Mathematics

1:05 AM

So if a derivative is how steep a function is at point on the graph, what is the second derivative?

How much the slope itself is changing?

1:52 AM

Does anyone here know what's the difference between a disconnected cake and connected cake? I need examples other than the technical definition because I can't appreciate it

Semiclassical

3:11 AM

@Startec That's right. For example, take f(x)=x^2. At the middle of the graph of y=x^2, the curve is flat; to the left and right it becomes increasingly steeper, but in the opposite directions. So the slope is more and more positive/negative the further out you go, and the rate at which that slope changes happens to be constant.

(Another way to put it: f'(x) is the rate at which f(x) changes with x, whereas f''(x) is the rate at which f'(x)---the rate of change---changes)

user76284

Any ideas on how to prove this is always positive for $\alpha^2 < 1$ and $\theta > 0$?

Silent

5:03 AM

Is the function given below, a counterexample to 'Suppose $f$ is a continuous bijection of a (not necessarily compact) metric space $X$ onto metric space $Y$, then inverse $f^{-1}$ continuous':
$$
g(x)=\left\{
\begin{array}{c}
\frac1y, \,\,\,1\le y<\infty \\
0,\,\,\,y=0
\end{array}
\right.
$$

I meant $$
g(x)=\left\{
\begin{array}{c}
\frac1x, \,\,\,1\le x<\infty \\
0,\,\,\,x=0
\end{array}
\right.
$$

can't edit now

Silent

5:22 AM

@AkivaWeinberger, will you please look at my question above?

Akiva Weinberger

So it's a map from $\{0\}\cup[1,\infty)\to[0,1]$?

I think works, actually

You could also do $[0,1)\cup[2,3]\to[0,2]$ defined by $f(x)=x$ if $x\in[0,1)$ and $f(x)=x-1$ if $x\in[2,3]$

or $[0,2\pi)\to S^1$ defined by $f(x)=e^{ix}$ (i.e. rolling it into a circle)

@Silent

Silent

Wow, @AkivaWeinberger, thank you very much! I, in fact had only that, rolling into circle example, which seemed a bit sophisticated. Thanks for this elementary example.

Startec

Thanks @sem

@Semiclassical

Akiva Weinberger

Continuous functions keep what's connected connected. Homeomorphisms also keep what's disconnected disconnected.

Alessandro Codenotti

6:06 AM

@Silent a very easy example: the identity from $\Bbb R$ with the discrete topology to $\Bbb R$ with the usual topology

Albas

6:23 AM

@BalarkaSen My project advisor says that most M.Math students from ISI just end up knowing a whole lot of fancy words without knowing anything about them. Didn't know how true that was

Adam

6:35 AM

awesome Pafnuty Lvovich Chebyshev fact of the day: he has family roots that trace back to a militant leader of the Tartar clan, who are more well known for an alliance they made with the Mongols lead by Genghis Khan

@WillHunting I hope global warming increases the iceberg separation and there is more shark food in the ocean for our prehistoric friends that are on the top of your food chain, basically the most necessary animal in the oceans ecosystem arguably

Secret

this post is soooo random

Silent

6:56 AM

@AlessandroCodenotti Wow! Awesome :-)

Thank you

Alessandro Codenotti

7:57 AM

I missed it but I guess you already solved the issue

Akiva Weinberger

8:12 AM

I guess "open functions" (functions which map open sets to open sets) would be functions that "preserve disconnectedness"

and continuous functions are ones that "preserve connectedness"

Does that make sense?

No it doesn't. Never mind.

$f:(-2,-1)\cup(1,2)$ defined by $f(x)=|x|$ doesn't "preserve disconnectedness" in any meaningful way.

Nor does the constant function (the function to a one-point space).

Liad

@AlessandroCodenotti nope

i still haven't figured out $\{(z,e^z) : z\in \Bbb C\}$ :/

Alessandro Codenotti

Did you think about the hint I gave you?

Liad

that $e^z$ is periodic?

Alessandro Codenotti

8:45 AM

Yes

Liad

hmm. im not sure how to use it. i think it can be shown that the set's closure is the whole space, but how to "project" (i know projection isn't good) it to $\Bbb A ^1$ is not clear to me @AlessandroCodenotti

maybe looking at $z$ in a circle of radius $2\pi$ ?

Alessandro Codenotti

The projection always works in the same, $(x,y)\mapsto x$ or $(x,y)\mapsto y$, you can think about this problem in terms of projections if you want, but which coordinate are you projecting on?

