Mathematics - 2024-03-20

psie

00:00

To show $\supset$, let $x\in [f^{-1}(a),f^{-1}(b))$, i.e. $f^{-1}(a)\leq x< f^{-1}(b)$. Since $f$ is increasing, the latter inequality holds iff $a\leq f(x)<b$, which shows that $x$ belongs to $f^{-1}([a,b))$.

Now, I'm unsure if I've done this correct because I have not used continuity anywhere...this feels not right.

psie

00:19

Upon further thinking, maybe continuity is required in saying that $f$ is injective...

psie

00:29

Upon even closer thought, what I said in my last message can't be right. A strictly monotone function needn't be continuous for it to be injective...so where in my argument, if it is correct, is continuity required? Hmm.

Thorgott

00:42

it isn#t required

psie

that's a relief...thanks for confirming

Jakobian

01:44

@psie its injective by definition

you also don't need that $f$ is continuous

being bijective and monotone, its continuous already

psie

👍

Daniel Donnelly

02:36

Hi, chat. Say you had two repeating sequences which we can write $\overline{1,2,3} := a$ and $\overline{2,1,2} := b$. What if you formed an operation on these that returns another repeating sequence. Here is one: the sum 1 + 2, and 2+1 match (from the first two entries of each respectively) so the first entry in the op output is $3$.

Then continue this way until what your doing starts to repeat itself

So the next would be 3+1 = 2+2 = 4

And teh next would be 2 + 3 = 2 + 2 + 2 = 5 and so on...

I mean 2 + 3 = 2 + 1 + 2 etc

Have you seen this operation before?

And how do we determine if it can terminate early? Meaning we already have determined the part that is going to repeat, without filling out $\text{LCM}(|a|, |b|)$ entries?

Believe it or not such operation's output values if you could somehow bound them by certain formulas has applications to NT problems.

Daniel Donnelly

02:55

1

Q: An exotic operation on repeating sequences of natural numbers such as $\overline{2,1,2} = 2,1,2, 2,1,2, \dots$

This is about repeating sequences of natural numbers such as $\overline{6} = 6,6,6, \dots$ or $\overline{2,1,2} = 2,1,2,2,1,2, \dots$. I am aware that these can be faithfully represented as the sequence of coefficients of a power series $f(x)(1 + x^k + x^{2k} + \dots)$ where $k = \deg f + 1$. ...

user 85795

03:31

@XanderHenderson ::savage::

Vivaan Daga

12:17

Is it possible for S4 to act on itself so that the orbits have length 1,3,4,4,4,4,4 ? @Thorgott do you have any idea for this question? Conjugation gets us 1,3,6,6,8

psie

14:12

I'm new to this and I'm not looking for big explanations (although I've tried to google the heck out of this, and yet I haven't been so successful). I'm wondering what the difference between a continuous and an absolutely continuous cumulative distribution function is?

In my textbook, the author simply says "In this book we are only concerned with two kinds: discrete distributions and (absolutely) continuous distributions." The author clearly has more to say, but unfortunately doesn't. If anyone could shed some light on what the difference is, I'd be grateful.

Thorgott

@Balarka daily fact (this one I can prove): given an element $\alpha\in\pi_{2n-1}(S^n)$ and $\beta_1,\beta_2\in\pi_n(X)$, we have $(\beta_1+\beta_2)\circ\alpha=\beta_1\circ\alpha+\beta_2\circ\alpha+H(\alpha)[\beta_1,\beta_2]$, where $H$ is the Hopf invariant and $[-,-]$ the Whitehead product. the Hopf invariant measures the failure of additivity of pulling back along the map in terms of the Whitehead product.

AliceD

@Moderators - mod from Psy & Neuro here: can we migrate this question to your site?

Jakobian

$Mathematics$ Math Mods' Office

For informal chat with the site moderators about moderation, s...

2

@AliceD

@psie Absolutely continuous means there exists a pdf. That is, a function $f(x)$ such that $P(X\in A) = \int_A f(x)dx$. If its merely continuous, then such pdf doesn't have to exist. We are only guaranteed that $t\mapsto P(X\leq t)$ is a continuous function.

