Mathematics - 2017-10-24 (page 3 of 5)

2:03 PM

@user193319 am sorry but how is the first set open?

(-infty,x) is not open in lower limit topology i suppose.

ugh, there is no hope to find an explicit well ordering of the reals. This binary tree, no matter how many partitions I do, just multiply itself

Anjani Gupta

Can you elaborate what u r trying to do?

Secret

Trying to do something that is probably impossible

-> Constructing an explicit well ordering of the reals

Anjani Gupta

hahaaha!!

Secret

I wish there is a proof for or against it, but it seems axiom of choice or its negations don't admit constructive proofs

and it is not clear which model of ZF I need to pick to get reals to give an explicit well ordering

Anjani Gupta

2:13 PM

@secret what are the open sets of lower limit topology?

Secret

lower limit are something of the form: $(-\infty,a)$, right?

then any open interval will be an open set $(a,b)$ since you can obtain these from the union of $(-\infty,a)$ and $(-\infty,b)$

Anjani Gupta

my memory says the are of this basis [a,infty)

sorry [a,b)

Secret

The open sets should be $(-\infty,b)$ (via countable unions of a decreasing sequence of a s) and $[a,\infty)$ (again, via countable unions) and disjoint unions of these

Anjani Gupta

yeah right!

To answer the user193319 question, don'y we need A to be open?

since we need intersections with A to be open.

Secret

Vitali set

In mathematics, a Vitali set is an elementary example of a set of real numbers that is not Lebesgue measurable, found by Giuseppe Vitali. The Vitali theorem is the existence theorem that there are such sets. There are uncountably many Vitali sets, and their existence depends on the axiom of choice. Assuming the existence of an inaccessible cardinal, Solovay constructed a model of Zermelo–Fraenkel set theory (ZF) without the axiom of choice, where all sets of real numbers are Lebesgue measurable. == Measurable sets == Certain sets have a definite 'length' or 'mass'. For instance, the interval [0...

I cannot believe it, it appears in the past I have been constructing vitali sets all the time without realising

user193319

2:28 PM

@AnjaniGupta No. We don't need $A$ open, only that those sets in the union are open in $A$.

Secret

$q+r,(q,r)\in \Bbb{Q}\times \Bbb{R}$

Anjani Gupta

@user193319 then how do u claim that the component of the union you mentioned are open in the bigger space?

congrats @Secret

Secret

nah, that's just a vtiali set, not a well ordering of the reals

user193319

@AnjaniGupta $(-\infty, x)$, $[x,y)$, and $[y,\infty)$ are all open in $\Bbb{R}_\ell$; so by definition their intersection with $A$ is open in $A$.

Anjani Gupta

Sorry my brain is stuck, but A need to be open for the intersection to be open

Simply Beautiful Art

2:35 PM

@AnjaniGupta Brain is stuck? Try surgery. :-)

Anjani Gupta

hahaa joining surgery chatroom :P

Secret

Uh, the reasoning on the non measurability of the vitali set is weird:

user193319

@AnjaniGupta I think you confusing a topology and subspace topology. I am not claiming that those intersections are open in $\Bbb{R}_\ell$; I am only claiming they are open in $A$, a claim which follows by definition.

Secret

Sure I can see why it is contained in said two intervals, but how can you prove that subintervals exists/does not exist in the vitali set

Anjani Gupta

ok got it @user193319 thanks for the patience.

user193319

2:37 PM

@AnjaniGupta Sure. No problem!

Anjani Gupta

@user193319 in general, connected sets and connected subspaces are different things.

Secret

Because if there are no intervals in the vitali set, then its lesbegue measure will always be zero, and then saying it is sandwiched between lesbegue measure of the intervals[0,1] and [1,-2] will be an invalid result?

Now if the vitali set does not contain intervals similar to the cantor set, and since all its translates by reals is disjoint, then even if one of its translate is located in [0,1] it will never give a lesbegue measure of 1

I think I need to read more carefully whether vitali sets contains intervals...

user193319

@AnjaniGupta Are you sure? I don't think it makes sense to talk of a connected set without reference to a topology.

From my understanding, a disconnected subspace is always separated by sets that are open in the subspace.

