Mathematics - 2012-06-27 (page 2 of 3)

Peter Tamaroff

2:00 AM

@BenjaLim I forgot how to do matrix multiplication!!!

BenjaLim

absolutely useless

robjohn

@PeterTamaroff Nooo!

Peter Tamaroff

@robjohn AAAAAAAAAAAAAAAAAAAAAAAA

@robjohn How do I multiply out what Ben wrote up there?

Eugene

@robjohn lol.

robjohn

@Eugene I wouldn't say that. I consider a 200 km trek to my friend's in San Diego a pretty big trip, though I have done the distance 3 times in one night.

Eugene

2:03 AM

@robjohn trek as in? because by car I've had friends who've done 2000 miles.

BenjaLim

@PeterTamaroff why do u want to multiply it out?

Peter Tamaroff

This one is in the first mid term of my algebra class starting in august:

robjohn

@PeterTamaroff $a=d+e$, $b=-e$, and $c=d$

Peter Tamaroff

@BenjaLim To find what $(x,y,z)$ are.

BenjaLim

WHAT KIND OF EXAM IS ONE WHERE YOU KNOW QUESTIONS BEFOREHAND

Peter Tamaroff

2:05 AM

@BenjaLim It is an old exam, silly!

robjohn

@BenjaLim A test in Prescience 101?

BenjaLim

OK

Peter Tamaroff

@BenjaLim Let $W=<(1,0,-1,1);(0,-1,-3,1)>$, $H_1 =\{ {\bf x} \in \Bbb R^4 : x_1+x_2-x_3-2x_4=0\}$, $H_2 =\{ {\bf x} \in \Bbb R^4 : x_1+x_2-x_3-x_4=0\}$

BenjaLim

ok

Eugene

@BenjaLim why not? there are some exams where they give you a list of 1000 questions and ask you maybe 10.

Peter Tamaroff

2:07 AM

Find a subpace $S$ of $\Bbb R^4$ such that

$(W \cap H_2) \oplus S=H_1$

And

$S \cap H_2 \neq \{0\} $

(if possible.)

BenjaLim

right

Peter Tamaroff

@BenjaLim Is it tough, or long?

BenjaLim

first can you identify $W \cap H_2$?

anon

you mean subspace S?

Peter Tamaroff

@BenjaLim Give me a sec.

@BenjaLim Should I find a system of equations for $W$ to make things easier?

BenjaLim

2:12 AM

you don't need to

well perhaps maybe yes

Peter Tamaroff

@BenjaLim So I just ask that the system is simultaneously satisfied...

Eugene

@BenjaLim where do you live?

melbourne? sydney? canberra?

robjohn

@Eugene Not in a day. 2000 miles takes at least 3 days switching drivers.

BenjaLim

@Eugene sydney now

@PeterTamaroff if you know what $W \cap H_2$ is

stuff should be easier

robjohn

@Eugene 2000 miles is across the entire US

Eugene

2:15 AM

@robjohn oh yeah. not in a day. i think 3 days would be sufficient too.

Peter Tamaroff

@BenjaLim I think it got it

Any vector of $W$ would be of the form

Eugene

@robjohn yup. my friend drove from wisconsin to Pasadena to see the rose bowl

@BenjaLim my cousins are in melbourne

robjohn

@Eugene Wow. Hopefully, they did more than just the Rose Bowl.

BenjaLim

@PeterTamaroff tell me

Eugene

@robjohn well i think they hit vegas on the way back. but wisconsin is such a big football state that it actually does happen that people travel all the way to california JUST for the rose bowl

Peter Tamaroff

2:17 AM

@BenjaLim I'm getting it together.

Eugene

@robjohn as in wisconsin was playing in the rose bowl for the first time in a decade

Peter Tamaroff

W can be the space such that

$2x_1-4x_2+x_3-x_4=0$

If I'm not talking crazy.

I just chose them appropriedly.

robjohn

@Eugene It's an hour drive to the Rose Bowl; what's the big deal? ;-)

Eugene

@robjohn not if you live in madison!

BenjaLim

@PeterTamaroff that space is three dimensioal

your $W$ is two dimensional :D

Peter Tamaroff

2:22 AM

@BenjaLim $W$? But that is not a base, it is a system of generators I suppose.

