Mathematics - 2015-12-04 (page 1 of 5)

Kevin Driscoll

00:14

@evinda I don't think so. Just taking derivatives gives you a mess that doesn't obviously cancel

More importantly, this is a first-order linear equation so it has at most 1 linearly independent solution. So adding arbitrary functions shouldnt be possible here

Well, after you seperate variables, I mean

@evinda I think you copied something wrong or something like that. I don't see how you could possibly get a Gaussian out of that because the resulting ODEs are just linear first-order. You can get periodic boundary condition, I think, by letting your exponentials be complex

That gets you some allowed imaginary eigenvalues

@TedShifrin I think $U(x,t) = e^{-2 \pi i x + 4 \pi i t}$ works

Rex

00:58

hello

I am a student and i have a question that involves induction, if anyone is willing to help me.

I will just post it up, because I don't see anyone active right now

Prove that for n in the set of natural numbers, n is greater thean or equal to 2: For all a belonging to the set of natural numbers, For all b belonging to the set of natural numbers, a is modular congruent(subscript n) to b -->a^2 (subcript n)b^2.

Sorry for writing in words, i am new to the site and am still leanring how to use symbols.

also what about this Prove by induction that for n in the set of natural numbers, n is greater thean or equal to 2: For all a belonging to the set of natural numbers, For all b belonging to the set of natural numbers, a is modular congruent(subscript n) t…

SAWblade

01:29

Well, if $a \equiv b \mod n$, then $a-b = nc$, where $c \in \mathbb{Z}$. Note then that $a^{2} - b^{2} = (a+b)(a-b) = (a+b)cn$, and as $(a+b)c \in \mathbb{Z}$, $a^{2} \equiv b^{2} \mod n$.

Ted Shifrin

@Kevin: How does that have the right initial condition?

L33ter

@anon why is $Q(\pi)/Q$ is purely transcendetal

I guess the elements of this thing looks like as follows

anon

why is pi transcendental?

Ted Shifrin

hi @anon :)

SAWblade

It's important?

anon

01:38

hello

L33ter

well, $\pi$ is transcendental because there doesn't exist a polynomial such that it evalutes to zero

i.e

Ted Shifrin

@Rex: Or, without factoring needed, write $a=b+nc$ and square both sides.

L33ter

the evaluation map has kernel zero

so put it more clearly we have the following $ev_\pi : R[x] \rightarrow Q(\pi)$ under evaluation by $\pi$ has kernel zero.

correct?

Ted Shifrin

$R=\Bbb Q$

So, $\Bbb Q(\pi)\cong \Bbb Q(x)$.

L33ter

I see

1 sec @TedShifrin just making some definitions clear in my head

SAWblade

01:46

@TedShifrin

Oops.

Do you know anyone I could get in touch with on the subject of self-avoiding walks?

Rex

thank you ted and sawblade

Ted Shifrin

I think @studentmath may know something about those, @SAWblade, but he's in the Israeli army and isn't around much.

Sure thing, @Rex. Keep learning!

SAWblade

Shame. Thank you, though. :)

Ted Shifrin

Other people around here may know things, but I don't.

L33ter

@TedShifrin so we when we say something is transcendatal over some field F we mean the following.. we Consider E to be an extension field of F that contain an element $\alpha$ and we consider the evaluation map $eva_{\alpha} : F[x] \rightarrow F(\alpha)$ where the mapping is defined as follows $f(x) \mapsto f(\alpha)$. Now $\alpha$ is said to be transcendental over F if kernel of that map is trivial right ?

Ted Shifrin

01:50

Sure thing.

But you also need the definition of a transcendental field extension.

L33ter

yeah

Ted Shifrin

Hi @Ali. Why in the world are you awake?

Hmm, I guess he isn't. Maybe it's like it happened with Huy the other night ... his computer just logs him in.

GBeau

02:10

putnam exam is saturday :o

I made a 2 last time

Ted Shifrin

That beats a 0. :)

0 is still the mean/median score, I believe.

Mike Miller

Hi @Ted. No, I wasn't here.

My phone's chat just doesn't work unless I take it out of mobile mode, which is unusable. Oh well.

PVAL

@TedShifrin Don't think its the mean. :)

Mike Miller

The median has not been zero for a very long time.

Guess I'm wrong, it was 0 in 2012, 2006, 2004.