Anjani Gupta

@user193319 I was going through this that generated the doubt math.stackexchange.com/questions/525830/…

i understand that lack of topology will be totally insane but its trying to say something in rather different setting

Secret

https://math.stackexchange.com/questions/137949/the-construction-of-a-vitali-set

hmmm... I already forgot, does lesbegue measure make sense under an uncountable union?

Anjani Gupta

2:46 PM

no @secret

user193319

@AnjaniGupta Hmm...I am not sure now...Perhaps Alessandro could help.

Unfortunately it doesn't look like he is on.

Balarka Sen

Hi @MikeMiller

Anjani Gupta

@Secret uncountable union isn't there in the definition of measure space.

Secret

ok

Lozansky

The change in internal energy $\Delta U$ must be zero for an isothermal process, no?

Balarka Sen

2:53 PM

Yes.

Secret

One random thought I had right now is that the lesbegue measure of the union of n vitali sets (assuming each of them don't contain any intervals (which I will try to figure out how to prove or disprove later)) can be zero or any number all the way up to infinity depending on whether the union contains intervals and how many.

Anjani Gupta

vitali set is non measurable. you claim their union is?

Secret

I am reading that non measurablility proof and wondering how to make sense of it: Specifically, have we ruled out the presence of intervals in a vitali set (because each interval (a,b) will have lesbegue measure a-b)

I knew that the union of all vitali set will give [0,1] since by definition it partitions that interval, but what is not clear to me is whether each disjoint vitali set contains intervals

Because if a vitali set contains only singletons similar to a cantor set, then its lesbegue measure has to be zero

Anjani Gupta

yeah it makes sense. intervals must not occur

see this proofwiki.org/wiki/Vitali_Theorem

Hercules Apergis

Hello, can I ask a question abut principal components?

Secret

3:06 PM

I have issues with this line. Let me give two examples to illustrate why:

Simply Beautiful Art

Hehe, thanks for starting the question about hyperoperations and ordinals @Secret

Anjani Gupta

@HerculesApergis sorry i dont know about it. Maybe a brief definition you would give.

Secret

Consider the sets $A=\{n\in \Bbb{N}, r \in [0,1)|n+r\}$ and $B=\{n,m\in \Bbb{N}, n+\frac{1}{m}\}$

Let each representative be $A_r$ and $B_m$ for fixed $r$ and $m$

Clearly, both are countable hence the Lesbegue measure is zero

It is also clear that all the $B_m$ s are disjoint hence their union is also a countable set, thus the lesbegue measure is also zero

However $A$ forms the whole of $\Bbb{R}$ hence the lesbegue measure is infinity

So since vitali sets are uncountable sets and they all union to form $\Bbb{R}$ what prevent them to be made entirely of singletons yet union together to form countably many intervals?

(Actually a better example will be cantor sets, and then take countable union of them somehow so that they give all reals (or at least a set formed by disjoint union of countably many intervals))

So I guess the question is then: Can the elements in a vitali set form intervals?

0

Q: If an unbounded Vitali set exists, then does it contain a set with positive measure?

Let $V$ be a Vitali set on $\mathbb{R}$ and suppose that $V$ is not bounded, ie the representatives $v$ of the cosets of $\mathbb{Q}$ are chosen in a such a way that $|v|\geq M$ for all $M\in\mathbb{R}$. If such a Vitali set exists, can it contain a set with positive measure? I know that all me...

Anderson Felipe Viveiros

3:35 PM

Does the flow determine uniquely its vector field? I.e., if $X$ and $Y$ are vector fields and $\varphi_X$ and $\varphi_Y$ are their flows, does $\varphi_X=\varphi_Y$ implies $X=Y$?

Balarka Sen

Look at the derivative of the flows

If the flows agree, their derivatives agree

Mike Miller

you beat me

Anderson Felipe Viveiros

Oh, nice. So easy

Balarka Sen

@MikeM better than getting sniped

Secret

0

Q: Does the vitali set contain only singletons?

A vitali set is one of the cosets of $\Bbb{R}$ quotient under $\Bbb{Q}$. That is, it is an element $V_r \in \Bbb{R}/\Bbb{Q}$ such that $\bigcup V_r =\Bbb{R}$. This MSE shows how a vitali set can only contain measure zero sets. Does it mean that vitali sets consists entirely of singletons and thu...

set-theory lebesgue-measure axiom-of-choice

Hopefully its a question that makes sense...

Tobias Kildetoft

3:39 PM

@BalarkaSen That sounds about right. It is better to get beaten than to get sniped.