The other two are three dimesional so I guess we're cool.

Eugene

@robjohn how long should i wait before i should read a preprint of a paper that's been submitted?

BenjaLim

@PeterTamaroff You described $W$ as the span of two vectors

not multiples of each other

so they are linearly independent

and so are a basis for $W$

Peter Tamaroff

@BenjaLim Oh, OK.

Fack!

robjohn

@Eugene why would you wait?

Eugene

@robjohn well errors? also since it's not verified it might not even be a proof

Peter Tamaroff

2:25 AM

@BenjaLim Wait!

The vectors of $W$ satisfy the equation of $H_1$

Si $W \subseteq H_1$.

BenjaLim

ah yes

Peter Tamaroff

Let me see what happens with $H_2$

Eugene

@robjohn should i wait at least until the first revision or just take a leap of faith?

Peter Tamaroff

@BenjaLim How should I take the intersection? What would you do?

BenjaLim

hold on

Eugene

2:27 AM

@robjohn it's by a very reputable author though. manjul bhargava

BenjaLim

@PeterTamaroff

Suppose $(a,b,c,d) \in W \cap H_2$

Then in particular $(a,b,c,d) \in W$

so that the components must satisfy $a + b - c - d = 0$

Also $(a,b,c,d) \in W$

So that there are real numbers $e,f$ such that

$(a,b,c,d) = e(1,0,-1,1) + f(0,-1,-3,1)$

so $a = e$, $b = -f$, $c = -e - 3f$

$d = e +f $

Peter Tamaroff

@BenjaLim I'm getting that $e+f=0$

BenjaLim

how?

@PeterTamaroff Can you wait for about 1 hour?

There is this stupid construction happening right outside of the window

it's so noisy

Peter Tamaroff

I assume that $v=c_1(1,0,-1,1)+c_2(0,-1,-3,1)$ so that $v=(c_1,-c_2,-c_1-3c_2,c_1+c_2)$

BenjaLim

and the excavator and drilling are driving me nuts

@PeterTamaroff I'm going to go to some café or something

Peter Tamaroff

2:34 AM

@BenjaLim I'll go to sleep in about $1$ hours :P

I have a midterm tomorrow.

BenjaLim

@PeterTamaroff What time in BA?

oh shit

really noisy here

Peter Tamaroff

@BenjaLim 11:35 PM

@BenjaLim Do you have earphones?

BenjaLim

no unfortunately

let's continue

mid term is more important than noise

Peter Tamaroff

Since $v=(c_1,-c_2,-c_1-3c_2,c_1+c_2)$ I use the equation from $H_2$ to get $c_1-c_2+c_1+3c_2-c_1-c_2=0$

Which gives $c_1+c_2=0$

BenjaLim

yes

So in fact $H_2 \cap W$ is just a line

Peter Tamaroff

2:36 AM

@BenjaLim It is in fact $v=c_1 (1,1,1,0)$

BenjaLim

yes

Peter Tamaroff

@BenjaLim Right.

@BenjaLim So $S$ must be a plane?

BenjaLim

@PeterTamaroff yes that's what I am guessing

because

if you call $H \cap W_2 = L$

then you want $L \oplus S = H_1$

Peter Tamaroff

I corrected the vector

BenjaLim

so that $\dim L + \dim S = \dim H_1$

Peter Tamaroff

2:38 AM

@BenjaLim Right.

BenjaLim

Hence $\dim S = \dim H_1 - \dim L = 3- 1 = 2$

Peter Tamaroff

@BenjaLim Wait

BenjaLim

@PeterTamaroff

Peter Tamaroff

They're asking that

$S \cap H_2 \neq \{ 0 \}$

BenjaLim

yes

Peter Tamaroff

2:42 AM

And that $S$ is the complement of $L$

That automatically means $S$ has dimesion two. So it must be spanned by two linearly indep vectors

Eugene

@BenjaLim throw another shrimp on the barbie!

