Mathematics - 2021-05-24 (page 3 of 3)

let's turn this email thing into a movement. nobody needs to look at or listen to anything. just send me text.

geocalc33

trust between parties will always be cultivated more effectively through in person interaction though

@copper.hat Ahem. Why is $(5)$ not correct? There are two fractional parts $\{j_1\pi\}\lt\{j_2\pi\}$ that are in the same $I_k$, which means that $0\lt{(j_2-j_1)\pi}\lt\frac1N$

As I state, this does not mean that $j_1\lt j_2$. It only means their fractional parts are ordered thus, and in the same $I_k$

5:14 PM

@robjohn My apologies. It is correct. I took it out of contex. Sorry, trying to balance a call and response.

Let me find and delete

@Koro What is true is that addition/subtraction holds modulo one. $\{x\}-\{y\} = \{x-y\} \mod 1$. So, if $\{x\} > \{y\}$ then $\{x-y\} = \{x\}-\{y\}$, otherwise $\{x-y\} = \{x\}-\{y\}+1$.

8 pi has a smaller fractional part than pi does.

Crap I cannot delete it.

@leslietownes $-7\pi$ has even smaller :-)

@robjohn Koro had made a stronger statement which I refuted, my (mis)understanding at the time was that it was a statement you had made and (5) was the conclusion. Sorry about that.

if you say so. i don't acknowledge negative numbers.

geocalc33

5:17 PM

How do you turn the positive real line into a complex manifold?

No problem. The density is not all that difficult to see, but not all that easy to write up.

Yeah, many straightforward ideas become lost in the transcription.

Not a comment on yours, which is good.

I think there is a clearer way, and I will modify that answer if I find it

One point is that sometimes you need to look at things like $\{-7\pi\}$, whose existence leslie denies.

Thorgott

@geocalc33 you dont

That needs $\{-x\}=1-\{x\}$ for $x\not\in\mathbb{Z}$

5:23 PM

Please explain why " this does not mean that and only implies that..."

:'(

i'm going to start a separate chat for people who deny the existence of "negative" "numbers."

we'll call it The REAL Math.SE chat and finance militant operations against people who contradict our principles.

Thorgott

but do you believe in "primes"?

@Koro $\{j_1\pi\}\lt\{j_2\pi\}$ does not mean that $j_1\lt j_2$

I agree with you Rob on this. But why the statement that $\{j_1\pi-j_2\pi\}$. I mean how could you club the fractional parts into one.

i believe in 2, 3, and 5. after that, i have to evaluate the cases according to their individual circumstances.

5:25 PM

$\{j_1\pi\}\lt\{j_2\pi\}$ and $\{j_1\pi\},\{j_2\pi\}\in I_k$ mean that $0\lt\{(j_2-j_1)\pi\}\lt\frac1N$

They need to be close

@robjohn It seems true . I have tried many examples. But then how does one prove that? I tried a lot yesterday :'(

I took

@koro if $\{x\} \le \{y\}$ then $\{y-x\} = \{y\} - \{x\}$.

$(j_2-j_1)\pi=floor (j_2-j_1)\pi +\{j_2\pi -j_1\pi\}$ and then tried to proceed to get some equality of the two things but failed miserably

@copper.hat Ahh, so that's the result

This is adding & subtraction modulo one.

Let me try to prove this one.

5:30 PM

@Koro Are you familiar/comfortable with quotient spaces?

$\{x-y\}=1-\{y-x\}$

@copper.hat You mean quotient groups? If yes, then yes.

@Koro only if $x-y\not\in\mathbb{Z}$

yes 😅. Forgot to put that. Sorry about that.

Let $x=y$.

Would you like to see an argument which doesn't require all this @robjohn?

5:33 PM

@Koro Sure, but make sure it doesn't wave hands.

@copper.hat Ok.

@robjohn It won't if agree upon that set $S=\{m+n\pi:m,n \in \mathbb Z\}$ has limit point 0

That is the first part of the answer I gave, so okay.

Ok. So we start from where trouble starts (5)

We first observe that $S$ is a subgroup of R under addition.

wait... $(5)$ is the conclusion that gives the limit point

Let's take any open interval $(p,q)$

We'll show that it contains an element from S

to prove density

5:39 PM

okay I'm thinking of $(p,q)=\left(\frac{157}{1000},\frac{158}{1000}\right)$

We choose an $\epsilon \lt (q-p)$. We know that there is one $s\in (0,\epsilon )$. We know that there is maximal integer $n$ such that $ns \le p$ so $(n+1)s\gt p$.

That is just what my argument says

I claim that: $(n+1)s\lt q$. Suppose not. It means that $(n+1)s\ge q$ but then $(n+1)s-ns\ge q-p\implies s\ge q-p$, which is a contradiction. So our claim is true. We are done!