The counter-part to absolutely continuous random variables are singular continuous random variables, for which there exists a Lebesgue measurable set $\Omega_0$ such that $P(X\in \Omega_0) = 1$ but $\lambda(\Omega_0) = 0$ where $\lambda$ is the Lebesgue measure

But nonetheless, $t\mapsto P(X\leq t)$ is still continuous

Every random variables decomposes into discrete, absolutely continuous, and singular continuous part

an example of a singular continuous random variable is the one whose CDF is based on the Cantor's devil staircase

Cantor function

In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous. It is a notorious counterexample in analysis, because it challenges naive intuitions about continuity, derivative, and measure. Though it is continuous everywhere and has zero derivative almost everywhere, its value still goes from 0 to 1 as its argument reaches from 0 to 1. Thus, in one sense the function seems very much like a constant one which cannot grow, and in another, it does indeed monotonically grow. It is also called the Cantor ternary function, the Lebesgue functio...

extended from $[0, 1]$ to the whole real line by setting $F(t) = 0$ for $t\leq 0$ and $F(t) = 1$ for $t\geq 1$

You should treat singular continuous random variables as the leftover part which no one really cares about

Vivaan Daga

14:33

am i missing something or is finding such an action hard?

where the orbits have length 1,3,4,4,4,4,4 and S4 is acting on itslef

is a computer search the best possible solution?

psie

@Jakobian does it need to be $1$ in $P(X\in \Omega_0) = 1$ or could it just be any other number in $(0,1)$? That would be the negation of absolutely continuous measure, right?

Jakobian

Also compare for the notion of absolute continuity of measures $\mu_0 < < \mu_1$ and singularity of measures $\mu_0\perp\mu_1$. Above names boil down to those concepts for the law of random variable $X$ and the Lebesgue measure.

AliceD

@Jakobian thanks for the tip!

Jakobian

@psie it needs to be the full measure

AliceD

@Jakobian You lost me after the second syllable in your reply, but I love your enthusiasm and admire your math skills :) I'll migrate the post :) Thanks!

Jakobian

14:39

Alright

@psie What do you mean by negation? If it were $P(X\in \Omega_0)\in (0, 1]$ that only tells you there is some singular part to $X$

Vivaan Daga

in general given orbit lengths is constructing an action easy? @Thorgott '

psie

@Jakobian yeah, I'm not really sure what I was talking about :) but what did you mean by "counter-part" to AC random variables? I interpreted "counter-part" as the "opposite", so that's why I thought, well, the opposite of AC, according the definition below, must be that there exists a $\Omega_0$ such that $\lambda(\Omega_0) = 0$ but $P(X\in \Omega_0) = 1\in (0,1]$ (since it's an implication)

the negation would be that the measure that is dominated just needs to be nonzero (like you negate, say, uniform continuity)

I have a typo above. I meant $P(X\in \Omega_0) \in (0,1]$.

Jakobian

15:14

> a person or thing that corresponds to or has the same function as another person or thing in a different place or situation.

how it is a counter-part I've explained in the same massage

@Jakobian .

Sahaj

15:51

@Jakobian hi!

Ive finally started reading the ebook you sent me

:)

Jakobian

@Sahaj hi

I've learned what an $F$-space is today

Ryder Rude

16:26

there is this idea that the notion of derivatives generalises to K algebras

becuz functions are essentially a K algebra becuz they also hav multiplication defined on them

Lucky Chouhan

@Jakobian what book??

Jakobian

16:50

@LuckyChouhan how should I know

Vivaan Daga

17:23

Any ideas for my Question I am extremely curious

If true it would make the proof of the Sylow theorems infinitely less mysterious

Thorgott

17:50

@VivaanDaga to the contrary, it seems an impossibly difficult. however, I don't have the time/interest to put effort into thinking about it

@RyderRude there are derivations

Vivaan Daga

@Thorgott even my specific case is hard without a computer?

1,3,4,4,4,4,4 and S4 I mean

Thorgott

18:28

dunno

Ryder Rude

@Thorgott does this mean that one can sort of take a derivative of just a complex number

these derivations must be 0 for multiples of the multiplicative identity element of the algebra

this means that all real numbers get mapped to 0 if we try to define a derivation on C

then D(a+bi)=D(a)+bD(i)=bD(i). so we can define D(i) to be anything?

we hav D((a+bi)(c+di))= (bc+ad)D(i). And we also hav D((a+bi)(c+di))= (a+bi)dD(i) + (c+di) bD(i)= (a+c + (b+d)i) D(i)

so (bc+ad)D(i)=(a+c + (b+d)i ) D(i). this means D(i)=0 is the only choice?

Thorgott

18:54

yes, there is only the trivial derivation on the $\mathbb{R}$-algebra $\mathbb{C}$

easier: $D(i^2)=2iD(i)$, but $D(i^2)=D(-1)=0$, so $D(i)=0$

VLC

19:08

Hi there!

Can somebody help me solve a quick equation:

(y dy - z dz)/dt = -z + 3

Can I just integrate every part ?

Xander Henderson

@VLC Is that how the problem was originally presented? In what context? In isolation, it doesn't seem like good notation...

