@BernardMeurer I would not have kicked you if this had been isolated within a larger conversation. However, I saw it as continuing the pattern of posting crap that had derailed the room earlier. And look, now we're having a moderation conversation instead of posting crap, so...goal somewhat accomplished? :P

Password is your sister's dog name

^basically us right now

@BernardMeurer She has a dog?

@BernardMeurer I'm not actually mad at you, please don't take this as a personal affront.

@ACuriousMind Did you delete my /b/ link?

@ACuriousMind I know, or well, thought but now know, you're not. My whole point is that unfair kicks suck balls. I'll take a punishment for what I've done wrong, but I refuse to not complain when I think I've been punished for something undeserving. Sure, you had a larger objective with that kick, but com on, Machiavelli is dead, the ends do not justify the means

Did someone get flagged?

Sigh, bloody foreign mods now

@ACuriousMind Sorry for the ruckus tonight

@BernardMeurer dog namespace has low bit-depth

@GPhys It's not exactly my bank password I'm protecting

@GPhys Good observation though :)

@BernardMeurer It's fine to complain, but I stand by what I did. Also, Machiavelli's Prince is a description of how the Medici ruled, not his advice on how rule should actually work (compare that work to his Discourses, and you'll see the blantant differences)

@ACuriousMind Wrong.

On the second part.

"The Prince" is Ezio Auditore da Firenze.

Not the Medici.

@0celo7 Shhh, we don't want the Templars to find this place

@ACuriousMind I'm using Machiavelli as the cultural figure for which he is, come on, look at the definition of "machiavellian" for example. And most importantly, you understood the argument. You stand by what you did, I stand in my belief that it was fundamentally wrong. I, for one, can't kick you, so I'll me removing myself from the room.

Forever?

@ACuriousMind Do you want to talk about something

I would like to discuss Fallout

@0celo7 NV or 4?

NV

Specifically the powder gangers

I killed a bunch of them at the beginning and got villified

So I never interacted with them outside of destroying them when they tried to ambush me

is there more to that story?

Yeah, I always felt that was kinda bugged because it felt like they, too, should have a story

But I never managed to talk to them

@MAFIA36790 tomorrow is a holiday here, and my sleep cycle is off anyway because I only have one day of classes this term

@ACuriousMind But re: 4, I started a new character the other day. The newspaper girl wandered too close to Swan's lake >.>

Swan is too OP

@0celo7 Piper?

@ACuriousMind Ah, right

And yes, the swan is crazy

Listening to music and typing a paper

I couldn't remember her name

@ACuriousMind I basically ran into the combat zone. He killed a bunch of raiders, got bored, and went back to his lake

/pond

I searched for nice mods for F4 the other day, but didn't really find much

@ACuriousMind Me too!

I couldn't find anything

I wanted to Frankenstein it :)

The scene seems to be much less active than Skyrim or NV

I should replay Skyrim

I never did play those fan made DLCs fully

If you manage to start it without crashing it, that is? :P

But a Skyrim playthrough is a commitment

@ACuriousMind Yeah, that's an issue. Also everyone is naked right now

I don't know why

reverse engineering my mod list and load order at this point will be impossible

@ACuriousMind did you play with Skyre?

Yep

Fun fact: Skyre + creature mods = invincible enemies from time to time

= very, very hard game

It was different, but I never actually figured out whether I actually liked it better

I have 100 smithing at level 40 or something but am still underpowered

100 enchanting

double enchants on everything

completely OP in the regular game --> destroyed by everything in Skyre

Well, SkyRe already amps up most enemies, together with creature mods it's to be expected it'll get impossibly hard

Also, I came to realize the combat system isn't really fun while playing Enderal

Oh shit

I forgot about that game

What's wrong with it?

It's...boring, at least to me, you don't do anything but chopping away at enemies

I got -3.9 for a wire radius.

Sigh.

Enderal's story and world were really nice, but I found myself wanting it to be in some other engine

-3.6, correction.

@ACuriousMind True, I guess.

Negative space wire, now that's engineering

I like archery though

I always end up leveling archery

Me too

and I kill Miraak early so I have access to the perk switcher

I've played Skyrim too much lol

I have so many playthroughs

@ACuriousMind I'll also switch between one-handed and two-handed and then axes, swords, greatswords, etc.

I've done a few mages but honestly they're terrible

End up going battle mage

Magic disappointed me in Enderal, just run-of-the-mill spells, nothing creative :/

There are lots of magic mods

But patching them with Skyre is a pain.

