Mathematics - 2019-04-01 (page 1 of 2)

12:04 AM

Oh man, this is confusing me so much...
From what I thought I knew, a radian is an angle-- a rotation!

But...

Steradians are "square angles," and I just don't understand that.

How can I orientate myself on a 3D sphere with only one number?

Is it a bit like how metres squared doesn't tell you where you are on the cartesian plane, just the area of the... square you made... ... uh
(I'm totally tired btw)

Basically my head is hopelessly lost and I'm asking how the concept of rotation/orientation can be translated from radians to steradians.

Tanner Swett

Well, steradians no longer give you enough information to orient yourself.

On the plane, an angle, all by itself, can represent two different things: a fraction of a circle, and a rotation.

A 90 degree angle, for example, can represent either a quarter-turn, or one fourth of a circle.

For a sphere, though, a "square angle" only represents a fraction of a sphere; it doesn't represent a rotation.

So, "is it a bit like how metres squared doesn't tell you where you are on the cartesian plane, just the area" - yes, it is.

Leaky Nun

12:29 AM

@TannerSwett do the inverse laplace transform to find $f$ :P

Tanner Swett

That's one way to do it. :D

Isa

How to integrate \int_0^1 x\sqrt{1-x^2} dx manually? any hint?

BananaCats

Need help understanding $e^{i2\pi x}$

Why isn't this always just $1^x$?

I mean $1$

Tanner Swett

12:44 AM

Well, what if x = 1/2?

Rithaniel

What is $e^{i2\pi(7/2)}$?

Tanner Swett

Then $e^{i 2 \pi x} = e^{i 2 \pi / 2} = e^{i \pi} = -1$.

BananaCats

Well if that's true then $e^{2\pi i} = e^{0i}$

So it always gets multiplied by $0$

when you add in $1/2$ to it

This is confusing the heck out of me

Tanner Swett

Well, it's true that $e^{2 \pi i} = e^{0 i}$.

Anyway, I think you're assuming that $e^{i 2 \pi x}$ must equal $(e^{i 2 \pi})^x$.

BananaCats

Well then $e^{2\pi i/7} = e^{0/7i} = 1$

Rithaniel

12:46 AM

Yeah, exponents don't behave well in the complex numbers

BananaCats

Oh it's not true that $e^{ab} = (e^a)^b$?

Rithaniel

$(z^x)^y\neq (z^y)^x$, in general.

BananaCats

Why did you commute them?

My goal is to solve this simply:

4

Q: A sequence of complex maps $a_n : \Bbb{N} \to \Bbb{C}, n \geq 1$ such that $a_i \cdot a_j = a_{\gcd(i,j)}$

We know that $\sum_{ab = n} f(a) g(b)$ is multiplicative in $n$ if $f, g$ are but what about $\sum_{\text{lcm}(a,b) = n} f(a) g(b)$. It associates because of associativity of $\text{lcm}$. Thanks @darij grinberg who says that this preserves multiplicativity. Where does it show up?: If the $a...

So I think that $\phi(n) = e^{i 2\pi (n) s}$ will work right?

Rithaniel

Just to highlight the main issue. You can't exchange the places of exponents. You have to be careful in the order you evaluate them as a result.

BananaCats

Oh cool

So $\phi(n)$ makes sense here and is not "always $1^s$?"

Rithaniel

12:50 AM

Potentially, yes.

(I say "potentially" because I'm not looking too deeply at the question at hand.)

BananaCats

But $\phi(n)$ is indeed equal to one value right?

or is it multivalued?

If it's multivalued that's something else and I can't really use it here

$(n) = $ an ideal

in $\Bbb{Z}$

See so it's not the usual way of looking at $e^{i2\pi n s}$

I've plugged in a full ideal of arguments

at once

Rithaniel

It might be multivalued. I'm just getting into ideals in my abstract algebra class, so I can't say for sure.

Tanner Swett

1:27 AM

All right, this Laplace analysis stuff is really easy. All I have to do is look at the behavior of

...which is less complicated than it looks; it's just a rational function in s.

BAYMAX

what is this fireworks thing on cursor pointer in old stack website known as?

AfronPie

I am trying to find an integer solution for a,b,c,d to (1) a+b+c+d=10 (2) 2a+5b+10c+20d=103 (3) a≥1 . I know these problems are known as diophantine-equations however I have never learned them, and for now I just need an answer quickly. I cannot find an online calculator. Does anyone know of a calculator or can provide me an answer for this.

