Mathematics - 2024-05-15

EE18

00:43

Am I going crazy or does this solution not really work?

5

Q: Proving $d_1=\frac{d(x,y)}{1+d(x,y)}$ is a metric equivalent with $d$, a given metric

Given a set $X$, define a metric by $d_1=\frac{d(x,y)}{1+d(x,y)}$ for all $x,y\in X$. I want to show that $d_1$ is a metric equivalent with $d$, a given metric. Here is my attempt so far: Given a set $U\subseteq X$ which is open in the $d$ metric and $x\in U$, there exists an $\varepsilon>0$ such...

I am using a slightly different strategy but I have no clue what's going on in this answer

Jakobian

@EE18 question or the answer

EE18

Answer

Will send my answer in two seconds, hopefully before i take off on this flight lol...

Jakobian

I don't think a question from 2016 needs another answer

unless perhaps the answer is wrong but let me see

EE18

No lol I'm not going to answer, I'm saying I'm working the same problem right now

and i feel like it's wrong

But also it has 5 upvotes so i doubt it lol

Jakobian

are you objecting that the OP proved $B_{d}(x, r)\subseteq B_{d_1}(x, r)$ part

EE18

00:54

Yes, no subscript 2 on $d$ though

Jakobian

so are you objecting that $d_1\leq d$

@EE18

EE18

No, the other direction is the challenging one right

Anyway I am taking off, apologies i have to run!

Jakobian

I don't care about your proof right now

I care about if you consider the answer true and if not then why

well its subtle but the answer on that question is indeed wrong

I left a comment on this

Jakobian

01:37

I also left a correct answer

with few additional side comments about the metric $d_1$ and that this actually shows more than just equivalence of metrics

EE18

07:05

Sorry, just landed

I agree with your comment and answer. I think in my solution I need to pay a little more attention to when we can rewrite d in terms of delta, with the d=0 case the one of concern

Tsundoku

08:07

@Derivative I'm not familiar with libgen. I am aware of several sites with free or creative commons maths books, e.g. openstax.org/subjects/math, but I don't know if libgen is similar.

S.M.T

09:27

Hi

I am having difficulty in understanding langrage interpolation theorem.

Like for starting point. Online it says with the first step. We have to multiply linear terms. (x-x1)(x-x2)(x-x3) and equate this to 0. Why do we do that ?

Soumik Mukherjee

09:45

@Tsundoku libgen is the Goat 🐐 of all those free sites

S.M.T

And a few more steps. Could someone help

Dannyu NDos

10:44

I've just noticed that linear logic is not paraconsistent.

Proof: From nothing, infer $\vdash 1$. Then dually, $\bot \vdash$.

Pizza

11:25

hi

Soumik Mukherjee

11:36

Hi

Pizza

@SoumikMukherjee whats up

user 70432

12:12

⬆️🆙🔝

nitsua60

@SoumikMukherjee Ooh, that's a nice one--thanks!

@XanderHenderson I do that, too, in my calc course, since I know they've seen PMI in precalc.

Soumik Mukherjee

@Pizza not much, what about you?

Jakobian

12:43

@EE18 your proof technically correct but there are steps which were done for no reason

Pizza

@SoumikMukherjee I'm quite well thanks

Jakobian

@S.M.T can you be more clear about the proof you are looking at

Pizza

12:58

@Jakobian can i you a question about mathexchange

if I publish a post on the homepage, after a few people publish their new posts, then mine disappears from the homepage, how do I make sure that it returns to the homepage?

Thorgott

you dont

user 70432

it is not polite to want your post always on the homepage

Pizza

yes but not having it on the homepage, can someone somehow see it?

user 70432

be patient

people search for good questions to answer

Pizza

I mean how do they find it, maybe through tags, I don't know much about this

user 70432

13:07

yes, tag it appropriately

Xander Henderson

@nitsua60 I teach induction in my precalculus classes, but the book we use here does not, so I don't think that my colleagues teach it, and I am sure that the high schools don't. So I do a short unit on induction in both precalc and calculus.

Pizza

@user70432 ok!

user 70432

Induction is definitely a topic for Algebra 2.

Xander Henderson

@user70432 "Algebra 2" is not a mathematical topic. It is the name of a class.

Pizza

$D$ is the domain contained in the first quadrant and delimited by the $x$ axis, the parabola $y = x^2-2$ and the bisector of the first and third quadrants.

$D = \{(x, y) \ | \ 0 \leq x \leq \sqrt{2}, \ x^2 - 2 \leq y \leq x \}$

is this correct? since I have to consider only the first quadrant, I take only $x = \sqrt{2}$

Xander Henderson

13:11

Huh... and "mathematical induction" doesn't seem to be in the Common Core. That's disappointing.

learning.ccsso.org/wp-content/uploads/2022/11/…

There is a reference to it at the bottom of page 64, in a footnote, noting that the binomial theorem can be proved via induction.

user 70432

Sad.