L33ter

@TedShifrin you want to laugh ?

our prof in the middle of lecture today in topology farted so loud

it was funny

GBeau

02:29

Yeah it's 0 sporadically these days @MikeMiller

I was happy about a 2 though; I was taking my first-ever proof class that semester, so I expected to make a 0

Kevin Driscoll

@TedShifrin AAAAAHHH Balls. I only saw your abbreviated boundary condition, not the original one. Didn't realize the original distribution was supposed to be Gaussian

Ted Shifrin

LOL @PVAL. Good point. I guess a mean of 0 would be difficult to achieve. Good thing I never taught statistics :P

Yeah, @Kevin, I'm somewhat puzzled by that question. But I think @evinda has an impossible problem there.

Kevin Driscoll

@TedShifrin I agree. Doesn't that make the problem have 3 boundary conditions, whne only 2 should be required?

or maybe the 'periodic' guy only counts as 1 condition, I'm not sure there

Ted Shifrin

@MikeM: You'd think I'd understand what you're talking about, but I don't. I chat (whether on the phone or on the desktop) through Chrome.

@Kevin: I haven't thought about it too carefully, but for a 1st order PDE, it seems that the initial condition should determine it uniquely.

Karim: You're still in middle school? :D

cap

In probability what is meant by "X is a zero mean unit variance random variable"? Is there a formula to describe such an X?

Kevin Driscoll

02:40

@TedShifrin Yes that's my intuition as well. Either you can have $u(t,0) = u(t,1)$ or you can have the initial distribution, but not both

Ted Shifrin

No, @cap. It means the mean is 0 and the variance is 1. There are zillions of such random variables. Of course, there's a unique normal random variable.

Kevin Driscoll

@cap I think it means the probability distribution function is a gaussian with 0 mean and standard deviation of 1

Oh they didn't say normal

So Ted's right

cap

sorry i meant to put normal. thank you

Ted Shifrin

Nowhere did it say normal/Gaussian, @Kevin! :P But you did scare me. Especially after my earlier screw-up :)

Oh, with normal, it's unique, and yes, there's a standard formula you can find any number of places.

Kevin Driscoll

@TedShifrin Haha! It seems my only recourse to besting you is to force you to answer with incomplete information which causes you to be retroactively overbroad

Ted Shifrin

02:42

something like $\dfrac 1{\sqrt{2\pi}} e^{-x^2/2}$ for the density.

No, @Kevin, I'm getting sloppier and stupider with my retirement.

Kevin Driscoll

I think its $\sqrt{\pi}$?

Ted Shifrin

Actually I think $\sqrt{2\pi}$.

I won't bother to look it up. @cap can find it or do the computation.

OK, I'm going back to cooking dinner.

Kevin Driscoll

Oh or maybe I can't put things into Mathematica properly now. Sounds like I need to retire.

Raphael

03:18

does anyone know the answer to this? What is the density of primes p such that p-1 has some fixed constant # of factors?

I have searched but can't find a reference if it is known

SAWblade

Man, primes are so strange.

Raphael

@SAWblade :)

SAWblade

They're strange in a good way, I guess. xD

I'm mostly just an idiot

03:51

@SAWblade I was about to recommend @PerplexedGuest help you, but you are the same person.

SAWblade

Indeed! I figured I would change my name to something I actually liked instead of a rushed moniker. xD

But I'm happy that I sprang into your mind as someone to talk to SAWs about. xD

Even though I know ... so little ...

I'm mostly just an idiot

Wow, it never even occurred to be that SAW was an acronym for that :D.

I thought the capitalization was a stylistic choice.

SAWblade

Hehe, I like math and wordplay, so SAWblade seemed like a good choice for a name. xD

I'm mostly just an idiot

Fair enough :).

SAWblade

My avatar image is a cube filled in by a self-avoiding walk made of only 90 degree turns. P:

Hamiltonian magic, I say.

Mike Miller

04:21

@DanielFischer: Glad to see people are hat-friendly.

I'm mostly just an idiot

04:31

I don't care for hats, but I don't care against them, but I must vote. ɪts ɔːl səʊ hɑːd

Rajesh Dachiraju

0

Q: Placing delta's at maxima, Is there any smart equation based expression?

Let $M$ be the set of maxima of the function $k:\mathbb{R}\to \mathbb{R}$. We define the function $$L(t) = \sum_{y\in M} k(y)\delta(t-y)$$ and there by the step function $$\Gamma(t) = \int_0^t L(\tau)d\tau$$. I'd like to know, if there is any single equation based (kind of closed form in terms o...