Liad

the second one would be better

because as we said $Im(e^z)$ is not closed in $\Bbb A ^ 1$

s.harp

Lie groups! Does every element of a connected Lie group admit a root? If the exponential map is surjective, then yes. But its not always surjective

Alessandro Codenotti

But the projection is not closed, so the image not being closed isn't very helpful

Liad

9:03 AM

i know, that's why im stuck.

Alessandro Codenotti

9:15 AM

So think about my hint instead

Another hint is that often the fibers of projections are more interesting than their images

Liad

what is fibers ?@AlessandroCodenotti

Alessandro Codenotti

Preimages of points

Liad

ok so it is enough to look at the preimage of $B_0(2\pi)$

maybe looking at the preimage of $0$ ? @AlessandroCodenotti

Alessandro Codenotti

I'm thinking of projecting on the first coordinate actually

Liad

that would give $\Bbb A^1$

Alessandro Codenotti

9:27 AM

Sure, but look at the fibers

Liad

the fibers are $\{(z,e^z)\}$

Alessandro Codenotti

What's the fiber over 0

Liad

$\emptyset$

Alessandro Codenotti

If the image of the projection is the whole of $\Bbb A^1$ how can it have an empty fiber?

Liad

i know its wrong.

its only $\{(0,1)\}$ as i wrote earlier?

Akiva Weinberger

9:38 AM

Whoa!

If $\epsilon_{12}=1$, $\epsilon_{21}=-1$, and $\epsilon_{11}=\epsilon_{22}=0$, then $P\epsilon P^\top=\det(P)\epsilon$

Is that just a weird coincidence?

Oh wait I see

Sort of

Alessandro Codenotti

Ah, wait, I got confused, projecting in the second coordinate was good, look at those fibers

mercio

that only works for size 2 matrices

Akiva Weinberger

Hold on, so the inverse of $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is $\frac1{\det(P)}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$

mercio

yes

something like that

Liad

@AlessandroCodenotti so here the preimage of $0$ is empty because $e^z$ is never zero

Akiva Weinberger

9:44 AM

and that last one has a superficial similarity to $P^\top$

Alessandro Codenotti

Ok let's look at the preimage of $1$ instead

Liad

so its $\{(z,e^z) : e^z =1 \}$

Akiva Weinberger

Is that actually $\epsilon P^\top\epsilon$?

Alessandro Codenotti

@Liad right, how many elements does this have?

mercio

no you are missing a minus sign because $\epsilon^2 = -I$

Akiva Weinberger

9:47 AM

Oh OK so hold on

Alessandro Codenotti

Hmm isn't that suspicious? Can that happen with polynomials?

Liad

it can't

because polynomials have finite number of zeros, this is how i proved it's closure is the whole space

(the closure of the set we started with)

Akiva Weinberger

So $P^{-1}=-\frac1{\det(P )}\epsilon P^\top\epsilon$?

mercio

for size 2 matrices only

Akiva Weinberger

Neat

Alessandro Codenotti

9:49 AM

Wait if proved the closure is the whole space then you're done, you already know it's not a variety

Akiva Weinberger

@mercio Right, because for 3x3 and above, the entries of the inverse matrix are quadratic instead of linear

(divided by the determinant)

Liad

my bad. @AlessandroCodenotti but itsn't it easier showing the closure is the whole space?

Alessandro Codenotti

It's the same argument

Liad

maybe i did something wrong

the closure is the zero set of a finite number of polynomials right ?@AlessandroCodenotti

Alessandro Codenotti

It is

mercio

9:51 AM

The entries of the inverse matrix still are degree (n-1) polynomials in the entries of the original matrix, divided by the determinant

Akiva Weinberger

Right, yeah

Alessandro Codenotti

So how did you show that the closure is the whole space?

Liad

i just going over it now, doubting my argument ^^

if $f$ vanish on the set then $f(z,e^z)=0$ for all $z$

that implies that $f$ has infinite number of zeros, doesn't it? @AlessandroCodenotti

Liad

10:14 AM

if we fix $z_0$ and look at $f(z_0,.)$ we get that this is a polynomial in $1$ variable

@AlessandroCodenotti with infinite number of zeros, so it must be the zero polynomial

Alessandro Codenotti

Yeah that works

Balarka Sen

12:38 PM

@Albas He's quite correct

M.Math in ISI is bullshit

Secret

1:02 PM

One thing I really want to happen if dedekind finite sets exists: Once a fool is smacked by them, they took a segmented shape that increasingly get segmented and then vanished like seeing an abstract artwork thining out of sight

Sounds like M.Math is a good test subject > : D

Secret

1:14 PM

in The h Bar, 2 hours ago, by Physics Meta

0

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