@user10478 you got 20t+4 because you thought R.R is (t+2)^2 + 9t^2 when in fact it is (t+2)^2 + 9t^4

THANK YOU! I worked it out 3 times looking for something just like that and didn't see it.

Nice clarification, @Leaky.

Screw the number 2, it's too confusing.

Cuz $2\times 2 =2$, @user10478?

Apparently

I'll remember that :P

Maybe it's a sequel to that political 2+2=22 video :P

Of that I know not.

What is $\operatorname{Hom}(\Bbb Q,S^1)$ (in the category of topological groups) when $\Bbb Q$ is given the "usual" topology?

What do you call a map $\Bbb R\to\Bbb R^n$ whose fourth derivative is zero? A cubic path?

In any case: given $x_0,v_0,x_1,v_1\in\Bbb R^n$, there exists a unique cubic path $f:[0,1]\to\Bbb R^n$ such that $f(0)=x_0$, $f'(0)=v_0$, $f(1)=x_1$, and $f'(1)=v_1$.

I think this is the one you get with Bezier cubics.

Conjecture: this path is the one that minimizes $\int_0^1\|f''\|^2dt$ given the boundary conditions.

Does that have any physical significance?

@AkivaWeinberger stationary points occur when the functional derivative is zero, i.e. when $f'''' = 0$, i.e. when $f$ is a cubic

Hello

I am currently reading Real Mathematical Analysis by Charles Pugh and it instructs me to try and visualize every concept.

$I[f+\varepsilon \xi] - I[f] = \int_0^1 [2 \varepsilon \xi'' f'' + \varepsilon^2 (\xi'')^2] \ \mathrm dt$, so the second variation is $\delta^2I[f,\xi] = \int_0^1 (\xi'')^2 \ \mathrm dt \ge 0$ so the stationary point is a minimum @AkivaWeinberger

that proves your conjecture

But the problem is I don't know how to visualize a limit point

the definition provided is

Let $M$ be a metric space, and let $S\subset M$. A point $p\in M$ is a limit of $S$ if there exists a sequence $(p_n)$ in $S$ that converges to it.

How am I supposed to visualize this?

@LeakyNun Is that $\delta$ times $f$, or is $\delta f$ one variable?

@AkivaWeinberger $\delta f$ is one variable

this is calculus of variations

@clathratus Imagine the set $S=\{0\}\cup\{1,\frac12,\frac13,\dots\}$

(here, $M=\Bbb R$)

Then $0$ is a limit point of $S$

A related example: $S=\{1,\frac12,\frac13,\dots\}$

$0$ is still a limit point of $S$ (even though it's not an element of $S$)

A final example: $S=(0,1)$, the open unit interval

The limit points of $S$ are everything in $[0,1]$ (the closed unit interval)

@clathratus Doesn't the definition say $p_n\ne p$?

'Cause otherwise, $1$ would also be a limit point of $\{1,\frac12,\frac13,\dots\}$ (take the constant sequence $(p_n)=(1,1,1,1,1,\dots)$)

Suppose set $X$ contains all $\text{n×n}$ matrices that can be formed using $-1$ and $1$. Find the probability of selecting an element of set $X$ whose $2×2$ sub matrix contain two $1$s and two $-1$s. Also find the probability of selecting an element $X$ whose each row and column has odd number of $1$s.

I asked for a general case. In the original question they ask for $n=4$ but I would like to derive it for any case.

@LeakyNun I wonder what curves result from the conditions $\|f'\|=1$ and $f''''\parallel f'$

(Are \| and \parallel the same symbol?)

$\|$

$\parallel$

Huh.

I don't get you

so f'=1 or -1?

$f:[0,1]\to\Bbb R^n$

like before

yes?

Conceptual Question: We say that there are only X groups of order Y "up to isomorphism." Does that mean that if you can write the multiplication table of the group out in two different ways, then you have written two different groups which are isomorphic to each other? Or would it be an automorphism to relabel variables and "isomorphism" are used when the underlying set is not the same in the image and preimage, even if the multiplication tables are the same?