Mike Miller

04:51

@I'mmostlyjustanidiot Well, the person with the most hats last year was from math.SE. It would be wrong to deprive the site the championship again.

I'm mostly just an idiot

@MikeMiller That does it, I am voting yes.

Done.

Mike Miller

:D

I'm mostly just an idiot

Who was our valiant champion?

Oh Normal Human won it, of course!

Mike Miller

No, some of the hats required one to upvote questions.

I'm mostly just an idiot

Was it you :D?

Mike Miller

04:59

ayup

pretty worried about getting another gold badge though.

i can get one on Academia.SE but hard to get one on MSE.

ugh no I can't I already got the electorate badge there

but can probably get it somewhere else pretty easily.

I'm mostly just an idiot

Oh, one of the hats requires obtaining another gold badge :O?

Mike Miller

one did last year; I got it then for steward, completing 1000 review tasks, which was possible in the given time frame

SAWblade

Hats?

Mike Miller

hats

I'm mostly just an idiot

@MikeMiller Raise 500 helpful flags is gold.

You've got the deputy for 80 helpful flags.

Copy editor is even easier probably.

" Edit 500 posts "

Mike Miller

05:10

yeah, winter bash is about 20 days long so if I spread those out it won't even be obnoxious, given the volume of the site

just careful editing of new posts

Ted Shifrin

05:22

Some of us are too lazy to be hat-motivated.

Mostly, what was your previous name?

I'm mostly just an idiot

05:37

**Question 8:** Why does it say that we define a submodule generated by $v\in V$ by $\mathfrak{g}(v):= \text{span}_{\Bbb F} \{u\in \{v,x_1x_2\cdots x_nv\}: x_i\in \mathfrak{g}, i=1,2,\cdots,n,\quad n\in \Bbb N\}$

Whereas I would expect it to be defined: $\mathfrak{g}(v) = \{ w\in V: w= x\cdot y, x\in \mathfrak{g}, y\in \text{span}{v}\}$?

Tobias Kildetoft

@I'mmostlyjustanidiot What context is that?

I'm mostly just an idiot

Ahhh, I guess modules of Lie algebras. Well this definition has $\mathfrak{g}$ as a Lie algebra.

I don't understand why in their definition they have $u\in \{v,x_1\cdot x_2\cdots x_n \cdot v\}$ instead of being able to apply any order of $x_i \cdot v$

Tobias Kildetoft

@I'mmostlyjustanidiot the submodule generated by $v$ is the smallest submodule containing $v$. The second one would not satisfy this.

@I'mmostlyjustanidiot Well, there might be no $x$ with $xv = v$ first of all

and there is also no reason why $x(y(v)) = zv$ for some $z$.

Mike Miller

@TedShifrin I think he's Alex.

I'm mostly just an idiot

05:52

Good point @TobiasKildetoft, that makes sense.

@MikeMiller Someone else said this. Is it the way I type?

Mike Miller

Something like that. Alternately, I thought you were previously someone named eg Lie algebras / Functional analysis / whatever, but I also thought that was Alex.

I'm mostly just an idiot

@JasperLoy You are black now. That is foreboding.

user174558

@I'mmostlyjustanidiot Are you Alex?

I'm mostly just an idiot

@JasperLoy I am not Alex, and I haven't went by any other name. Is that so normal here?

@TobiasKildetoft They previously defined $\mathfrak{g}$ to be of dimension $n$, but it was a page ago. Is their definition above consistent with it being of dimension $n$?

(Or are we taking any length of $x_1x_2\cdots x_k$ in that span?)

Tobias Kildetoft

@I'mmostlyjustanidiot Any length

I'm mostly just an idiot

06:03

@TobiasKildetoft Ahhh that makes more sense to me.

user174558

@I'mmostlyjustanidiot I am now an orange.

I'm mostly just an idiot

@JasperLoy You are still black on my screen, but I look forward to that beautiful orange returning.

user174558

@I'mmostlyjustanidiot You only need to refresh your browser.

I'm mostly just an idiot

@JasperLoy Fixed, thank you.

user174558

@I'mmostlyjustanidiot The orange makes me happy.

I'm mostly just an idiot

06:11

@JasperLoy Me also. Or is it 'me too' or 'me as well'?

user174558

@I'mmostlyjustanidiot All are OK, but 'me too' is more common.

user174558

@I'mmostlyjustanidiot It should be 'haven't gone'. You should use the participle form of the verb 'go' there.

I'm mostly just an idiot

@JasperLoy Thanks.