BenjaLim

@PeterTamaroff What is true is that $(W \cap H_2) \cap S = \{0\}$

this is because your sum is direct

Peter Tamaroff

@BenjaLim Yes. You mean $\{ 0 \}$ =D

BenjaLim

@PeterTamaroff too lazy

FYI in most algebra books, e.g. like AM you will just see 0

Peter Tamaroff

@BenjaLim But one thing

BenjaLim

2:44 AM

@PeterTamaroff wait

Peter Tamaroff

Since $L =<(1,1,2,0)>$

BenjaLim

you have a problem

Peter Tamaroff

We have taht $L \subset H_2$

BenjaLim

wait

what is $L$?

wasn't it $(1,1,1,0)$?

Peter Tamaroff

@BenjaLim I had messed a sum. hehehe

BenjaLim

2:45 AM

what is $L$ now?

crap all this arithmetic we're messing up

It is $(1,1,2,0)$

ok

So it is in $H_2$

ophewww

I was like

Peter Tamaroff

Since $x_1+x_2-x_3-x_4=0$

BenjaLim

2:46 AM

crap it is not in $H_1$ then what does the internal direct sum make sense?

Peter Tamaroff

@BenjaLim It is also in $H_1$

BenjaLim

@PeterTamaroff You just need to find two other vectors $v_1$ and $v_2$ that are in $H_1$

Peter Tamaroff

@BenjaLim Yay!

BenjaLim

$v_1,v_2$ and $(1,1,2,0)$ are linearly independent

Peter Tamaroff

@BenjaLim But we want $L \oplus S =H_1$ not $H_2$

BenjaLim

2:47 AM

this should be pretty easy

sorry

got the $H's$ mixed up

@PeterTamaroff Is this ok now?

Peter Tamaroff

Yeah. That is part of first mid term in Algebra (like half the semester), which hasn't even started yet, hehehehe.

BenjaLim

@PeterTamaroff In fact let me prove to you rigorously now why $H_1$ has dimension 3

Peter Tamaroff

@BenjaLim I'm all eyes.

BenjaLim

$H_1 $ is the orthogonal complement of the one dimensional subspace spanned by $(1,1,-1,-2)$

so since $\Bbb{R}^4 = H_1 \oplus H_1^{\perp} = H_1 \oplus (1,1,-1,-2)$

we have that $\dim \Bbb{R}^4 = \dim H_1 + 1$

so that $\dim H_1 = 3$

@PeterTamaroff satisfied?

Peter Tamaroff

@BenjaLim I don't know what the orthogonal complement is! AHHHH

BenjaLim

2:50 AM

useless

Peter Tamaroff

@BenjaLim I have like $50$ excersices from the book I got. It is from a Spaniard professor, which has his books TeX pdfs (they are only available online as far as I know)

BenjaLim

@PeterTamaroff crazy

Peter Tamaroff

But so far I know up to direct sum.

BenjaLim

linear algebra is crazy

that is good enough

but at the moment you are only dealing with the internal direct sum

knowing the direct sum

there is such a thing as the external direct sum

Peter Tamaroff

@BenjaLim Hm, OK.

BenjaLim

2:53 AM

@PeterTamaroff You should go to bed

Vamos Vamos Peter Tamaroff!!!

Peter Tamaroff

@BenjaLim Interestingly enough, the $50^{th}$ excersice if from Fibonacci.

BenjaLim

for your exam tomorrow

Peter Tamaroff

He asks to get Binet's Formula from some algebraic considerations.

BenjaLim

@PeterTamaroff yes

Peter Tamaroff

@BenjaLim Hahah thanks!

@BenjaLim It is calculus, though.

BenjaLim

2:54 AM

The proof I know is using generating functions

Peter Tamaroff

And I have 3hs

BenjaLim

@PeterTamaroff No problem

Peter Tamaroff

@BenjaLim I really like those

BenjaLim

Go to bed now

you need enough sleep for an exam

Peter Tamaroff

@BenjaLim Yes, true. But it isn't that early. It is at 1 PM

Let's look at one more, and I'll go, OK?

BenjaLim

2:55 AM

doesn't matter

ok

hit em up

Peter Tamaroff

I need to find a basis for $\mathcal P(\{a,b,c\})$ with $K=\Bbb Z_2$

The action is simply $1 \cdot S = 1$ and $0 \cdot S = \emptyset$

anon

Using $A+B:=(A\cup B)\backslash(A\cap B)$ again.

