Mathematics - 2015-01-29 (page 1 of 4)

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12:07 AM

Hello =) Could someone help me at the multiplication: math.stackexchange.com/questions/956859/… ???

@Fundamental I don't know how often you pull this query, but right now you're the #2 meta user by (meta)"rep."

12:43 AM

Hi all can someone answer me simple question that baffels me

@Karlo Baffles?

I can try.

yes :D

here is the post

ops

I am not quite sure whether I can follow you. But here is hint which may guide you to the solution: Let $W$ be the image of $\phi$. Let $W'$ be the vector space generated by the columns of $A$. If you can show that $\dim W=\dim W'$, you are done. (Do you know why?) — russoo 13 mins ago

here it is

What's your question?

did you get the link?

oh ok

here is my question

You pasted a comment.

12:46 AM

Here is my question.

The problem and solution seem to be isomorphism but it is not said that $\phi$ is isomporphism

@Karlo Don't you define the rank of a linear transformation $\phi:V\to V'$ as the dimension of the image of $\phi$?

r(phi)=dimIm phi

Yes.

yes

If you pick a basis for $V$, say $e_1,\ldots,e_n$ then $\phi(e_1),\ldots,\phi(e_n)$ generate the image of $\phi$.

12:51 AM

yes

and the rank will be maximum

now let us look at the V' the other space

If you consider the matrix of $\phi$ from the basis $(e_i)$ to some base $(f_1,\ldots,f_m)$ then it's columns are the coordinates of $\phi(e_j)$ in the base $(f_i)$.

The rank of this matrix is the dimension of the space in $k^n$ generated by the vectors corresponding to the $\phi(e_j)$, say $\phi(e_j)=\sum a_{ij}f_i$, then the vectors are $(a_{i1},\ldots,a_{im})$ for $i=1,\ldots,n$.

But "taking coordinates" is an isomorphism so it preserves dimension.

This means that ${\rm rk}(\phi)=\dim {\rm im} \;\phi $ equals the dimension of this "column space."

@Karlo Do you understand?

not quite sure but

if we set basis $e_n$ on V

dimIm$phi$ will be $\phi(e_n)$

that is easy

but we have to be careful what Im will be

because V' has basis all vectors of the basis e_n which are in Im$\phi$!?

Sayan Chattopadhyay

Hi

Hi Sayan are you from aops?

:D

Sayan Chattopadhyay

Aops?

12:57 AM

one forum

Sayan Chattopadhyay

Hi@Chris'ssis

@Karlo A dimension cannot equal a vector.

yes but

if we pick e_2 ... e_n

for basis

then clearly the dim = n-1

Sayan Chattopadhyay

Hey karlo what are you guys talking about

@Karlo No. If the map is the zero map then the dimension is $0$ regardless of the basis.

I have to go, sorry.

1:00 AM

Sayan

you here

Sayan Chattopadhyay

Yes

Suppose we have linear operator

and we map from V -> V'

if we pick basis for V e_n

the dimension of the matrix in V will be n right

Sayan Chattopadhyay

Well I don't know what's. Linear operator

ah ok

Sayan Chattopadhyay

You can ask me anything in number theory

Well what's your profession karlo

1:02 AM

1st year

in university

Anyone that understands linear algebra here ;P

Sayan Chattopadhyay

Oh.....which topic in maths do u like the most

I am competitor

I used to go to olympiads 12 years

Sayan Chattopadhyay

Well this room is pretty good then for you

yes :)

Sayan Chattopadhyay

This is a good room

1:06 AM

could have gone to IMO

but too lazy :D

Sayan Chattopadhyay

imo

international math olympiad

Sayan Chattopadhyay

Do u like the Riemann hypothesis

:D :D and you bet

I always joke : I am at starbucks sloving Riemann hyop

hypo

1:24 AM

@SayanChattopadhyay Hehe.. this is awesome and shows exactly the image an non-mathematician has about math. It's so widespread that they don't even understand each-other :-)

1:46 AM

just proved $\prod_{i=1}^\infty \mathbb{Z}$ isn't free as a $\mathbb{Z}$-module

that was quite something

Does there exist a field with 1, 2, 4, 6, 7, 15 elements??

