Mathematics - 2013-12-09 (page 4 of 6)

Don Larynx

07:00

how is it not injective?

Karl Kronenfeld

It has to do with finding a map from an ideal of the form $(p,x)\subset\mathbb Z[x]$ to $\mathbb Q(x)/\mathbb Z[x]$ which does not extend to a map from $\mathbb Z[x]$.

John

I need to prove that sinc(t) is bandlimited while sinc(t^3) is not using the Paley Wiener Theorem and quite frankly don't know how to go about this

robjohn

@Potato It is being handled.

Potato

07:28

@robjohn Thanks.

usukidoll

I saw it too...wow I agree with Potato on this one. LIke if you're going to help others, don't be selfish and delete

Potato

@anon Questions got bulk-deleted?

anon

some sort of contest/exam and a single user IIRC

supposed to be temporary. might have been just closed not deleted, don't remember.

Potato

Huh. I guess there's all sorts of MSE inside baseball I'm missing.

@anon Do you know which old user Post No Bills is?

anon

no

Potato

07:36

This is clearly not his/her first rodeo.

robjohn

07:53

@Potato Thank Martin Sleziak and Arthur Fischer; they've done most of the work.

Jasper Loy

08:10

I saw on the news that many people in the Minnesota died because of the snow.

@John You should ask on the main site. Paley Wiener is nontrivial.

Now that I have 300 points, I may retire, lol.

Chris's sis

Greetings

I just saw a ghost and I need to be sure this is real!

Is Marvis gone?

I saw this on his account "This account is temporarily suspended for rule violations. The suspension period ends in 7 days ".

Please delete account

1 12 125 267

As far as I know, Marvis had a great contribution on MSE, his answers were always very good.

user1876508

08:28

If I remove the coordinate axes from R^3, do I get six circles which wrap a cube?

homotopically?

usukidoll

yeah but he threw a tantrum, went nuts, and decided to act selfish @Chris'ssis

it's sad that someone that smart would stoop low by deleting answers and editing a lot of things.

badass

@Chris I see the same thing :-(

Chris's sis

@badass Well, I think he should make a decision when he's not angry.

badass

I agree

He was a great helper in this room also

usukidoll

I wonder what caused him to snap?

badass

08:39

Why doesn't he just invite each mod to a separate room and talk it out?

Divide and conquer

Bill D tried to take them all on at one time and it cost him one year

usukidoll

huh?

robjohn

@Chris'ssis from the comments on this answer and his meta post, it would seem he had some problems with downvotes.

badass

"Useless moderators and thankless users" are his last words

Chris's sis

08:55

@robjohn thanks, I understand now.

usukidoll

I'm going to bed night guys

badass

Later

Chris's sis

@badass Sometimes the words are simple words, not necessarily meaningful. (depending on the circumstances)

@badass I say many things when I'm angry ... :-)

badass

Yes context/circumstances mean everything

robjohn

@Chris'ssis He was sort of on the edge before, but then Shog9's answer pulled the plug.

badass

09:00

And down the tube he went....

...sad to see

Chris's sis

@robjohn Yeah, that guy was a bit tough. ... Nothing new under the sun, this usually happens in the real world.

robjohn

@Chris'ssis I never noticed any signs of a problem until recently

Chris's sis

@robjohn Everybody makes mistakes.

robjohn

@Chris'ssis He was making 200+ points regularly and a few downvotes shouldn't have even affected him. I mean not at all, any downvotes would be completely erased by the opvotes.

Chris's sis

@robjohn Maybe he was reguraly downvoted and this can affect you. I remember when I was regurarly downvoted by an user ... it was pretty annoying.

badass

09:08

He must have been a perfectionist

Trying to please all of the people all of the time

Chris's sis

@badass Also on this site, it depends on your aim here. For instance, if you wanna have more rep than some users it's one thing, but if you want to learn and enjoy math is a different thing.

@badass no matter how good you are, there is always something to learn from the others.