$f'$ is a vector on the unit sphere, i.e. $f$ has constant speed

oh lol

@Rithaniel Automorphisms are a type of isomorphism, no?

so $f'''' = f' + C$?

Indeed they are.

@LeakyNun More like $f''''=C(t)\cdot f'$

why?

doesn't parallel mean same slope everywhere

Two vectors are parallel if one is a scalar multiple of the other

so f''''' = f''

we must have different notions of parallel :P

@Rithaniel I'm not sure I understand the question 100%. I mean, you have an isomorphism either way, it's just a question of whether the isomorphism is the identity map or not

I think

and I don't like yours :P

@LeakyNun I just mean that, for all $t$, $f''''$ is perpendicular to all the vectors that $f'$ is perpendicular to

perpendicular now?

Read it again

right, sorry, not focusing

@LeakyNun Under your definition, is $(1,2)$ parallel to $(2,4)$?

this doesn't make sense under my definition

As vectors in $\Bbb R^2$

I'm more concerned about the two curves being parallel

Arright well I dunno

What's the fourth derivative called

Snap?

The snap vector is parallel to the velocity vector at all times.

what's the significance

No, jounce

Oh, both are used

@LeakyNun I want to minimize $\int_0^1\|f''\|^2dt$ among all constant-speed paths with given boundary positions and velocities

Trying to pin down a definition with a friend. It's a concept of being "invariant under automorphism." But the question is kind of "In practice, what is an isomorphism and what is an automorphism? When are two groups automorphic and when are they isomorphic. Is the only thing that separates an isomorphism from being an automorphism the set that the group is operating upon?"

is that even possible though

you're just travelling around a circle if n=2

wait no

but f' is a circle

@Rithaniel I don't think "two groups are automorphic" is a concept that makes sense

An automorphism is a map from a group to itself, you don't have two groups to compare

well, it's either false or it's a tautology

if they aren't the same group, it's false

I agree with that. I suppose it would be better to say "You start with an isomorphism. When can you say that it's mapping to and from itself?"

if they are the same group, then it's trivially true

@Rithaniel An isomorphism is an automorphism iff the domain and codomain have the same underlying set and the same group operation

Normally though you don't care about whether the underlying sets overlap unless they're contained in some bigger set

like, two subgroups of a group

perks of a set-theoretic foundation

hmm. is it possible to have two different group operations on the same set of elements?

and the group operation being the same is usually show via the existence of an isomorphism?

otherwise we don't really care because you could just replace them with isomorphic copies

@Semiclassical of course

@Semiclassical Take any two groups of the same cardinality

oh, of course

same number of elements, but different structure

@Rithaniel No, it's shown by showing that $a\circ_1b=a\circ_2b$ for all $a,b\in G$

(where the groups are $(G,\circ_1)$ and $(G,\circ_2)$)

another version of that question, I guess, would be whether there's a group which "naturally" has two different group operations. I imagine that's also true though, despite how vaguely-formulated that statement is

So, suppose if you have a isomorphism which takes $b$ to $a$ and $a$ to $b$, but the structure is the same after this "relabeling"

@Semiclassical {e,a,b,c} :P

I guess the quaternions, since you could make $ij=k$ or $ij=-k$ (and $i^2=j^2=k^2=-1$ and fill in the rest of the multiplication table accordingly) and have not much change

The former has $ijk=-1$ and the latter has $kji=-1$

(and all even permutations, for each)

"Quaternions with the right-hand rule" and "quaternions with the left-hand rule" would be another way to describe that

that sounds legit

Okay, that's actually what I was thinking about.

going back to what leaky suggested, you can view Z/2 x Z/2 as {(0,0), (0,1), (1,0), (1,1)} equipped with component-wise binary addition

So, those two groups are the same, just "written differently"

@Rithaniel As an example, take $G=\{e,a,b,c\}$ with an operation $\circ$ defined by $a\circ b=c$, $~b\circ c=a$, $~c\circ a=b$, anything $\circ$ itself is $e$, and $e$ $\circ$ anything is that thing

That's just the Klein four group

Indeed.