Peter Tamaroff

@anon I suppose

BenjaLim

symmetric difference

@PeterTamaroff what is the identity element in $\mathcal{P}(\{a,b,c\})$?

Peter Tamaroff

@BenjaLim $\emptyset$

anon

2:58 AM

empty set

BenjaLim

no the multiplicative identity

that is an order

anon

multiplication in a vector space?

Peter Tamaroff

@BenjaLim Oh, $A \cap A =A$

I gues....

But I doesn't work for every set in $\mathcal P$

BenjaLim

@anon Forget it I was thinking field extensions :D :D

Peter Tamaroff

Hm....

anon

3:00 AM

{a} {a,b} {a,b,c} ?

BenjaLim

@PeterTamaroff You can bash out and check that the above is linearly independent :D

there are only $2 \times 2 \times 2 $ possibilities

taylor

hello world

Peter Tamaroff

@BenjaLim Hahaha true.

Well, I'll go get some rest!

BenjaLim

yes

I should go too

anon

@PeterTamaroff Surely you should at least test {{a},{b},{c}}?

BenjaLim

3:02 AM

need to get started on algebraic topology

Peter Tamaroff

@anon Well, that would work, right?

@BenjaLim GO, GO!

Figure it out :)

@anon I'm confident!

bye all !

later

(also, my first guess is also a legit basis)

@PeterTamaroff It's impossible for any nonzero element to be a scalar multiple of another in a $\Bbb F_2$ vector space, so you merely needed to pick two arbitrary ones and figure out the third. (And you can find $\dim V=3$ by noting $|P(\{a,b,c\})|=2^3$.)

Peter Tamaroff

3:07 AM

@anon What is an $\Bbb F_2$ vector space?

anon

$\Bbb F_2$ is the field with two elements, i.e. $\Bbb Z/2\Bbb Z$.

Peter Tamaroff

@anon the or a?

anon

how many fields with two elements do you know of?

Peter Tamaroff

@anon I don't know!

Hahahha

anon

I'll give you a hint: every field has distinct additive identity and multiplicative identity. Can F2 have any other elements?

Peter Tamaroff

3:08 AM

@anon I know that one is a field, because 2 is prime.

@anon What I don't understand is if you're saying $\Bbb F_2 := \Bbb Z_2$ or $\Bbb F_2$ is a notation for a field with two elements.

anon

There is only one field with two elements (up to isomorphism). It happens to be $\Bbb Z/2\Bbb Z$.

Peter Tamaroff

@anon Ok, I'll catch up on that later. I'll go sleep now.

@BenjaLim Bye!

Eugene

@BenjaLim throw another shrimp on the barbie!

Jeff

3:58 AM

hey folks. anyone in here?

would like some help with basic linear algebra if possible

anon

k

Jeff

$A=([0,1,1],[1,0,1],[1,1,0])$ and one of the evals $\lambda=-1$. Solve $(A-\lambda I)v=0$. Therefore, $([1,1,1],[1,1,1],[1,1,1])v=0$. Row reduce matrix to get $([1,1,1],[0,0,0],[0,0,0])v=0$. Then the answer is that $E_\lambda=span{(-1,0,1),(0,-1,1)}$. How that last step? I get the components of v: $v_1+v_2+v_3=0$. what then?

(i don't remember how to make a matrix, those are rows between the [])

anon

you just need two arbitrary LI vectors that generate the subspace of v with components adding to 0, which is equivalent to finding two LI vectors perpendicular to (1,1,1). they picked two arbitrarily.

Jeff

oh

anon

(obviously (-1,0,1) and (0,-1,1) are LI vectors in the given subspace, and the subspace is dim 2 so this is a basis)

Jeff

4:06 AM

bottom line: any two vectors such that the components of each vector equal zero and the two vectors must be LI.

anon

sum of components

Jeff

yes, the algabraic multiplicity was 2, and the text was showing that the geometric one is 2, too. thx anon

robjohn

@Eugene you could be the first to find the errors :-)

Jeff

@anon right. i meant sum

@anon: did they pick two vecs because the alg. mult. was 2?

anon

does $E_\lambda$ stand for eigenspace, or generalized eigenspace? If the former, because of algebraic multiplicity, if the latter, geometric multiplicity. (I think I got that right.) Of course they're equal in this case.

