Mathematics - 2017-09-26 (page 1 of 4)

@Jasper morning

morning nitsua60

10 Replies

12:25 AM

wtf... (13-5x)^2 = (5x-13)^2 0_0

Damn negative numbers...

Yes. (a)^2= (-a)^2.

10 Replies

Never had seen it applied with equations as "a" before.

Its not as visible because the "-" is distributed throughout the equation

CausingUnderflowsEverywhere

12:55 AM

Morning Ted

1:39 AM

What should I do if I have to find the x coordinate of a quadratic function that is greater than a given x coordinate such that the average rate of change from the given x coordinate to the one I need to solve for is 1 ? I'm stumped.

Jeff

Suppose $f$ is continuous on $[a,b]$, then there is a point $c$ in $(a,b)$ such that $f(c)=[f(a)+f(b)]/2$. Textbook says this statement is false. But isn't $[f(a)+f(b)]/2$ halfway between $f(a)$ and $f(b)$. And if $f$ is continuous, then it must take on all values between $f(a)$ and $f(b)$. It's true, no?

@Jeff The example which comes to mind is this. Let $f(x)=x(1-x)$ and $[a,b]=[0,1]$. Then $f$ is continuous, but there's no point in $(0,1)$ such that $f(c)=[f(a)+f(b)]/2=0.$

Jeff

@Semiclassical OK. Good example. I guess the IVT says point $c$ is in $[a,b]$ (not $(a,b)$).

It was a good homework question ;)

@Semiclassical TY

3:06 AM

@Semiclassic @Jeff The point is that the intermediate value theorem specifically requires $y$ strictly between $f(a)$ and $f(b)$. When $f(a)=f(b)$, the hypotheses of the theorem cannot hold.

@Faust: Morning.

3:18 AM

its almost aftermorning now Ted

You made me come back for that? Bah.

3:33 AM

Hey chat.

hi gian

@TedShifrin, I was actually meaning to ask you something.

Remember the other day when we were computing the homology of the torus? But instead of row reducing a bunch of matrices, we used the first isomorphism theorem. You said that was a major tool in homological algebra...

So what's the major tool? :D

We knew that $H_2 = \Bbb Z = H_0$.

Right.

We also knew the ranks of the chain groups. The Hopf lemma says that $\sum (-1)^i \rank H_i = \sum (-1)^i \rank C_i$.

3:39 AM

I searched it and it took me to a lemma about harmonic functions.

there are lots of Hopf things.

oi ted whats an example of a group G and a subset A such that, for all g in G
and a in A, $ gag^{−1} \in A$
but A is not a subgroup of G

any chance u know of one off the top of ur head?

i wanted to pick a left coset of something

I've never thought about that phenomenon, Faust.

hmm ok

i have lol

learned the first isomorphism theorem today =)

@TedShifrin, is there another name for that lemma?

3:44 AM

Not that I know of, @gian. I think it's named that in Hatcher (and that's how I learned it in Spanier years ago).

It's a repeated application of nullity/rank to prove it, @gian. A neat argument.

Oh, I have an example @Faust

O.o

i spent like an hr trying to figure it out

What is one the main examples you know of groups?

cyclic

DN

d_n

D_n

Non abelian ... keep going

GL_n(R)

S_n

3:49 AM

There you go.

So think about $S_n$ ... or even $S_3 = D_3$.

k gimmie a minute

I found it Ted.

It is a beautiful result, @gian.

So the Euler Characteristic is just an alternating sum of the ranks?

Wow never would've seen that coming.

Yes.

And knowing $V-E+F$ is the alternating sum of the ranks of the homologies is a special case of this result.

3:53 AM

oh hey, betti numbers

heya @Semiclassic ... you saw my stoopid note earlier?

which?

oh, the jeff one?

Yup.

@TedShifrin H= the rotations and A= the reflections without the identity?

kinda annoyed at myself for pointing that out, tbh

getting thanked for providing a HW answer is not something I'm fond of.

3:54 AM

What's $H$, @Faust?

But, yeah, take the set of reflections. It's invariant under conjugation (for various reasons, algebraic or geometric) but it is not a subgroup.