Balarka Sen

indeed

Secret

O c****, the f*** is an outer measure...

Akiva Weinberger

Hlello

Balarka Sen

Hey

Secret

4:01 PM

Interestingly, vitali sets are a lot more concrete than the sets in ZF-C. It's just a coset of $\Bbb{R}$ under $\Bbb{Q}$. I can "kinda" visualise that

Tobias Kildetoft

@Secret Isn't it a representative of each coset?

Secret

It is, I just tend to take the cosets in full, because then I can see how they partition the reals

just as how when thinking about $n\Bbb{Z}$ I like to think of the whole thing instead of its representative nm

Tobias Kildetoft

But if you take the whole thing, you don't get a vitali set.

GFauxPas

For my prof's proof of Caught Goursat theorem, he first proves it for triangular paths, is that a standard way of teaching it? It's cool but my textbooks dont do it that way

Balarka Sen

Yes, it is one of the standard ways to do it without further assumptions

In particular you avoid assuming C^1 holomorphic

(In which case you can use Green's theorem to directly arrive at Cauchy-Goursat)

GFauxPas

4:17 PM

Thanks, he didn't finish it, he extended it to all polygonal paths, then said that if we wanted to we could on our own time prove that the sequence of integrals along the polygonal path converges to the integral along the smooth path

Which i didn't try yet

Balarka Sen

Well, actually you can completely avoid working with polygonal paths. You can use the Goursat's theorem on triangular contours to construct an anti-derivative of your holomorphic function

GFauxPas

And then use the ftoc, interesting

Balarka Sen

Right

GFauxPas

But why does an antideriv on the triangle imply an antideriv on the arbitrary closed curve?

Balarka Sen

Well, suppose $f$ is a holomorphic function on a domain $\Omega$ bounded by a circular contour $\partial \Omega$

GFauxPas

4:22 PM

Simply connected or not necessarily?

Or does "domain"mean sc

Balarka Sen

Thanks, edited

GFauxPas

K

Listening:)

Secret

Ok I mistaken, taking all of them only forms the cosets in $\Bbb{R}/\Bbb{Q}$. A Vitali set consists of only one member for each coset

Balarka Sen

Actually I'm going to end up constructing a local antiderivative, so $\Omega$ can be whatever. It's a whatever open set.

Anyway, pick some point $p \in \Omega$

GFauxPas

Okay

Balarka Sen

4:28 PM

Define the following thing. $F(z) = \displaystyle \int_\gamma f(z) dz$ where $\gamma$ is a path going from $p = p_1 + i p_2$ to $z = z_1 + iz_2$ by going along the x-axis first, and then along the y-axis. Namely, you go from $(p_1, p_2)$ to (keeping the y-component fixed) $(z_1, p_2)$ to (keeping the x-component fixed) $(z_1, z_2)$

So we have $\gamma(0) = p$ and $\gamma(1) = z$.

It's an "L" curve, if that makes sense

GFauxPas

Okay

Like a taxicab shortest path

Balarka Sen

Exactly.

Notice that $F$ is well defined only a small disk neighborhood of $p$ (it may not be well defined on $\Omega$ because the lines might get outside if $\Omega$ is eg not convex)

I claim on such a neighborhood, $F$ is an antiderivative of $f$

GFauxPas

Right

Well its almost a triangular path

So we can make the hypotenuse in the other orientation ?

Adeek

hey everyone I was wondering I had a student who is taking a lab with me we are looking at the following limit $lim_{x \rightarrow 0} 2ax^2 sin(1/x^2)$

Balarka Sen

The idea, IIRC, is to think about $F(z + h) - F(z)$ as $\int_C f(z) dz$ where $C$ is union of such L paths $\gamma_1$ and $\gamma_2$ joining $p$ to $z$ and $z + h$, along with a line going from $z$ to $z + h$

That makes up a closed contour

Adeek

4:35 PM

x approaches 0 from the negative side

if we look at squeeze theorem we see that limit is 0

but if we replace z = 1/x^2 and use sin z / z = 1 we get 2a

does anyone have the reason why this is true ?

Balarka Sen

The integral would be $F(z + h) - F(z) + \int_{z}^{z+h} f(z) dz$ where the last integral is taken along the straightline segment from $z$ to $z + h$

Adeek

why we get different answer ?