L33ter

07:30

@anon do you know how can I make like a map diagram ?

I need to write the universal property of quotient space in latex for my presentation

is it possible to do that in latex?

user134177

hey. if X is a topological space, then the singular homology of Xx[0,1]^2 and Xx[0,1] with coefficients in Z are the same, right?

robjohn

@L33ter I know that there are packages to help do such things. Whether you can do them in MathJax is not so certain.

L33ter

oh

robjohn

@L33ter But using LaTeX, you should be able to do it.

L33ter

alright thank you @robjohn I will try it out

Blue Bug

07:48

Hi new to #mathmatics :D

iwriteonbananas

@BalarkaSen Right, we're on the same page I think. I'll see if I can come up with something.

Overflowh

08:14

Good morning

Chantry Cargill

08:29

Just finished thinning apple trees for 8 hours in 30+ degree weather

Overflowh

09:18

What is the proof that for a given endomorphism, to each eigenvector is associated only one eigenvalue? Does it requires the use of the Fundamental Theorem of Linear Transformations?

Tobias Kildetoft

@Overflowh No, it requires just looking at the definitions

(though I am not actually sure what that fundamental theorem would be)

Overflowh

@TobiasKildetoft This one:
Let $V$ and $W$ linear subspaces.
Given an ordered basis $R=[e_1, e_2, ..., e_n]$ of $V$ and a system $S = [w_1, w_2, ..., w_n]$ of $n$ vectors.
$\exists !$ linear transformation $f: V \to W / f(e_i) = w_i$ with $i = 1..n$

Tobias Kildetoft

@Overflowh Yeah, that has nothing to do with this

Overflowh

Then I have no idea :/

Tobias Kildetoft

Just write up what it would mean to have two distinct eigenvalues

Overflowh

09:27

Mh...
Well, given $k, h \in K$ (with $K$ being a scalar field and $k \notequal h$), it would mean that...
$$f(v) = kv$$
and
$$f(v) = hv$$

...it doesn't make sense to me honestly

Tobias Kildetoft

@Overflowh Which part?

I'm mostly just an idiot

$Av=\lambda_1 v$, $Av=\lambda_2 v$

Overflowh

Saying that $f(v)$ is equal to two different things. I mean, who tells us that it can't be equal to two different things?

I'm mostly just an idiot

$\lambda_1 v= \lambda_2 v\implies (\lambda_1 - \lambda_2)v = 0$

Overflowh

Intuitively I know that it's not possible, but practically? @TobiasKildetoft

Tobias Kildetoft

09:29

@Overflowh By the definition of function

I'm mostly just an idiot

Let $v\ne 0$ in an integral domain

Tobias Kildetoft

@I'mmostlyjustanidiot this $v$ does not live in any integral domain, but in a vector space

I'm mostly just an idiot

We get the integral domain property?

We get if for the scalars?

Which lie in a field?

$K\times V\to V$

Tobias Kildetoft

@I'mmostlyjustanidiot integral domain is for rings. There is no multiplication of the elements here, only by scalars

the word you are looking for is faithful action, but that will not help the asker here

Overflowh

@TobiasKildetoft I know several definitions of "functions" (actually, linear applications) to be honest.

Tobias Kildetoft

09:31

@Overflowh There is only one (up to equivalence) definition of function.

Overflowh

@TobiasKildetoft Pardon, what is it? "An operation that preserves the operations of vector addition and scalar multiplication"?

Tobias Kildetoft

@Overflowh No, that is what it means for a function to be linear. You need to know what a function is first.

Overflowh

@TobiasKildetoft Wait, I have to go google for this since I can't manage to find it in my notes...

Tobias Kildetoft

@Overflowh There is a good chance it will not have been defined in the course on linear algebra, as it is way more basic than that

Overflowh

@TobiasKildetoft The one I know from my first Analysis course is "An operation that, to distinct elements of the domain, associate different elements of the codomain." But this one is for injective functions. I remember there are also surjective functions that basically do the opposite and bijections that do both things.

Tobias Kildetoft

09:43

@Overflowh You really need to look this stuff up properly then. If you don't know the basic properties of functions, then there is no way you can understand linear functions.

Overflowh

@TobiasKildetoft Any resource?

(apart from wikipedia)

Tobias Kildetoft

@Overflowh Wikipedia is a fine place to start

I'm mostly just an idiot

I thought the point was that $(\lambda_1-\lambda_2)v=0$ for nonzero $v$ means that $\lambda_1=\lambda_2$?