Pride in mathematics can simply "kill" you.

When you think there is nothing to learn from the others, then you're really done.

badass

Indeed!

Chris's sis

@badass I like to know things but I'm aware that I need to learn as long as I live - - this is not true only for math but for any other area of life.

If I have 100 solutions for a question (supposing that) and someone wants to teach me another solution, I'm very willing and humble to learn it.

user97303

This is probably not the best place to ask, but I was wondering if anyone knows of a intuitive and simple tutorial of MFC? Particularly if you just want to drag a command line program in there where do you put it?

badass

The many proofs of the Pythagorean theorem is a great model of this @chris

Chris's sis

09:23

@badass Indeed :-)

user97303

@Chris'ssis I read your last comment and immediately thought of the 100 prisoner's problem

user97303

ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf

Chris's sis

@ReilaLee Interesting. :-)

user97303

Actually, don't read the paper

user97303

see if you can come up with some ways of solving it

user97303

09:27

The riddle:
One hundred prisoners have been newly ushered into prison. The warden tells
them that starting tomorrow, each of them will be placed in an isolated cell,
unable to communicate amongst each other. Each day, the warden will choose
one of the prisoners uniformly at random with replacement, and place him in
a central interrogation room containing only a light bulb with a toggle switch.
The prisoner will be able to observe the current state of the light bulb. If he
wishes, he can toggle the light bulb. He also has the option of announcing that

badass

Thanks for sharing @ReilaLee

user97303

anytime, @badass

Karl Kronenfeld

09:44

@ReilaLee I suppose the simplest solution is to have one prisoner keep track of how many prisoners entered the interrogation room.
Something like having the other prisoners toggle the light if it is off and it has never been off before when they were in the room; otherwise do nothing. The prisoner keeping count will do nothing if the light is off or add one and untoggle the light if it is on. (We can have him/her count to 100 excluding him/herself to account for the possibility of the light initially being on)

Arturo

To solve the riddle:
Wouldn't a possible solution be that given that there are 100 prisoners and they are randomly selected uniformly, then at $t< \infty$, they will all at least be interrogated once. The strategy would consist of knowing the maximum potential of the current in the lightbulb, or to avoid electric circuits technicalities, "the voltage" or brightness. Suppose the value is 100V. They are only allowed to lower it 1V the first time they are interrogated. This means that it will undoubltey arrive to 0V, and the lightblub will be off when that happens, and the prisoner will turn t…

user97303

@KarlKronenfeld, yep, that's about the simplest solution. The caveat is that we're waiting so many days between successive counts that this solution takes much longer than the statistical 99% confidence of a normal distribution

Martin Sleziak

Neither meta nor the main site works for me. Am I the only one, or do other people have problems too?

Chris's sis

@robjohn Isn't this limit $\log(11)$? This is what I got. math.stackexchange.com/questions/599203/… 99231

Martin Sleziak

It seems that other sites in SE network do not work for me either.

user97303

09:50

@Arturo the lightbulb is an ideal lightbulb only capable of depicting 1 bit of information (otherwise we could have prisoners sign the bulb when they go into the room or something :) )

Arturo

@ReilaLee oops! ok thanks

Martin Svanberg

Can anyone here explain what the "rank" of an elliptic curve means?

I can't wrap my head around the concept

anon

do you know that an elliptic curve has the structure of an abelian group?

if you have an abelian group isomorphic to Z^n x torsion-part, then the rank is n

it's kind of like the dimension of a vector space, but Z-linear instead of K-linear

user97303

10:08

Does anyone know how to stick a c++ function in MFC so you can use it from there?