Balarka Sen

But that becomes arbitrarily small as $h \to 0$

GFauxPas

A desk, then the limit changes

Balarka Sen

@Adeek What? If $x \to 0$, $z \to \infty$

Adeek

4:37 PM

ohh rightt

GFauxPas

Lim as x->?

Adeek

right lol

Balarka Sen

This is basic high school algebra error

Adeek

haha

See @BalarkaSen dealing with this abstract sh!t I make stupid mistakes now

lol

GFauxPas

We all make mistakes :)

Balarka Sen

4:39 PM

@GFauxPas In any case, you'll end up integrating $f$ over the triangle with vertices $z, z + h, z + i\text{Im}{h}$ I think.

That's zero, by Goursat

GFauxPas

I'll play when I have paper thanks Balatka

Balarka Sen

Sure.

Ted Shifrin

@GFauxPas: Did you work out that $f$ injective problem?

hi Balarka, Karim

Balarka Sen

Hi @Ted

GFauxPas

No I did something else but im going to the college library today to work on it

Ted Shifrin

4:42 PM

I expect you to figure it out!

GFauxPas

:)

Secret

Ok, lets take your intuition and slap it around a bit. Think of a perfect nowhere dense set, either the Cantor set which has measure zero or a "fat" Cantor set in $[0,1]$ which has measure as close to 1 as you like. They all look alike. Talk about "disconnected!"-- between any two points there is a genuine gap, not like the irrationals. Your intuition is telling you that there is no way these sets are not small. But facts are facts. So tell your intuition not to confuse measure statements with topological statements and vice versa. — B. S. Thomson Nov 21 '15 at 6:15

Fine I am not touching Vitali sets until I get my head through measure theory!

Clearly, I need to stop using my intuition on uncountable sets

You're conflating positive measure and intervals. The irrational numbers contain "only singletons" but have a full measure. — Asaf Karagila 13 mins ago

I think measure theory is those things you cannot rely on intuition even a single moment

... I had a feeling I will be question permabanned by someone very soon if I keep asking set theory stupid questions...ack

Maneesh Narayanan

-4

Q: What is the radius of convergence of given series?

Let $f(x) = {1\over 1+x^2}$ .Consider its Taylor expansion about a point a ∈ R given by $f(x) = \sum_{n=0}^{∞} a_n(x − a)^n$. What is the radius of convergence of this series ? My attempt: I take $A(a,0)$ and $B =(0,0)$ and $C(0,\pm i)$ where a is real . $AC$ is the distance of hypotenuse of the...

complex-analysis power-series

I tried by completing the square and applied the known series.

ie

Secret

This chat is having an increased frequency of people of a certain type recently which gives me the urge to permaspam it to oblivion. I wonder if the exam period is near...

Not to mention my own mistakes frustrates me too

not to mention, decreasing frequency of regulars that are used to the weirdness thus providing an outlet for it to go somewhere

Maneesh Narayanan

4:58 PM

$\frac{1}{1+x^2}=\frac{1}{1+(x-a)^2 -2ax-a^2}$ How to procede further?

Please help me.

Answer given in the answer key is $\sqrt{1+a^2}$

@Secret What is this?

@Alen K chen

Alex K Chen

Sorry if that sounded harsh, but stop spamming this room

2

Maneesh Narayanan

type error

@AlexKChen I am a human being. Not a super computer. sorry for my mistakes.

Simply Beautiful Art

5:21 PM

@ManeeshNarayanan Please don't be so spammy.

Edit/delete comments to avoid creating confusion.

Secret

Suddenly a huge number of people in the chat :O

Erik the Outgolfer

well, what happens with flags

Nog Shine

Flags.

Simply Beautiful Art

>.O

Phase

@0celo7 did you flag it

Simply Beautiful Art

5:22 PM

@Secret Also why

0celo7

@Phase no, why would you assume I did?

Phase

@0celo7 revenge for the universe flagging you all the time

0celo7

@Phase I am a peaceful person

2

Simply Beautiful Art

Also, don't run ChatJax with Secret's mess

0celo7

I do not try to exact revenge or other such petty things

Simply Beautiful Art

5:24 PM

Yeah thanks

Froze my page >:O

0celo7

@Loong why did you remove that?