Semiclassical

Most obvious thing that comes to mind is to subtract (1) from (2) twice

That gives you 3b+8c+18d==83, with no constraints on b,c,d

actually, maybe it's worth looking for solutions with a=1?

in that case you'd have b+c+d=9, 5b+10c+20d=101. subtracting the first equation 5 times from the second gives 5c+15d=101-45=56

oh, but that won't work since the LHS is odd.

AfronPie

1:43 AM

Is there a calculator that can compute this (I can't figure how to do it in Wolfram)

I mean this problem was originally: A person pays 10 bills in a store with total value 103. They use $2,$5,$10 and $20 bills. I then wrote two equations. I know the problem probably wanted you to guess the solution but I was hoping I didn't have to

Rithaniel

a=3, b=1, c=3, d=3

AfronPie

Thanks @Rithaniel

Semiclassical

Are you trying to find one such solution, or count all the solutions?

AfronPie

Just find one

Rithaniel

Wait, no, I messed up

Semiclassical

1:47 AM

(The latter is actually pretty easy using the concept of generating functions in combinatorics. Turns out to be 6300 of them)

Rithaniel

a=4, b=1, c=1, d=4

AfronPie

Btw my equation (3) was a hint for the problem and am not sure if its actually necessary

Ok @Rithaniel Ill check it

Thanks again @Rithaniel and @Semiclassical

Rithaniel

The approach was to try and find ways to get something with a "ones" digit of 3 and work backwards from there.

This won't work for larger problems, though, keep that in mind.

AfronPie

Ok, and is there a calcualtor that can compute to diophantine-equations?

Rithaniel

Not that I know of. Diophantine equations are a fairly difficult topic in math.

Partly because it's difficult to get computers to work with them, in most cases.

AfronPie

1:51 AM

Oh that is interesting

Rithaniel

The field of study that would best equip a person with handling diophantine equations is number theory, if you'd like to read up some more.

Secret

Linear Diophantine equations with two variables can be solved using the Euclidean algorithm. Anything else need hardcore number theory

Semiclassical

By contrast, the question of counting the number of solutions is pretty straightforward , at least if you disregard negative integers

BananaCats

Can you think of a way such that $\phi(a) \phi(b) = \phi(\gcd(a,b)), \phi : \Bbb{N} \to \Bbb{C}$?

Semiclassical

huzzah for generating functions

Rithaniel

1:58 AM

There are probably a few more cases than just that where a computer would be well equipped to solve the problem, but yeah.

BananaCats

Anyone?

I'm working with $e^{i 2\pi (n \Bbb{Z})}$ but I think that's multi-valued...

Rithaniel

I don't know of any functions satisfying that property.

BananaCats

@Rithaniel addition of ideals

$(n) + (m) = (\gcd(m,n))$ in $\Bbb{Z}$ where $(n)$ is ideal notation

So I was thinking combine that with Euler's formula somehow

Rithaniel

Ah, yeah, that does ring a bell.

BananaCats

$e^{i(n)}e^{i(m)} =e^{i(n,m)}$

Maybe requires a summation then the cyclicity will do something neat...

Semiclassical

2:15 AM

@Rithaniel Interestingly, that seems to be the only solution (with no negative integers anyways)

Rithaniel

Hmmmm, yeah, I can imagine that being the case. There's definitely infinite solutions if you include negative integers, though.

Semiclassical

yeah

oh hey, WolframAlpha can do it: wolframalpha.com/input/…

Rithaniel

wolfram fails a lot when finding integer solutions, though.

I'm actually kind of surprised that it found this one.

Semiclassical

well, over the positive integers you need a,b,c,d at least one

and since a+b+c+d=10

that limits you to at most 7 in any one of them

so there's only 7^4 options to check

not that hard

Rithaniel

True

BananaCats

2:38 AM

true true

:D

user10478

Given a vector field, is there a way to find the curve a water particle would travel as a result (this sort of thing youtu.be/VJ2ZDLQk3IQ?t=102) analytically (in cases where the curve can be expressed with functions)?

Semiclassical

Those would be called the streamlines of the vector field

user10478

Ahh, so it is possible then

Semiclassical

Whether you can actually compute those streamlines analytically will almost certainly depend on your vector field, tho

user10478

I'm guessing it has to be conservative?