Well, @Semiclassic, I don't mind helping/hinting with homework.

all the rotations like (123)

Yeah.

Rotations form a subgroup, @Faust.

(with the identity of course)

yeah it has e in it

intresting

Have you learned what conjugation does to elements of $S_n$?

3:57 AM

leaves them alone?

had I thought about the possibility of it being a HW question, I'd not have given the counterexample.

Nooo, @Faust.

What happens if you conjugate the reflection $(12)$ in the triangle group?

sorry could you state that diffrently?

In $S_3=D_3$, what happens when you conjugate $(12)$?

like write (21)(g)(12) or $a(12)a^{-1}$?

4:00 AM

Yes.

to which? Lol

i thought those are two different things

whenever I see questions about S_n, my reflex is usually to think in terms of string diagrams

Oh, conjugating $(12)$ is different from conjugating by $(12)$. [That's the first]

oh

so $a(12)a^{-1}$

sorry lanaguage i understand now

does it equal $a^2$?

Try it :)

You could do it physically with a triangle. Like do $(123)(12)(123)^{-1}$.

4:04 AM

(123)(12)(321)

string diagraaaams

Hush, @Semiclassic.

me and $S_n$ dont get along ok :P

Oh, well, then all the more reason to cut out an equilateral triangle and play with it.

I made my students do triangle, square, and usually pyramid.

goes to (132)

4:06 AM

Noooo.

A reflection can't turn into a rotation.

oh

im reading my thing wrong

i becomes a diffrent reflection

(32)

no (23)

well same diffrence

Same thing.

Right.

In general, any time you conjugate a $k$-cycle you get a different $k$-cycle. Very powerful.

So that's an easy example for your question.

some of theses exercies are too hard

as it must, since conjugation definitely preserves the group operation

ALannister

Heya

4:08 AM

Hi @ALannister

i really need to elarn $S_n$ better

Anyway Ty Ted

Gabriel Romon

6:03 AM

@Did can you please have a look at this math.stackexchange.com/questions/2445077/… ? I've seen this sigma-algebra defined in your lecture notes as well

Q: Must probability density be continuous?

6:46 AM

[Random]

10

From other materials that I've read, the probability density of a continuous random variable must itself be continuous. Is this correct? If it is, I don't understand why that would be so, why can't the probability change abruptly?

probability

I don't think it is possible to have a nowhere continuous probability density function as that will mean that function is not Lebesgue integrable

@Secret There are nowhere continuous functions that are Lebesgue integrable

(just take the indicator function for the rationals on a bounded interval)

I thought lesbegue integrability will need the number of discontinuities to be restricted to countably many?

or I might have confused that requirement with some other property of functions

@Secret see math.stackexchange.com/questions/1079172/… for that specific function (the answer mentions Lebesgue)

Carl Mummert's answer covered my query, it turns out that criteria I was stating is about whether the function is riemannian integrable not lesbegue integrable

8:09 AM

@TobiasKildetoft have you received my email?

@LeakyNun Yeah

Vrouvrou

9:44 AM

hello

Little Rookie

9:57 AM

Is significant digit the same as significant figure?

11:20 AM

@LittleRookie yes

11:37 AM

Hi @AkivaW

Hello

@BalarkaSen @AkivaWeinberger Hi

Hi Tobias!

What's up, @Akiva?

Not much to be honest

On the train ride to school

exciting :P

11:48 AM

@TobiasKildetoft how is my work?

@LeakyNun I haven't had time to look through it yet. Busy with grant applications still

ok

Hi, does someone know how to simplify $\sum_{j=1}^N \mathbf r_j^T (\sum_{i=1}^N \mathbf r_i \mathbf r_i^T)^{-1} \mathbf r_j $

@user86234 $i$ is unused?

Thank you, I edited the equation

I think it should be something beautiful, like $1$ or $N$, but I dont see how to get it :)

11:56 AM

what is $\textbf r$?

$\mathbf r_i \in \mathbb{R}^n$, $n > 1$

Well that thing in the parentheses is a scalar, right?