@robjohn I'm sure it does but unfortunately, I didn't understand those fractional part things. :'(

Copper was suggesting me a proof above: Let x=y and then ...? copper

I'm a bit lost as to what is happening now.

I think that he was talking about the $\{x-y\}=1-\{y-x\}$

5:45 PM

Copper: what happened was that: I stated a proof which didn't require fractional parts too much.

if you let $x=y$, you get $0=1$

it's a theorem, but there are exceptions and counterexamples.

So we may get back to the point of difficulty that is fractional parts thing

I have a question, So if I knew something will take say, 30 days but I won't do it on weekends how would I use a mathematical equation to remove two days for every 7 there are?

If $\{x \} \le \{y\}$ then $\{y-x\} = \{y\} - \{x\}$.

5:46 PM

@CiurkitboyN it will take $30\cdot\frac75$ days then

I understand from Rob and you @copper that: {x-y}={x}-{y}, if {x}>= {y}

That is what you need for Robjohn's proof.

Yes. I requested a proof of this only.

@robjohn What?

You were suggesting the proof starting with let x=y ,

Then what

5:47 PM

No, I was giving a counterexample to a statement you made.

@CiurkitboyN If something will take $30$ days and you only work $5$ of the $7$ days, it will end up taking $30\cdot\frac75$ days.

It might be better if you focused on the proof.

why are you putting those dollar signs?

what is cdot?

mathjax

ah, you are new here...

5:48 PM

mathjax?

Ahh. Okay. copper. I forgot mentioning $x-y\notin \mathbb Z$ when I mentioned that.

@CiurkitboyN MathJax allows LaTeX to be rendered here

what is ltex

latex?

ah. hang on...

it will take 30 x 7/5 days if you don't work weekends

ok

Q: MathJax basic tutorial and quick reference

5:51 PM

3445

(Deutsch: MathJax: LaTeX Basic Tutorial und Referenz) To see how any formula was written in any question or answer, including this one, right-click on the expression and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next point...

ChatJax installation page

so it would be x*7/5=y
x being days I do the task or whatever and y being days it will be done?

yes. That is because you are working only 5/7 of the time

ok thanks

ok got it working thanks for the help

here is what I used it for if you care at all

planning what needs to be done?

It's actually for my school

Oh I see now that I have mathjax I can see what yall were actually typeing

or whatever yall called it

6:17 PM

@CiurkitboyN It comes in very handy in a math chat room

30 mins ago, by robjohn

@CiurkitboyN If something will take $30$ days and you only work $5$ of the $7$ days, it will end up taking $30\cdot\frac75$ days.

now you can hopefully read that without dollar signs

man-hour rate problems are funny. i remember in middle school being set some problem, and i thought. there's no way that putting an unlimited number of people on baking this cake will get it out in under the time it takes to bake a cake. baking doesn't work like that. ten thousand degrees for X seconds is not 10000/Y degrees for YX seconds. sometimes you just gotta watch the oven.

all ten thousand of you just have to sit down and wait for it.

@leslietownes you disappoint a lot of people that way

also, has nobody thought of the fire inspector and building codes. you definitely do not want to see more than a dozen people in this kitchen.

are they wearing masks?

they are NOT 6 feet apart

"what would the fire marshal say?" is a necessary step in any applied math problem.

6:40 PM

@robjohn In the proof that Koro was looking at, after (5) its says "that there are integers $p,q>0$", did that mean an integer $p$ and an integer $q>0$?

Jairo A. del Rio

Hi everyone. This may be a weird question, but does anybody know how to construct the tangent circles (those next to the diameter on the left) in this figure? I ask for an exact construction, in case it helps.

@robjohn Actually, you could note that as long as $k \{ (j_2-j_1) \pi \} < 1$ (for $k \in \mathbb{N}$, of course) you have $\{ k (j_2-j_1) \pi \} = k \{ (j_2-j_1) \pi \}$ which sidesteps the issue and shows $<{1 \over N}$ 'densenesss' without any more work.

@copper.hat yes. Since I am looking at $|p-q\pi|$.

@robjohn I often write $p,q>0$ as meaning both which I now realise is ambiguous.

@copper.hat that might work. Let me look.

@copper.hat but we don't know that $j_2-j_1\gt0$, so something might need extra care

6:55 PM

he wrote realise with an s. he's a spy.

@robjohn You don't need that here, you are just showing that $\{ \mathbb{Z} \pi \}$ is dense.

Finally I have understood it. Thanks a lot @robjohn and copper

@copper.hat at some point, we may need to use $\{-x\}=1-\{x\}$ to get $p,q\gt0$

but let me see if that simplifies things

@robjohn Are you looking for both $p,q $ to be positive?

