_5
yes
Since all the groups of order 10 are either cyclic or dihedral. Clearly the first option is impossible.
 
2:35 AM
then sure, the subgroup of order 10 must be generated by an elt of order 2 and of order 5; the latter is necessarily a cycle, disprove the first is a 2-cycle (so it must be a product of two 2-cycles)
 
my argument was the following take a 5 cycle WLOG suppose is p =(12345) since tpt=p^-1 so the t that makes the work is (15)(24) which is in A5
something like that using the property of the conjugates in any cycle.
Because without that the computation is a nightmare
t(12345)t=(54321) iff t=(15)(24) since the t is in A5 and the 5 cycle are, therefore the entire group is in A5
 
@anon: Hi, blue!
 
hi
 
@anon: I've decided to annoy you in this way: When you're blue, I'll call you anon and when you're anon, I'll call you blue.
 
k
 
2:47 AM
@anon: So, how are things?
 
fun
 
@Hippalectryon: There are many kinds of high.
 
By the way, some days ago, talking with friends someone thought in the following what functions can live in C_b(Q) continuous and bounded functions from Q to R . Seems to be very pathological ones. Does someone knows a good place to check or study this types or problems?
 
@anon: void fun(){int f = 0; cout<<"How much fun from 1 to 10"; cin>>f; }
 
When I asked this question here ,everyone told me that is a futile work to try of characterizes this weird functions. But seems to be very fun.
 
3:00 AM
@IceBoy: Hi skull.
 
Hi @anon @Nick
 
hi
 
Teaching still good? :)
 
mmhmm
 
Great ... I give my first prob test Wednesday ... Bet it'll be very bimodal. When's your first test?
 
3:09 AM
hiya'
 
@TedShifrin in intermediate/college algebra we had Test 1 last week (we have four tests + final in those classes over the semester). in the abstract algebra class I T/A the takehome portion is due this Thursday I believe, and so I'll be grading with the teacher on Friday afternoon.
 
Hi @Nick
 
@TedShifrin: Is the Gauss's Law in math something different to the one I learned in my Electrostatics class?
 
Wow ... Great experience for you heading to grad school, @anon. How did they do on test 1 (the college alg kids)?
 
the results weren't out yet at our monday morning meeting.
but I could look up my own students right now I suppose
 
3:22 AM
ah, computer testing or something?
 
yes
 
@Nick it is the same
 
Gotcha ... I always followed my precalc students when they had computer testing.
@Nick: What Gauss's Law in math? It's Gauss's Theorem = Divergence Theorem.
 
Oh!'
 
Gauss's Law in physics is basically cohomology + Divergence Thm in math :)
 
3:26 AM
@IceBoy: I was talking about $$\Phi = \int{ E \cdot dS} = \frac{Q}{\epsilon_0}$$
 
It's the Divergence Theorem applied to the appropriate one of Maxwell's Equations, @Nick.
 
@TedShifrin: Mhh. i should look into that. Thanks.
 
Sure ... I cover that in my course that's on YouTube, but I use fancier notation. @Nick
I discuss gravitation, not electrostatics, though.
 
@TedShifrin: $$F_e \propto \frac{q_1 q_2}{r} \sim F_m \propto \frac{m_1 m_2}{r}$$
 
@anon: Take-home exam, huh? How many students got help from MSE?
Precisely, @Nick.
 
3:33 AM
@TedShifrin so far as I can tell, none.
 
Good ... I would never give a takehome in this day and age ...
 
my combinatorics class is open notes/book/internet - you can use any source as long as you cite it.
 
Night, all!
 
night
 
@TedShifrin: We won't misbehave without you.
Night
 
3:38 AM
I need help with my linear algebra homework
I have the polynomials over $F$
with addition and scalar multipliaction as usual
how do I prove they are generated by $1,x,x^2\dots $?
 
every polynomial is by definition a linear combination of those powers
 
really
well the example is actually different
instead of polynomials consider sequences
and sequences are added and multiplied as usual
how can I justify they are generated by the secuences which are 1 at i and 0 elsewhere
?
I can't do it
 
do you have any restriction on the sequences?
 
they are over a field $F$
 
if you don't have any restrictions, then the things of the form (0,...,1,0,...) do not generate the space
since for instance (1,1,1,...) is not a finite linear combination of those things
 
3:44 AM
yeah, that's what i thought
so what would be a base for that vector space?
or basis
 
one needs the axiom of choice to know there is one; you won't be able to write a basis for it down
 
oh crap
so what should I write<'
?
I was told to find a basis
 
what was the original question
 
it's from friedberg
it says find a basis for it and justify
 
give me a verbatim quote
don't paraphrase
 
3:47 AM
find a basis for the vector space in example 5 of section 1.2 . justify your answer.
example 5:
oh damn
it say they have a finite number of nonzero terms
sorry for bothering you
 
and there you have it
well I am going to bed now
 
4:10 AM
You need Zorn's lemma.
Ok a finite number, no problem.
 
 
2 hours later…
6:14 AM
Really dumb question. Whenever we talk about the functional limit of a function between metric spaces (X,d) and (Y,w) , does the limit of the function as its dependent variable approaches some point in the domain, have to be in Y ?
 
 
1 hour later…
7:37 AM
Hey guys
 
8:16 AM
Greetings
@robjohn nice approach there that I upvoted.
@robjohn could you possibly get some information from UCLA about the asymptotic of theta function partial sum? Then we finish the job pretty easy.
That's because we may use integrals and immediately all gets reduced to some theta function partial sum.
 
Greetings
r
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e.
t
i
n
g
s
 
 
1 hour later…
9:46 AM
@IceBoy Hi.
 
10:10 AM
@ParthKohli Hello
 
@robjohn I have an indirect proof that Cleo is Tunk-Fey ..
Question: what is the probability that Cleo and Tunk-Fey both make such a horrible mistake and get the same answer?
21
A: A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$

Tunk-FeyIn the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operat...

Just asking ... nothing more ...
This formula does not seem to work $$ \sum_{n=1}^\infty \frac{H_n x^n}{n^3}=&\frac12\zeta(3) \ln x-\frac18 \ln^2 x\ln^2(1-x) + \frac12 \ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{ Li}_3(x)\right]\\&+ \operatorname{Li}_4( x)- \frac{ \pi^2}{12} \operatorname{ Li}_2(x)-\frac12\operatorname{Li}_3(1-x) \ln x+ \frac{\pi^4}{60}$$
 

« first day (1511 days earlier)