I miss the days of Morrowind when levitation and damage strength was a thing. Horribly unbalanced, but also rather fun

@ACuriousMind If I do another Skyrim playthrough I'll disable the damage and difficulty scaling of Skyre. I bet half of my problems are from reproccer and all those damn patches

but I like the spawn changes a lot

more creatures in the overworld is nice

It's clearly a dude's operator.

@ACuriousMind Quick! What's the operator norm of a general matrix given by?

@0celo7 largest singular value

@ACuriousMind Eigenvalues?

@0celo7 basically, yes, but in general the singular values are the square roots of the eigenvalues of $A^\dagger A$.

How does one prove such a thing?

hmm

@0celo7 You prove $\lvert\lvert A^\dagger A \rvert\vert = \lvert\lvert A\rvert\rvert^2$ and then prove that the operator norm of a self-adjoint operator equals its largest eigenvalue

@ACuriousMind yeah, apparently so

but how does one prove the first thing

@ACuriousMind $||A^\dagger A||=\sup \{||A^\dagger Av||\mid ||v||=1\}$.

What the heck are you supposed to do with that?

@0celo7 note that $\lvert\vert Av\rvert\rvert^2 = \langle Av,Av\rangle = \langle v, A^\dagger A v\rangle \leq \lvert\lvert v\rvert\rvert\cdot \lvert\lvert A^\dagger A v\rvert\rvert$.

@ACuriousMind Lol, forgot I had an inner product.

What was the dagger supposed to be, then? ;P

Conjugate transpose, clearly

You don't get the conjugate transpose as a map on the same vector space without an inner product

I know

Did I need a ":P" there?

Maybe

Go to sleep

I'm fading

how are you alive?

Yes, I can read.

Is it possible to map a resistivity or conductivity value to a 0..1 range representing voltage drop over a specific length and cross section?

@0celo7 barely

Should I ask a question on this stack exchange for something like that?

@ACuriousMind that's not an answer

go to sleep dude

@MAFIA36790 are you ready to learn Arzela-Ascoli?

@Rodolvertice I don't understand the question. Resistivity is by definition resistance times cross section per unit length.

@MAFIA36790 Fitting for Halloweeen, isn't it? :P

@MAFIA36790 of Banach spaces?

Prove that a linear map on a normed vector space is continuous iff it is bounded.

Also show that on finite-dimensional spaces they are all bounded.

I should probably do the second one, actually.

I mean write it out on a paper

I probably have it somewhere

@0celo7 Singular value has a maximum, done ;)

@ACuriousMind I don't understand functional analysis.

@ACuriousMind It's much simpler.

The singular value of (finite-dimensional) matrices is not functional analysis :P

This line of proof would also fit elegantly into the "largest singular value is operator norm" we just discussed

$$||T(v)||=||\sum x_iT(e_i)||\le \sum|x_i|||T(e_i)||$$

@ACuriousMind I don't understand any of that though

So it might as well be

So we get $||T(v)||\le M\sum|x_i|$

@ACuriousMind I'm a game developer looking to have an accurate representation of electrical components. How can I use a value that can range from 62 million S/m for silver, all the way to 0.0000000001 S/m for bromine? S/m seems to be a measure of amps per volt meter, and i have no idea how to interpret such a value.

I'm sure that last factor is $\le ||x||$, maybe need a factor of 1 or something

Or not

Triangle

Yeah, so $||T(v)||\le M||v||$

@ACuriousMind Ok, what were you proposing?

Damn vector components/notation.

@0celo7 hmm? Once you know that the largest singular value is the operator norm, you just use that there are only finitely many singular values on a finite-dimensional space, and take the maximum, which is finite, so the operator is bounded.

But I don't know that

@Rodolvertice What's S? And what do you actually want to do?

Siemens

@ACuriousMind Suppose I have a long wire, and I put a current through it, and measure the magnetic field at some distance r

now suppose I have a short wire, same type, and put the same current through it

what happens to the magnetic field at r?

@Rodolvertice Ah, so that's conductivity. Well, since there's no such thing as a maximum conductivity, you can't expect to map it into a 0..1 range. Of course, you could just set the maximum conductivity that appears in your setting to 1.

You'll be working in rather weird units, then, but I guess that's never stopped anyone writing a simulation ;)

resistivity is just the reciprocal of conductivity, no?

I was looking for some sort of mapping where 0 is no conductance, and 1 is perfect conductance

@Rodolvertice yep

@ACuriousMind wait what

How is there no such thing as maximum conductivity?

German efficiency

Superconductors arent the maximum?

@Rodolvertice Superconductors have effectively zero resistance, so they have "infinite" conductivity.

@GPhys what's wrong?

@ACuriousMind What would be considered as non-weird units then, may I ask?