Martin Svanberg

i understand that it's an abelian group. but i don't see how it is isomorphic to Z^n x torsion-part

i'm pretty new to this, so it's quite confusing to me. what corresponds to Z^n in the elliptic curve group?

anon

10:25

there is no unique "torsion-free part"

it doesn't work like that

Given an abelian group G with torsion subgroup T, one may write G=TxA and G=TxB, with A and B distinct subgroups of G both isomorphic to Z^n

Mordell's theorem states that the group of an elliptic curve is finitely generated

since the group is also abelian this forces it to be isomorphic to a direct product of a torsion group and a free abelian group - this is classical structure theory for abelian groups

Martin Svanberg

hm, I think I may lack some background here on torsion groups.

anon

http://en.wikipedia.org/wiki/Finitely-generated_abelian_group

a f.g. torsion abelian group is just a finite abelian group btw

Martin Svanberg

so a torsion group is a finite group that generates the rational points on the elliptic curve?

anon

huh?

the adjective "torsion" used to modify the word "group" just means every element in the group has finite order

by itself it needn't refer to elliptic curves or anything like it

and saying "the group generates the group" is fairly redundant btw

but no, the torsion elements of the elliptic curve need not generate all rational points

the fact that one speaks of rank at all means generally elliptic curves have elements of infinite order

and they cannot be generated by any torsion elements in an abelian group

so far it seems your questions are about group theory, not elliptic curves

Martin Svanberg

yeah, that's probably where i'm weak

Anonymous

11:19

Hello?

agent154

13:21

If anybody happens to be here at this hour: When performing the euclidean algorithm on a pair of Gaussian integers, is it done when the remainder is a unit or 0? In the case of normal integers, a remainder of 1 tells that the two integers are relatively prime... but in the Gaussian integers, there's also $i$ to contend with.

Also, if asked to find all GCDs $\gamma$ of two Gaussian integers, does it suffice to calculate $-\gamma$, $i\gamma$ and $-i\gamma$?

Huy

@agent154: It depends on when you consider it to be "done". If the remainder is a unit, that remainder is the GCD since the norm of the remainder strictly decreases with each step, so you are basically done already. Of course you can still do one more step and will get remainder 0 and are done then at the latest.

agent154

Is it convention to consider only the "positive" GCDs? So in the case of normal integers, we never seem to say that $-1$ is a GCD, though it is an associate of $1$...

Huy

And yes, the GCD of two Gaussian integers is unique up to multiplication by a unit (i.e. $\pm 1$ or $\pm i$).

I don't know about convention but I imagine saying $1$ is the GCD looks a lot nicer than saying $-1$ is.

agent154

OK, thanks

Romiox

13:36

Hi there, I am having a little trouble understanding power series. I know that e^x = sum(0 to inf) (x^n)/(n!) for (presumably, cause that's where I struggle) x_0 = 0. What would it be for x_0 = 1 ?

Chris's sis

@robjohn I was wondering the following question: for instance, is it fair someone that knows nothing about calculus downvotes someone that knows calculus? I mean there should be some kind of filters such that one that didn't ask or answer calculus questions cannot downvote someone that posts answer/questions related to calculus.

agent154

@Huy Do you know where I can read up on how to do prime factorization on Gaussian integers?

Huy

@agent154: I don't know if I can recommend anything to you, my Algebra course was in German and so was the textbook I used for it.

agent154

ah

One of the only things I have here is saying that if the norm of a gaussian integer is a rational prime, then that gaussian integer is a gaussian prime

Huy

@Romiox: The Taylor series of a smooth function $f$ around a point $x_0$ is defined as $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$. Just plug in $f$ and $x_0$ and you will find your power series for $e^x$ around your desired point.

@agent154: Try your textbook or just googling. For example, this came up as one of the first results when googling: math.ou.edu/~kmartin/nti/chap6.pdf

@Romiox: Did you figure it out?

Romiox

13:49

@Huy I believe so, sec

robjohn

@Chris'ssis looking at his reputation in his profile, it seems that the downvotes started around Nov 23. It was a fairly recent occurrence.

Romiox

@Huy sum(0 to inf) (e^x)/(n!) (x-1)

Huy

You forgot a few things.