So it commutes with everything

@AkivaWeinberger it is a rank-$N$ matrix

Oh. You're right. I was confusing it with $\mathbf{ r^\top r}$

It depends on convention right

whether we are working with column or row vectors

11:59 AM

These are column vectors, I assume

Thank you @BalarkaSen, these are column vectors, @LeakyNun in standard basis

@user86234 are they the standard basis?

well then one only has to manually check some cases :P

Ah, excuse me, they are not basis vectors themselves.

It's not linear in them, is it?

That expression looks like a quadratic form of sorts...

12:00 PM

@user86234 oh ok

so they are random vectors?

finally finished the assingment except for the last question.

To be precise they are the residuals formed from some iterative process, and they are not generally linearly dependent.

Does the bracket term seemed to resemble some kind of trace of a matrix?

Hi, Nate

since only the diagonal entries were summed

12:02 PM

@dodsy thumbs

@user86234 did you mean independent?

@Secret $\mathbf r_j$ is a vector in $\mathbb R^n$

@Secret I made an error yesterday: you need AC to go from "exists surjective f:A->B" to "exists injective g:B->A"

So let $B$ be an infinite Dedekind finite set and $S \subsetneq B$

@LeakyNun yes: they are likely linearly independent

easily exists surjective $f:B \to S$ defined by $f(x) = \begin{cases}x & x \in S \\ s & x \notin S\end{cases}$ with any $s \in S$

[you don't need AC to pick out one element]

injective $g:S \to B$ defined by $f(x)=x$ I guess

wait, in this case both exist without AC

12:07 PM

hm, I was expecting this to be easy, but I might be wrong :) maybe it makes a good math.stackexchange.com question

@AkivaWeinberger I want to look back at some of the covering space theory I learnt at some point. Would you be interested to go through stuff with me?

I don't see how one can simplify this quadratic form expression if the matrix does not have some kind of special structure

@BalarkaSen Sure why not

Unrelated

You know what's easier than knot tying?

Not tying

Are there theorems that talk about matrices produced by sum of outer products of possibly arbitrary vectors?

lol

12:10 PM

@Secret Thank you, let me analyse this for a bit

@Secret did you take into account that we consider the inverse of $\sum_{i=1}^N \mathbf r_i \mathbf r_i^T$?

@AkivaW Let me know when you want to do this and we could just pick up Hatcher chapter uh 1.3? and start reading.

I mostly want to recall the subtleties with non-regular covers etc

There's also basepoint issues removing which we get Galois-like correspondences but upto conjugacy

I have forgotten shit

@user86234 I am not very sure whether the existence of an inverse of a given quadratic form will imply it is not indefinite

It might also help to check for its eigenvalues so that one can decompose it with some kind of diagonalisation procedure

and since you mentioned it is the residue of some iterative process, I will imagine the eigenvalues will give some information on how the data is distributed or something

@Secret I can put the assumption that $N > n$, then the inverse exist. I think this matrix must be semi-definite but I might be wrong

@Secret that is a good idea, the SVD might work here

@LeakyNun Yeah and note that without countable choice nor a countable proper subset, there is no way to define a map to shift elements in $S$ to make room for elements in $B -S$ thus preventing an injective and surjective, hence bijective map to exist

12:28 PM

so axiom of countable choice already destroys infinite dedekind finite sets?

Finite set

In mathematics, a finite set is a set that has a finite number of elements. Informally, a finite set is a set which one could in principle count and finish counting. For example, { 2 , 4 , 6 , 8 , 10 } {\displaystyle \{2,4,6,8,10\}} is a finite set with five elements. The number of elements of a finite set is a natural number (a non-negative integer) and is called the cardinality of the set. A set that is not finite is called infinite. For example, the set of all positi...