The field $\mathbb{F}_{p^n}$ has $p^n$ elements.

So, a field with
$2$ elements is : $\mathbb{F}_2$
$4$ elements is : $\mathbb{F}_2^2$
$7$ elements is : $\mathbb{F}_7$

Is this correct?? What about $1, 6, 15$?? Are there fields that have so many elements??

depending on your definition {0} could be a field with 1 element

Shouldn't a field contain the identity element of the addition and the multiplication?? @SamuelYusim

that's why I said depending on your definition

So, doesn't it exist a field with one element??

@SamuelYusim

1:54 AM

some people don't require rings to have a multiplicative identity, and so a field is just a "ring" of this sort where every nonzero element is invertible. this is vacuously true, as they say, of {0}.

also there's only one ring with one element up to isomorphism and you know it's not a field so there you go

if you require fields to have at least 2 elements, there certainly can't be one with one element

Ok... @SamuelYusim And what's with $6$ and $15$?? Are there fields with $6$ or $15$ elements??

What's your gut feeling, @Mary? :)

@MaryStar A finite field can only have cardinality a power of a prime.

Both of them, $6$ and $15$, cannot be written as $p^n$, where $p$ is prime and $n \in \mathbb{N}$, so that we can say that the field id $\mathbb{F}_{p^n}$... So, I don't know... How could we check this @AlexWertheim

Are you familiar with the characteristic of a ring, @Mary?

2:02 AM

@PedroTamaroff Since neither $6$ nor $15$ can be written as $p^n$, where $p$ is prime and $n \in \mathbb{N}$, there are no such fields??

@MaryStar Yes.

Pedro

$n$ is the characteristic, so $n$ is smallest natural number such that $n \cdot a=0$, where $a \in R$, right?? @AlexWertheim

yeah

Ok!! Thanks!! @PedroTamaroff

2:04 AM

now why must it be prime in a finite field

@PedroTamaroff remember my question?

@Mary: yes. For any ring $R$, the characteristic of $R$, $n$, can be thought of as the generator of the kernel of the unique ring homomorphism $\mathbb{Z} \rightarrow R$.

@AlexWertheim can you help with linear algebra question

Perhaps, @Karlo. I'll try. What's the question?

if we have linear operator that maps V->V'

and we set basis for V e_n then the matrix of e_n has dim=n

ok?

2:08 AM

You're saying $V$ is $n$ dimensional, and you fix a set of basis vectors $e_{n}$. Then the $n \times n$ matrix formed by the taking the $e_{n}$'s as row/column vectors has rank $n$, sure.

okey now

If we have the basis e_1 ... e_n and

e_1 is in Ker and others are in Im

for V' we have that each vector is represented as a linear combination

of the vectors in im

Im

but they are lineary indipendant

sk V' has basis e_2 to e_n

ops i mean

V' has basis $\phi(e_2)$ to $\phi(e_n)

of the matrix

then it's dimension of matrix is n-1

right?

I'm assuming $\phi$ is the linear operator?

yes

If you're saying that the matrix with rows $\phi(e_{2}), \ldots,\phi(e_{n})$ has rank $n-1$, sure.

of V' yes

so it looks like the dimension of the matrix of V' depends on the basis vectors in Imphi

right?

2:14 AM

@Fundamental What brings you here? =)

I'm not sure what you're trying to say. Are you saying the dimension of $V'$ depends on the image of the basis vectors under $\phi$?

yes

@PedroTamaroff I'm not here, it's an illusion. The "Rejoin favorite rooms" button.

Yes, I believe so. Well, we can draw explicit conclusions, assuming $\phi$ is surjective, which it sounds like it is in this case.

btw why if a matrix has solution $(x_1,x_2 .... x_n )$ and all $x_i$ are diffrent from 0 the determinant is 0

2:16 AM

@Fundamental Hehe, that happens to me too.

hello?