Tobias Kildetoft

@I'mmostlyjustanidiot It is

Chantry Cargill

@Overflowh If you're taking an analysis course, you would have an analysis textbook, correct?

Tobias Kildetoft

09:49

@I'mmostlyjustanidiot But without knowing basic properties of functions, we have not yet gotten to the part where $\lambda_1 v = \lambda_2 v$.

Overflowh

@ChantryCargill I *took* an Analysis course. Around two years ago. And yes, I *had* a textbook, but it's not where I live now, unfortunately.
I'm checking wikipedia.

Chantry Cargill

@Overflowh Ah, well in that case, google/wikipedia is your friend.

or should that be are your friends? good thing Jasper isn't here :s.

Overflowh

@ChantryCargill I always interpreted the use of the / as you did like an "or". So I think is your friend looks fine.

I'm mostly just an idiot

I thought it was not an exclusive-or though, as in, google and/or wikipedia, in that case it should be: google/wikipedia is/are your friend/s

@TobiasKildetoft I am somewhat confused about linear maps between $\mathfrak{g}$-modules. Are these $\mathfrak{g}$-linear, in the $R$-linear map sense?

Overflowh

@TobiasKildetoft I checked wikipedia and it honestly didn't tell me anything new. I know about composition, images, injective and surjective functions, identity and inverse functions. I know that a function transforms something from the domain into something else from the codomain and I know that for equals values of this something it will always produce the same value of something else (which also means that for different inputs you'll get different outputs).

Question is: how do you take this and put it in mathematical symbols to write a proof? Because as I told you, I get it intuitively.

Tobias Kildetoft

09:59

@I'mmostlyjustanidiot if they are just required to be linear, then that usually means $k$-linear where $k$ is whatever field (or ring) you work over

which is weaker than being homomorphisms of $\mathfrak{g}$-modules.

I'm mostly just an idiot

Because that requires $\phi([a,b])=[\phi(a),\phi(b)]$?

Tobias Kildetoft

@Overflowh what you wrote first does not mean that different inout results in different output

I'm mostly just an idiot

(like a lie algebra homomorphism, or something else)

Tobias Kildetoft

@I'mmostlyjustanidiot no, that would not make sense for modules

it requires that $\varphi(g.v) = g.\varphi(v)$

I'm mostly just an idiot

Oh, isn't that required of a module homomorphism?

Tobias Kildetoft

10:01

@Overflowh The only thing we needed here though was that since $f(v)$ equals two things, those things are equal

I'm mostly just an idiot

I thought $R$-module homomorphism $\cong$ $R$-linear map?

Tobias Kildetoft

@I'mmostlyjustanidiot For modules over a ring, yes. These are not modules over a ring (at least not obviously so)

I'm mostly just an idiot

So we have that acting $g$ on $v$ before or after mapping is equivalent, that isn't multiplication up there

Tobias Kildetoft

@I'mmostlyjustanidiot Right, I just like to write $g.v$ for the action because it is easier

Overflowh

@TobiasKildetoft Okay, I get the logic here since I know examples of functions that do this ($x^2 = 4$ both for $x = 2$ and $x = -2$). But how would you do this without examples? I mean, at the abstract level, are there "logic" rules to infer things from other things? I'm starting to think that I have problems with basic logic more than Math...

Tobias Kildetoft

10:07

@Overflowh I am really not sure what you are confused about. We know that $f(v) = \lambda_1 v$ and $f(v) = \lambda_2 v$ and we just need to conclude (for now) than this means that $\lambda_1 v = \lambda_2 v$.

Overflowh

@TobiasKildetoft And this because we know that, for functions, to equal inputs correspond equal outputs, right?

Tobias Kildetoft

@Overflowh Right

Overflowh

@TobiasKildetoft Well, my confusion starts here. What if at this point someone asks: why? Why is it true that to equal inputs correspond equal outputs? I mean, is saying "because this is the definition" an acceptable answer?

Tobias Kildetoft

@Overflowh Yes, it is pretty much the only acceptable answer

I'm mostly just an idiot

@Overflowh Please write 'two' :D.

Overflowh

10:19

@TobiasKildetoft Oh. I though it went deeper than this. Nice, this makes things a lot easier.

@I'mmostlyjustanidiot Umh? Where? o.o

I'm mostly just an idiot

@Overflowh t(w)o equal inputs correspond equal outputs, right?

@Overflowh that t(w)o equal inputs correspond equal outputs?