VLC

@XanderHenderson

The problem was to find the first integral of a PDE. I got a characteristic system (t is the parameter) and then by applying some functions and operations I got to the above equation. Basically the system is linear and homogeneous

Pizza

$M_{m*n}(\mathbb{R})$
$A\subseteq M_{m*n}(\mathbb{R})$
$A=\{a_{ij},1<i<m \land 1<j<n\}$
$W=\{A:a_{ij}=a_{ji} , \forall i\neq j\}$
Prove that A is a subspace of M, pls help

VLC

@XanderHenderson The original equation that I got was:

(dy/dt)y- (dz/dt) = -z + 3

Sha Vuklia

19:52

Ted hasn't been online for over a week, which is not typical for him. Does anyone know if he is ok?

Pizza

@ShaVuklia 3 days ago I saw on the home page that he had replied to someone

But I don't know why he isn't writing here anymore :(

Sha Vuklia

ah, then the "last seen 8 days ago" that I saw must have been last seen in the chat

Pizza

Yes I think so, then 11 days ago he wrote his last message

Thorgott

yeah, Ted is active on the site still, just hasn't been in chat

I don't know if he's busy, taking a break or something else, but at least there's no acute reason to worry

psie

20:57

Just want to double check something. If a function is continuous, then $X$ connected implies $f(X)$ is connected. The contrapositive of this must be; if $X$ is connected but $f(X)$ isn't, then $f$ isn't continuous. Is this right? The reason I'm asking is because this is a claim made here.

leslie townes

i would push back a little on "the" contrapositive and "must be". one always has some choice as to how you phrase logically equivalent things in english, if not also in symbols. whatever uniqueness there is, is not as strict as you might expect.

psie

ok 👍

leslie townes

so here, e.g., if your P is "f: X to Y is continuous and X is connected" and your Q is "f(X) is connected," one way of phrasing the contrapositive of P => Q is "if f(X) is not connected, then either f: X to Y is not continuous, or X is not connected"

and maybe you're unconsciously using even further logical rewrites to add in "okay, so if f(X) is not connected and X is connected, then f: X to Y is not continuous." logic books will give interesting latin names to what is going on here.

so, it is a logically sound thing to say, and it comes from "if X is connected then f(X) is connected," but i don't know that there is a machine where if you fed in that sentence and pressed one button labeled "contrapositive," that it would spit out what that comment is using.

psie

@leslietownes hmm, interesting, so would you say the claim in the link is correct? The answerer says that the inverse of $f:(-1,0]\cup[1,2]\to[0,4]$ given by $x\mapsto x^2$ is not continuous because $[0,4]$ is connected but $(-1,0]\cup[1,2]$ isn't...

I guess the claim is correct, rereading your version of the contrapositive :)

i.e. "if f(X) is not connected, then either f: X to Y is not continuous, or X is not connected"

leslie townes

yes, it is correct. i am not sure i would choose that way of explaining that example to a classroom full of people who have just learned about continuity, but it's fine

e.g. the last time i taught real analysis, students first encountered continuity as a pointwise condition that was only later extended to sets, and connectedness came a few pages later than continuity in the textbook. in that kind of setting i might first "point" out (ha ha) that the function f^{-1} is not continuous at 1 in that example.

but that way of looking at it is a good one, because it shows that you don't need to engage with the specifics of what f^{-1} is, or where its discontinuities might be in some example, at all.

psie

21:22

the theorem I'm used to is:

Theorem Suppose $f:X\to Y$ is continuous. Then $f(X)$ is connected if $X$ is.

Here the contrapositive wouldn't negate continuity I think

The (or "a") contrapositive would be:

leslie townes

one thing you might be wrestling with is that often a theorem is stated as list of hypotheses or even just a narrative story that sets a context, followed by "if P then Q," where the context is not explicitly part of the "if-then" statement.

psie

Contrapositive Suppose $f:X\to Y$ is continuous. Then $X$ is disconnected if $f(X)$ is.

leslie townes

and there's the question if, if you want to symbologize it, whether you symbologize just the P and Q, or if you also take the story and symbologize that and add it with an "and" to the P, or do something else logically equivalent with it.

psie

true

leslie townes

and different people and textbooks will resolve that in different ways, depending on context. i don't think there's one machine way that does it, any more than there's one way in english of saying the underlying logical content.

in "intro to proof" books they would often symbologize something like "Let a and b be integers. If a and b are even, then a+b is even" such that the contrapositive is just "if a+b is not even, then either a is not even or b is not even," and not take the hypothesis about a and b being integers and shoehorn that into the statement, to arrive at something like "if a + b is not even, then not (a is even and b is even) or not(a and b are integers)."

maybe something similar is going on here. as you've stated the theorem in English, it's "about" continuous functions, and it feels weird to turn it into something that might also be "about" functions that are not continuous. even though logically it is also about that.

all of those expository choices seem to affect mainly which pieces of your setting get explicit expression in symbols and which don't, i.e., they are about symbolic formalism and not what the meaning is. and they are genuine choices, not just the output of a mapping that takes in English and spits out The One True Formalism.