Semiclassical

2:50 AM

i don't remember that much, unfortunately

user10478

Okie, I'll look it up, thanks

BananaCats

$\varphi(\gcd(m,n)) \varphi(\text{lcm}(m,n)) = \varphi(m) \varphi(n)$

Is close to what i need

So ive deduced that $\varphi(\text{lcm}(m,n)) = \varphi(m)\varphi(n/\gcd(m,n)) = \varphi(m/\gcd(m,n)) \varphi(n)$

By coprime multiplicativity

of totient

is there any way to "distribute that symmetrically" over the RHS of the first line above?

Silent

4:09 AM

Please confirm if I am getting this right:

Let $γ$ be a simple closed contour that contains the
distinct points $z_1, z_2, ··· , z_n$ in its interior. We have to find value of $\displaystyle{\int_{\gamma}}\,\frac{dz}{(z-z_1)(z-z_2)\cdots(z-z_n)}$.

If $n=1$, then the value is $2\pi i$, but for $n>1$, it is zero. Am I right?

Semiclassical

I think you're thinking in terms of $1/(z-z_1)^2$, which would indeed be zero

But that's a bit different.

@silent I think you may be right, though I'm a bit unsure as to why

(As in, I checked the case of $n=2$ by hand and it seems to indeed be zero.)

My guess right now would be to consider the substitution $w=1/z$

One thing to note: I'm pretty sure that the integral would not be zero for $n>1$ if your numerator were a function of $z$. for instance, it wouldn't vanish if your numerator were $z^{n-1}$ instead of $1$

(not totally sure about that example)

user76284

4:48 AM

https://en.wikipedia.org/wiki/Residue_theorem#Statement

Consider $n = 2$. The residue at $z_1$ is $1/(z_1-z_2)$. The residue at $z_2$ is $1/(z_2-z_1)$. Their sum is zero. Similarly, consider $n = 3$. The residue at $z_1$ is $1/(z_1-z_2)/(z_1-z_3)$. The residue at $z_2$ is $1/(z_2-z_1)/(z_2-z_3)$. The residue at $z_3$ is $1/(z_3-z_1)/(z_3-z_2)$. Their sum is again zero. And so on.

Semiclassical

yeah, that works. but it's a bit tedious to compute all the residues within the unit circle

especially when it can be done directly by considering the residue at infinity

Silent

5:19 AM

thanks, @Semiclassical and @user76284

Silent

6:06 AM

Let $\gamma$ be circle of radius $r$, namely $a+re^{it}$. My book claims that $\int_{\gamma}dz=2\pi r$, but if I go by definition, $\int_0^{2\pi}f(z(t))z'(t)=\int_0^{2\pi}ire^{it}\,dt=0$

Semiclassical

@Silent Do they possibly mean $\int_\gamma |dz|$? That's the only way I could fathom their claim.

Silent

@Semiclassical I don't think so. It is from here, see P127, after line 'Hence, from Theorem 13.1, we find'

sorry! i should have looked at 13.1 before ranting about it.

Semiclassical

yeah, all they're claiming is that the length of contour $\gamma_r$ is $2\pi r$

Silent

6:23 AM

silly me. I went for all that computation

piece_and_love

7:47 AM

When restricting a function $f$ to a subset $A$ of its domain and saying the value of $f$ under such values is, say, $1$, is it proper notation to write $f|_A=1$?

Leaky Nun

sure

piece_and_love

Cheers. One follow up question; is it always true that for a set $B$ that the limit of

oops lemme type it out in full

is it always true that for a set $B$ that the limit of $b_n$ ($b_n \in B$) as $n$ goes to infinity is defined and is an element of the closure of $B$?

Leaky Nun

no

0,1,2,3,... has no limit for example

piece_and_love

Oh, sorry, I maybe should have stated that B is a non-empty connected subset of a metric space haha. I don't suppose it would be true then?

Leaky Nun

R is connected

piece_and_love

7:54 AM

Though I guess you just take the metric to be R

ye

Cheers for the answers!

Secret

8:36 AM

[Random]

$y = {}^{\infty}x$
$\int y dx = \int x^{y} dx$

$y^2=yx^y$

$W(y^2) = x$

error

$y^2=ye^{y\ln x}$

$y^2 \ln x = (y \ln x)e^{y\ln x}$

$W(y^2\ln x) = y \ln x$

$W (x^{2y} \ln x) = x^{y} \ln x$

$yW(x^{2y} \ln x) = (y \ln x) e^{y \ln x}$

$W(yW (x^{2y} \ln x)) = y \ln x$

$yW^{\circ^{\infty}} (x^{2y} \ln x) = y \ln x$

$yW^{\circ^\infty} (y \ln x) = y \ln x$

$yW^{\circ^{\infty}} (y \ln x) = \ln x$

$e^{yW^{\circ^{\infty}} (y\ln x)} = x$

$(yW^{\circ^{\infty}} (y\ln x))' e^{yW^{\circ^{\infty}} (y\ln x)} = dx$

error, too much recursions, calculation terminated

lush

9:01 AM

hey everybody :-)

it's been a while since my last visit

Secret

Overview of differential equations | Chapter 1

►

Silent

@Secret wow! thanks for sharing. this channel has helped me a lot with calculus and linear algebra