Overflowh

No that wasn't I typo, I really meant "to". Is it not correct? I think it's the use of "equal" that messes things up. I meant equal in the sense: "that to the same inputs correspond the same outputs". @I'mmostlyjustanidiot

Tobias Kildetoft

@I'mmostlyjustanidiot No, what he wrote is correct

Overflowh

For once :P

I'm mostly just an idiot

@TobiasKildetoft You think it is correct English to say "Why is it true that to equal inputs correspond equal outputs?"

Tobias Kildetoft

10:24

@I'mmostlyjustanidiot Yes, while replacing "to" with "two" would make it very incorrect

I'm mostly just an idiot

You can either get rid of 'to' completely, or change it to two

Two equal inputs correspond equal outputs is wrong?

Tobias Kildetoft

@I'mmostlyjustanidiot You need a "to" somewhere in there to bind to the "correspond"

very wrong. They correspond to equal outputs

I'm mostly just an idiot

Would it be fine to say "Two equal inputs corresponds to two equal outputs"?

Overflowh

I think "Two equal inputs correspond to two equal outputs" would sound correct too. But as @TobiasKildetoft said, you need a "to" somewhere.

I'm mostly just an idiot

Sounds good to me @Overflowh!

Corresponds, or correspond?(in our latter phrasing)

Tobias Kildetoft

10:30

@I'mmostlyjustanidiot correspond (as the subject is plural)

I'm mostly just an idiot

and the same holds usually with regard vs regards?

(with subject plural)

Overflowh

@I'mmostlyjustanidiot Input**s** --> correspond (plural subject)
Input --> correspond**s** (singular subject)

Damn markdown, how I hate it

I'm mostly just an idiot

I think any forced spaces break **stars**

Test

Yep

@Overflowh Thanks!

Overflowh

You're welcome @I'mmostlyjustanidiot. I would have never imagined to turn out useful to someone in this channel :P

I'm mostly just an idiot

@Overflowh Haha :D, everyone has something to teach you :).

Overflowh

10:36

@I'mmostlyjustanidiot I agree. But after having clamorously failed at simple logic my stocks went down pretty hard :P

I'm mostly just an idiot

in English Language & Usage, 1 min ago, by Matt E. Эллен

so, to rephrase the question "why does one have to correspond equal outputs to equal inputs?"

That's the English master take on it, I like it!

Tobias Kildetoft

@I'mmostlyjustanidiot That is not the same phrase. It has a completely different subject

I'm mostly just an idiot

@TobiasKildetoft How do you use correspond?

Tobias Kildetoft

@I'mmostlyjustanidiot What do you mean?

I'm mostly just an idiot

I just mean, I read that rephrasing to mean $x=y \implies f(x)=f(y)$, which was what he originally asked right?

Tobias Kildetoft

10:42

@I'mmostlyjustanidiot Right

I'm mostly just an idiot

@TobiasKildetoft What do you mean 'subject' here, maybe this is an English studies word?

Tobias Kildetoft

@I'mmostlyjustanidiot The subject of a sentence is the thing in the sentence which acts

I'm mostly just an idiot

So the subject in the first was equal inputs, and in the second phrasing was equal outputs?

Tobias Kildetoft

@I'mmostlyjustanidiot No, in the first it was indeed "equal inpits", but in the rephrasing, it was "one"

which actually also changes the meaning of "correspond" from a passive to an active verb

but this is getting rather technical on the grammar side, so this is not really the place for it.

I'm mostly just an idiot

True, thanks though.

Overflowh

10:49

@TobiasKildetoft Just for the sake of completeness (and to thank you since I didn't yet), this is how I wrote the final thing:

*Prove that to each eigenvector is associated an unique eigenvalue.*
Given $f: V \to V$ endomorphism.
Given $k, h \in K$ with $K$ being a scalar field, $k \neq 0$ and $h \neq 0$.
Given a vector $v \in V$.

By the definition of *function* we know that
$f(v) = kv$ and $f(v) = hv \Rightarrow kv = hv \Rightarrow k = v$

If (absurdly) $k \neq h \Rightarrow kv \neq hv \Rightarrow f(v) \neq f(v)$ which is clearly a contraddiction.

Tobias Kildetoft

@Overflowh There are a few things wrong here. First, you can't assume that neither $k$ nor $h$ is $0$.

and secondly, $kv = hv$ does not imply that $k=v$.

Overflowh

@TobiasKildetoft The second makes sense. For the first one I thought it was part of the definition of eigenvalue (imposing that an eigenvalue can't be zero), but I checked and it seems I was wrong.