Xander Henderson

21:35

@psie I wouldn't phrase it that way. The theorem really should be read "If $f$ is continuous and the image of $X$ is connected, then $X$ is connected."

leslie townes

which i guess is a long way of restating my first comment on this :)

Xander Henderson

The contrapositive is then "If $Y$ is the image of a function $f$ and $Y$ is disconnected, then either $f$ is not continuous, or $X$ is not connected."

leslie townes

this is your brain on symbols math.stackexchange.com/questions/4884590/…

psie

ok, that's how it is stated in Rudin's as well.

leslie townes

oh, they edited their post to make it less an illustration of what i was talking about. the title was "Can someone explain this to me? I for the life of me cannot understand this" and the post was "if X, then X" where X was the exact same symbolic expression.

of course it was a typo and they fixed it, but it illustrates my point better in its original state.

Simd

22:04

I have two arrays A and B of length 6 each. I want to multiply A[i:6] by B[i] for each i (indexing from zero). Is this operation called something?

Xander Henderson

22:18

I don't understand. An array is usually an $m\times n$ arrangement of numbers. Does this mean that $A$ and $B$ are both simply lists of $6$ numbers?

And what is $A[i:6]$?

I suspect that you are a programmer of some sort---we are mathematicians, not programmers. You need to speak math, not C (or python, or FORTRAN, or whatever language that is).

Jakobian

We need to speak goblin

leslie townes

simd: this may or may not be relevant but if A is a 6x1 "column vector" and B is a 6x1 "column vector" with entries (b_1,b_2,b_3,b_4,b_5,b_6), then the columns of the 6x6 matrix A B^T are, in order, A scaled by b_1, then A scaled by b_2, then A scaled by b_3, and so on. i.e. the columns of this 6x6 matrix product encode the results of scaling the vector A by each one of the entries of B, in order

i'm now silently wondering about "length 6" and "indexing from 0" and the whether the 6 in "i:6" (whatever that is) should be a 5

user10478

Does Von Neumann's Minimax Theorem apply to zero-sum games only if they consist of a single move made by each player, or also in zero-sum games with several alternating moves? If the latter, then why does it not imply that in Chess neither white nor black has an edge? If the former, then why does it appear when players use multiplicative weights to play zero-sum games over a sequence of days (Example: youtube.com/watch?v=-_pIUD5Jzl0)?

Xander Henderson

@leslietownes That, too.

Semiclassical

22:55

@Jakobian the Cantor distribution is one of those things which still weirds me out

formally it makes sense

Xander Henderson

@Semiclassical How does it weird you out? It is a perfectly nice distribution...

Semiclassical

being neither continuous nor discrete

it does mostly just come down to CDFs being more fundamental than PDFs

Semiclassical

23:11

i do like that it's not hard to sample from the Cantor distribution. (or, at least, it's not hard to sample up to some precision, but that goes for the uniform distribution too)

Jakobian

@leslietownes help? I don't have any inspiration to do math. I'm just recycling topics someone already been through.

Its all bland, boring, overdone

Xander Henderson

@Jakobian Switch to cocaine?

...

Oh! You said MATH... not METH.

nvm

Semiclassical

@XanderHenderson hey, if it was good enough for Erdos... (technically that was amphetamines not methamphetamines but joke > facts)

Jakobian

23:29

Math is like a mass delusion. There is literally no point to it. Its not even made for anything. Sure maybe that was the purpose at the beginning. Right now? Bizzarre things NO ONE needs

@XanderHenderson could as well be doing that

Thorgott

that's the cool thing about it

meth sounds fun too, but sadly it's not good for your health

leslie townes

@Jakobian alwayshasbeen.jpg

Xander Henderson

@Jakobian How do you feel about ballet? opera? hip-hop? swing? oil-on-canvas? literature?

Lots of people have argued that these things are pointless and serve no purpose, yet they seem, as a group, to be rather important to human flourishing.

Jakobian

23:52

Because there seems to be no other point to stay alive than enjoy some of those form of entertainment, I'm guessing

leslie townes

jakobian you should set a goofy meta mathematical goal, a kind of stunt, and try to achieve it. e.g. "publish a paper that contains some number worked out to 20 decimal places in it" in a non pay to play journal, or "publish a paper in a journal that has the funniest [to you] name that you can manage, without paying for the privilege"

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