@Karlo: It means the columns of the matrix are linearly dependent.

which column is 0?

No column has to be zero. If one column is zero, then the columns have to be linearly dependent.

yes and that is what you say

if all are diffrent from 0 then one col has to be 0 => det=0

oh wait you can get det=0 if we have entries that are not zero

2:30 AM

That's not what I said. I said if a matrix $A$ has a solution $x$ such that $Ax = 0$ with $ \neq 0$, then its columns are linearly dependent. I did not say that one of the columns was $0$.

yes yes

and how to prove this any idea

Yes, I'm quite aware of how to prove it. Think about elementary column operations, and the multiplicativity of the determinant.

I've got to run though.

2 hours later…

4:14 AM

Hey guys I need to prove this fact (Ax C) cup(B × D) ⊂ (A ∪ B) × (C ∪ D). When does equality occur? I proved it is a subset and I also got the case of equality when A = B and C = D but is this the only case ??

4:35 AM

> i dont know why it dowsn't let me publish my question (it says something about quality standards...). so i'll write know some crap so it will let me post it. ignore it.

Source

I don't understand people... "Hm, I can't post my question because it doesn't meet quality standards. I got it! I'll append some crap to get around the restriction." :\

That's called "gaming" the system.

What function can I use to represent the next sequence: 2,2,2,2,2,3,3,2,2,2,2,2,3,3,2,2,2,2,2,3,3…?

5:01 AM

@Ilya_Gazman $$f(n)=\left\{\begin{array}{} 3&\text{if }7\mid n^2+n\\ 2&\text{otherwise}\end{array}\right.$$

or using Iverson Brackets $$2+[7\mid n^2+n]$$

1 hour later…

6:14 AM

@MikeMiller: Good morning.

@MikeMiller: HA! No way to beat me today.

Good morning, @Huy

@MikeMiller: What's up. Ready to sleep?

No, @Huy, it's only 10PM (in the morning)

I see.

It's 7am here.

And I'm already at uni.

Let's look at some more differential geometry then.

No, I'm going to watch an episode of Dr. Katz, Professional Therapist.

Maybe even two.

6:17 AM

Pf.

7:02 AM

@MikeMiller: If, by any chance, you're still here: For a smooth submanifold $N$ of $M$ we defined an atlas by $$\mathcal{A}_N:= \{(V,\chi) | \, V = N \cap U, \, \chi = \psi|_{N \cap U}: N \cap U \to \mathbb{R}^k \text{ where } (U,\psi) \in \mathcal{A}_M\}.$$ Is that already a maximal atlas?

@SayanChattopadhyay Hi

I'll take a break from doing math for a short period of time, I need to recover myself. Far too exhausted.

7:53 AM

@BalarkaSen: Why is $\mathbb{R}P^2$ compact?

@Huy Do you know any measure theory?

@Anthony: I should, but I don't.

Awh.

Thanks anyway.

@Huy Quotient space of compact spaces are...?

You should have a good picture of RP^n in mind when you're talking about them. Projective spaces can be befuddling.

The way I think of RP^2 is as [0, 1] cross [0, 1] with opposite edges identified in the opposite orientation.

From that, compactness is more or less obvious.

8:10 AM

@BalarkaSen: I didn't remember quotient spaces of cpct spaces are cpct as well. Should have properly studied topology.

You don't have to remember it. Just visualize.

A space is compact if it can be finitely covered. You identify like mad inside your space. It can still be finitely covered.

:P

yooooooooo

ok

I usually don't prove point-set stuff if it's not too necessary.

anyone know matlab? I'm stuck at trying to introduce a new parameter so I can find the value of the initial condition and time step

8:12 AM

RP^n is actually rather easy to visualize. On the other hand, I am struggling with CP^n right now.

8:22 AM

Why are you thinking so much about RP^n lately, @Huy?

@BalarkaSen: Just coming up a lot as examples in my DG notes. I have an exam in 7 minutes. See you later.

Ah, ok. Good luck.