If you map $a\mapsto b$ and $b\mapsto a$ (and keep the other two constant) then the structure stays the same

That's an automorphism.

but you can also add them like binary numbers mod 4, in which case 11+01 = 00 rather than 10

So that seems like a pretty natural example: same group elements, but two different group operations and (in this case) two different groups

@Rithaniel $(\Bbb H^\times,\times_L)$ is isomorphic to $(\Bbb H^\times,\times_R)$

Alright, and the right quaternions versus left quarternions are another form of this. The structure remains the same even as you rearrange elements.

Right, but it's not an automophism because ${\times_L}\ne{\times_R}$.

Aaaaah, okay.

So $(\Bbb H^\times,\times_L)\ne(\Bbb H^\times,\times_R)$.

Even though the underlying set is the same.

Those two groups would still be isomorphic, though, wouldn't they?

Alright, that's what I wanted to know.

@Semiclassical Yeah

figured

Yes, they're the same "up to isomorphism"

So it's an automorphism up to isomorphism? :3

I wonder if there is a concept of counting groups "up to automorphism"

@Semiclassical Negate a basis imaginary quaternion, or swap two of them

Sounds unpleasant.

@Semiclassical Actually, do any orientation-reversing isometry on the 3D subspace of imaginary quaternions

Ted, why is $||x|| = \sup_{||y||=1} \langle x, y \rangle$?

which means that the ones that are automorphisms are the orientation-preserving ones

rotations

@JoeShmo draw a picture

$\langle x, y \rangle \le ||x||$ by the Schwartz inequality, but what's the other direction?

it's maximized when y and x are parallel

which makes sense 'cause those are all defined by conjugation

i.e. y = x/|x|

$z\mapsto p^{-1}zp$ is a rotation of $\Bbb R^3$

oops

no i knew that

@Leaky told you the obvious answer.

as long as x != 0 anyways

thanks :)

Cauchy-Schwarz in fact tells you precisely when equality holds.

(As a sidenote, the rotations of $\Bbb R^4$ are $z\mapsto pzq$, yeah?)

it's great when an inequality tells you the conditions for equality

stupid question

What is a rotation of $\Bbb R^4$, DogAteMy?

(which is why $SO(4)$ is $S^3\times S^3/(x\sim -x)$)

Any element of $SO(4)$?

Is there a term for groups for which every isomorphism is an automorphism?

(well, I guess C-S is really the inequality + the conditions for equality)

Yeah

I need to go now

Bye

bye

@Rithaniel: I don't know what that means. Every isomorphism $G\to G$ is an automorphism of $G$.

@Rithaniel you really need to read the definition of everything

Hates to agree with @Leaky, but ...

yay :P

Well, above we concluded that there are two relabeling the quaternion groups and that they're not the same group, necessarily, but that they are the same group "up to isomorphism"

yes so they're isomorphic

that isomorphism is not an automorphism

because the two groups are not equal

something seems inconsistent here

So you're asking if there are groups with $G\cong G'$ if and only if $G=G'$?

But imagine a group where no matter what isomorphism you constructed, let's say "the underlying multiplication table looks the same"

What does $=$ mean? So $\Bbb Z_2\times \Bbb Z_3 \cong \Bbb Z_6$, but they're not equal.

Hmmm, a good question.

A cyclic group of order $n$ under addition is not equal to a cyclic group of order $n$ under multiplication, but ... I'm not sure I like this question.

One really should think about groups up to isomorphism.

Indeed, this is a murky topic.

Not really.