@copper.hat for a positive irrational $\pi$, they will be

to make $|p-q\pi|$ small

7:01 PM

Of course.

Either way, you can skip that whole part.

Rob: for $\{j_1\pi\}\lt \{j_2\pi\}$, we'll have $1-\frac 1N \lt \{(j_2-j_1)\pi\}\lt 1$

Typo: it should be "for $\{j_2\pi\}\lt \{j_1\pi\}$..."

7:16 PM

okay I think I'm going nuts. My calculus textbook defines the sine and cosine as the only pair of functions $\mathbb{R}\to\mathbb{R}$ which satisfy the conditions:
1. $\sin(0)=0$
2. $\cos(0)=1$
3. $\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)$
4. $\cos(a-b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
5. There exists an $r>0$ so that $0<x<r$ implies $0<\sin(x)<x<\sin(x)\cos(x)$

but don't the pair of functions $\sin(2x)$, $\cos(2x)$ also satisfy these?

which would mean they're underdefined

seems like it

do 3 and 4 hold with the doubling? mind where you double and where you don't.

they do. I verified it

oh what about 5

it also does but the $r$ is smaller

0.9something

7:23 PM

wait which is the r for 5 in the usual one?

often in characterizing trig functions by functional equations there is something on where "pi" is. i don't see that in your list of 1-5.

Hello, I have question. How can we check if the topological spaces $(\mathbb{R},\tau_d)$ and $(\mathbb{R},\tau_{eucl})$ - that is discrete and euclidean topology - are homeomorphic?

just thinking out loud.

@Onir $\pi$

sorry

$\pi/2$

then the tangent changes sign

7:24 PM

so the value that works is $\pi/2$?

larry, connectedness might be one way of differentiating the spaces.

@Koro In @robjohn's proof, note that as long as $k \{ (j_2-j_1) \pi \} < 1$ (for $k \in \mathbb{N}$, of course), you have $\{ k (j_2-j_1) \pi \} = k \{ (j_2-j_1) \pi \}$. This shows that every point in $[0,1)$ is less than ${ 1 \over N}$ away from one of the points $\{ (j_2-j_1) \pi \} , \{ 2(j_2-j_1) \pi \}, ...$. Since $N$ was arbitrary this shows that $\{\mathbb{Z}\pi \}$ is dense in $[0,1)$.

and for the version with $2x$ which one do you propose?

@leslietownes he defines it a little bit later with the root of the cosine

R is a connected topological space if it is given the euclidean topology. it is not a connected topological space if it is given the discrete topology. connectedness is preserved by homeomorphism.

7:26 PM

Thanks @leslietownes

@Onir whatever the smallest positive solution of $\sin(2x)=x$ is

it may also be checkable directly. heuristically it is easy to be continuous when mapping from a fine topological space to a coarser one, but hard to be continuous the other way around. so you might look at what happens if you map euclidean R to discrete R

Good point

but what about $tan(x)/x $?

2x beats tanx in (0,\epsilon) I think.

I think sin(ax) and cos(ax) cant work because of 5 unless a=1

if a<1 you use the first part of the inequality and if a>1 the other

i think i agree with you.

Vrouvrou

hello

7:31 PM

good morning

Vrouvrou

why spectral values = Eigenvalues in finite dimensional vectorial space ?

@copper.hat @Koro: I have simplified my proof to not force $p,q\gt0$ and replaced $\varepsilon$ by $p+q\pi$ everywhere. See if that looks easier to understand

Because a matrix is invertible if and only if its kernel has dimension 0

kernel*

because finite dimension vector spaces can't have subspaces of the same dimension

@robjohn Looks good to me. I wouldn't even bother with the upper bound on $j$ :-). Ohh, I guess you have "in $M$".

@copper.hat then it is no longer restricted to $I_k$

7:37 PM

Yeah, catching up.

I still havent been suspended for eoqs

What a great hooman being I must be

I could put something about that covers all of the $I_k$ because one more $j$ would put the number over $1$

I don't understand 5) to 6). I'm sorry 5) to 6) is not obvious to me

@Koro set $q=j_2-j_1$

Do you see that $\{x\}=p+x$ for some $p\in\mathbb{Z}$?

@JairoA.delRio You might try messing around with inverting in some circles. i probably wont do this as this might set me off on a few days of circle drawing as i have no self control, but i might start by inverting the top inscribed circle and tracking the biggest circle, the diameter that the circle you want to construct it tangent to and incidence relations

7:44 PM

Ah I see so 6) is basically re-writing 5) differently

yes re-writing it differently in another way

Rob: I think that somewhere you should also mention that $q\ne 0$ and the reason is that, just before 6) you have mentioned a condition which is violated when q=0

we know that q is >0 as it is j2-j1

such that |q|$\le$N...