Secret

It's seriously lagging the chat

Loong

@0celo7 because of the flags

0celo7

it wasn't for me. Linux ftw

Phase

Weird, maybe it's browser dependent? I didn't experience anything from it

Secret

5:24 PM

I did not expect the bomb is that strong. I just want "insert suitable person" to shut up

0celo7

@Loong people flagged Secret's message? Jeez

Erik the Outgolfer

if you have a userscript...

Simply Beautiful Art

@ManeeshNarayanan Doesn't mean you should spam.

Phase

I run ChatJax

Bálint

Wait, you guys get LaTex in the chat

Phase

5:25 PM

I just saw a long chain of infinities

Bálint

Common

Phase

Didn't have any lag tho

0celo7

me too

Phase

physics masterrace

0celo7

@Phase we have superior computers

Phase

5:25 PM

We've developed an immunity

that too

0celo7

I just ran @Secret 's LaTeX on my GPU

Simply Beautiful Art

@Bálint We get chatjax

VatsalSura

Can some1 please help me with my question. math.stackexchange.com/questions/2487816/…

Alex K Chen

Spamming started again. I guess the room should be put to rest/some users shoud be kickmuted/some messages should be deleted[cc: @Loong ]

Simply Beautiful Art

-_-

Phase

5:27 PM

Isn't that a little extreme

Secret

Is it seriously close to the exam period. Both h bar and math are recently quite flooded with these people?

0celo7

@Loong feel free to kick me

Simply Beautiful Art

@Phase Probably. I've never seen it happen in this chat before.

0celo7

@Loong if you know fortran hmu in the physics room tho

Alex K Chen

Hi @SimplyBeautifulArt ! Remember we used to play Super Tic Tac Toe (and Gale occasionally) when we and @EriktheOutgolfer were kids ?

Abcd

5:28 PM

@Secret No, isn't the month of march the "exam period"?

Simply Beautiful Art

@AlexKChen yes I do.

0celo7

@Abcd I have exams this week

I have exams every week

(roughly)

Abcd

Oh.

VatsalSura

https://math.stackexchange.com/questions/2487816/number-of-square-submatrices-of-a-matrix-and-occurrence-of-each-value-in-the-new

Please help me in this question.

Secret

Does anyone want another dose of the infinity bomb. If not STOP HELP VAMPIRING!!

2

Simply Beautiful Art

5:32 PM

@VatsalSura Please do not ask twice without giving people time.

Phase

Nothing wrong in asking chat for help but if someone doesn't respond to your first link, posting it 5 minutes after probably won't achieve much

Secret

though to be fair, he had show effort. It's just combinotorics I cannot help much because I suck at it

Simply Beautiful Art

Also note that questions on main may take more than an hours to be answered. Some will go days, even weeks without being answered. Some simply don't get answered.

Liad

i got a quesiton : $f(z) = z \ ^ 2 + \lambda$ for fixed $\lambda \in \Bbb C$.
Define $V = \{x \in \Bbb C : |x| \gt \dfrac{1}{2} + \sqrt{\dfrac{4}{5} + |\lambda|} \}$
i need to show that if $x \in V$ then $|f(x)| \gt |x| +1$ , someone can help ?

Alex K Chen

lol

Simply Beautiful Art

5:33 PM

@Secret You're not helping.

Secret

in The h Bar, 22 mins ago, by Secret

I have a very $very^{very^{very^{very}}}$ bad mood today

Alex K Chen

Click on the link, please :D

Secret

death by rick rolled

mixedmath

@VatsalSura It's fine to ask for help, but don't do it too often. Users will react negatively towards "noise" in the channel.

Phase

Predictabo

Liad

5:35 PM

i got a quesiton : $f(z) = z \ ^ 2 + \lambda$ for fixed $\lambda \in \Bbb C$.
Define $V = \{x \in \Bbb C : |x| \gt \dfrac{1}{2} + \sqrt{\dfrac{5}{4} + |\lambda|} \}$
i need to show that if $x \in V$ then $|f(x)| \gt |x| +1$ , someone can help ?

mixedmath

Aw, someone is flagging the rickroll

Simply Beautiful Art

@Secret perhaps sifting through these ordinals may brighten your day up.

Phase

Ok @Liad for gods sake

I literally just wrote that it's pointless to repeatedly ask it so soon

And you defy expectation by doing it in like 2 seconds

Your previous one isn't even off the page yet

Nog Shine

@Secret Flags again.