So, do you don't think there can exist a group for which every isomorphism gives the same inherent structure? What about $\mathbb{Z}/2\mathbb{Z}$?

this is a stupid question

There is only one isomorphism.

What does same inherent structure mean?

Under isomorphism, that's precisely what you have.

I quit.

@TedShifrin I already quitted way earlier :P

yes, i guess my question was why was it the supremum

but thats also a stupid question..

its right there in cauchy schwartz

it's the supremum because it's the maximum

its the supremum because its the supremum?

but that question is more significant than you might think

what

isn't that what you wrote?

no, maximum and supremum are not the same thing

so a compact space will contain its supremum

hence its a maximum

sure...

but you were saying something else

?

never mind...

no really. i wasn't following then

To note: a maximum doesn't necessarily exist even if the supremum does.

I think that's what Leaky wanted to get across

a compact space will always have a maximum

Indeed. I think your logic is correct. But Leaky was trying to say something and he didn't think it was coming across. I could be wrong about what he wanted to say, though.

@JoeShmo anyway it shows that $x \mapsto \langle x,-\rangle$ is an isometry $\Bbb R^n \cong L(\Bbb R^n,\Bbb R)$

which fact shows that?

that $\|x\| = \sup_{\|y\|=1} \langle x,y \rangle$

how do you get your nice norm typeset?

\|

thanks

merp. $\|$

nice!

Learning latex is an ongoing process. It wasn't until last week that I learned about \not.

Indeed

who would need to use \not

@Rithaniel I was going to tease with the indeed's by just typing a bunch of "Indeed"s

Well, I typed up a $\not\cong$ earlier today. Is there an easier way?

but the forum thinks it's a mistake. meaning it won't let you type the same thing over and over

:P

@JoeShmo Indeed

Indeed, Indeed

$\ncong$ \ncong

test: $\not{p}$

So, Ted said "there is only one isomorphism" when I mentioned the cyclic group of order 2 just before he quit. Maybe that's what I'm really wondering about.

bah, doesn't even work as Feynman slash

But I'll think about that stuff another day.

An isomorphism is, almost by definition, something that preserves the inherent structure

Our example from earlier was $\Bbb H^\times_L:=(\Bbb H\setminus\{0\},\times_L)$ and $\Bbb H^\times_R:=(\Bbb H\setminus\{0\},\times_R)$

yes. from greek -- shape/form preservation

i.e. an isomorphism preserves structure

Yeah, it's difficult to refer to what I mean. The best way I can explain it is to imagine the "action table" for the group and how an isomorphism might change the appearance of the table.

that is, preserves how things compose

$\Bbb H^\times_L\ne\Bbb H^\times_R$, but $\Bbb H^\times_L\simeq\Bbb H^\times_L$

I'm thinking about groups where every isomorphism leaves the "action table" alone.

Notice that the identity map is not an isomorphism

(in this specific case where the two groups have the same underlying set but different operation)

@Rithaniel Give me an example where an isomorphism changes the "action table"

Well, the quaternions with the left rule and the quaternions with the right rule.

I'd try typing up the tables to show what I mean, but two 8x8 tables sound unpleasant in chat.

Oh I see

so basically you want a group where all permutations of the elements are automorphisms?

Yes, that sounds correct.

Yes, that's exactly correct.

(Well, hopefully all permutations that fix the identity element)

(Naturally)

@Rithaniel it's not so bad

$(\Bbb Z_2)^n$ and $\Bbb Z_3$ I think

and that's it

Really? That is a very short list.

Think about it. Take a non-identity element $a$

(I'm assuming this isn't the trivial group)

Also, how would you type them up, Leaky? I use tabu in my latex documents and don't know how I'd do it in chat.

Either $a^2=e$, or $a^2=b\ne e,a$

In the former case, since you can permute $a$ with any other element, all elements square to $e$, and we get $(\Bbb Z_2)^n$

Hmmm, I think I see what you mean.