Quick question: in working with matrices that have complex eigenvalues, I notice that the eigenvectors produced from a complex eigenvalue and its complex conjugate appear to have some relation. For example, (1, 2-i) is an eigenvector for lambda=3i and (1, 2+i) is an eigenvector for lambda=-3i.

In this case, the vector whose components are complex conjugates of another eigenvector's components is also an eigenvector. Is this always the case, or is the relationship less straight-forward, or is there no relationship at all?

8:37 AM

I think this is always the case: consider Av = kv (the vector v is an eigenvector of matrix A). Then this implies (Av)* = (kv)*, where * indicates the complex conjugate. If A is a real matrix, then I think this is equivalent to Av* = k*v*, and so the eigenvector corresponding to k* is just v*.

Sayan Chattopadhyay

9:32 AM

Hi @BalarkaSen

A quick question:what do u call log to the base of pi

@Hippalectryon you might like to know that what we did is far from being complete.

Sayan Chattopadhyay

@Chris'ssis what do u call log to the base of pi

@SayanChattopadhyay Not sure I get your point.

Sayan Chattopadhyay

See log to the base of e of any number is called its natural logarithm so what do u call log to the base of pi for any number

@SayanChattopadhyay I don't know if there is such a name. I'm not aware of it.

9:46 AM

@SayanChattopadhyay Hello.

iwriteonbananas

@Huy quotient maps are always continuous...image if a compact space under continuous map is compact

balarka

@DanielFischer @robjohn do you see a nice way of computing $$\lim_{\epsilon \to 0}\int_{\epsilon}^{\infty} \frac{\sin(x)}{x}(\log(\epsilon+x)-\log(x)) \ dx$$? The limit should be $0$.

bananas

iwriteonbananas

why is $SL_2(\mathbb{R})$ generated by the following matrices

fuck i dont wanna write down the matrices

( t 0 \\ 0 t )

( 1 s \\ 0 1 )

and

( 0 1 \\ -1 0 )

?

where s,t are not 0

well think about it

Sayan Chattopadhyay

9:49 AM

Hi @BalarkaSen

iwriteonbananas

the claim means that any matrix with determinant 1 can be written as a product of those 3 matrices right?

just check that every matrix in SL2(R) can be written as a product of those three

iwriteonbananas

lol

i've tried

@iwriteonbananas no

Sayan Chattopadhyay

So there is no name for log to the base of pi

9:50 AM

product of those generators raised to some exponent

iwriteonbananas

no ?

@SayanChattopadhyay no

iwriteonbananas

yeah

i dont know how to do it

Sayan Chattopadhyay

Oh hi@iwriteonbananas

iwriteonbananas

hey sayan

whats your power level today?

Sayan Chattopadhyay

9:51 AM

Amazing name @iwriteonbananas

iwriteonbananas

its my real name

Sayan Chattopadhyay

Power level?

iwriteonbananas

yeah

i suck at homology. gah.

iwriteonbananas

balarka give me a hint

Sayan Chattopadhyay

9:52 AM

Even me

no

iwriteonbananas

prick

@SayanChattopadhyay you know homology?

Sayan Chattopadhyay

I was watching few videos on it yesterday

Its basically algebraic topology

oh. that's not really knowing.

@SayanChattopadhyay it's a tool. algtop is a lot more than just homology.

Sayan Chattopadhyay

9:55 AM

well I have started liking group theory which is used in this

used in where?

Sayan Chattopadhyay

Algebraic topology

oh. sure. but you haven't studied algebraic topology yet, have you?

Sayan Chattopadhyay

Are u reading any books currently @BalarkaSen

Nope not yet

Maths books

yes. Hatcher, for one. and side-reading Atiyah-MacDonald.

Sayan Chattopadhyay

9:57 AM

Names of the books

Hatcher's Algebraic Topology and A-M's Commutative Algebra

Sayan Chattopadhyay

About maths

Fine

Why do you ask?

Sayan Chattopadhyay

I want to read something about algebraic topology

OK, but be advised that it takes background on point-set topology and basic algebra.

How much algebra are you familiar with?