@ACuriousMind I wasn't following the entire conversation, but certainly such a bijection does exist

@Rodolvertice SI units, of course

@GPhys yes, but not a continuous one.

Ohms per meter?

@GPhys Even in light of superconductors having zero resistance?

$[0,1]$ is compact, but $\Bbb R$ is not.

Sure there's a bijection between an interval and $\mathbb{R}$, as @0celo7 points out, but it's not gonna be useful.

oh I see what you want now

I lied slightly. There might be a continuous map $\Bbb R\to [0,1]$ (not the other way around)

Not sure how you would get such a map

But certainly no homeo

Isn't there some way to assume a square meter cross section, then obtain a number that when raised to the power of the length of the sample represents the total power drop?

@ACuriousMind Seriously, what does the length of the wire do?

Ampere's law seems to only care about the radius of the wire, not its length

@0celo7 inverse tangent?

@GPhys Certainly not to the closed interval

IIRC that's a diffeomorphism.

oh I was thinking open

@GPhys then arctan is the standard one, yeah

tanh works too

there's a few

@Rodolvertice Hm, you could take $1 - \exp(-\sigma)$ for $\sigma$ the conductivity and it would map zero conductivity to zero and infinite conductivity to 1.

I think there's a continuous one that's of the form $f/g$ for $f,g$ polynomials but I cba to work it out

@GPhys Not for the closed interval!

still talking about open

Actually it might not be possible

@GPhys Yes, there's a diffeomorphism by rational functions

@0celo7 I can prove it

there is no surjection from a closed interval to an open interval by IVT

closed to open*

No surjection? Lie.

@ACuriousMind Thank you! Should $e$ be the base? What would the number physically represent?

Stop editing :P

@0celo7 (continuous) surjection

@GPhys The proof is that a closed interval is compact, hence has compact image.

An open interval isn't compact, hence is not the image of a closed interval.

@Rodolvertice The base can be anything you want, we have $x^0 = 1$ and $\lim_{n\to\infty} x^{-n} = 0$ for any real $x$.

are you saying there is a continuous surjection from a closed interval to an open interval?

I'm afraid I have nothing to offer for the physical interpretation of that number

Ok.

@GPhys I just proved there isn't.

I'm tired and about to head to bed; I"m not sure what you're getting at

@0celo7 I know, that's what I was pointing out

Yeah, I don't know what you want to do with IVT.

continuous function maps closed to closed

@GPhys What's more interesting is: is there a bijective continuous map $(0,1)\to[0,1]$.

@GPhys that's not IVT

@GPhys Not true.

@GPhys Careful, the open interval is closed in its own (subspace) topology, so that statement probably doesn't do what you want it to do.

e.g.: take a deformation retract $\Bbb R\to(0,1)$.

closed interval to closed interval

not closed, sorry

Also, continuous functions don't map closed sets to closed sets, that's what closed maps to. Continuity is that the preimage of closed sets is closed.

@GPhys that's true

but that's exactly what I said (twice)

the other direction is more interesting

unless I'm missing something @ACuriousMind any thoughts?

Thought: I should go to bed before the sun comes up.

Night

Or

Morning

Whichever you prefer :)

@0celo7 I know Q_Q

IVT was what made it obvious to me, but maybe this doesn't follow so obviously from IVT itself as from its proof in NSA I was thinking of

NSA? Are you a spook?

my undergraduate mathematics thesis was practically a short textbook on internal set theory (approach to NSA)

what is NSA?

nonstandard analysis

ew

internal set theory being the more obscure of the approaches

can you give me a good theorem?

the proof of IVT is amazing

What is IVT?

intermediate value theorem

image of connected sets is connected

Oh ok.

not very amazing

@0celo7 the NSA proof is you divide up the interval into a finite number of points and pick the point satisfying the theorem

@GPhys sounds like some shit to me

and the assumptions just naturally let you do so

finite number of points?

indeed

are you sure you didn't make an error

in internal set theory there's a finite set containing every standard element of $\mathbb{R}$ for example

(or any other set)

(standard)

that does that even mean

how is R not uncountable

IST adds axioms to ZFC that form a conservative extension (so everything true before is still true)

but enriches your vocabulary to talk about new things, so-to-speak

the other thing it adds is the unary predicate standard, so that every set is standard or nonstandard

every set that can be uniquely defined without IST axioms is standard (provable with IST axioms)

If you send me your thesis I will peruse it

there exists a finite set containing every standard element of any set (infinite or otherwise)

in the case of, e.g. $\mathbb{R}$, this finite set is necessarily nonstandard

in fact, there is a finite set containing every standard set