Secret

This channel is also Leaky's favourite. His number theory videos often inspires a lot of ideas

Silent

9:16 AM

number theory videos? will you please share a link? i have seen none!

@Secret

Secret

All possible pythagorean triples, visualized

►

for example

Silent

Thank you !! never saw that before

Suppose $a_n\ge0$ and $\sum a_n$ convergent. Show that $\sum \frac1{n^2a_n}$ is divergent.

I have a proof that uses cauchy-schwarz

is there some other proof? for examle, the one which uses $a_n\to0$?

piece_and_love

10:09 AM

Quick question: if $f$ maps a set $A$ to the point $1$ then you say the the pre-image of $1$, $f^-1(1)$ contains $A$, not that $A$ contains $f^-1(1)$, right?

Secret

yes, the preimage is larger than or equal to the set $f$ maps

piece_and_love

Then I must be misunderstanding what's mean in a proof from a book about Metric spaces

am I allowed to post images?

meant *

Secret

use the upload button

piece_and_love

I'm trying to find where that is hahaha

I'm either blind or don't have permission to upload I believe

I can just write out the relevant parts I guess

'$f$ is continuous and maps $B$ to the discrete space ${0,1}$, suppose there exists some $b \in B$ s.t. $f(b)=1$ then since ${1}$ is open in {0,1}$ and f is continuous we have that $f^-1(1)$ is open in $B$'

Secret

10:40 AM

yup that's correct

$f^{-1}(1)$ can be $\{b\}$ or some open superset of $\{b\}$, which is what I meant in the previous statement

piece_and_love

does the statement '$f^-1(1)$ is open in $B$' not imply that $f^-1(1)$ is contained in $B$? this is probably where I'm misunderstanding

Secret

11:03 AM

Ah right, the largest the preimage of $f$ can be is the whole domain, thus $f^{-1}(1) \subseteq B$

sorry for the mistake

piece_and_love

no worries, thanks for the explanation!

Secret

11:29 AM

9

Q: Does the concept of infinity have any practical applications?

I know what you're thinking: "of course it has, for example, it can be used to tell you how many times you can go around a circle". But that isn't really true, now is it? You'd be dead or the world would go under long before an infinite amount of loops had been reached. Are there any practical a...

infinity applications

30

Q: Math without infinity

Does math require a concept of infinity? For instance if I wanted to take the limit of $f(x)$ as $x \rightarrow \infty$, I could use the substitution $x=1/y$ and take the limit as $y\rightarrow 0^+$. Is there a statement that can be stated without the use of any concept of infinity but which un...

soft-question infinity

Defining infinity. Attempt 4 (💥(Infinity)):

Let $S$ be some set, let $f$ be an injective function. Then $S$ is infinite if:

$$f(S) \subseteq S$$

and for all $s,t \in S$:

$$f(s)=t, f(t) \neq f(s), f(t) \neq s$$

and there exists some natural number $n, m$ such that there exists bijection between all fixed points and cycles of $f$ and $n, m$:

In English, $S$ is infinite if there exists some injective function $f$ such that its orbit never revisits itself nor can be listed out on a piece of paper of some fixed size $x$

More generally, infinity:

1. Is often larger than a set of interest, which we considered as finite

2. Is invariant under some mappings

3. It enforces a given process and pattern to continue without interruption and without repetition

In other words: Infinity = Invariant, nonstop and repeating process each with different outcomes

user193319

11:49 AM

If $G$ is a topological group, how is $\ell^1(G)$ defined?

usukidoll

How does this even work?
http://prntscr.com/n5upr6
omg this is the last one and I just hate this >:(
It's like a special definition of efficency UGH!

Secret

group algebra with functions in $\ell^1$ space?

user193319

From my understanding, it consists of all functions (continuous?) $f : G \to \Bbb{C}$ such that $\sum_{g \in G} |f(g)| < \infty$. But what does $\sum_{g \in G} |f(g)| < \infty$ mean? It should mean something like, "For every $\epsilon > 0$, there is some...(?)...such that $\sum_{g \in ?} |f(g)| < \epsilon$."

usukidoll

uh oh

when Latex and regular words get mashed together

user193319

But what goes in the question mark?