Maneesh Narayanan

@SimplyBeautifulArt I have been using this chat for 2 days. I don't know much about this chat. sorry, if I did anything wrong.

VatsalSura

5:36 PM

@mixedmath I am seriously looking for some help here, but everyone here is just spamming instead of discussing a question or something.

Semiclassical

I take it I missed some drama

Liad

@Phase it is suppposed to be 5/4 not 4/5 , i couldnt edit it. relax

Phase

Vatal please don't slander someone with the same brush that applies to you

Secret

@VatsalSura We know you are legitmate, it's just at this time there are too many help vampires around and we get a bit stirred...

Semiclassical

I’d look back at the transcript but uh

I don’t want to

Phase

5:37 PM

@BalarkaSen nice ad

Maneesh Narayanan

How to delete a chat?

Secret

Semiclassical: You missed the detonation of the infinity bomb in response to my emotions to Maneesh and a couple of random things

Maneesh Narayanan

please help me.

mixedmath

@BalarkaSen We shall see if anyone bites at this one.

Nog Shine

@BalarkaSen Flagged already. ;)

Semiclassical

5:38 PM

Not sure counter-spam is a productive response to spam

Balarka Sen

@mixedmath Well, the flagger is going to get to enjoy some nice jam at least.

Secret

It's not, but awe induced silence does work like a charm

Balarka Sen

@NogShine I was going to compliment you on your gravatar, but actually I also compliment you on your username.

Simply Beautiful Art

@Semiclassical You definitely don't.

Phase

I just realised.

Semiclassical

5:39 PM

I figured

Phase

God damn it.

Riker

@BalarkaSen can people not? this flag noise is getting annoying

8

Nog Shine

@BalarkaSen Thanks.

Phase

@NogShine are you he who's name is written by the ancients?

Balarka Sen

lmao that was flagged wtf

VatsalSura

5:40 PM

@Secret If they can't help, they can atleast star it as interesting so that someone can help

Secret

@VatsalSura It need to be really interesting to get a star (and there is no control on what the chat thinks is interesting)

Phase

@VatsalSura that's implying anyone finds it interesting. If no-one responded, they don't. Just stop asking chat for now. If you don't have an answer tomorrow, and there's more people in chat then then maybe consider asking again

If you want it to get starred, just write something that you really dont want a mod to see

people will star it and you'll be shot into space as punishment

Simply Beautiful Art

@VatsalSura No, that's not how starring (usually) works. Stars are for interesting material, not for your own personal questions.

Secret

The maths chat is really crowded today

mixedmath

@VatsalSura People are here to chat. You've advertised your question. If someone is interested, great! If not, then you'll just have to hope that someone on the main site responds.

Secret

5:41 PM

Is it really because of the infinity bomb have shattered reality :P?

Simply Beautiful Art

I suppose we didn't need to all respond lol

mixedmath

@SimplyBeautifulArt (sorry)

Simply Beautiful Art

@Secret Man, it wasn't that big. It was just $\varepsilon_{\infty+1}$ at most.

Semiclassical

I’d also remind people of their ability to ignore other users messages

2

Secret

Well it is big enough to lag some computers, so...

This is the first time I saw a 3rd row appeared in this chat

Semiclassical

5:43 PM

Not everything you find annoying needs moderator intervention

28

Riker

@Semiclassical flags are also visible to anybody with over 10k rep, which is a lot more than just mods

Simply Beautiful Art

@Secret On a brighter topic, I may have found a nicer way to deal with ordinal collapsing functions

Secret

hmm?

Simply Beautiful Art

I assume you haven't figured out how normal OCF's work yet?

Semiclassical

@Riker yeah. I’d be seeing it myself were I not here by mobile

Balarka Sen

5:44 PM

They seriously need to modify the flag system

This is hilarious

Secret

@SimplyBeautifulArt Well, the epsilon level fixed point behaviour are still trickly, but I can at least imagine because I now have a more solid grasp on how the epsilon and above numbers behave

Riker

quick question regarding my calc 1 homework: is it fair to say that taking the implicit derivative of a expression, say y^3 from x + y^3 = 4 is the same as continuously taking the derivative of y^3?

Semiclassical

I’d prefer it if a flag required at least a short message saying why it’s flagged

4

Balarka Sen

@Riker Only on a small enough neighborhood in the domain of definition of $y(x)$.