It is $f^{(n)}(x_0)$, not $f^{(n)}(x)$ and also it is $(x-x_0)^n$, not just $(x-x_0)$.

robjohn

@Chris'ssis yes. I upvoted your comment. I doubt the OP will correct it, however.

Romiox

@Huy "it is $(x-x_0)^n$, not just $(x-x_0)$." I see, thanks. "$f^{(n)}(x_0)$" not sure what that translates to

Huy

13:54

Well, $f^{(n)}(x)$ for $f(x) = e^x$ is just $e^x$. Now you plug in $x_0$ for $x$.

Chris's sis

@robjohn I hope he changes his mind. I'd like to go through his answers.

Romiox

@Huy So... e? oô

robjohn

@Chris'ssis I have said this countless times before. Voting is capricious. It is impossible keep voting fair while encouraging knowledgeable people to vote, yet discouraging others.

Huy

@Romiox: Exactly.

Romiox

hmm, that's nice

Huy

13:58

That's because $1$ is a nice number. :P

Romiox

And e^x does play its part, too :D

Chris's sis

@robjohn Indeed.

Huy

Yes, indeed.

Charlie

@Chris'ssis GREAT CHRISSY! :)

Chris's sis

@Charlie GREAT CAT! :-)

Charlie

14:00

@Chris'ssis hehe how are you?

Chris's sis

I was preparing to leave for some shopping. How about you? :D

Charlie

@Chris'ssis I'm fine :D

Romiox

@Huy Thanks a lot, I'll go and put that knowledge to work :)

Huy

@Romiox: You're welcome, I'm glad I could help.

Chris's sis

@Charlie hehe, glad to hear that! :D

Charlie

14:03

@Chris'ssis buy me a gift ;)

Chris's sis

@Charlie Well, I'd prefer to create a question to you! (a kind of mathematical gift) :D

Mike

Morning @TedShifrin

Charlie

14:51

@Chris'ssis fair enough :)

Chris's sis

@Charlie This one is really great!

Let $(a_n)_{n\ge1}$ be an increasing sequence such that $\lim_{n\to\infty}(a_{2n}-a_{n})=1$. Compute
$$\lim_{n\to\infty} \frac{a_n}{\log(n)}$$

Adam

How can I see that $4*p_1*p_2..p_m-1$ contains a factor of the form $4*q+3$?

Chris's sis

@Charlie it's not created by me, but I can use it as a gift. :D

Huy

@Adam: 4-1=3.

Adam

@Huy: are you refering to the fact that we can write the former in the form 4(p_1*p_2*..p_m-1)+3 ?

Huy

15:01

Something like that. Did I misunderstand your question?

Adam

Well, I already know this fact. Maybe I am thick-headed or something, but does that imply a factor of the form $4q+3$?

Huy

Well, if you can write it as such then clearly one if its factors is $4q+3$?

Ted Shifrin

@Adam, you're in the middle of a proof by contradiction, aren't you?

Hi @Mike.

Adam

Well, yes :-)

Ted Shifrin

So, say explicitly what you're assuming and use the Fundamental Thm of Arithmetic.

Adam

15:06

I assumed that there are finitely many primes of the form $4n+3$

Ted Shifrin

And used them all in your product above. Go on.

Mike

@Chris'ssis Cute question.

Adam

Now I want to show that this product has to contain a factor that is of the form $4q+3$, distinct from all $p_i$ and prime.

Ted Shifrin

So what does FTA tell you?

Well, your $N$ is the product minus $1$. :)

Adam

For one thing, I know that any factor of the N is distinct from all $p_i$.

Mike

15:16

Cute question with a cute answer

Ted Shifrin

Ok, @Adam, so that leaves what for prime factors?

Adam

They have to be of the form $4q+1$?

Ted Shifrin

So when you multiply such numbers, what is the mod 4 remainder?

Adam

1, I guess.

Ted Shifrin

Yes ... Prove that. So there's your contradiction.