@Secret I don't think so. I believe $C_0(G)$, all the continuous complex-valued functions vanishing at $\infty$, is the Group Algebra of $G$ (but maybe I am wrong).

Secret

11:56 AM

hmm, your definition makes it clearer to me. I will suspect $\sum_{g \in G} |f(g)| < \infty$ means for every $H \subseteq G$, there always exists some real number $M$ such that $\sum_{g \in H} |f(g)| < M$

meaning every such series $|f(g)|$ must be bounded regardless of what and how many $g$ s you taken from the group $G$

user193319

Why not for every $\epsilon > 0$, there exists...etc. I am trying to parallel the definition of what it means for $\sum_{n = 1}^\infty |x_n| < \infty$ to be true. For that to be true, it means that for every $\epsilon > 0$, there is some $N \in \Bbb{N}$ such that $\sum_{n=N}^\infty |x_n| < \epsilon$.

Perhaps I should ask on the main.

Secret

Ah I see, so you want to define convergence in terms of the tail, in that case I think you need something like the tail of the sequence or net in question, which I don't recall that notation, but that is the $?$ you need to substitute into $g \in ?$

user193319

Filters may help, but it isn't clear what the "tail" of a group is.

Secret

if the set $f(G)$ forms a partial order, then the tail can be described using the limit notion of nets

Otherwise I have no idea, though there should be some ordering induced since $G$ has a topology

@usukidoll I don't have background on this but based on the very limited knowledge I have about Fisher information $I$ and variance $V$, you will expect something estimates a quantity very well if the variance it returns is no larger than the information needed to specify it

that is, an estimator is good if it has low amount of noise in what it thinks $\theta$ is

user193319

12:18 PM

0

Q: $\ell^1$ Space of a Group $G$

If $G$ is a topological group, how is $\ell^1(G)$ defined (is it necessary to require that $G$ be a topological group?)? From my understanding, it consists of all functions (continuous?) $f : G \to \Bbb{C}$ such that $\sum_{g \in G} |f(g)| < \infty$. But what does $\sum_{g \in G} |f(g)| < \infty$...

functional-analysis fourier-analysis harmonic-analysis

Leaky Nun

@user193319 1. you want $G$ to be a Hausdorff topological group and then use the Haar measure 2. $< \infty$ just means it is finite

user193319

Doesn't $G$ also have to be locally compact to have a Haar measure?

Leaky Nun

oh yeah sure

user193319

As I said above, I know that $\sum_{n=1}^\infty |x_n| < \infty$ means for every $\epsilon > 0$, there is some $N \in \Bbb{N}$ such that $\sum_{n=N}^\infty |x_n| < \epsilon$, so I figured that $\sum_{g \in G} |f(g)| < \infty$ would means something similar.

Leaky Nun

no it just means that it is finite

although what you said should be equivalent

but when we say $< \infty$ we just mean finite

Secret

12:29 PM

But surely there is a way to elaborate what it means to be finite. As user 193319 said, for complex number sequences, there exists some $M$ such that the tail of the sum is smaller than $\epsilon$, but what about any net in a topological group in general?

Leaky Nun

it means there is a real number $R$ that is the value

user193319

Someone actually just posted a nice answer on my question.

Leaky Nun

ok fair

Secret

$$\sup \{\sum_{\{g \in A\}} |f(g)|:A \subset G \, \text {finite}\} <\infty$$

hmm... why are $A$ being finite is sufficient

cause I can e.g. pick all $f(g)$ such that it forms a harmonic series, then taking the supremum of this will always be $< \infty$. It is only when you sum the harmonic series then you diverge

Leaky Nun

this ensures that $f$ has countable support (why?)

Secret

12:38 PM

Hmm... the above supremum compute all possible finite sums of $f(g)$, which means we have at least countably many real numbers forming a sequence that converges, thus ensuring $f$ has nonzero image when fed a countable set

the special case being take $A$ to be singletons, and take countably many of these to produce a sequence of $f(g)$. That converges meaning that all possible sequences are bounded

In fact, we can take arbitrarily many of such $A$ s under some preorder to produce a net, and this net converges, thus ensuring every sum is bound

I do not understand why I keep underestimating the power of sup

user193319

1:04 PM

Problem: Let $f : \Bbb{R} \to \Bbb{R}$. Prove that the set of irrational numbers cannot be the set of discontinuities $f$...