Sayan Chattopadhyay

10:00 AM

How much do u need to be familiar with

a fair bit. a thorough knowledge of groups theory, rings and fields a little.

do you know any topology?

Sayan Chattopadhyay

Nope then I have to start from scratch

Yes, you have to.

Sayan Chattopadhyay

Any book recommended for that?

Simmons. Introduction to modern analysis.

Sayan Chattopadhyay

10:03 AM

Oh......it has topology

?

It has topology in the first section.

Sayan Chattopadhyay

Well I want to prove the Riemann hypothesis that is why I am learning allghis

All this

Haha. You do realize that it's been unsolved for 200 years, right?

Sayan Chattopadhyay

Yes but even Fermat's last theorem was unsolved for500 years

Fair enough. Interesting goal.

@SayanChattopadhyay Not at all a bad goal since even understanding the statement of Riemann hypothesis would push you to learn a lot of mathematics.

I presume you don't understand what RH says yet?

Sayan Chattopadhyay

10:08 AM

It says that the zeros of the zeta function have a real part equal to 0

not entirely correct.

Sayan Chattopadhyay

Well but I don't understand one thing

you mean real part equal to 1/2.

but anyway, do you understand the zeta function?

Sayan Chattopadhyay

Why is it called zeta function why not phi function theta function or something else

Just a name. But tell me the definition of zeta function.

Sayan Chattopadhyay

10:12 AM

A function of the form 1|1^s+1|2^s..........1|n^s

You mean $\sum_{n = 1}^\infty 1/n^s$, right?

Sayan Chattopadhyay

Then

What does underscore show

In your equation

@SayanChattopadhyay use this

Sayan Chattopadhyay

How to

well drag the ChatJax link to your bookmark tab

Sayan Chattopadhyay

10:18 AM

Wait then let me switch on my pc

yes i do mean that

But then RH doesn't make sense, see.

Since that thing converges only for $\Re[s] > 1$

And talking about zeros being of real part $1/2 < 1$ is nonsense.

Sayan Chattopadhyay

it should converge for 1 also right

The statement of the Riemann hypothesis might not be as trivial as you think it is ;)

@SayanChattopadhyay no.

$\sum_{n = 1}^\infty 1/n$ diverges, for example.

Sayan Chattopadhyay

yes but cant we use complex numbers

i mean we use

i am not sure what you mean

$\sum_{n = 1}^\infty 1/n^s$ diverges if $\Re[s] = 1$

you can prove that if you know enough calculus.

Sayan Chattopadhyay

10:25 AM

yes u are right...

so your definition of zeta doesn't make it clear what RH says.

Sayan Chattopadhyay

well u know i think we have to link RH to some other theorem as in to prove it

it takes work to make sense of RH, @SayanChattopadhyay. You need to be familiar with complex analysis to make sense of it.

@SayanChattopadhyay but you don't understand RH yet.

first try to understand it, then you can worry about proving it :P

Sayan Chattopadhyay

i will try my best and and atleast take one step forward in proving it

wait for one year

haha. there have been many steps made towards proving RH by many great mathematicians.

Sayan Chattopadhyay

10:28 AM

well i was thinking that twin prime conjecture,goldbach conjecture and RH might have a link

no they don't.

i will repeat myself : first try to understand what RH says before trying to find links or fancy stuff

Sayan Chattopadhyay

well before knowing the link between taniyama shimura conjecture and fermats last theorem even u would have said there is no link

well but RH has been by dream since 7th standard

i'd really really prefer you don't talk about big names you are not familiar with

first study calculus, complex analysis. then think about what it says and what the links might be.

Sayan Chattopadhyay

well yes but u see iam still in 10th standard

But i will try my best

you know taniyam shimura conjecture

yes, but i barely understand it. i have little background on modular forms.

Sayan Chattopadhyay

10:32 AM

i know just this much that it says that every modular forms is equal to an elliptical equation

but you don't even know what a modular form is :P

Sayan Chattopadhyay

But u cant show a modular form right i mean an actually

don't throw big names if you don't understand them, a wise person once said.