Adam

15:22

So the whole number has to have a mod 4 remainder 1, and a mod 4 remainder 3 at the same time - contradiction, right?

Ted Shifrin

Yup.

Mike

How do you write ~ in LaTeX?

Chris's sis

@Mike thanks :-)

Charlie

15:39

@Chris'ssis thank you, chris :D

badass

Hi

Chris's sis

@badass Hello

@Charlie Welcome :D

Charlie

@Chris'ssis :D

badass

:-)

Charlie

@Chris'ssis I think I'll change my avatar

@badass hi mr.Patrol

Chris's sis

15:41

@Charlie Won't you be a cat anymore? :D

Adam

@Mike: \sim is one possibility.

Charlie

@Chris'ssis for a day, maybe

Chris's sis

@Charlie ok :D

badass

Hi @charlie how are you?

Mike

Thanks, @Adam!

Charlie

15:46

@badass fine, and you?

badass

Fine thanks @charlie this Mobile experience is so miniature

Charlie

@badass yup, sometimes quite annoying

badass

Indeed

And sometimes quite interesting @charlie

Charlie

@badass yes, internet everywhere mwahahaha

badass

16:05

Mwahahaha

Charlie

16:16

@badass will you wear christmas hat this year?

Ah, another christmas with mse guys, how cute :)

badass

@charlie the hats at stack exchange have to be earned starting Dec 15

Charlie

@badass eh?

badass

Something about Christmas Bash @charlie

Charlie

@badass hmmm

badass

16:32

@Charlie I am not sure if math SE is going to take part in it

Jasper Loy

@badass Not all sites are involved?

badass

Correct @JasperLoy

Jasper Loy

@badass Mind telling me how old you are?

badass

Last year MSE did not take part

Jasper Loy

ELU definitely took part.

Charlie

16:41

@badass :/

badass

@charlie :-)

Don Larynx

back from algebra final

brb studies 2 days straight ideals and polynomial congruences and i get questions about irreducible polynomials for 5 questions out of 8. brb killing self, brb hope professor gets fired.

omg one of the questions was find a prime factor of $2^{35} - 1$

wtf?

didnt do it

Jasper Loy

@DonLarynx What is brb?

badass

Be right back

Don Larynx

@jasper.........be right back............

Jasper Loy

16:53

But it does not make sense in your sentence

Don Larynx

@Jasper: stop taking it literally and it will make sense...

anon

hey jasps

Jasper Loy

@anon Hey anon, I am feeling very bad. I am afraid I may never recover...

anon

@DonLarynx by geo sum formula it's divisible by 2^5-1, which is easily computable by hand as 31

@JasperLoy aww

Don Larynx

oh geometric sum formula, something we never learned in class. oh how cool

i better get at the very least a B

in that class

Pedro Tamaroff

16:56

@JasperLoy Dude.

Jasper Loy

The worst I got for math classes was a B+, so not too bad.

@PedroTamaroff Hi Pedro.

Don Larynx

@Jasper: Did you ever have a bad professor?

Jasper Loy

@DonLarynx Yes, about half of them were bad.

Don Larynx

@Jasper then I am assuming you read 99% of the time

Jasper Loy

@DonLarynx Well, to be honest, I think I did OK despite my mental problems because the course content was not too hard.

Don Larynx

16:58

@Jasper: Have you been thrown in a course where you don't know what to study?

That is precisely this course.

With this teacher.

And this book.

Jasper Loy

@DonLarynx Ah, yes, and I think I got B+

Don Larynx

@Jasper: Did you ever have a sucky book + professor

??

@Jasper: If you have...I praise you.

@Jasper: I have a book published by pearson

thats so lame

Jasper Loy

@DonLarynx Yes, definitely. I just went into the exam hall and bluffed my way.

Don Larynx

i want to kill all that is evil in the world

Jasper Loy

@DonLarynx Tell me what book it is.

Don Larynx

16:59

like this book

i want it gone

Transcript for

$Mathematics$ Mathematics