Should there be more hypotheses on $f$?

Secret

Thomae's function

Thomae's function, named after Carl Johannes Thomae, has many names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function, the Riemann function, or the Stars over Babylon (John Horton Conway's name). This real-valued function of a real variable can be defined as: f ( x ) = { 1 q ...

I will admit I still don't really understood that proof, but the short answer is you do not need anything else other than the axiom of choice

Daminark

1:46 PM

@user193319 nope, that's enough. So, equivalently to your statement, and easier to process, the set of rational numbers cannot be the set of points of continuity of $f$

There's a trick here, turns out the set of points of continuity of a function is a set of a very certain type, and the rational numbers are not that type

I'm sorta deciding whether to say which type. It's not easy to figure out, but there isn't really that much work left (though I guess you have to use a definition of continuity that you probably never heard of unless you read Spivak Calc on Manifolds)

Secret

I wonder what will happen in choiceless universe where baire category fails, do such weird functions can exist...

Daminark

"choiceless universe"

There will be darkness, and time will stop. For it will violate logic itself

Secret

lol

user193319

2:28 PM

@MatheinBoulomenos A while back you were helping me with the problem of classifying all noncommutative $p^3$ groups (with $p \ge 3$). The hint the book I am using suggested showing that there exists a normal subgroup $N$ of order $p^2$ and an element $c$ of order $p$ outside of $N$. You said this was false and gave the following example:

$G= \left\{ \begin{pmatrix}1+pa & b \\ 0 & 1 \end{pmatrix} \mid a,b \in \Bbb Z/p^2 \Bbb Z \right\}$

My professor just went over this problem today in class, but maintained that there does in fact exist such an element $c$. When I asked him to prove this, he said it was a simple counting argument...and now I am confused...How does one show that $G$ is in fact a counterexample to the hint?

@MatheinBoulomenos Who is right? You or my professor? I'm inclined to believe you, especially since Balarka also agreed (and both of you are brilliant).

Larry Eppes

3:39 PM

hello

anybody can help my previous question? here it is: math.stackexchange.com/questions/3168201/…

amanuel2

Here I know your suppose to check for remainder 1 for case 1 and remainder 2 for case 2.

But what about if it asked if an integer is disvisble bye 2 or 4 or 5(Something Greater or lower than 3). For which remainders do i check for the cases.

Larry Eppes

just calculate the remainder of square of 1^2, 2^2,...,n^2 modula mod m...

this question is more basic in elementary number theory...

"calculate the remainder of square of 1^2, 2^2,...,n^2 modula mod n" for my typo

amanuel2

You have to proof using proof by cases method. My question is how do i know the max number of the remainder. For example the max remainder for 3 is 2

Larry Eppes

oh, that is a difficult prob to me at a first sight... sorry for misunderstand the question..

??? the max remainder for n is n-1, this is obvious or not?

Lucas Henrique

@LarryEppes it's not.

Larry Eppes

3:54 PM

or you want the maximum remainder of 1^2, 2^2, ..., n^2 mod n?

Lucas Henrique

@amanuel2 suppose $x$ and $y$ are both remainders such that $x^2 \equiv y^2$. So $(x-y)(x+y) \equiv 0$. Then $x \equiv \pm y \in \Bbb Z_n$ if $n$ is prime, i.e, $\Bbb Z_n$ is an integral domain.

amanuel2

If N and D are positive integers, and if N divided by D is equal to Q with remainder R. Im asking for the possible values for R

Larry Eppes

D===Q mod R, then R|(D-Q)

Lucas Henrique

@amanuel2 Well, those are the values ranging from $0$ up to $n-1$ (included)

Larry Eppes

N=DQ+R? sorry for my poor english..

amanuel2

3:59 PM

Ok 0 -> (n-1). Thats all i needed . Thanks @LucasHenrique @LarryEppes

anakhro

4:11 PM

Hey hot cats.

Secret

reddit.com/r/dataisbeautiful

datasplosion

actually, I wonder how does one could define a paranormal distribution

googles paranormal activity in pure mathematics

anakhro

paranormal concerns a fundamental lack of empirical evidence for an empirical claim.

Since math doesn't touch empirics, you will be hard-pressed.

Secret

make sense

Ghost algebra ?_?

anakhro

$user image$

Ted Shifrin

Is that your April Fools present, @anakhro?

anakhro

4:21 PM

@TedShifrin it's not really foolery.

Perturbative

Hey everyone!

anakhro

Happy April Fools to you, Ted!