In the latter case, since you can permute $a$ with $b$, we have $b^2=a$

so $a^4=a$ or $a^3=e$

We can't have anything that's not $e$, $a$, or $b$, because if we did (say $c$), we could swap $b$ and $c$ and get $a^2=c$

contradiction

so we're left with $\{e,a,b\}$ with $a^2=b$ and $b^2=a$

which is $\Bbb Z_3$

@Rithaniel $\begin{pmatrix}1&2\\3&4\end{pmatrix}$ \begin{pmatrix}1&2\\3&4\end{pmatrix}

It always feels a little bad to see an easy proof to something that you were thinking was a very interesting topic.

Ah, danke Leaky.

$\begin{array}{c|c}a&b&c\\\hline d&e\end{array}$ \begin{array}{c|c}a&b&c\\\hline d&e\end{array}

Aw, sorry

Bit funny though

Ah, it's fine. You never learn without asking. :P

Also I didn't realize that $\Bbb Z_3$ was so unique 'til I tried answering your question

We have an infinite family, plus that

Yeah, that is an interesting stand-out.

Maybe we could generalize this and ask "for a given number n, what groups G exist, if any, for which the permutations of the elements of G result in n possible isomorphism?"

Notice by the way that the Klein four group is $(\Bbb Z_2)^2$

so it also has the property

I guess you could phrase it as ${\rm Aut}(G)=S_{|G|-1}$ iff $G$ is one of the things we listed

Indeed.

Danke schon, again, for the help, Akiva

TextTranslation["You're welcome", "German"]

why ?

Test

video.twimg.com/tweet_video/Co7eYOvXYAAtaM_.mp4

Aw, doesn't embed

@CroCo This seems wrong… you sure it's not $\int_0^\infty|e^{-t}|dt$?

@AkivaWeinberger, this is the definition of the norm L_1

The book I'm reading states

Strange

Fibonacci

Mornin' all

Hello! How to determine if bell-shaped distribution is too skewed to the right/left to be used as approximate normal distribution?

I oughta learn statistics at some point

Of all the areas of math I want to learn, that's probably the one that'll be most useful for jobs

(Well, that and computer science)

wow most of complex analysis is working with the cartesian product of the real numbers with themselves along with some "additional structure"

@isaac9A Isn't all of (1-variable) complex analysis just that?

'Death sentence for women': Alabama proposes law to make abortion punishable by up to 99 years in prison

america at it again

@TobiasKildetoft yep!

@isaac9A I can't tell if this is satire

@LeakyNun it is if you want it to be

that's even more confusing

@isaac9A It's surprising you get so much for so little, isn't it?

can anyone show why this ?

@CroCo Just do the integral as you always would. It is a straightforward one

@TobiasKildetoft I did it and ended up with infinity.

ohh, and it is. So the claim is wrong.

Maybe they wanted the numerical signs to be around the $t$ instead

Not sure but the book I'm reading states

I'm not sure how the norm is equal to unity.

Hmm, looks like they only care about the function on the positive reals or something like that

that is hard to tell since nothing stated in the book. I can say that the book is about nonlinear systems and usually time starts at zero. This is a possible guess.

All I know is $\int_0^\infty|e^{-t}|dt$ equals $1$, and $\int_{-\infty}^\infty|e^{-t}|dt$ doesn't.

I suspect that they only care about functions defined on $[0,\infty)$.

The only way to find out for sure is to read on in the book and see from context.

@TobiasKildetoft, @AkivaWeinberger indeed you are right. I've looked at the definition in the book about how they define the norm and got this

they start from zero.

Oooops

@CroCo errare humanum est

hello,

if f_k\to f strongly in L_{loc}^2(R^n), does this means for any compact subset K of R^n, f_k strongly converges to f in L^2(K)?