@SayanChattopadhyay i am not sure what you mean by "showing a modular form"

Sayan Chattopadhyay

i wont just trying to collect info

math is not collecting info

it's about understanding

Sayan Chattopadhyay

10:33 AM

i will

well in which standard are u @BalarkaSen

10th.

Sayan Chattopadhyay

what

what then how do u know this much

amazing

unbelievable

The next terrence tao of INdia

nothing amazing

Sayan Chattopadhyay

really i feel like saluting you

a lot of my classmates know a lot more than me

10:35 AM

hello, please is the condition $\lim\sup_{n\rightarrow+\infty} |\nabla u_n|^N_N<\frac{1}{8^N}$ impliese that $(u_n)$ is bounded ? thank you

Sayan Chattopadhyay

where do u study man

ok, i gotta go. i recommend you study mathematics than collecting information @Sayan

Sayan Chattopadhyay

where r u going

@BalarkaSen i will study maths and will start from algebra.......thanks for inspiration

@MikeMiller have you an idea ? please

please

Sayan Chattopadhyay

What idea@Vrouvrou

10:46 AM

is the condition $\lim\sup_{n\rightarrow+\infty} |\nabla u_n|^N_N<\frac{1}{8^N}$ impliese that $(u_n)$ is bounded ? thank you

@SayanChattopadhyay

Sayan Chattopadhyay

11:15 AM

@BalarkaSen my only question to you

@BalarkaSen now since i am going to start higher lecel mathematics where should i start from

Can anyone give me answer for the question i asked just now

@Vrouvrou where did u start from

11:29 AM

@Chris'ssis Of course :-)

@Hippalectryon This morning, very early morning, I began to put things on paper.

@Hippalectryon This limit is crazy $$\lim_{\epsilon \to 0}\int_{\epsilon}^{\infty} \frac{\sin(x)}{x}(\log(\epsilon+x)-\log(x)) \ dx$$

@Hippalectryon but you know what? I have something even more crazier.

$$\lim_{\epsilon \to 0}\int_{\epsilon}^{\infty} |\sin(x)|(\log(\epsilon+x)-\log(x)) \ dx$$

Why am I not even surprised :D

@Chris'ssis I think we can use the fact that $$\left|\int_{2k\pi}^{2(k+1)\pi}\frac{\sin(x)}{x}\mathrm{d}x\right| \le\frac1{\pi k(2k+1)}$$

@Hippalectryon The first limit came from the problem we discussed.

@robjohn Yeah, that would be a good idea.

@Chris'ssis I don't think that converges: the log part is $\sim\frac\epsilon x$

11:34 AM

hi @hippa !

@robjohn There I have sine in absolute value.

@Chris'ssis I know, so there is no cancellation. The integral diverges at $\infty$

@robjohn Mathematica suggests that limit goes to $0$ too. That is wrong.

@Chris'ssis The integral over any inverval $[n\pi,(n+1)\pi]$ is $\ge\frac{2\epsilon}{n\pi}$

@robjohn Yeah.

11:49 AM

hi @Chris'ssis @robjohn

@Ramanewbie Hi

@Chris'ssis are you ok with your book ? Will it be ready for sale soon ?

@Ramanewbie No. I already have many hours of work on a single question just to make it beautiful.

@Chris'ssis Still bugs to fix maybe ?

@Ramanewbie There is no bug in my work.

11:51 AM

@Chris'ssis so what ? why isn't it ready then

@Ramanewbie Try to write such a book and see why. I don't wanna publish a book for the sake of publishing.

@Chris'ssis I bet it's not easy but... I don't have this experience !

I planned not to work on math for a couple of days ... but ...

but ?

I'm still working. I'll have to take a break of a few days (it's hard to do that though).

11:55 AM

I though you already had all you wanted to be in it, right ? @Chris'ssis

@Ramanewbie Yeap, but in the meantime I found other things I'd like to add to it.

@Chris'ssis So will it be a really big book ? @Chris'ssis

@Ramanewbie Will see.

ok

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