Ted Shifrin

Thanks, @anakhro. Hi, @Perturb.

Perturbative

Hey @TedShifrin :)

anakhro

Maybe too general of a question: if I have a smooth family of surfaces $\{S_i \mid i\in[0,1]\}$, and a generic property $P$ each $S_i$ could satisfy, I can't necessarily assume each $S_i$ satisfies $P$, can I?

Ted Shifrin

4:24 PM

That doesn't make sense.

"could satisfy"?

anakhro

Sorry I had a huge problem with that sentence, maybe I fixed it for you, @TedShifrin

Ted Shifrin

"Generic" is meaningless here.

So if you say each $S_i$ could satisfy $P$ and then you ask if each $S_i$ must satisfy $P$, it's pretty obviously wrong.

anakhro

Yeah, I am wondering if I can ``generically'' assume that the family satisfies a generic property.

Rithaniel

Heya Ted.

anakhro

My intuition leads me to refuse to believe you can "fix" any singular $S_i$.

Ted Shifrin

4:27 PM

If you mean "generic" in a technical way, that should mean that $S_i$ should satisfy it for most — sometimes an open dense subset of — $i$ values.

anakhro

Without messing up some other $S_j$ possibly.

Ted Shifrin

hi @Rithaniel

anakhro

@TedShifrin that is what I am using. The set of surfaces which satisfy $P$ form a dense subset of the set of all surfaces.

So almost every $S_i$ is generically $P$.

Ted Shifrin

OK. If the property is generic in that technical sense (for every family ....), then you've answered your own question.

anakhro

But the question I am hung up on is whether you can always fix those that are not generic so that they are, while maintaining the genericity of the rest of the family.

Because it is a smooth family.

Ted Shifrin

4:30 PM

This is too vague a discussion.

anakhro

I agree it is vague, but I fear that if I make it precise, you won't be able to understand it because it deals with objects you are unfamiliar with.

Shall I just try?

Ted Shifrin

No.

anakhro

lol, love you, Ted.

Ted Shifrin

Hi, demonic @Alessandro.

Alessandro Codenotti

Hi @Ted

anakhro

4:33 PM

Actually, I can phrase it in terms of the Poincare-Bendixson theorem.

Ted Shifrin

It sounds like you probably need to study the transversality theorems.

Alessandro Codenotti

The new semester began today, it's nice to have classes again

anakhro

I know of them already, but I don't think I can apply them here.

At least to fix.

Ted Shifrin

Congrats, Alessandro.

anakhro

@AlessandroCodenotti what classes are you taking?

Alessandro Codenotti

4:36 PM

Models of set theory, type theory, a set theory seminar, a practical project in logic and noncommutative geometry

I'll also go to topology II without taking the exam most likely

Ted Shifrin

Way too much logic stuff!

Alessandro Codenotti

I'm also doing a "geometry" course!

anakhro

topos theory, right

Type theory is super cool.

Alessandro Codenotti

(noncommutative geometry is actually counted as analysis credits, it will all be functional analysis apparently)

anakhro

Got a chance to do a lambda calculus and categorical logic course and it was great.

Ted Shifrin

4:39 PM

Yeah, that typically is mostly analysis, hardly geometry.

anakhro

What's geometry if not placing wooden shapes through similarly shaped holes?

Alessandro Codenotti

After the algebraic geometry course I learned I shouldn't trust "geometry" in the name of courses

anakhro

@AlessandroCodenotti definitely check out the classical approach to algebraic geometry.

It redeems it.

Semiclassical

hi chat

anakhro

Great notes on it: math.lsa.umich.edu/~idolga/CAG.pdf

Ted Shifrin

4:43 PM

@Alessandro: And your algebraic topology course was barely what I would call topology. Says something about your professors ;P

Semiclassical

boring probability question

Alessandro Codenotti

@TedShifrin I actually think that had the right balance of algebra and topology, but the topology was all in the exercise sheets

Ted Shifrin

Blah.

Semiclassical

Suppose I've got a random vector $X$. I want the individual components $X_1,X_2,\ldots,X_n$ to be uniformly distributed (on {-1,0,1}, for instance) but I don't want to assume independence

anakhro

What was the algebraic topology course? Category theory 101?

Mine was like that and I retained nothing. :(

Semiclassical

4:45 PM

e.g. I don't want $X_1,X_2,...,X_n$ to be $n$ independent samples of the uniform distribution

So I guess the question is whether saying "each $X$_k$ is uniformly distributed" assumes anything about their joint distribution?