pretty sure it's just eigenspace, so yes algebraic multiplicity

Jeff

4:11 AM

well, $E_\lambda$ was my shorthand notation. they use notation $E_{\lambda=-1}$, where $-1$ is a eigenvector with algebraic multiplicity of $2$.

anon

it's pretty easy to see the multiplicitis are the same from the rred form

Jeff

what is the giveaway in the rred form?

anon

two of the rows are zero

well, hmm

Jeff

ok. thx

Martin Sleziak

BTW egreg is the first 100k user at TeX.SE.

I guess Math.SE and TeX.SE have non-trivial intersection, so I thought it might be interesting information for some folks here too.

Eugene

4:40 AM

@robjohn or the first one i know of. lol.

robjohn

4:55 AM

@Eugene You don't need to worry about errors, you can just find them :-)

Eugene

@robjohn cool. i'll just go ahead and read it then. thanks.

Frank Science

6:04 AM

@Eugene Yes, my problem.

Frank Science

6:52 AM

@Eugene There's no doubt that my problem is extrememly elementary, but there seems nobody interested in such hard elementary problem.

leo

7:28 AM

hi there

I'm having troubles

anon

aren't we all

leo

:-)

I suppose so...

robjohn

@leo what sort of troubles?

leo

I'm trying to see $L^2[a,b]$ as the completion of the pre-Hilbert space $C[a,b]$.

I understand how goes all the construction.

In this setting $L^2[a,b]$ is a set of equivalence classes

@robjohn My problem is that I can not realize $\chi_{\Bbb Q\cap [a,b]}$ as an equivalent class

To be more specific I'll sketch the construction

robjohn

@leo Isn't the equivalence class, functions which are constant on that set?

anon

7:40 AM

Well, what is the $L^2$ (semi?) norm of that characteristic function?

@robjohn Let $f$ be a function that is $0$ everywhere on $\Bbb Q\cap [a,b]$ except a specific point (assume $a\ne b$). Then isn't $f\sim\chi$?

My understanding is that $L^2$ is made up of functions with $\|f\|_2<\infty$, modulo the space of functions $g$ such that $\|g\|_2=0$.

leo

1. In $C[a,b]$ (continuous functions over $[a,b]$) with the Riemann integral, define $(f,g)=\int_a^b f$
2. Let $\cal{C}$ the set of Cauchy sequences in $C[a,b]$ (on the norm induced by the inner product).
3. Define $f\sim g\iff \lim_{n\to\infty} |f_n-g_n|^2=0$
4. Define $L^2[a,b]=C[a,b]/\sim$

@anon is this construction

user19161

@leo So you don't understand the last step - completion of a metric space?

leo

And the inner product defined in $L²[a,b]$ is given by $(f,g)=\lim(f_n,g_n)$

anon

Hmm. You require the functions be continuous on $[a,b]$. How is $\chi_{\Bbb Q\cap[a,b]}\in C[a,b]$ then?

leo

@JasperLoy I understand All the construction.

anon

7:50 AM

Oh, we need a Cauchy sequence to serve as an avatar for $\chi$. Hmm.

leo

@anon Exactly!

Gigili

@anon: Did you get my email?

anon

let me check my email

user19161

@Gigili Did you meet Mahnax?

leo

The candidate is something like $f_n=\chi_{r_1,\ldots, r_n}$ where $\{r_i\}$ is an enumeration of the rationals of $[a,b]$

but this doesn't works.

user19161

7:52 AM

The latex is confusing me, haha. You guys go ahead!

user19161

I actually wrote 20 pages on this in my undergrad days!

leo

Because $\int_a^b |\chi-f_n|^2$ doesn't make sense.