Riker

i.e. apply chain rule on d/dx(y^3), which gives 3y^2*d/dx(y^3)?

I'm not entirely sure

Phase

5:46 PM

I don't get why it doesn't send the notification in waves. Send the flag to like 2 moderators online on the site and if there's no response in a couple minutes send it forward

Erik the Outgolfer

huh

how you send a flag to "2 moderators"

Secret

that could be dangerous...

Riker

@BalarkaSen can you ELI<taking calc 1 this year and have no clue what you mean>

@EriktheOutgolfer you don't

Semiclassical

It’d have to be an implementation change

Riker

@Phase sometimes it needs more immediate attention, like browser-crashing unicode

@Semiclassical mod flags do, but isee what you mean

@BalarkaSen what's "neighborhood" mean in this context?

Evinda

5:48 PM

Hello!!! By taking spherical coordinates, why is $\theta$ in the interval $[0,2 \pi]$ and $\phi$ in in the interval $[0,\pi]$ ?

Erik the Outgolfer

@Riker well, if you cast a blue flag then you supposedly flag because the message is "spam, inappropriate or offensive"

Riker

/shrug

Semiclassical

Look at a globe and the lines of latitude/longitude

Phase

Because every point is fully specified by those co-ordinates without covering the same point twice.

Balarka Sen

@Riker I'm sorry. What I mean is this; say you have the expression $x^2 + y^2 = 1$. Then if you look at the point $x = 1$, you'd run into trouble doing $dy/dx$

Simply Beautiful Art

5:51 PM

Define a function as follows:
$$\psi(0)=\varepsilon_0\\ \psi(z)=\sup\{~^\omega (\psi(z[n,z])),n<\omega\}\\ (x+y)[n,z]=x+(y[n,z]) \\(x\cdot y)[n,z] = x\cdot(y[n,z]) \\(x^y)[n,z] = x^{y[n,z]} \\1[n,z]=0\\ \omega[n,z] = n \\ \Omega[n,z]=\begin{cases}\omega,& n=0\\ \psi(z[n-1,z]),&n>0\end{cases}$$

Riker

@BalarkaSen I see

Secret

@SimplyBeautifulArt It does look a lot cleaner. Also [n,z] are two dimensional fundemental sequeneces?

Balarka Sen

@Riker But yes, away from the problematic points, your method is 100% correct and is called implicit differentiation.

Riker

hm ok

yeah, my chapter is indeed called "implicit differentiation", I just wanted to clarify I was doing it right

thanks for the help

Simply Beautiful Art

@Secret One dimensional. The z serves as a place-holder so that the Ω can do it's job correctly.

These are more or less the fundamental sequences I used to write a program who's output exceeds TREE(3).

Secret

5:55 PM

ah, so they are like nested functions such as $\Omega(\Omega,0))$ ?

Simply Beautiful Art

@Secret No no no....

Single Ω's

e.g. $\psi(Ω\cdot\omega+3)$

No such $\psi(\Omega_\Omega)$ type stuff yet

Or $\Omega_3$ or anything of that sort.

Secret

This new OCF is easier to understand then the $C$ functions, and you also explicitly construct an define the fundemental seqences

Simply Beautiful Art

Huehuehue yeah, you can credit @user21820 for the idea, got it from him.

Secret

I see

Simply Beautiful Art

Unfortunately, he doesn't seem to be planning on extending it further, so I took it up in some places.

Wanna try denesting some OCFs?

Perhaps, for insight, start with simple old $\psi(\Omega)$

Balarka Sen

5:59 PM

@Riker Here's a small remark: The issue with doing this with $x^2 + y^2 = 1$ at the point $(1, 0)$ is at that point the curve (which is a circle) cannot be defined as a graph of a single function of $x$; you don't know what to choose, $y(x) = \sqrt{1 - x^2}$ or $y(x) = -\sqrt{1 - x^2}$. So $y$ is not even well-defined as a function of $x$.

Secret

Well the issue is that $\omega_1^{CK}$ is impredicative as far we have tried, there seemed to be no natural way to find a binary function that can well order all computable ordinals. user21820 is interested in predicative mathematics which means ordinals below $\omega_1^{CK}$

Hmm sure

Abcd

4 rows...1st time I guess.

Riker

@BalarkaSen yeah, I see how that would be a problem

I think my homework questions avoid that though

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$Mathematics$ Mathematics