Tobias Kildetoft

@Overflowh Yeah, that is a common mistake. It is the eigenvector that cannot be $0$.

Overflowh

@TobiasKildetoft About the second one: if we know that $v$ is the same for both $f(v)$, why we can't say that $kv = hv$ implies $k = h$? I mean, $k$ and $h$ would be the only two things that change, so the only two things that determinate the equality or inequality, right?

I'm mostly just an idiot

See $v=0$(you just have to state that $v\ne 0 \vee (h-k)v=0$)

$102(0)=124120(0)$

Tobias Kildetoft

11:01

@Overflowh We can, but that was not what you wrote (you wrote $k = v$).

I'm mostly just an idiot

OH haha

Tobias Kildetoft

But you do need an argument for this.

Overflowh

@I'mmostlyjustanidiot Well you can't have $v = 0$ because an eigenvector can't be zero by definition.

@TobiasKildetoft Oh damn. I'm very very sorry, that was a typo :/

I'm mostly just an idiot

You wrote given a vector $v\in V$, not an eigenvector though

Huy

does anyone know of actual (popular) scientific experiments conducted where I could apply the normal distribution? for example, every high schooler knows about Mendel's experiment on inheritence.

according to my lecture notes in probability, he conducted his experiment with $8023$ plants with $2001$ successes, where the predicted probability of success was $\frac{1}{4}$. approximating this binomial distribution with a normal distribution, we can quickly see that a deviation at least as bad as this was very likely, namely $90\%$. are there similar actual data available for such popular experiments?

Overflowh

11:02

@I'mmostlyjustanidiot Right, I should have made that clear too probably.

Tobias Kildetoft

@Huy You mean where one can easily see that the numbers were probably rigged?

I'm mostly just an idiot

I mean I guess it because an eigenvector by definition after $Av=\lambda v$

Huy

@TobiasKildetoft: why were the numbers rigged? too likely?

Tobias Kildetoft

@Huy I thought it was a common agreement that Mendel almost certainly rigged the numbers, as they fit the model way too well

Huy

I don't know about that kind of stuff

and no, that's not what I meant, but it's interesting to know

just similar experiments where on high school level we can compute "how likely was it to get such a good/bad result, theoretically"

Balarka Sen

11:05

@TobiasKildetoft really? I thought it fit the model because the model was correct? :P

Tobias Kildetoft

@BalarkaSen The model was indeed correct, but the numbers were too good to fit the variance build into it (as far as I recall the story)

Huy

yeah but he probably wanted to prove his point so much that he changed the numbers slightly to be even more correct

Tobias Kildetoft

It was a long time ago that I did any statistics

Huy

@TobiasKildetoft: do you maybe have anything in mind similar to that experiment?

Tobias Kildetoft

@Huy Sorry, no

Huy

11:07

what about you @BalarkaSen, you seem to be enjoying non-math things too these days :P

Balarka Sen

@Huy I always enjoyed them. I simply realized that acknowledging that I do doesn't make me bad at math (otherwise stated, I realized I need not be a nerd to be interested in math).

Huy

good for you

you're growing up

Balarka Sen

besides, Mendel's theorem/law is very mathematical.

one of the few mathematical things in biology I have seen.

Huy

I taught normal distribution today

I was surprised they understood a lot more of it than I expected

so now I'm looking for cool applications which are familiar to them

like the Mendel thing

Balarka Sen

what's normal distribution?

Huy

11:16

just a probability distribution which can very often be applied

for example for binomially distributed random variables for "large" $n$

Tobias Kildetoft

@Huy And that is of course something one can show mathematically to be true

Huy

well yea

Balarka Sen

what's a probability distribution

Huy

but I think de moivre is a bit overkill for high school

Tobias Kildetoft

but that is usually beyond the first course in which students see the distribution

Huy

11:17

also I need to teach them complex numbers and differential equations too before they graduate

so I'll just do the most important concepts and show applications

Balarka Sen

teach them the ricci flow

Huy

ok

Balarka Sen

and hence, ultimately, perelman's proof of poincare conjecture

Huy

sure

Newb

How do you find the intersection of two cylinders in four dimensions?

Suppose they are given by $C_1 = \{(w,x,y,z) \in \R^4 | y^2 + z^2 \leq 1\}$

$C_2 = \{(w,x,y,z) \in \R^4 | w^2 + x^2 \leq 1\}$

I'm totally stumped

I've read about them in three dimensions, mathworld.wolfram.com/SteinmetzSolid.html

Newb

12:00

I'm thinking that the volume of the duocylinder would be equivalent?

iwriteonbananas

12:19

Harro.