May be I'm wrong.

@Koro $(6)$ forces $q\ne0$. If $q=0$, we have an integer between $0$ and $\frac1N$

Ok :)

besides the fact that $j_2-j_1\ne0$

7:51 PM

Yeah that I observed already j2 not equal to j1

I didn't want to break up the flow by mentioning that, since the whole idea of the rewrite was to attempt to make things easier to understand

Sir I also want to know how you came up with the idea of 7b from 7a. I understand that in 7a, we used 5) to get 7b)

@robjohn Alright :)

The whole idea was to show that $\{\mathbb{Z}\pi\}$ was dense in $[0,1]$, so since $0\lt p+q\pi\lt\frac1N$, we know that $p+q\pi=\{q\pi\}$

because it is in $[0,1)$ which is where $\{x\}$ lives

Wow!! Phenomenal, Flamboyant, Amazing, magical, extraordinary, wonderful

Thanks a lot for this @robjohn You are fantastic :)

Indeed, $\{ \mathbb{N} \pi \}$ is dense in $[0,1]$. Another day.

8:00 PM

I'm trying to answer all of my old questions to get them out of the unanswered pile

is this frowned upon?

@copper.hat No way

Just saying intuitively

:-). It is true.

:)

@copper.hat that follows from $\{-x\}=1-\{x\}$ for $x\not\in\mathbb{Z}$

but that was removed for simplification

@Koro That was the idea of the first draft

the only thing missing is that we don't know the sign of $q$

Since $I_k=1-I_{N-k}$ all we need is $\{-x\}=1-\{x\}$ for $x\not\in\mathbb{Z}$

Is it possible for a continuous bijection that isn't a homeomorphism to exist between homeomorphic toplogy space?

8:11 PM

a continuous bijection that isn't a homeomorphism?

@robjohn :)

yes, i guess it means not open/closed

yes it is

ah, not bicontinuous

Take the discrete topology on an infinite set X

Then take an equivalence class on X such that each equivalence set has finite size

consider the topology in which the open sets are the ones that are made of unions of equivalence sets.

8:16 PM

okay thanks

And now let your function be the identity between the discrete topology and the new one (both on the infinite set X).

Thanks!

The second topology is also a discrete topology

only our "points" are artificial

I don't feel like anything I said made a lot of sense

but you're welcome

8:35 PM

In fact you just need that the set of equivalence sets has the same cardinality as X, asking each one has finite size is a way to do it.

I think mapping R to R via identity does it also. for instance, if the source R has discrete top and the target euclidean, it is a continuous bijection. if the source is euclidean and the target discrete, it is not continuous.

but we need that both topologies are homeomorphic

ahh im not reading lol ignore that

it is all gucci bass

11:02 PM

@Onir what is the homeomorphism between those two spaces?

11:18 PM

there is none

in one of them there are points that aren't open

:/

haha

Let's find an example that actually works

a classic example is $[0,1) \to \{z \in \mathbb{C}: |z| = 1\}$ given by $t \mapsto e^{2\pi i t}$

oh wait there is an easy example.

Take the right order topology

or use goofy topologies. remember that a continuous bijection from a compact space to a hausdorff space will have a continuous inverse.

and consider a right shift

right order topology on $\mathbb R$

oh nevermind

it's not easy

Q: Are continuous self-bijections of connected spaces homeomorphisms?

11:23 PM

scrolled up and saw the issue. i can't think of a simple example.

i do like the idea of using an infinite product and playing with infinities.

25

I hope this doesn't turn out to be a silly question. There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms. But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnec...

general-topology

A: Are continuous self-bijections of connected spaces homeomorphisms?

If we only make the open sets be $\mathbb R$ and $\varnothing$ and the sets $[a,\infty)$ with $a\geq 0$ it's a topology?

I think with that topology in the domain and the codomain the function $x=x-1$ is continuous

well I took the risk and added my ramblings as a solution

0

Consider the topology on $\mathbb R$ in which the open sets are $\varnothing,\mathbb R$ and $(a,\infty)$ such that $a\geq 0$. Is this a topology? $\varnothing$ and $\mathbb R$ are open. If we take $(a,\infty)$ and $(b,\infty)$ without loss of generality $a<b$ and so $(a,\infty) \cap (b,\infty) = ...

I may have made a mistake tho

Im predicting I did something wrong

as the accepted answer has some circles and lines

which is more advanced than what I did

Oh wait my solution is just an off-brand version of what qiaochu did

well, I guess it must be correct at least

vitamin d

11:41 PM

@Onir I posted a message in our chat a few days ago. I suppose you haven't seen it yet. An answer would be extremely kind of you. I'd really like to know if the author was not telling the truth.