It appears so, the axiom of countable choice provide a choice function over a countable domain to destroy amorphous and infinite dedekind finite sets

However, be warned that I am not sure if my intuition on why it is so is correct, I might need to do some formal proof soon

well I have a proof sketch in my head

(To be added to Mathworks) The strength of axioms related to infinity, and how they change the available type of infinite sets:
1. Axiom of finite choice (is a theorem in ZF) : Allow picking a finite number of elements from any set
2. Axiom of infinity: Existence of the naturals and $\omega$
3. Axiom of countable choice: All dedekind finite sets are finite
4. Axiom of dependent choice: (No idea yet)
5. Axiom of choice: Existence of nonmeasurable sets

proofwiki.org/wiki/…

@Secret look at this proof

12:45 PM

Interesting, this statement seemed to be in 2nd order logic

$A \left({n}\right) = \left\{{h: P \to T}\,\middle\vert\,{h \left({0}\right) = a \land \forall m < n: h \left({s \left({m}\right)}\right) = g \left({h \left({m}\right)}\right)}\right\}$

since it has a map in it as an object

it does seem to be second order, but it can actually be done in first order

yes, by writing as a schema similar to how the axiom schema replacement is done, which tend to confuse me because there are too many letters

A: How is exponentiation defined in Peano arithmetic?

no that isn’t how it is done

29

The bare bones answer is something like what Hagen has said. The idea is this: Exponentiation is understood to be a function defined recursively: $y=2^x$ iff there is a sequence $t_0,t_1,\dots,t_x$ such that $t_0=1$, $t_x=y$, and For all $n<x$, $t_{n+1}=2\times t_n$. In this respect, exponen...

12:59 PM

@Secret

It surprised me that defining exponentiation in peano arithmetic need the chinese remainder theorem

and as a result it is not as comprehensible atm since I have a very poor number theoric intuition

it’s just a way to encode sequences

Morning

@Faust top of the morning to you

Having a good day?

1:06 PM

sure

Thats good =)

Im having a good semester i got 5 classes and i am not dieing at all =)

Whats $\frac{\mathbb{R}}{\mathbb{Z}}$ look like?

$\Bbb R/\Bbb Z$

it looks like a circle

we don't write it as a fraction

yeah what u did not what i did

$\Bbb R/\Bbb Z$ that's it

a circle?\

1:11 PM

yes, among other representations

a satisfactory answer would depend on what you mean by "look like"

what would i google to find an image?

The follow concept is crucial in understanding infinity:

cantorsattic.info/Infinity

Q: Show that $\mathbb{R}/\mathbb{Z}$ is isomorphic to $\{e^{i\theta} : 0 \le \theta \le 2\pi \}$

@Faust r/z quotient group

3

This question is asking to prove that the quotient group $\mathbb{R}/\mathbb{Z}$ is isomorphic to the group of complex numbers with modulus 1 (under multiplication). It's hard for me to visualize the structure of $\mathbb{R}/\mathbb{Z}$, so I can't think of an isomorphism. I feel like the first...

group-theory

The infinity that appeared in the usual induction is a potential infinity, while those of infinite ordinals and transfinite induction are actual infinities

@Secret I'm not aware of any infinity that appears in induction

1:14 PM

@LeakyNun it must be isomorphic to something on the complex numbers if its a circle?

If P holds in the base case and induction holds, then P holds for the whole inductive set

@Secret "infinity"?

@Faust $\{z \in \Bbb C ~:~ |z|=1\}$ is a circle for all intends and purposes

uh... I am thinking about induction using the natural numbers, where P holds in step n, thn it holds in step n+1 for all $n\in \Bbb{N}$

so lets assume it maps to a circle in C whats the operation in C ?

multiplication?

@Secret I don't connect that to infinity

1:16 PM

what is being shown in such induction is that P holds for all finite n in $\Bbb{N}$ but does not show whether P holds for some infinite case

@Faust indeed.

cause anthing |z|=1 *|z'|=1 =1

intresting

@SimplyBeautifulArt hi

Simply Beautiful Art

:o hi

Or consider recursively constructing the naturals from the empty set, no matter which step you stop at, you have constructed some natural number that is finite ,however large, but you cannot construct the set of naturals itself without the axiom of infinity

1:17 PM

@Secret I do connect this to infinity

Another example is that you can apply the successor function to some finite ordinal endlessly and never reach $\omega$

sure

so induction and recursive definition are like using potential infinities

Another example of it is in calculus, when we wrote $\lim_{x\to \infty}$, $(a,\infty)$ etc.

here the symbol $\infty$ means go up without bound

This, I think, is why $\infty$ is not a number

I see

Endless Stairs 10 Hours - Super Mario 64

because it is more like a process, than an object

More figuratively, any operation that involves induction and $\infty$ seemed to be something like this...