Balarka Sen

hi @iwriteonbananas

iwriteonbananas

Learned about Baire's category theorem today.

Newb

@iwriteonbananas any idea about the geometry problem I posed above?

Balarka Sen

really? but that's usually taught in point-set courses.

Newb

I'm thinking that the solution is the volume of the duocylinder, which for radii 1 is 4pi^2.

iwriteonbananas

12:22

@BalarkaSen We did it in my functional analysis course, not alg top :P

Balarka Sen

oh, aha

@iwriteonbananas Here's a fun problem solution of which uses BCT : no space filling curve $\Bbb R \to \Bbb R^2$ is injective.

iwriteonbananas

@Newb Well, the intersection is $\{(w,x,y,z):w^2+x^2\leq 1 \text{ and } y^2+z^2\leq 1\}$

Are you trying to visualize it?

Newb

Nope, just trying to obtain the volume

iwriteonbananas

@BalarkaSen Hehe, fun.

Balarka Sen

Try to prove it when you have time :)

Did you see my message? I proved your $a^2 = 0$ thing when $m/2$ is even.

We just had to stare at the diagram, which we were too lazy to do.

iwriteonbananas

12:28

@Newb Oh, ok. I suck at these things and I gotta leave in 5 minutes, sorry.

@BalarkaSen Yeah, I saw the message, and I responded

Balarka Sen

Hmm, weird, I didn't get the message.

I have no idea how to prove that cup square map is mult. by $m/2$ when it's odd, though. Something cohomology operations ...

iwriteonbananas

@BalarkaSen Yeah, that seems like a rather nasty thing to prove. Didn't Hatcher give the hint to mimic the proof of thm 3.12?

Balarka Sen

We can't mimick that proof, it's strictly restricted to $\Bbb Z/2$ coefficients.

At least, I'll shave my head if it's not. It's Poincare duality one way or another, which is valid for nonorientable things only mod 2

iwriteonbananas

@BalarkaSen Yeah, he said to mimic it with $\Bbb Z_m$ instead of $\Bbb Z_2$ coeff's

Balarka Sen

Yes, and I doubt very much whether it can be done.

iwriteonbananas

12:32

mhm :(

Balarka Sen

I mean, that construction with transverse copies of projective spaces inside $\Bbb P^n$ inherently points towards some sort of Poincare duality hiding behind the window.

iwriteonbananas

Right, transversal intersection of submanifolds yada yada

Balarka Sen

I'll see if it can be done though. Haven't thought about it.

I'll be surprised if it can be done.

@iwriteonbananas how dare you yada such a beautiful construction

iwriteonbananas

@BalarkaSen hahah. by the way, where does hatcher discuss the correspondence between cup product and transversal intersection of submfds?

Balarka Sen

he doesn't.

iwriteonbananas

12:35

T__T

Balarka Sen

I have a proof, assuming a hard theorem, which Mike and prof told me.

But you never listened to it...

iwriteonbananas

I want to know the precise formulation. And I want to see the proof.

Balarka Sen

I already gave you the precise formulation.

iwriteonbananas

Yeah I suck...I forget almost everything.

Balarka Sen

Subamnifolds of cco manifolds represent cycles, which are Poincare dual to cocycles. Cup product of the corresponding cocycles gives me a cocycle. Poincare dualizing gives me a cycle which is in fact represented by a submanifold which is transverse intersection of the two submanifolds.

iwriteonbananas

12:37

I gotta go now. got a lecture in a few minutes.

Balarka Sen

That's it.

iwriteonbananas

sorry to leave hastily

Balarka Sen

@iwriteonbananas Have fun.

Let me know when you want to hear the proof.

iwriteonbananas

laters.

yeah, hopefully i'll be back in chat tonight.

Balarka Sen

cool.

Actually, to be honest, my proof is only for dual dimensional submanifolds, I'd guess you can generalize that with great care.

Anubhav.K

12:58

@BalarkaSen sorry to join here lately, it seems an interesting discussion is going on over here. Could you please just explain the main problem of your discussion

Balarka Sen

@Anubhav.K You're certainly welcome to join this chat, there's no need to apologize for that :) We were discussing Poincare duality between cup product and transverse intersection in cohomology of smooth manifolds.

The statement of the theorem is described in the starred message above.

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$Mathematics$ Mathematics