:(

Jakobian

13:23

I can see why someone is struggling with induction

we're doing things on intuitive level and induction is completely intuitive

so people might think, why do we even state it

and I agree, why do people even state induction, its not like people are getting introduced to basic propositional logic

not formally at least

but its like, no, induction is special

and has to be introduced formally for some reason

user 70432

The reason is clarity of thought.

Soumik Mukherjee

@Pizza no

The bound on $x$ is incorrect, also you need to add x,y>=0

Pizza

mm wait

im trying to solve this

Calculate the following double integral (directly and with the Gauss Green formula)
$$\int\int_D xy \ dx \ dy$$ Where $D$ is the domain contained in the first quadrant and delimited by the $x$ axis, the parabola $y = x^2-2$ and the bisector of the first and third quadrant.

is it necessary to add x,y >= 0?

$\int_{0}^{\sqrt{2}} \int_{x^2 - 2}^{x} xy \ \, dy \, dx$

so it's not like that?

@SoumikMukherjee

Xander Henderson

13:59

@Pizza You should use \iint for double integrals: \iint_D xy\,\mathrm{d}x\,\mathrm{dy} renders as $$\iint_D xy\,\mathrm{d}x\,\mathrm{dy}.$$

Pizza

oh ok thanks!

but do I have to integrate with $x$ first? it is given as $\mathrm{dx}$ $\mathrm{dy}$ so for this reason I have to do $x$ first?

or can I also integrate first for $y$?

Xander Henderson

Well, if you want to compute it directly, you can start with either variable. That is Fubini-Toneli, n'est-ce pas?

Boy, "the bisector of the first and third quadrant" is awkward phrasing. Why not just give the equation for the line, i.e. $y=x$?

Pizza

@XanderHenderson I honestly copied and pasted it, it was written like this

i think i fixed my domain wait

Xander Henderson

@Pizza I didn't say that you had made a mistake. Only that this is a very awkward way of phrasing things (in my opinion).

Pizza

ah yes, no I thought you were reading the sentences where I said if the domain was correct

Xander Henderson

14:10

I haven't looked at your math at all.

Pizza

okok

Pizza

14:22

i think i fixed it

so

the parabola meets the line at point (2,2) by the way not sqrt2

I had done $x^2-2=0$ before, but it is $x^2-2=x$ because that will be where the parabola meets the line , since that is basically the corner of the domain
other corner being the origin , the line is y = x and the parabola is $y = x^2 -2$ so $x^2 -2 = x$

right?

so for y first D is

$$D={(x,y)|y\le x \le \sqrt{y+2}, 0\le y \le 2}$$

for x first

$$x^2-2\le y\le x,0\le x\le 2$$

so it is $\int_{0}^{2} \int_{x^2 - 2}^{x} xy \ \mathrm{d}x\,\mathrm{dy}$

i can do it in y because the integral will be easier i think

Pizza

15:24

@SoumikMukherjee Thanks anyway, yes the x part was wrong

you were right

I calculated the double integral directly and got $\frac{2}{3}$, now using gauss green I should get the same result right?

Xander Henderson

15:40

@Pizza However you compute the integral, you should get the same thing.

Pizza

@XanderHenderson do I need to define the vector field?

I'm having a bit of trouble honestly

Xander Henderson

@Pizza I haven't taught Gauss/Green/Stokes since 2017. It is not at all top of mind for me.

Pizza

okay don't worry, I'll see better later

Jakobian

17:16

When I see n'est-ce pas I just blank out

Definitely not a word in my dictionary

Well, word. Sentence or whatever. I don't know how to call it

Tanner Swett

17:30

Suppose we have a category with all finite products and coproducts. Given an object A, can we create the selection function f : (1 + 1) * A * A -> A, satisfying f(i1, g, h) = g and f(i2, g, h) = h?

robjohn

@Pizza the inner integral is with respect to $x$ and the limits are in terms of $x$. This can be done, but it makes the result a function of the $x$ in the limits. Perhaps you intended $\mathrm{d}y\,\mathrm{d}x$?

Pizza

@robjohn ah yes sorry

i write it wrong

Tanner Swett

If the category were also cartesian closed, it would be easy: take the two projections A * A -> A and curry them into 1 -> hom(A * A, A), then take the coproduct of those to obtain a single morphism 1 + 1 -> hom(A * A, A), then uncurry.

Thorgott

@TannerSwett it depends, e.g. the answer is yes if the category is distributive and no if it is pointed (and non-trivial)

cartesian closed categories are distributive, so that generalizes your observation

Tanner Swett

Good to know! Now I'm trying to think if I can name a particular counterexample, like the category of pointed sets or the category of monoids.