►

note that whenever you decide to stop, you never reached a step call infinity

1:21 PM

indeed

Simply Beautiful Art

@Secret lol

Therefore, this might explain why the infinty in hyperreals are so different from the concept $\infty$

because those are actual infinities, elements that are greater than any real number

Therefore hereby I conclude with the following poetic statement:

$\textit{Just as Art is both a process and the result, Infinity is both an object and a process}$

Therefore, the reason infinite sets are so weird is that we might be not supposed to reach them

but whether our universe allow existence of infinite configuration of energy momentum is an open question, and this metaphysical discussion will be continued later in favour of set theoric stuff

Btw, I now have a more detailed reread of that chinese remainder theorem MSE. It does seems the reason the coding works is because chinese remainder theorem kinda allow a unique number to represent n equations hence n numbers in a sequence

Now logically, the next step of this thought process is...

"Chinese remainder Theorem for countable congruence"

Now googling...

countable congruence... what??

it only works for finite numbers

Q: Chinese remainder theorem for infinite equations & equations of infinities

2

The Chinese remainder theorem is a very well-known theorem of arithmetic and algebra. Inspired by the following paper in model theory, Li Rong Yu, Li Bo Luo - The Generalization of the Chinese Remainder Theorem, Acta Mathematica Sinica, July, 2002, Vol.18, No.3, pp. 531–538 which proves a...

number-theory reference-request ordinals chinese-remainder-theorem

O.o

TACO

1:31 PM

So... this is the kind of ASMR that's hip nowadays?

Good stuff

How is that 40 minutes?

Okay so the quesiton I couldn't answer was:

$f(a,b)=a^2+b^2$

Give examples to show that $f(a,b)$ is neither one-to-one or onto.

lol

ok

can u think of something that isnt a squared number?

in the whole real line

Q: Generalization of Chinese Remainder Theorem to infinite ideals

2

I'm looking for any (obviously weaker) generalization of this famous theorem in the special case that the family of ideals is not finite.

abstract-algebra reference-request ring-theory commutative-algebra

@Faust No earthly idea.

1:34 PM

@Dodsy which part?

lol

also let $D=\{(a,b) \in \mathbb{N}\times\mathbb{N}: a \leq b)$ and define $f:D \rightarrow \mathbb{N}$

thats an important part dodsy

is it?

very

1:35 PM

Hmm, so that's basically a parabloid with the domain a triangular natural number lattice

I couldn't figure out if this was the equation for a circle $y=x^2 + y^2$

Lmfao @Secret

what?

a = 1, b = 2.

$a^2+b^2=6 $ ?

@LeakyNun I did none of the problem.

@Dodsy ok

Well I already handed it in without the answer completed @Faust how about you enlighten me.

1:39 PM

you cant

so its not onto

no for 1-1

$3^2 +4^2 =5^2$ yes?

can you think of any other example that also equal $ 5^2$?

So really he just tried to make it more complicated than it was?

I was caught up thinking about how it's a 3 dimensional object.

assuming your N has 0

in the end its just a function

well

not sure its a function by definition

nvm it is

it's a function

but its not one-one because you can find two things that map to the same place

and its not onto because some things dont get mapped to

@Dodsy you just need to apply the definition of "one-to-one" and "onto"

1:43 PM

@Secret if you happen to be interested: here is someone with an interesting idea math.stackexchange.com/questions/2445901/…

you should of asked for a hint man

im sure you could of got it with alittle help ^^

i always get the definition of function half confused with one to one

@LeakyNun I just thought it was more complicated than that.

@Dodsy don't complicate things :)

I'll try not to. :)

and seriously stareing at a piece of paper until you have some godly inspiration is overated ask for a hint if your stuck for more than an hr on a problem

unless your in a dif geo class then you can give it 6

1:47 PM

@Faust he is not allowed to ask us

im not suggesting he asks for the answer

it's the same

go talk to your prof in his office hours if your really worried about it