Mathematics

Ted Shifrin

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.

Karl Kronenfeld

You should have stopped him.

ccorn

@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.

Balarka Sen

@TedShifrin you were the guy

only in a different space-time.

Pedro Tamaroff

@ccorn Oh, OK.

Ted Shifrin

You're asleep, @BalarkaSen

Balarka Sen

12:08 AM

he is.

i am his cat.

robjohn

1:18 AM

@Chris'ssis I was at least able to get an asymptotic series that verifies my computational claim.

Jose Antonio

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?

Pedro Tamaroff

Maybe they meant order $2$?

Jose Antonio

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.

Pedro Tamaroff

2:10 AM

@MikeMiller Hello there.

anon

@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.

Pedro Tamaroff

@anon HARRO!

anon

how's things and stuff?

Jose Antonio

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.

Pedro Tamaroff

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.

anon

@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.)

Jose Antonio

The exercise is to show that any group of order 10 in S_5 are in A_5.

anon

A_5 has order 60

Pedro Tamaroff

I'm off to sleep. Cheers @anon

anon

night

Pedro Tamaroff

2:31 AM

Adiós, Jose.

Jose Antonio

Good night

Yes, I know

anon

so, you're trying to prove groups of order 10 is a particular group of order 60?

Jose Antonio

No no

anon

oh, are in A_5

Jose Antonio

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.

anon

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)

Jose Antonio

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

Nick

@anon: Hi, blue!

anon

hi

Nick

@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.

anon

k

Nick

2:47 AM

@anon: So, how are things?

anon

fun

Nick

@Hippalectryon: There are many kinds of high.

Jose Antonio

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?

Nick

@anon: void fun(){int f = 0; cout<<"How much fun from 1 to 10"; cin>>f; }

Jose Antonio

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.

Nick

3:00 AM

@IceBoy: Hi skull.

Ted Shifrin

Hi @anon @Nick

anon

hi

Ted Shifrin

Teaching still good? :)

anon

mmhmm

Ted Shifrin

Great ... I give my first prob test Wednesday ... Bet it'll be very bimodal. When's your first test?

Nick

3:09 AM

hiya'

anon

@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.

Ice Boy

Hi @Nick

Nick

@TedShifrin: Is the Gauss's Law in math something different to the one I learned in my Electrostatics class?

Ted Shifrin

Wow ... Great experience for you heading to grad school, @anon. How did they do on test 1 (the college alg kids)?

anon

the results weren't out yet at our monday morning meeting.

but I could look up my own students right now I suppose

Ted Shifrin

3:22 AM

ah, computer testing or something?

anon

yes

Ice Boy

@Nick it is the same

Ted Shifrin

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.

Nick

Oh!'

Ted Shifrin

Gauss's Law in physics is basically cohomology + Divergence Thm in math :)

Nick

3:26 AM

@IceBoy: I was talking about $$\Phi = \int{ E \cdot dS} = \frac{Q}{\epsilon_0}$$

Ted Shifrin

It's the Divergence Theorem applied to the appropriate one of Maxwell's Equations, @Nick.

Nick

@TedShifrin: Mhh. i should look into that. Thanks.

Ted Shifrin

Sure ... I cover that in my course that's on YouTube, but I use fancier notation. @Nick

I discuss gravitation, not electrostatics, though.

Nick

@TedShifrin: $$F_e \propto \frac{q_1 q_2}{r} \sim F_m \propto \frac{m_1 m_2}{r}$$

Ted Shifrin

@anon: Take-home exam, huh? How many students got help from MSE?

Precisely, @Nick.

anon

3:33 AM

@TedShifrin so far as I can tell, none.

Ted Shifrin

Good ... I would never give a takehome in this day and age ...

anon

my combinatorics class is open notes/book/internet - you can use any source as long as you cite it.

Ted Shifrin

Night, all!

anon

night

Nick

@TedShifrin: We won't misbehave without you.

Night

Jorge Fernández

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 $?

anon

every polynomial is by definition a linear combination of those powers

Jorge Fernández

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

anon

do you have any restriction on the sequences?

Jorge Fernández

they are over a field $F$

anon

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

Jorge Fernández

3:44 AM

yeah, that's what i thought

so what would be a base for that vector space?

or basis

anon

one needs the axiom of choice to know there is one; you won't be able to write a basis for it down

Jorge Fernández

oh crap

so what should I write<'

?

I was told to find a basis

anon

what was the original question

Jorge Fernández

it's from friedberg

it says find a basis for it and justify

anon

give me a verbatim quote

don't paraphrase

Jorge Fernández

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

anon

and there you have it

well I am going to bed now

Jorge Fernández

gnight

Jose Antonio

4:10 AM

You need Zorn's lemma.

Ok a finite number, no problem.

nerdy

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 ?

Partly Putrid Pile of Pus

7:37 AM

Hey guys

Chris's sis

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.

Ice Boy

Greetings

r

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t

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Parth Kohli

9:46 AM

@IceBoy Hi.

Ice Boy

10:10 AM

@ParthKohli Hello

Chris's sis

@robjohn I have an indirect proof that Cleo is Tunk-Fey ..

4

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 $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$

In 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 ...

I mean $$\sum_{n=1}^\infty \frac{H_nx^n}{n^3}$$ formula doesn't work ...

OK, I'll recheck that again when I'm back

bbl

Moron

@Chris'ssis does that matter?

Will Hunting

10:33 AM

@Moron Hi!

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$Mathematics$ Mathematics