Thorgott

17:40

those work, as does any abelian category

Tanner Swett

Thanks!

The reason I ask is that I'm trying to figure out the smallest amount of structure I need to build functions involving things like booleans, natural numbers, and tuples. I was wondering if, for booleans, it's sufficient to have finite products and two arrows 1 -> 2 with a universal property stating that a pair of arrows 1 -> A can be factored into a single arrow 2 -> A.

Now I know that that's not sufficient, and I should probably instead use a universal property with a parameter, asserting that a pair of arrows B -> A can be factored into a single arrow 2 * B -> A.

Pizza

18:11

Calculate the following double integral (directly and with the Gauss Green formula)
$$\iint_D xy\,\mathrm{d}x\,\mathrm{dy}$$Where $D$ is the domain contained in the first quadrant and delimited by the $x$ axis, the parabola $y = x^2-2$ and $y=x$.

$$\text{For x:} \ \ \ \ D = (x,y)|x^2-2\le y\le x,0\le x\le 2$$ $$\text{For y:} \ \ \ \ D={(x,y)|y\le x \le \sqrt{y+2}, 0\le y \le 2}$$

One thing isn't clear to me, can I choose the domain?

the thing is, I don't have a solution, so I don't know if I'm doing it correctly

could someone help me a moment pls

Sine of the Time

did you draw the domain $D$?

Pizza

em wait i dont have enough points to share

wait

cdn.discordapp.com/attachments/903481101744484362/…;

Frieren

Hi, in the theorem: "A function $\alpha:S\to T$ is bijective iff there exists $\beta:T \to S$ such that $\alpha\beta=1_T$ and $\beta \alpha =1_S$", I am not sure if I proved one thing correctly. In the implication $\implies$, from the bijectivity hypothesis I deduced that for each $t \in T$ there exists a unique $s_t \in S$ such that $\alpha(s_t)=t$, and so I defined $\beta$ as $\beta(t):=s_t$ and so $(\alpha \beta)(t)=\alpha(s_t)=t=1_T$.

However, I am not sure how to show that $\beta \alpha=1_S$. My approach is this: let $s \in S$. Hence, $\alpha(s)\in T$ and so there exists $t\in T$ such…

Pizza

something like that

psie

Let $C$ be the Cantor set. Suppose $x\in C$, so that $x=\sum_1^\infty a_j3^{-j}$ where $a_j=0$ or $2$ for all $j$. Let $$f(x)=\sum_1^\infty b_j2^{-j}\text{ where }b_j=a_j/2.$$ Then the claim in Folland's book is that $f$ maps $C$ onto $[0,1]$ and therefor $C$ has the cardinality of the continuum. I see how the map is surjective, but it also needs to be injective, right? How can I see that it is injective?

Sine of the Time

18:19

@Pizza correct

Pizza

@SineoftheTime it's not clear to me when I go to choose whether to integrate for y or x, which domain to take

Sine of the Time

the domain "for y" doesn't look correct

@Pizza do you know Fubini's theorem?

Pizza

yes wait

$$\iint_R f(x, y) \, dx \, dy = \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx$$

or $$\iint_R f(x, y) \, dx \, dy = \int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy$$

is this

Sine of the Time

yes, so you can choose to integrate first with respect to x and then with respect to y or viceversa

here is convenient to use what you wrote in "for x"

Pizza

but can I choose it?

for $y$ i mean to solve $x^2-2=y$ , for $x$ i mean $x^2-2=x$

is correct?

Sine of the Time

18:30

@Pizza yes

Pizza

@SineoftheTime where did i go wrong

Sine of the Time

wait, it's correct maybe

psie

@psie I guess injectivity follows from the fact that we choose the unique ternary expansion of some number in $[0,1]$. So to each sequence $(a_j)$ and consequently to $(b_j)$, there is exactly one number $x\in[0,1]$, i.e. $f$ is injective.

Sine of the Time

for x it's wrong

since when $x=0$ for example, $y$ is negative

but $y\ge 0$ in $D$

Pizza

isnt for x,$x_1 = -1$ $x_2 = 2$

so i consider only 2 bc we are looking for the right-most x value

@SineoftheTime

Sine of the Time

18:44

what are $x_1$ and $x_2$?

for x is wrong since you're including negative values of y

Pizza

the solution of $x^2-2=x$

Sine of the Time

try to draw for x and you'll understand why it's wrong

Pizza

@SineoftheTime do i have to include y >= 0?

Sine of the Time

in D y is bigger than 0

you're not integrating over D

you're integrating over this: desmos.com/calculator/3zabdb3vgt

Pizza

what is this

i dont understad

Thorgott

18:52

@psie The function is not injective, nor does it need to be.