@Semiclassical The dependent, pairwise independent variables thing reminds me of one-time pads

the cypher where, we agree on a random key (the "one-time pad") in secret, and I transmit the sum of the message and the key

(these are taking place in some group of some form)

and you decode it by subtracting out the key

Order plus randomness is indistinguishable from randomness, so this is unbreakable

(the weakness is in transmitting the key)

(The key needs to be roughly the same size as the message)

The connection with the probability thing is, I'm giving you a random variable X in private, and then later I give you a random variable Y in public

and given X and Y, the variable Z isn't uniformly random, but given only X or only Y, it is

@LarryEppes what is $L_{loc}^2(\Bbb R^n)$?

There's no similar analogy that's possibly for public-key cryptography, I think

Probability doesn't work well when we talk about computational intractability

"The probability that the tenth digit of $\sqrt{17}$ is 1" isn't 10%, it's either 0% or 100%, we just don't know which until we figure it out

it's 0% @AkivaWeinberger

Thanks

however there is an interpretation of probability that would make it 10%, I believe

The problem is, the way probability is classically defined, we have $P(A)=P(B)$ if $A\Leftrightarrow B$, no matter how difficult the proof of $A\Leftrightarrow B$ is

Like, $P(p\land(q\lor r))=P((p\land q)\lor(p\land r))$, etc

so we can just take the proof that the tenth digit of $\sqrt{17}$ isn't $1$ step-by-step and turn it into a (long) string of equalities

I'm not aware that probability is classically defined

I'm sure it's possible to make a formalism that takes computational difficulty into account, I just don't know what it would look like

Like, in such a system, is it 10% until you compute it?

yes

our subjective assignment probability changes with the introduction of more data

that's why r/dataisbeautiful

Is it 50% (it's a yes-no question after all) until you realize that a random digit has probability 10% of being a 1?

@LeakyNun r/subredditsashashtags

The thing is, we get no new data. There's no hidden information.

well to someone who hasn't computed the tenth digit of sqrt(17), that is hidden information

If I flip a coin and hide it under a cup and ask you if it's heads or tails, you're missing information, and no amount of thinking will change that.

probability is a subjective assignment of your belief

probability doesn't exist outside of your mind, according to de Finetti

But you see how this is fundamentally different from a coin hidden under a cup

if you hide it then to me the probability of it being heads is 50%

btw heads or tails :P

Right, and you'll never know what side it is on unless you look under the cup.

I don't know what you mean :P

and upon observation I gain data which changes my assignment of probability

Whereas with $\sqrt{17}$, you don't need to manipulate the world or anything.

Would you say $P(p\land(q\lor r))$ doesn't equal $P((p\land q)\lor(p\land r))$ until you retrieve De Morgan's law from memory?

(or derive it from first principles or whatever)

That's not De Morgan

no idea

$P(\lnot(p\land q))=P((\lnot p)\lor(\lnot q))$ is De Morgan

The other one is just distributivity

and in quantum mechanics it really is 50%

Is this a quantum coin we're flipping?

anyway how does this relate to computational difficulty?

Well if I do a pubic-key cryptography scheme, where the encryption key and the encrypted message are both public,

from an eavesdropper's perspective, is the message a uniform probability distribution from all possible messages?

Well, no - it's at most a uniform probability distribution of all possible messages that the eavesdropper hasn't tested yet

('cause you can guess at a message, encrypt it with the public encryption key, and compare it with the encrypted message)

If it's a good public-key scheme, the eavesdropper can't change the probability distribution much more than that — that's another way of saying the eavesdropper can't do anything faster than naïve guess-and-check

It's an open problem whether or not RSA can be cracked efficiently

@LeakyNun this is something the “QBists” in base a lot of their interpretation of quantum mechanics around, and specifically with reference to de Finetti

@Semiclassical what is a QB-ist?

en.m.wikipedia.org/wiki/Quantum_Bayesianism

aha

Hello! How to determine if bell-shaped distribution is too skewed to the right/left to be used as approximate normal distribution?

Meanwhile trying to compatify a dense set, you get spocks everywhere