Martin Sleziak

@JasperLoy you know that you can display math in chat using robjohn's or Zhen's bookmarlet, right?

user19161

@MartinSleziak Yes, thanks!

leo

@robjohn did you see what I wrote

Rajesh D

10:43 AM

What is the derivative of $\frac{sin((n+\frac{1}{2})x)}{sin(\frac{x}{2})}$, and what is the value of the derivative at $x = 0$. In my computations the derivative is $\infty$ at $x = 0$ But I know it is $0$. I don't know where I have done the mistake.

anon

hard to tell where the mistake is if you don't write your computations out. in any case, alternative methods might be Taylor expansions (you only need to go partial) and complex exponentials

Rajesh D

@anon : How to compute the value of a function at a point, and also draw a proper graph (unlike the contour plots here which I do not understand) in Wolfram math

anon

10:59 AM

f(x) at x=a will compute the value of a function at a point, and drawing a graph you just do y=f(x) (the word "plot" seems to be optional). however, you have a parameter $n$; unless you plug a specific value for $n$ in, it can't give you a definitive graph, so it won't really work...

Rajesh D

I am getting $\infty$ for any $n$ in my calculation, just want to make sure it is wrong, I'll enter some value for $n$

anon

definitely wrong

pretty sure that's like a Dirichlet kernel or something...

yep, dirichlet kernel

Rajesh D

This is what I've got : $\frac{(2n+1)sin(\frac{x}{2})cos((n+\frac{1}{2})x) - sin((n+\frac{1}{2})x)cos(\frac{1}{2}x)}{4(sin(\frac{x}{2}))^2}$

by hand computation of derivative

@anon

anon

how did you get a sine of a cosine?

brb bathroom

Rajesh D

sorry mistake

i've edited

anon

11:09 AM

k, now how do you plug x=0 into that?

Rajesh D

the denominator goes to zero as $x^2$ and the numerator goes to zero as $x$, hence I get $\infty$

Frank Science

@RajeshD What do you want to do?

anon

what makes you think the numerator goes to zero as x^1?

@FrankScience he wants to see where his computational error is

Frank Science

Derivation?

Rajesh D

$sin(x) = x$ for small x @anon

anon

11:13 AM

differentiation of sin((n+1/2)x)/sin(x/2) at x=0

Frank Science

You can do it by Taylor series.

Rajesh D

Is my differentiation itself is wrong I guess, Wolfram shows different one

anon

I'm too inebriated for this sort of arithmetic.

Frank Science

Well, I'll compute it. Wait for a moment.

Rajesh D

here

anon

11:16 AM

@FrankScience It's 0, as w|a confirms, Raj just wants to know where his error lies.

I still think it's with the numerator.

Frank Science

@RajeshD Let $f(x)=\sin\alpha x/\sin\beta x$.

Rajesh D

ok

fine

Frank Science

@RajeshD We have $f'(x)=(\alpha\cos\alpha x\cdot\sin\beta x-\beta\sin\alpha x\cdot\cos\beta x)/\sin^2\beta x$ for $x>0$.

Rajesh D

yes fine

robjohn

@anon Oh, I missed the point of null sets.

Frank Science

11:27 AM

@RajeshD $f(0)=\alpha/\beta$. I've omited.

Rajesh D

How @Frank

How did you compute $f(0)$?

OHHHHH

Frank Science

@RajeshD No, it's only definition. Otherwise, $f'(0)$ never exists.

Rajesh D

OMG

Frank Science

@RajeshD Because $f'(0)=\lim_{x\to0}(f(x)-f(0))/x$.

Rajesh D

So the usual derivative does not work for $x = 0$, as the definition is different. Is it right @Frank

Frank Science

11:30 AM

@RajeshD To be rigorously, we should prove some theorems.

Rajesh D

the now how do we get $f'(0)$

L' Hopital rule??

Frank Science

@RajeshD For example, if $f(x)$ is coutinuous on $-H\le x\le H$, and it's differentiable on $0<|x|\le H$, we have $f'(0)=\lim_{x\to 0}f'(x)$. Can you prove it?

@RajeshD You should prove it first.

Rajesh D

ok let me look at it for a moment

Frank Science

@RajeshD Well, you seems not very familiar with such stuff. We can compute it without that theorem, only use the definition of derivation.

$f(x)=\sin\alpha x/\sin\beta x$ for $x\neq0$, and $f(0)=\alpha/\beta$. Let's compute $f'(0)$.

Rajesh D

Has it got to do with the Darboux Theorem ? @Frank

I'll take it granted for the moment, i'll think of it later

Frank Science

11:38 AM

@RajeshD I'm not sure.