Sine of the Time

draw for x

psie

@Thorgott Are you sure that it's not injective? Which theorem/proposition are you relying on when you say it does not need to be? I think Folland establishes that there is a bijection between the Cantor set $C$ and $[0,1]$, the latter which has the cardinality of the continuum. So they have the same cardinality.

Pizza

@SineoftheTime do i have to modify this ?

Sine of the Time

use "for y" to integrate, you don't need both

Thorgott

Yes, I'm sure it's not injective. Note also Folland never says that it is.

If $A$ surjects onto $B$, then the cardinality of $B$ is $\le$ the cardinality of $A$. This is basic set theory.

Alessandro Codenotti

18:58

@Thorgott Are you assuming my axioms?

Pizza

desmos.com/calculator/cxolzrlcgg?lang=it

@SineoftheTime

you mean this?

Sine of the Time

yes

it's wrong

Thorgott

yes, and with pleasure

psie

@Thorgott right, but here we want to show the cardinality is the continuum, so an inequality is not enough?

Pizza

@SineoftheTime you mean the red part?

Sine of the Time

19:02

confront this: desmos.com/calculator/3zabdb3vgt with D

Thorgott

@psie the other inequality is obvious

also, re: injectivity: $f$ maps $0.2$ in base $3$ to $0.1$ in base $2$ and $0.022\dotsc$ in base $3$ to $0.011\dotsc$ in base $2$

the two base $3$ expansions represent different numbers, but the two base $2$ expansions represent the same number

psie

ok :) thanks (the other inequality I assume follows from $C$ being a subset of $[0,1]$)

Thorgott

yeah

Pizza

@SineoftheTime desmos.com/calculator/3zabdb3vgt?lang=it

you mean there are points out D?

Sine of the Time

yeah....

desmos.com/calculator/f67czrplck?lang=it: this is the correct link @pizza

so when you use "for x" you're including the region with $y\le 0$

Pizza

19:18

yes as you told me when $x$ = 0, then $y$ = -2

i can see it in your graph

desmos.com/calculator/igzjoojq7p?lang=it

so for y is correct here

Sine of the Time

19:34

right

Sha Vuklia

Would anyone know how to show the two inclusions at the bottom? I think it's enough to know here that $\Omega\in\operatorname{End}(V)$ is invertible and commutes with $\rho(x)$ for each $x\in\mathfrak g$ to argue that $V^+(\Omega)\subset V_{\operatorname{eff}}$ and $V^\mathfrak g\subset V^0(\Omega)$. However, I fail to show this. Any idea?

I should add that $V^+(\Omega)=\bigcap_n \Omega^n(V)$ and $V^0(\Omega)=\bigcup_n\ker\Omega^n$

Sha Vuklia

19:54

I was thinking of using that invertibility of $\Omega$ and $\Omega\rho(x)=\rho(x)\Omega$ yields $\rho(x)=\Omega\rho(x)\Omega^{-1}$

but it leads to nowhere

ohh

I see now that the second inclusion can be shown from the definition of $\Omega$ (it is the Casimir element belonging to a certain ideal of the Lie algebra)

yea, and same for the other inclusion

ok, my problem is solved then

leslie townes

you're welcome

Sha Vuklia

2

Sine of the Time

@SoumikMukherjee @AlessandroCodenotti Do you know duck chess?

Alessandro Codenotti

yes

Sine of the Time

I discovered it a couple of days ago watching Eric Rosen

Derso

20:06

@SineoftheTime I know it. It's pretty fun

Sine of the Time

so many chess players is chat:)

Pizza

20:28

@SineoftheTime i need to put in ${0 \le y }$

?

Sine of the Time

why are you trying to compute this?

Pizza

@SineoftheTime i mean for x its wrong, i need to fix it?

Sine of the Time

you don't have to do both, one sufficies

Alessandro Codenotti

@SineoftheTime Indeed

I started playing OTB recently

Sine of the Time

"for y" is correct so compute the integral using it

@AlessandroCodenotti really? what time control?

Soumik Mukherjee

20:32

@SineoftheTime yes

@AlessandroCodenotti nice

Fun fact : there is something called omega delta gambit in chess

Sine of the Time

imagine if it was called epsilon delta

Soumik Mukherjee

I was thinking the same

Alessandro Codenotti

@SineoftheTime 10+0 mostly, but just informal games so far

I hadn't played on a real board in years

Sine of the Time

me too

7years

Soumik Mukherjee

Do you have a fide rating?

Alessandro Codenotti

20:42

Nope

Sine of the Time

yes, around 1930 with the "correction"

Soumik Mukherjee

Correction?

Sine of the Time

yeas basically who was under 2000 elo had a correction

Soumik Mukherjee

Okay

Sine of the Time

now that I'm studying differential geometry, @ted left the chat :'(

Pizza

20:53

yes, I also noticed that he is no longer in the chat

...

Bml