I don't think it should, but I'm rusty

Rithaniel

Today I ~~wasted~~ spent an hour defining custom symbols to use for variables in Latex documents.

Semiclassical

Good reframing.

Perturbative

Okay so I just wanna recap about change of coordinates on manifolds again.

So say I have some manifold $M$ of dimension $n$ and I take two charts $(U, \varphi)$ and $(V, \psi)$ with nonempty intersection and represent them via local coordinates by $(U, (\theta^i))$ and $(V, (y^i))$. If I choose a point $p \in U \cap V$, then I can represent a tangent vector $X_p \in T_pM$ either as $$X_p = c^1 \frac{\partial}{\partial \theta^1} \big|_{p} + \dots + c^n\frac{\partial}{\partial \theta^n} \big|_{p} $$

or as $$X_p = d^1 \frac{\partial}{\partial y^1} \big|_{p} + \dots + d^n\frac{\partial}{\partial y^n} \big|_{p} $$.
Apparently we can relate the two expressions by $$d^j = \sum_{i = 1}^n \frac{\partial \tilde{x}^j}{\partial r^i} (\varphi(p))c^i$$

where $\tilde{x}^i$ is the $i$-th coordinate function of the function $\psi \circ \varphi^{-1} : \varphi[U \cap V] \to \psi[U \cap V]$. In other words $\psi \circ \varphi^{-1}(a) = (\tilde{x}^1(a), \dots, \tilde{x}^n(a))$.

Or I can write it (more to my liking) as

$$d^j = \sum_{i = 1}^n \frac{\partial \left(\pi_j \circ \psi \circ \varphi^{-1}\right)}{\partial r^i} (\varphi(p))c^i$$

Where in both cases $r^1, \dots r^n$ are just the standard coordinates on $\mathbb{R}^n$

Is that correct?

anakhro

n i c e

Ted Shifrin

@Semiclassic: No, that certainly doesn't suggest independence.

Semiclassical

4:52 PM

good

Ted Shifrin

@Perturb: I can't begin to unwind all the notation in your "more to my liking."

Semiclassical

I guess I should also be able to restate what I'm saying in terms of the marginal distributions

Ted Shifrin

I don't understand the "apparently" statement. There should be a direct chain rule application with derivatives of $\theta^i$ with respect to $y^j$.

Where are this excessive notation and all the unnecessary letters coming from?

Semiclassical

i.e. I should be able to say that $X=(X_1,X_2,\cdots,X_n)$ has uniform marginal probabilties

Ted Shifrin

I don't know nothing about no marginal probability.

anakhro

4:55 PM

double negative, bro

Semiclassical

I'm not great on them either, but the wiki definition is simple enough: "In probability theory and statistics, the marginal distribution of a subset of a collection of random variables is the probability distribution of the variables contained in the subset."

Ted Shifrin

triple @anakhro

anakhro

I didn't know you were a constructivist.

I forget, is there a natural identification for triple negatives

or is that only some places.

Semiclassical

I don't not not know

Ted Shifrin

@Semiclassic: So it's basically a joint probability distribution in however many variables?

Semiclassical

4:59 PM

Yeah

anakhro

@Semiclassical u rock

Are source/sink nodes commonly distinguished from normal source/sinks in dynamical systems?

Perturbative

@TedShifrin I only said apparently in case I was wrong :p

The $\pi_j$ is just the $j$-th projection map

Ted Shifrin

I don't have the patience to unwind it all.

anakhro

@Perturbative what exactly are you looking for? That everything you said is correct? In that case, just go check against your textbook.

From Tu's book for you.

Semiclassical

anyways, I think the problem I've been working on lately can summed up like this. Suppose I have random vector $X=(X_1,X_2,X_3)$ such that $X\stackrel{D}=-X$ and $X_1,X_2,X_3$ are each uniformly distributed on $\{-s,-s+1,\cdots,s\}$ where $s=1/2,1,3/2,2,\ldots$.

anakhro

5:08 PM

Then you can just feed the partials through the definitions.

Semiclassical

And the question is what values of $(\mathbb{E}X_1 X_2,\mathbb{E}X_2 X_3,\mathbb{E}X_1 X_3)$ can be produced by such distributions.

main thing that's nice is that, at least in principle, you can actually visualize that

Semiclassical

5:45 PM

Huh, turns out there’s a nice name for the construction I’m articulating:

en.m.wikipedia.org/wiki/Copula_(probability_theory)

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$Mathematics$ Mathematics